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#### Question 1:

Test the continuity of the function on f(x) at the origin:

Given:

We observe
(LHL at = 0) =
=â€‹

(RHL at x = 0)â€‹ =
=â€‹$\underset{h\to 0}{\mathrm{lim}}\frac{h}{\left|h\right|}=\underset{h\to 0}{\mathrm{lim}}\frac{h}{h}=\underset{h\to 0}{\mathrm{lim}}1=1$

Hence, $f\left(x\right)$ is discontinuous at the origin.

#### Question 2:

A function f(x) is defined as,

Show that f(x) is continuous that x = 3.

Given:

We observe
(LHL at = 3) =
=â€‹

And, (RHL at = 3)â€‹ =
=â€‹

Also, $f\left(3\right)=5$

Hence, $f\left(x\right)$ is continuous at $x=3$.

#### Question 3:

A function f(x) is defined as

Show that f(x) is continuous at x = 3

Given:

We observe
(LHL at x = 3) = $\underset{\mathrm{x}\to {3}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(3-h\right)$
=   â€‹$\underset{h\to 0}{\mathrm{lim}}\frac{{\left(3-h\right)}^{2}-9}{\left(3-h\right)-3}=\underset{h\to 0}{\mathrm{lim}}\frac{{3}^{2}+{h}^{2}-6h-9}{3-h-3}=\underset{h\to 0}{\mathrm{lim}}\frac{{h}^{2}-6h}{-h}==\underset{h\to 0}{\mathrm{lim}}\frac{h\left(h-6\right)}{-h}=\underset{h\to 0}{\mathrm{lim}}\left(6-h\right)=6$

(RHL at x = 3) = $\underset{\mathrm{x}\to {3}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(3+h\right)$
=   â€‹

Given:
$f\left(3\right)=6$

$\therefore \underset{\mathrm{x}\to {3}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{\mathrm{x}\to {3}^{+}}{\mathrm{lim}}f\left(x\right)=f\left(3\right)$

Hence, $f\left(x\right)$ is continuous at $x=3$.

#### Question 4:

If
Find whether f(x) is continuous at x = 1.

Given:

We observe
(LHL at x = 1) = $\underset{\mathrm{x}\to {1}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(1-h\right)$
=   â€‹$\underset{h\to 0}{\mathrm{lim}}\frac{{\left(1-h\right)}^{2}-1}{\left(1-h\right)-1}=\underset{h\to 0}{\mathrm{lim}}\frac{1+{h}^{2}-2h-1}{1-h-1}=\underset{h\to 0}{\mathrm{lim}}\frac{{h}^{2}-2h}{-h}=\underset{h\to 0}{\mathrm{lim}}\frac{h\left(h-2\right)}{-h}=\underset{h\to 0}{\mathrm{lim}}\left(2-h\right)=2$

(RHL at x = 1) = $\underset{\mathrm{x}\to {1}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(1+h\right)$
=   â€‹$\underset{h\to 0}{\mathrm{lim}}\frac{{\left(1+h\right)}^{2}-1}{\left(1+h\right)-1}=\underset{h\to 0}{\mathrm{lim}}\frac{1+{h}^{2}+2h-1}{1+h-1}=\underset{h\to 0}{\mathrm{lim}}\frac{{h}^{2}+2h}{h}=\underset{h\to 0}{\mathrm{lim}}\frac{h\left(h+2\right)}{h}=\underset{h\to 0}{\mathrm{lim}}\left(2+h\right)=2$

Given:
$f\left(1\right)=2$

$\therefore \underset{\mathrm{x}\to {1}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{\mathrm{x}\to {1}^{+}}{\mathrm{lim}}f\left(x\right)=f\left(1\right)$

Hence, $f\left(x\right)$ is continuous at $x=1$.

#### Question 5:

If
Find whether f(x) is continuous at x = 0.

Given:

We observe
(LHL at x = 0) = $\underset{\mathrm{x}\to {0}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(0-h\right)=\underset{h\to 0}{\mathrm{lim}}f\left(-h\right)$
=   â€‹$\underset{h\to 0}{\mathrm{lim}}\frac{\mathrm{sin}\left(-3h\right)}{-h}=\underset{h\to 0}{\mathrm{lim}}\frac{-\mathrm{sin}\left(3h\right)}{-h}=\underset{h\to 0}{\mathrm{lim}}\frac{3\mathrm{sin}\left(3h\right)}{3h}=3\underset{h\to 0}{\mathrm{lim}}\frac{\mathrm{sin}\left(3h\right)}{3h}=3·1=3$

(RHL at x = 0) = $\underset{\mathrm{x}\to {0}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(h\right)$
=   â€‹$\underset{h\to 0}{\mathrm{lim}}\frac{\mathrm{sin}3h}{h}=\underset{h\to 0}{\mathrm{lim}}\frac{3\mathrm{sin}3h}{3h}=3\underset{h\to 0}{\mathrm{lim}}\frac{\mathrm{sin}\left(3h\right)}{3h}=3·1=3$

Given:
$f\left(0\right)=1$

It is known that for a function $f\left(x\right)$ to be continuous at xa,
$\underset{\mathrm{x}\to {a}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{\mathrm{x}\to {a}^{+}}{\mathrm{lim}}f\left(x\right)=f\left(a\right)$

But here,
$\underset{\mathrm{x}\to {0}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{\mathrm{x}\to {0}^{+}}{\mathrm{lim}}f\left(x\right)\ne f\left(0\right)$

Hence, $f\left(x\right)$ is discontinuous at $x=0$.

#### Question 6:

If find whether f is continuous at x = 0.

Given:

We observe
(LHL at x = 0) = $\underset{\mathrm{x}\to {0}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(0-h\right)=\underset{h\to 0}{\mathrm{lim}}f\left(-h\right)$
=   â€‹$\underset{h\to 0}{\mathrm{lim}}{e}^{\frac{-1}{h}}=\underset{h\to 0}{\mathrm{lim}}\left(\frac{1}{{e}^{\frac{1}{h}}}\right)=\frac{1}{\underset{h\to 0}{\mathrm{lim}}{e}^{\frac{1}{h}}}=0$

(RHL at x = 0) = $\underset{\mathrm{x}\to {0}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(h\right)$
=   â€‹$\underset{h\to 0}{\mathrm{lim}}{e}^{\frac{1}{h}}=\infty$

Given:
$f\left(0\right)=1$

It is known that for a function $f\left(x\right)$ to be continuous at x = a,
$\underset{\mathrm{x}\to {a}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{\mathrm{x}\to {a}^{+}}{\mathrm{lim}}f\left(x\right)=f\left(a\right)$

But here,
$\underset{\mathrm{x}\to {0}^{-}}{\mathrm{lim}}f\left(x\right)\ne \underset{\mathrm{x}\to {0}^{+}}{\mathrm{lim}}f\left(x\right)$

Hence, $f\left(x\right)$ is discontinuous at $x=0$.

#### Question 7:

Let
Show that f(x) is discontinuous at x = 0.

Given:

Consider:
$\underset{x\to 0}{\mathrm{lim}}f\left(x\right)=\underset{x\to 0}{\mathrm{lim}}\left(\frac{1-\mathrm{cos}x}{{x}^{2}}\right)\phantom{\rule{0ex}{0ex}}⇒\underset{x\to 0}{\mathrm{lim}}f\left(x\right)=\underset{x\to 0}{\mathrm{lim}}\left(\frac{2{\mathrm{sin}}^{2}\frac{x}{2}}{{x}^{2}}\right)\phantom{\rule{0ex}{0ex}}⇒\underset{x\to 0}{\mathrm{lim}}f\left(x\right)=\underset{x\to 0}{\mathrm{lim}}\left(\frac{2{\mathrm{sin}}^{2}\frac{x}{2}}{4\left(\frac{{x}^{2}}{4}\right)}\right)\phantom{\rule{0ex}{0ex}}⇒\underset{x\to 0}{\mathrm{lim}}f\left(x\right)=\underset{x\to 0}{\mathrm{lim}}\left(\frac{2{\left(\mathrm{sin}\frac{x}{2}\right)}^{2}}{4{\left(\frac{x}{2}\right)}^{2}}\right)\phantom{\rule{0ex}{0ex}}⇒\underset{x\to 0}{\mathrm{lim}}f\left(x\right)=\frac{2}{4}\underset{x\to 0}{\mathrm{lim}}{\left(\frac{\mathrm{si}\mathrm{n}\frac{x}{2}}{\frac{x}{2}}\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒\underset{x\to 0}{\mathrm{lim}}f\left(x\right)=\frac{1}{2}·{1}^{2}=\frac{1}{2}$

Given:
$f\left(0\right)=1$

∴ â€‹$\underset{x\to 0}{\mathrm{lim}}f\left(x\right)\ne f\left(0\right)$

Thus, f(x) is discontinuous at x = 0
.

#### Question 8:

Show that
is discontinuous at x = 0.

The given function can be rewritten as:

â€‹$⇒$

We observe
(LHL at x = 0) = $\underset{x\to {0}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(0-h\right)=\underset{h\to 0}{\mathrm{lim}}f\left(-h\right)$$=\underset{h\to 0}{\mathrm{lim}}\left(-h\right)=0$

(RHL at x = 0) = $\underset{x\to {0}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(0-h\right)=\underset{h\to 0}{\mathrm{lim}}f\left(h\right)$$\underset{h\to 0}{\mathrm{lim}}0=0$

And, $f\left(0\right)=2$
∴ â€‹

Thus, f(x) is discontinuous at x = 0
.

#### Question 9:

Show that
is discontinuous at x = a.

The given function can be rewritten as:

â€‹$⇒$

$⇒$

We observe
(LHL at x = a) = $\underset{x\to {a}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(a-h\right)$$=\underset{h\to 0}{\mathrm{lim}}\left(-1\right)=-1$

(RHL at x = a) = $\underset{x\to {a}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(a+h\right)$$\underset{h\to 0}{\mathrm{lim}}\left(1\right)=1$

∴ â€‹$\underset{x\to {a}^{-}}{\mathrm{lim}}f\left(x\right)\ne \underset{x\to {a}^{+}}{\mathrm{lim}}f\left(x\right)$

Thus, f(x) is discontinuous at x = a.

#### Question 10:

Discuss the continuity of the following functions at the indicated point(s):
(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

(i) Given:

We observe

$⇒\underset{x\to 0}{\mathrm{lim}}f\left(x\right)=f\left(0\right)$
Hence, f(x) is continuous at x = 0.

(ii) Given:

We observe

$⇒\underset{x\to 0}{\mathrm{lim}}f\left(x\right)=f\left(0\right)$
Hence, f(x) is continuous at x = 0.

(iii) Given:

Putting xa = y, we get

$⇒\underset{x\to a}{\mathrm{lim}}f\left(x\right)=f\left(a\right)=0$
Hence, f(x) is continuous at x = a.

(iv) Given:

We observe
$\underset{x\to 0}{\mathrm{lim}}f\left(x\right)=\underset{x\to 0}{\mathrm{lim}}\frac{{e}^{x}-1}{\mathrm{log}\left(1+2x\right)}\phantom{\rule{0ex}{0ex}}⇒\underset{x\to 0}{\mathrm{lim}}f\left(x\right)=\underset{x\to 0}{\mathrm{lim}}\frac{{e}^{x}-1}{\frac{2x\mathrm{log}\left(1+2x\right)}{2x}}\phantom{\rule{0ex}{0ex}}⇒\underset{x\to 0}{\mathrm{lim}}f\left(x\right)=\frac{1}{2}\underset{x\to 0}{\mathrm{lim}}\frac{\left(\frac{{e}^{x}-1}{x}\right)}{\left(\frac{\mathrm{log}\left(1+2x\right)}{2x}\right)}\phantom{\rule{0ex}{0ex}}⇒\underset{x\to 0}{\mathrm{lim}}f\left(x\right)=\frac{1}{2}×\frac{\left(\underset{x\to 0}{\mathrm{lim}}\frac{{e}^{x}-1}{x}\right)}{\left(\underset{x\to 0}{\mathrm{lim}}\frac{\mathrm{log}\left(1+2x\right)}{2x}\right)}=\frac{1}{2}×\frac{1}{1}=\frac{1}{2}$
And, $f\left(0\right)=7$
$⇒\underset{x\to 0}{\mathrm{lim}}f\left(x\right)\ne f\left(0\right)$

Hence, f(x) is discontinuous at x = 0.

(v) Given:

Here, $f\left(1\right)=n-1$

$\underset{x\to 1}{\mathrm{lim}}f\left(x\right)=\underset{x\to 1}{\mathrm{lim}}\frac{1-{x}^{n}}{1-x}\phantom{\rule{0ex}{0ex}}⇒\underset{x\to 1}{\mathrm{lim}}f\left(x\right)=\underset{x\to 1}{\mathrm{lim}}\left[{\left(1-x\right)}^{n-1}+{}^{n}C_{1}{\left(1-x\right)}^{n-2}x+{}^{n}C_{2}{\left(1-x\right)}^{n-3}{x}^{2}+...+{}^{n}C_{n-1}{\left(1-x\right)}^{0}{x}^{n-1}\right]$
$⇒\underset{x\to 1}{\mathrm{lim}}f\left(x\right)=0+0...+{\left(1\right)}^{n-1}=1\ne f\left(1\right)$

Thus, .

(vi) Given:

We observe
(LHL at = 1) = $\underset{x\to {1}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(1-h\right)=\underset{h\to 0}{\mathrm{lim}}-\left(1-h\right)-1=\underset{h\to 0}{\mathrm{lim}}-2+h=-2$
And, $f\left(1\right)=2$

$⇒\underset{x\to {1}^{-}}{\mathrm{lim}}f\left(x\right)\ne f\left(1\right)$

Hence, f(x) is discontinuous at x = 1.

(vii) Given:

We observe

(LHL at x = 0) = $\underset{x\to {0}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(-h\right)=\underset{h\to 0}{\mathrm{lim}}\left[-h-2\right]=-2$
(RHL at x = 0) = $\underset{x\to {0}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(h\right)=\underset{h\to 0}{\mathrm{lim}}\left(2+\mathrm{h}\right)=2$

$⇒\underset{x\to {0}^{-}}{\mathrm{lim}}f\left(x\right)\ne \underset{x\to {0}^{+}}{\mathrm{lim}}f\left(x\right)$

Hence, f(x) is discontinuous at x = 0.

(viii) Given:

We observe

(LHL at x = a) = $\underset{x\to {a}^{-}}{\mathrm{lim}}f\left(x\right)=\left(-a+a\right)\mathrm{sin}\left(\frac{1}{-a+a}\right)=0$
(RHL at x = a) = $\underset{x\to {a}^{+}}{\mathrm{lim}}f\left(x\right)=\left(a-a\right)\mathrm{sin}\left(\frac{1}{a-a}\right)=0$

$⇒\underset{x\to {a}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{x\to {a}^{+}}{\mathrm{lim}}f\left(x\right)=f\left(a\right)$

Hence, f(x) is continuous at x = a.

#### Question 11:

Show that is discontinuous at x = 1.

Given:

â€‹
We observe
(LHL at x = 1) = $\underset{x\to {1}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(1-h\right)$$=\underset{h\to 0}{\mathrm{lim}}\left(1+{\left(1-h\right)}^{2}\right)=\underset{h\to 0}{\mathrm{lim}}\left(2+{h}^{2}-2h\right)=2$

(RHL at x = 1) = $\underset{x\to {1}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(1+h\right)$$=\underset{h\to 0}{\mathrm{lim}}\left(2-\left(1+h\right)\right)=\underset{h\to 0}{\mathrm{lim}}\left(1-h\right)=1$

∴ â€‹$\underset{x\to {1}^{-}}{\mathrm{lim}}f\left(x\right)\ne \underset{x\to {1}^{+}}{\mathrm{lim}}f\left(x\right)$

Thus, f(x) is discontinuous at x = 1.

#### Question 12:

Show that

Given:

We observe
(LHL at x = 0) = $\underset{x\to {0}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(0-h\right)=\underset{h\to 0}{\mathrm{lim}}f\left(-h\right)$
$=\underset{h\to 0}{\mathrm{lim}}\left(\frac{\mathrm{sin}3\left(-h\right)}{\mathrm{tan}2\left(-h\right)}\right)=\underset{h\to 0}{\mathrm{lim}}\left(\frac{\mathrm{sin}3h}{\mathrm{tan}2h}\right)=\underset{h\to 0}{\mathrm{lim}}\left(\frac{\frac{3\mathrm{sin}3h}{3h}}{\frac{2\mathrm{tan}2h}{2h}}\right)\phantom{\rule{0ex}{0ex}}=\frac{\underset{h\to 0}{\mathrm{lim}}\left(\frac{3\mathrm{sin}3h}{3h}\right)}{\underset{h\to 0}{\mathrm{lim}}\left(\frac{2\mathrm{tan}2h}{2h}\right)}=\frac{3\underset{h\to 0}{\mathrm{lim}}\left(\frac{\mathrm{sin}3h}{3h}\right)}{2\underset{h\to 0}{\mathrm{lim}}\left(\frac{\mathrm{tan}2h}{2h}\right)}=\frac{3×1}{2×1}=\frac{3}{2}$

(RHL at x = 1) = $\underset{x\to {0}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(0+h\right)=\underset{h\to 0}{\mathrm{lim}}f\left(h\right)$
$=\underset{h\to 0}{\mathrm{lim}}\left(\frac{\mathrm{log}\left(1+3h\right)}{{e}^{2h}-1}\right)=\underset{h\to 0}{\mathrm{lim}}\left(\frac{3h\frac{\mathrm{log}\left(1+3h\right)}{3h}}{\frac{2h\left({e}^{2h}-1\right)}{2h}}\right)\phantom{\rule{0ex}{0ex}}=\frac{3}{2}\underset{h\to 0}{\mathrm{lim}}\left(\frac{\frac{\mathrm{log}\left(1+3h\right)}{3h}}{\frac{\left({e}^{2h}-1\right)}{2h}}\right)=\frac{3}{2}\frac{\underset{h\to 0}{\mathrm{lim}}\left(\frac{\mathrm{log}\left(1+3h\right)}{3h}\right)}{\underset{h\to 0}{\mathrm{lim}}\left(\frac{\left({e}^{2h}-1\right)}{2h}\right)}=\frac{3×1}{2×1}=\frac{3}{2}$

And, $f\left(0\right)=\frac{3}{2}$

∴ â€‹$\underset{x\to {0}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{x\to {0}^{+}}{\mathrm{lim}}f\left(x\right)=f\left(0\right)$

Thus, f(x) is continuous at x = 0.

#### Question 13:

Find the value of 'a' for which the function f defined by
$f\left(x\right)=\left\{\begin{array}{ll}a\mathrm{sin}\frac{\mathrm{\pi }}{2}\left(x+1\right),& x\le 0\\ \frac{\mathrm{tan}x-\mathrm{sin}x}{{x}^{3}},& x>0\end{array}\right\$is continuous at x = 0.

Given:
We have

(LHL at x = 0) =

(RHL at x = 0) =

#### Question 14:

Examine the continuity of the function

Also sketch the graph of this function.

The given function can be rewritten as:

We observe

(LHL at x = 0) = $\underset{x\mathit{\to }{0}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\mathit{\to }0}{\mathrm{lim}}f\left(0-h\right)=\underset{h\mathit{\to }0}{\mathrm{lim}}f\left(-h\right)$ = â€‹$\underset{h\mathit{\to }0}{\mathrm{lim}}3\left(-h\right)-2=-2$

(RHL at x = 0) =  $\underset{x\mathit{\to }{0}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\mathit{\to }0}{\mathrm{lim}}f\left(0+h\right)=\underset{h\mathit{\to }0}{\mathrm{lim}}f\left(h\right)$ = â€‹$\underset{h\mathit{\to }0}{\mathrm{lim}}\left(h+1\right)=1$

$\therefore \underset{x\mathit{\to }{0}^{-}}{\mathrm{lim}}f\left(x\right)\ne \underset{x\mathit{\to }{0}^{+}}{\mathrm{lim}}f\left(x\right)$

Hence, $f\left(x\right)$ is discontinuous at $x=0$.

#### Question 15:

Discuss the continuity of the function f(x) at the point x = 0, where

Given:

(LHL at x = 0) = $\underset{x\mathit{\to }{0}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\mathit{\to }0}{\mathrm{lim}}f\left(0-h\right)=\underset{h\mathit{\to }0}{\mathrm{lim}}f\left(-h\right)$ = â€‹$\underset{h\mathit{\to }0}{\mathrm{lim}}-\left(-h\right)=0$

(RHL at x = 0) =  $\underset{x\mathit{\to }{0}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\mathit{\to }0}{\mathrm{lim}}f\left(0+h\right)=\underset{h\mathit{\to }0}{\mathrm{lim}}f\left(h\right)$ = â€‹$\underset{h\mathit{\to }0}{\mathrm{lim}}\left(h\right)=0$

And, â€‹$f\left(0\right)=1$

$\therefore \underset{x\mathit{\to }{0}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{x\mathit{\to }{0}^{+}}{\mathrm{lim}}f\left(x\right)\ne f\left(0\right)$

Hence, $f\left(x\right)$ is discontinuous at $x=0$.

#### Question 16:

Discuss the continuity of the function f(x) at the point x = 1/2, where
$f\left(x\right)=\left\{\begin{array}{c}x,\\ 1/2,\\ 1-x,\end{array}\begin{array}{c}0\le x<1/2\\ x=1/2\\ 1/2

Given:

We observe

(LHL at x = $\frac{1}{2}$) = $\underset{x\mathit{\to }{\frac{1}{2}}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\mathit{\to }0}{\mathrm{lim}}f\left(\frac{1}{2}-h\right)$ = â€‹$\underset{h\mathit{\to }0}{\mathrm{lim}}\left(\frac{1}{2}-h\right)=\frac{1}{2}$

(RHL at x = $\frac{1}{2}$) = $\underset{x\mathit{\to }{\frac{1}{2}}^{\mathit{+}}}{\mathrm{lim}}f\left(x\right)=\underset{h\mathit{\to }0}{\mathrm{lim}}f\left(\frac{1}{2}+h\right)$$\underset{h\mathit{\to }0}{\mathrm{lim}}\left(1-\left(\frac{1}{2}+h\right)\right)=\frac{1}{2}$

Also, â€‹$f\left(\frac{1}{2}\right)=\frac{1}{2}$

Hence, $f\left(x\right)$ is continuous at $x=\frac{1}{2}$.

#### Question 17:

Discuss the continuity of

Hence, f(x) is discontinuous at x = 0.

#### Question 18:

For what value of k is the following function continuous at x = 1?

Given:
If $f\left(x\right)$ is continuous at x = 1, then

$k=2$

#### Question 19:

Determine the value of the constant k so that the function

Given:

If $f\left(x\right)$ is continuous at x = 1, then,

$k=-1$

#### Question 20:

For what value of k is the function

Given:

If $f\left(x\right)$ is continuous at x = 0, then

$k=\frac{5}{3}$

#### Question 21:

Determine the value of the constant k so that the function

Given:

If $f\left(x\right)$ is continuous at x = 2, then
...(1)

Now,

And, $f\left(2\right)=3$

From (1), we have

$4k=3\phantom{\rule{0ex}{0ex}}⇒k=\frac{3}{4}$

#### Question 22:

Determine the value of the constant k so that the function

Given:

If $f\left(x\right)$ is continuous at x = 0, then

$k=\frac{2}{5}$

#### Question 23:

Find the values of a so that the function

Given:

We observe
(LHL at x = 2) = $\underset{\mathrm{x}\to {2}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(2-h\right)$ =  â€‹$\underset{h\to 0}{\mathrm{lim}}a\left(2-h\right)+5=2a+5$

(RHL at x = 2) = $\underset{\mathrm{x}\to {2}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(2+h\right)$ =  â€‹$\underset{h\to 0}{\mathrm{lim}}\left(2+h-1\right)$ = $1$

And, $f\left(2\right)=a\left(2\right)+5=2a+5$

Since $f\left(x\right)$ is continuous at x = 2, we have
$\underset{\mathrm{x}\to {2}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{\mathrm{x}\to {2}^{+}}{\mathrm{lim}}f\left(x\right)=f\left(2\right)$
$2a+5=1$
$2a=-4$
$a=-2$

#### Question 24:

Prove that the function

remains discontinuous at x = 0, regardless the choice of k.

The given function can be rewritten as:

We observe
(LHL at x = 0) = $\underset{x\to {0}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(0-h\right)=\underset{h\to 0}{\mathrm{lim}}f\left(-h\right)$
=$\underset{h\to 0}{\mathrm{lim}}\frac{1}{-2h-1}=-1$

(RHL at x = 0) = $\underset{x\to {0}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(0+h\right)=\underset{h\to 0}{\mathrm{lim}}f\left(h\right)$
=$\underset{h\to 0}{\mathrm{lim}}\frac{1}{2h+1}=1$

So, â€‹$\underset{x\to {0}^{-}}{\mathrm{lim}}f\left(x\right)\ne \underset{x\to {0}^{+}}{\mathrm{lim}}f\left(x\right)$ such that are independent of k.

Thus, f(x) is discontinuous at x = 0, regardless of the choice of k.

#### Question 25:

Find the value of k if f(x) is continuous at x = π/2, where

Given:

If f(x) is continuous at x = $\frac{\mathrm{\pi }}{2}$, then

$\underset{x\to \frac{\mathrm{\pi }}{2}}{\mathrm{lim}}\frac{k\mathrm{cos}x}{\mathrm{\pi }-2x}=3$     ...(1)

Putting $\frac{\mathrm{\pi }}{2}-x=h$, we get

From (1), we have

$⇒k×1=6$
$⇒k=6$

Hence, for $k=6$ , f(x) is continuous at x = $\frac{\mathrm{\pi }}{2}$.

#### Question 26:

Determine the values of a, b, c for which the function
is continuous at x = 0.

The given function can be rewritten as:

We observe
(LHL at x = 0) = $\underset{x\to {0}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(0-h\right)=\underset{h\to 0}{\mathrm{lim}}f\left(-h\right)$

(RHL at x = 0) = $\underset{x\to {0}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(0+h\right)=\underset{h\to 0}{\mathrm{lim}}f\left(h\right)$
=$\underset{h\to 0}{\mathrm{lim}}\left(\frac{\sqrt{1+bh}-1}{bh}\right)=\underset{h\to 0}{\mathrm{lim}}\left(\frac{bh}{bh\left(\sqrt{1+bh}+1\right)}\right)=\underset{h\to 0}{\mathrm{lim}}\left(\frac{1}{\left(\sqrt{1+bh}+1\right)}\right)=\frac{1}{2}$

And, $f\left(0\right)=c$

If $f\left(x\right)$ is continuous at x = 0, then

â€‹$\underset{x\to {0}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{x\to {0}^{+}}{\mathrm{lim}}f\left(x\right)=f\left(0\right)$

$⇒a=\frac{-3}{2}$

Now, $\frac{\sqrt{1+bx}-1}{bx}$ exists only if $bx\ne 0⇒b\ne 0$.

$b\in R-\left\{0\right\}$

#### Question 27:

If

Given:
If $f\left(x\right)$ is continuous at x = 0, then
...(1)

Consider:

$⇒\underset{x\to 0}{\mathrm{lim}}f\left(x\right)=\frac{2{k}^{2}}{4}×1=\frac{{k}^{2}}{2}$

From equation (1), we have
$\frac{{k}^{2}}{2}=f\left(0\right)$
$⇒\frac{{k}^{2}}{2}=\frac{1}{2}\phantom{\rule{0ex}{0ex}}⇒k=±1$

#### Question 28:

If is continuous at x = 4, find a, b.

Given:

We observe
(LHL at x = 4) = $\underset{x\to {4}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(4-h\right)$
$=\underset{h\to 0}{\mathrm{lim}}\left(\frac{4-h-4}{\left|4-h-4\right|}+a\right)=\underset{h\to 0}{\mathrm{lim}}\left(\frac{-h}{\left|-h\right|}+a\right)=a-1$

(RHL at x = 4) = $\underset{x\to {4}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(4+h\right)$
$=\underset{h\to 0}{\mathrm{lim}}\left(\frac{4+h-4}{\left|4+h-4\right|}+b\right)=\underset{h\to 0}{\mathrm{lim}}\left(\frac{h}{\left|h\right|}+b\right)=b+1$

And $f\left(4\right)=a+b$

If f(x) is continuous at x = 4, then

â€‹$\underset{x\to {4}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{x\to {4}^{+}}{\mathrm{lim}}f\left(x\right)=f\left(4\right)$
$⇒a-1=b+1=a+b$

#### Question 29:

For what value of k is the function

continuous at x = 0?

Given:

If f(x) is continuous at x = 0, then

â€‹$\underset{x\to 0}{\mathrm{lim}}f\left(x\right)=f\left(0\right)$
$⇒\underset{x\to 0}{\mathrm{lim}}\frac{\mathrm{sin}2x}{x}=k$
$⇒\underset{x\to 0}{\mathrm{lim}}\frac{2\mathrm{sin}2x}{2x}=k$
$⇒2\underset{x\to 0}{\mathrm{lim}}\frac{\mathrm{sin}2x}{2x}=k$
$⇒2×1=k$
$⇒k=2$

#### Question 30:

Let $f\left(x\right)=\frac{\mathrm{log}\left(1+\frac{x}{a}\right)-\mathrm{log}\left(1-\frac{x}{b}\right)}{x}$, x ≠ 0. Find the value of f at x = 0 so that f becomes continuous at x = 0.

Given:

If f(x) is continuous at x = 0, then

â€‹$\underset{x\to 0}{\mathrm{lim}}f\left(x\right)=f\left(0\right)$
$⇒\underset{x\to 0}{\mathrm{lim}}\left(\frac{\mathrm{log}\left(1+\frac{x}{a}\right)-\mathrm{log}\left(1-\frac{x}{b}\right)}{x}\right)=f\left(0\right)$
$⇒\underset{x\to 0}{\mathrm{lim}}\left(\frac{\mathrm{log}\left(1+\frac{x}{a}\right)}{\frac{ax}{a}}-\frac{\mathrm{log}\left(1-\frac{x}{b}\right)}{\frac{bx}{b}}\right)=f\left(0\right)$
$⇒\frac{1}{a}\underset{x\to 0}{\mathrm{lim}}\left(\frac{\mathrm{log}\left(1+\frac{x}{a}\right)}{\frac{x}{a}}\right)-\left(-\frac{1}{b}\right)\underset{x\to 0}{\mathrm{lim}}\left(\frac{\mathrm{log}\left(1-\frac{x}{b}\right)}{\frac{-x}{b}}\right)=f\left(0\right)$

#### Question 31:

If is continuous at x = 2, find k.

Given:

If f(x) is continuous at x = 2, then

â€‹$\underset{x\to 2}{\mathrm{lim}}f\left(x\right)=f\left(2\right)\phantom{\rule{0ex}{0ex}}⇒\underset{x\to 2}{\mathrm{lim}}\frac{{2}^{x+2}-16}{{4}^{x}-16}=f\left(2\right)\phantom{\rule{0ex}{0ex}}⇒\underset{x\to 2}{\mathrm{lim}}\frac{4\left({2}^{x}-4\right)}{\left({2}^{x}-4\right)\left({2}^{x}+4\right)}=k\phantom{\rule{0ex}{0ex}}⇒\underset{x\to 2}{\mathrm{lim}}\frac{4}{\left({2}^{x}+4\right)}=k\phantom{\rule{0ex}{0ex}}⇒\frac{4}{\left({2}^{2}+4\right)}=k\phantom{\rule{0ex}{0ex}}⇒\frac{4}{8}=k\phantom{\rule{0ex}{0ex}}⇒k=\frac{1}{2}$

#### Question 32:

If is continuous at x = 0, find k.

Given:

If f(x) is continuous at x = 0, then

â€‹$\underset{x\to 0}{\mathrm{lim}}f\left(x\right)=f\left(0\right)\phantom{\rule{0ex}{0ex}}⇒\underset{x\to 0}{\mathrm{lim}}\frac{{\mathrm{cos}}^{2}x-{\mathrm{sin}}^{2}x-1}{\sqrt{{x}^{2}+1}-1}=k\phantom{\rule{0ex}{0ex}}⇒\underset{x\to 0}{\mathrm{lim}}\frac{1-{\mathrm{sin}}^{2}x-{\mathrm{sin}}^{2}x-1}{\sqrt{{x}^{2}+1}-1}=k\phantom{\rule{0ex}{0ex}}⇒\underset{x\to 0}{\mathrm{lim}}\frac{-2{\mathrm{sin}}^{2}x}{\sqrt{{x}^{2}+1}-1}=k\phantom{\rule{0ex}{0ex}}⇒\underset{x\to 0}{\mathrm{lim}}\frac{-2\left({\mathrm{sin}}^{2}x\right)\left(\sqrt{{x}^{2}+1}+1\right)}{\left(\sqrt{{x}^{2}+1}-1\right)\left(\sqrt{{x}^{2}+1}+1\right)}=k\phantom{\rule{0ex}{0ex}}⇒\underset{x\to 0}{\mathrm{lim}}\frac{-2\left({\mathrm{sin}}^{2}x\right)\left(\sqrt{{x}^{2}+1}+1\right)}{{x}^{2}}=k\phantom{\rule{0ex}{0ex}}⇒-2\underset{x\to 0}{\mathrm{lim}}\frac{\left({\mathrm{sin}}^{2}x\right)\left(\sqrt{{x}^{2}+1}+1\right)}{{x}^{2}}=k\phantom{\rule{0ex}{0ex}}⇒-2\underset{x\to 0}{\mathrm{lim}}{\left(\frac{\mathrm{sin}x}{x}\right)}^{2}\underset{x\to 0}{\mathrm{lim}}\left(\sqrt{{x}^{2}+1}+1\right)=k\phantom{\rule{0ex}{0ex}}⇒-2×1×\left(1+1\right)=k\phantom{\rule{0ex}{0ex}}⇒k=-4$

#### Question 33:

Extend the definition of the following by continuity
at the point x = π.

Given:

If f(x) is continuous at x = $\mathrm{\pi }$, then

â€‹
$\underset{x\to \mathrm{\pi }}{\mathrm{lim}}f\left(x\right)=f\left(\mathrm{\pi }\right)\phantom{\rule{0ex}{0ex}}⇒\underset{x\to \mathrm{\pi }}{\mathrm{lim}}\frac{1-\mathrm{cos}7\left(\mathrm{x}-\mathrm{\pi }\right)}{5{\left(\mathrm{x}-\mathrm{\pi }\right)}^{2}}=f\left(\mathrm{\pi }\right)\phantom{\rule{0ex}{0ex}}⇒\frac{2}{5}\underset{x\to \mathrm{\pi }}{\mathrm{lim}}\frac{{\mathrm{sin}}^{2}\left(\frac{7\left(\mathrm{x}-\mathrm{\pi }\right)}{2}\right)}{{\left(\mathrm{x}-\mathrm{\pi }\right)}^{2}}=f\left(\mathrm{\pi }\right)\phantom{\rule{0ex}{0ex}}⇒\frac{2}{5}×\frac{49}{4}\underset{x\to \mathrm{\pi }}{\mathrm{lim}}\frac{{\mathrm{sin}}^{2}\left(\frac{7\left(\mathrm{x}-\mathrm{\pi }\right)}{2}\right)}{\frac{49}{4}{\left(\mathrm{x}-\mathrm{\pi }\right)}^{2}}=f\left(\mathrm{\pi }\right)\phantom{\rule{0ex}{0ex}}⇒\frac{2}{5}×\frac{49}{4}\underset{x\to \mathrm{\pi }}{\mathrm{lim}}\frac{{\mathrm{sin}}^{2}\left(\frac{7\left(\mathrm{x}-\mathrm{\pi }\right)}{2}\right)}{{\left(\frac{7}{2}\left(\mathrm{x}-\mathrm{\pi }\right)\right)}^{2}}=f\left(\mathrm{\pi }\right)\phantom{\rule{0ex}{0ex}}⇒\frac{2}{5}×\frac{49}{4}\underset{x\to \mathrm{\pi }}{\mathrm{lim}}{\left[\frac{\mathrm{sin}\left(\frac{7\left(\mathrm{x}-\mathrm{\pi }\right)}{2}\right)}{\left(\frac{7}{2}\left(\mathrm{x}-\mathrm{\pi }\right)\right)}\right]}^{2}=f\left(\mathrm{\pi }\right)\phantom{\rule{0ex}{0ex}}⇒\frac{2}{5}×\frac{49}{4}×1=f\left(\mathrm{\pi }\right)\phantom{\rule{0ex}{0ex}}⇒\frac{1}{5}×\frac{49}{2}×1=f\left(\mathrm{\pi }\right)\phantom{\rule{0ex}{0ex}}⇒\frac{49}{10}=f\left(\mathrm{\pi }\right)$

Hence, the given function will be continuous at $x=\mathrm{\pi }$, if $f\left(\mathrm{\pi }\right)=\frac{49}{10}$.

#### Question 34:

If x ≠ 0 is continuous at x = 0, then find f (0).

Given:

If f(x) is continuous at x = 0, then

â€‹$\underset{x\to 0}{\mathrm{lim}}f\left(x\right)=f\left(0\right)\phantom{\rule{0ex}{0ex}}⇒\underset{x\to 0}{\mathrm{lim}}\frac{2x+3\mathrm{sin}x}{3x+2\mathrm{sin}x}=f\left(0\right)\phantom{\rule{0ex}{0ex}}⇒\underset{x\to 0}{\mathrm{lim}}\frac{x\left(2+3\frac{\mathrm{sin}x}{x}\right)}{x\left(3+2\frac{\mathrm{sin}x}{x}\right)}=f\left(0\right)\phantom{\rule{0ex}{0ex}}⇒\underset{x\to 0}{\mathrm{lim}}\frac{\left(2+3\frac{\mathrm{sin}x}{x}\right)}{\left(3+2\frac{\mathrm{sin}x}{x}\right)}=f\left(0\right)\phantom{\rule{0ex}{0ex}}⇒\frac{\underset{x\to 0}{\mathrm{lim}}\left(2+3\frac{\mathrm{sin}x}{x}\right)}{\underset{x\to 0}{\mathrm{lim}}\left(3+2\frac{\mathrm{sin}x}{x}\right)}=f\left(0\right)\phantom{\rule{0ex}{0ex}}⇒\frac{2+3\underset{x\to 0}{\mathrm{lim}}\left(\frac{\mathrm{sin}x}{x}\right)}{3+2\underset{x\to 0}{\mathrm{lim}}\left(\frac{\mathrm{sin}x}{x}\right)}=f\left(0\right)\phantom{\rule{0ex}{0ex}}⇒\frac{2+3×1}{3+2×1}=f\left(0\right)\phantom{\rule{0ex}{0ex}}⇒\frac{5}{5}=f\left(0\right)\phantom{\rule{0ex}{0ex}}⇒f\left(0\right)=1$

#### Question 35:

Find the value of k for which
is continuous at x = 0;

Given:

If f(x) is continuous at x = 0, then

â€‹

#### Question 36:

In each of the following, find the value of the constant k so that the given function is continuous at the indicated point;
(i) at x = 0

(ii) at x = 1

(iii) at x = 0

(iv) at x = π

(v) at x = 5

(vi) at x = 5

(vii) at x = 1

(viii)

(ix)

(i) Given:

If f(x) is continuous at x = 0, then
$\underset{x\to 0}{\mathrm{lim}}f\left(x\right)=f\left(0\right)\phantom{\rule{0ex}{0ex}}⇒\underset{x\to 0}{\mathrm{lim}}\frac{1-\mathrm{cos}2kx}{{x}^{2}}=8\phantom{\rule{0ex}{0ex}}⇒\underset{x\to 0}{\mathrm{lim}}\frac{2{k}^{2}{\mathrm{sin}}^{2}kx}{{k}^{2}{x}^{2}}=8\phantom{\rule{0ex}{0ex}}⇒2{k}^{2}\underset{x\to 0}{\mathrm{lim}}{\left(\frac{\mathrm{sin}kx}{kx}\right)}^{2}=8\phantom{\rule{0ex}{0ex}}⇒2{k}^{2}×1=8\phantom{\rule{0ex}{0ex}}⇒{k}^{2}=4\phantom{\rule{0ex}{0ex}}⇒k=±2$

(ii) Given:

If f(x) is continuous at x = 1, then

(iii) Given:

We have
(LHL at x = 0) = $\underset{x\to {0}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(0-h\right)=\underset{h\to 0}{\mathrm{lim}}f\left(-h\right)=\underset{h\to 0}{\mathrm{lim}}k\left({h}^{2}+2h\right)=0$
(RHL at x = 0) = $\underset{x\to {0}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(0+h\right)=\underset{h\to 0}{\mathrm{lim}}f\left(h\right)=\underset{h\to 0}{\mathrm{lim}}\mathrm{cos}h=1$

$\therefore \underset{x\to {0}^{-}}{\mathrm{lim}}f\left(x\right)\ne \underset{x\to {0}^{+}}{\mathrm{lim}}f\left(x\right)$

(iv) Given:

We have
(LHL at x = $\mathrm{\pi }$) = $\underset{x\to {\mathrm{\pi }}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(\mathrm{\pi }-h\right)=\underset{h\to 0}{\mathrm{lim}}k\left(\mathrm{\pi }-\mathrm{h}\right)+1=k\mathrm{\pi }+1$
(RHL at x = $\mathrm{\pi }$) = $\underset{x\to {\mathrm{\pi }}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(\mathrm{\pi }+h\right)=\underset{h\to 0}{\mathrm{lim}}\mathrm{cos}\left(\mathrm{\pi }+h\right)=\mathrm{cos\pi }=-1$

If f(x) is continuous at x = $\mathrm{\pi }$, then
$\underset{x\to {\mathrm{\pi }}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{x\to {\mathrm{\pi }}^{+}}{\mathrm{lim}}f\left(x\right)$
$⇒k\mathrm{\pi }+1=-1\phantom{\rule{0ex}{0ex}}⇒\mathrm{k}=\frac{-2}{\mathrm{\pi }}$

(v) Given:

We have
(LHL at x = 5) = $\underset{x\to {5}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(5-h\right)=\underset{h\to 0}{\mathrm{lim}}k\left(5-\mathrm{h}\right)+1=5k+1$
(RHL at x = 5) = $\underset{x\to {5}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(5+h\right)=\underset{h\to 0}{\mathrm{lim}}3\left(5+\mathrm{h}\right)-5=10$

If f(x) is continuous at x = 5, then
$\underset{x\to {5}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{x\to {5}^{+}}{\mathrm{lim}}f\left(x\right)\phantom{\rule{0ex}{0ex}}⇒5k+1=10\phantom{\rule{0ex}{0ex}}⇒k=\frac{9}{5}$

(vi) Given:

If f(x) is continuous at x = 5, then
$\underset{x\to 5}{\mathrm{lim}}f\left(x\right)=f\left(5\right)\phantom{\rule{0ex}{0ex}}⇒\underset{x\to 5}{\mathrm{lim}}\left(x+5\right)=k\phantom{\rule{0ex}{0ex}}⇒k=5+5=10$

(vii) Given:

We have
(LHL at x = 1) = $\underset{x\to {1}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(1-h\right)=\underset{h\to 0}{\mathrm{lim}}4=4$
(RHL at x = 1) = $\underset{x\to {1}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(1+h\right)=\underset{h\to 0}{\mathrm{lim}}k{\left(1+\mathrm{h}\right)}^{2}=k$

If f(x) is continuous at x = 1, then
$\underset{x\to {1}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{x\to {1}^{+}}{\mathrm{lim}}f\left(x\right)\phantom{\rule{0ex}{0ex}}⇒k=4$

(viii) Given:

We have
(LHL at x = 0) = $\underset{x\to {0}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(0-h\right)=\underset{h\to 0}{\mathrm{lim}}k\left({\left(-h\right)}^{2}+2\right)=2k$
(RHL at x = 0) = $\underset{x\to {0}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(0+h\right)=\underset{h\to 0}{\mathrm{lim}}3h+1=1$

If f(x) is continuous at x = 0, then
$\underset{x\to {0}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{x\to {0}^{+}}{\mathrm{lim}}f\left(x\right)\phantom{\rule{0ex}{0ex}}⇒2k=1\phantom{\rule{0ex}{0ex}}⇒k=\frac{1}{2}$

(ix) Given:

If f(x) is continuous at x = 2, then
$\underset{x\to 2}{\mathrm{lim}}f\left(x\right)=f\left(2\right)\phantom{\rule{0ex}{0ex}}⇒\underset{x\to 2}{\mathrm{lim}}\left(x+5\right)=k\phantom{\rule{0ex}{0ex}}⇒k=2+5=7$

#### Question 37:

Find the values of a and b so that the function f given by
is continuous at x = 3 and x = 5.

Given:

We have
(LHL at x = 3) = $\underset{x\to {3}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(3-h\right)$$=\underset{h\to 0}{\mathrm{lim}}\left(1\right)=1$

(RHL at x = 3) = $\underset{x\to {3}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(3+h\right)$$=\underset{h\to 0}{\mathrm{lim}}a\left(3+h\right)+b=3a+b$

(LHL at x = 5) = $\underset{x\to {5}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(5-h\right)$$=\underset{h\to 0}{\mathrm{lim}}\left(a\left(5-h\right)+b\right)=5a+b$

(RHL at x = 5) = $\underset{x\to {5}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(5+h\right)$$=\underset{h\to 0}{\mathrm{lim}}7=7$

If f(x) is continuous at x = 3 and 5, then

∴ â€‹

On solving eqs. (1) and (2), we get

#### Question 38:

If . Show that f is continuous at x = 1.

Given:

We have
(LHL at x = 1) = $\underset{x\to {1}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(1-h\right)$$=\underset{h\to 0}{\mathrm{lim}}\frac{{\left(1-h\right)}^{2}}{2}=\frac{1}{2}$

(RHL at x = 1) = $\underset{x\to {1}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(1+h\right)$$=\underset{h\to 0}{\mathrm{lim}}\left[2{\left(1+h\right)}^{2}-3\left(1+h\right)+\frac{3}{2}\right]=2-3+\frac{3}{2}=\frac{1}{2}$

Also, $f\left(1\right)=\frac{{\left(1\right)}^{2}}{2}=\frac{1}{2}$

∴ â€‹

Hence, the given function is continuous at $x=1$.

#### Question 39:

Discuss the continuity of the f(x) at the indicated points:
(i) f(x) = | x | + | x − 1 | at x = 0, 1.
(ii) f(x) = | x − 1 | + | x + 1 | at x = −1, 1.

(i) Given: $f\left(x\right)=\left|x\right|+\left|x-1\right|$

We have
(LHL at x = 0) = $\underset{x\to {0}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(0-h\right)$$=\underset{h\to 0}{\mathrm{lim}}\left[\left|0-h\right|+\left|0-h-1\right|\right]=1$

(RHL at x = 0) = $\underset{x\to {0}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(0+h\right)$$=\underset{h\to 0}{\mathrm{lim}}\left[\left|0+h\right|+\left|0+h-1\right|\right]=1$

Also, $f\left(0\right)=\left|0\right|+\left|0-1\right|=0+1=1$

Now,

(LHL at x = 1) = $\underset{x\to {1}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(1-h\right)=\underset{h\to 0}{\mathrm{lim}}\left(\left|1-h\right|+\left|1-h-1\right|\right)=1+0=1$

(RHL at x =1) = $\underset{x\to {1}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(1+h\right)=\underset{h\to 0}{\mathrm{lim}}\left(\left|1+h\right|+\left|1+h-1\right|\right)=1+0=1$

Also, $f\left(1\right)=\left|1\right|+\left|1-1\right|=1+0=1$

∴ â€‹

Hence, $f\left(x\right)$ is continuous at .

(ii) Given: $f\left(x\right)=\left|x-1\right|+\left|x+1\right|$

We have
(LHL at x = −1) = $\underset{x\to -{1}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(-1-h\right)$$=\underset{h\to 0}{\mathrm{lim}}\left[\left|-1-h-1\right|+\left|-1-h+1\right|\right]=2+0=2$

(RHL at x = −1) = $\underset{x\to -{1}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(-1+h\right)$$=\underset{h\to 0}{\mathrm{lim}}\left[\left|-1+h-1\right|+\left|-1+h+1\right|\right]=2+0=2$

Also, $f\left(-1\right)=\left|-1-1\right|+\left|-1+1\right|=\left|-2\right|=2$

Now,

(LHL at x = 1) = $\underset{x\to {1}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(1-h\right)=\underset{h\to 0}{\mathrm{lim}}\left(\left|1-h-1\right|+\left|1-h+1\right|\right)=0+2=2$

(RHL at x =1) = $\underset{x\to {1}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(1+h\right)=\underset{h\to 0}{\mathrm{lim}}\left(\left|1+h-1\right|+\left|1+h+1\right|\right)=0+2=2$

Also, $f\left(1\right)=\left|1+1\right|+\left|1-1\right|=2$

∴ â€‹

Hence, $f\left(x\right)$ is continuous at .

#### Question 40:

Prove that is discontinuous at x = 0

The given function can be rewritten as

â€‹$⇒$

We have
(LHL at x = 0) = $\underset{x\to {0}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(0-h\right)=\underset{h\to 0}{\mathrm{lim}}f\left(-h\right)=\underset{h\to 0}{\mathrm{lim}}2=2$

(RHL at x = 0) = $\underset{x\to {0}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(0+h\right)=\underset{h\to 0}{\mathrm{lim}}f\left(h\right)=\underset{h\to 0}{\mathrm{lim}}0=0$

∴ â€‹$\underset{x\to {0}^{-}}{\mathrm{lim}}f\left(x\right)\ne \underset{x\to {0}^{+}}{\mathrm{lim}}f\left(x\right)$

Thus,  f(x) is discontinuous at x = 0
.

#### Question 41:

If , then what should be the value of
k so that f(x) is continuous at x = 0.

The given function can be rewritten as

â€‹
We have
(LHL at x = 0) = $\underset{x\to {0}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(0-h\right)=\underset{h\to 0}{\mathrm{lim}}f\left(-h\right)=\underset{h\to 0}{\mathrm{lim}}-2{\left(-h\right)}^{2}+k=k$

(RHL at x = 0) = $\underset{x\to {0}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(0+h\right)=\underset{h\to 0}{\mathrm{lim}}f\left(h\right)=\underset{h\to 0}{\mathrm{lim}}\left(2{h}^{2}+k\right)=k$

If $f\left(x\right)$ is continuous at $x=0$, then

k can be any real number.

#### Question 42:

For what value of λ is the function

continuous at x = 0? What about continuity at x = ± 1?

The given function f is

If f is continuous at x = 0, then

Therefore, there is no value of λ for which f(x) is continuous at x = 0.

At x = 1,

f (1) = 4x + 1 = 4 × 1 + 1 = 5

Therefore, for any values of λ, f is continuous at x = 1

At x = -1, we have

f (-1) = $\lambda \left(1+2\right)=3\lambda$

$\underset{x\to -1}{\mathrm{lim}}\lambda \left(1+2\right)=3\lambda \phantom{\rule{0ex}{0ex}}\therefore \underset{x\to -1}{\mathrm{lim}}f\left(x\right)=f\left(-1\right)$

Therefore, for any values of λf is continuous at x = -1

#### Question 43:

For what value of k is the following function continuous at x = 2?

Given:

We have
(LHL at x = 2) = $\underset{x\to {2}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(2-h\right)$$=\underset{h\to 0}{\mathrm{lim}}\left(2\left(2-h\right)+1\right)=5$

(RHL at x = 2) = $\underset{x\to {2}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(2+h\right)$$=\underset{h\to 0}{\mathrm{lim}}3\left(2+h\right)-1=5$

Also, $f\left(2\right)=k$

If f(x) is continuous at x = 2, then
â€‹

Hence, for k = 5, $f\left(x\right)$ is continuous at $x=2$.

#### Question 44:

Let If f(x) is continuous at x = $\frac{\mathrm{\pi }}{2}$, find a and b.

Given:

We have
(LHL at x = $\frac{\mathrm{\pi }}{2}$) = $\underset{x\to {\frac{\mathrm{\pi }}{2}}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(\frac{\mathrm{\pi }}{2}-h\right)$

$=\underset{h\to 0}{\mathrm{lim}}\left(\frac{1-{\mathrm{sin}}^{3}\left(\frac{\mathrm{\pi }}{2}-h\right)}{3{\mathrm{cos}}^{2}\left(\frac{\mathrm{\pi }}{2}-h\right)}\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\underset{h\to 0}{\mathrm{lim}}\left(\frac{1-{\mathrm{cos}}^{3}h}{3{\mathrm{sin}}^{2}h}\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\frac{1}{3}\underset{h\to 0}{\mathrm{lim}}\left(\frac{\left(1-\mathrm{cos}h\right)\left(1+{\mathrm{cos}}^{2}h+\mathrm{cos}h\right)}{\left(1-\mathrm{cos}h\right)\left(1+\mathrm{cos}h\right)}\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\frac{1}{3}\underset{h\to 0}{\mathrm{lim}}\left(\frac{\left(1+{\mathrm{cos}}^{2}h+\mathrm{cos}h\right)}{\left(1+\mathrm{cos}h\right)}\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\frac{1}{3}\left(\frac{1+1+1}{1+1}\right)=\frac{1}{2}$

(RHL at x = $\frac{\mathrm{\pi }}{2}$) = $\underset{x\to {\frac{\mathrm{\pi }}{2}}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(\frac{\mathrm{\pi }}{2}+h\right)$
$=\underset{h\to 0}{\mathrm{lim}}\left(\frac{b\left[1-\mathrm{sin}\left(\frac{\mathrm{\pi }}{2}+h\right)\right]}{{\left[\mathrm{\pi }-2\left(\frac{\mathrm{\pi }}{2}+h\right)\right]}^{2}}\right)\phantom{\rule{0ex}{0ex}}=\underset{h\to 0}{\mathrm{lim}}\left(\frac{b\left(1-\mathrm{cos}h\right)}{{\left[-2h\right]}^{2}}\right)\phantom{\rule{0ex}{0ex}}=\underset{h\to 0}{\mathrm{lim}}\left(\frac{2b{\mathrm{sin}}^{2}\frac{h}{2}}{4{h}^{2}}\right)\phantom{\rule{0ex}{0ex}}=\underset{h\to 0}{\mathrm{lim}}\left(\frac{2b{\mathrm{sin}}^{2}\frac{h}{2}}{16\frac{{h}^{2}}{4}}\right)\phantom{\rule{0ex}{0ex}}=\frac{b}{8}\underset{h\to 0}{\mathrm{lim}}{\left(\frac{\mathrm{sin}\frac{h}{2}}{\frac{h}{2}}\right)}^{2}\phantom{\rule{0ex}{0ex}}=\frac{b}{8}×1\phantom{\rule{0ex}{0ex}}=\frac{b}{8}$

Also, $f\left(\frac{\mathrm{\pi }}{2}\right)=a$

If f(x) is continuous at x = $\frac{\mathrm{\pi }}{2}$ , then

#### Question 45:

If the functions f(x), defined below is continuous at x = 0, find the value of k.

Given:

We have
(LHL at x = 0) = $\underset{x\to {0}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(0-h\right)$

$=\underset{h\to 0}{\mathrm{lim}}\left(\frac{1-\mathrm{cos}2\left(-h\right)}{2{\left(-h\right)}^{2}}\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\underset{h\to 0}{\mathrm{lim}}\left(\frac{1-\mathrm{cos}2h}{2{h}^{2}}\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\underset{h\to 0}{\mathrm{lim}}\left(\frac{2{\mathrm{sin}}^{2}h}{{h}^{2}}\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\frac{2}{2}\underset{h\to 0}{\mathrm{lim}}\left(\frac{{\mathrm{sin}}^{2}h}{{h}^{2}}\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\frac{2}{2}\underset{h\to 0}{\mathrm{lim}}{\left(\frac{\mathrm{sin}h}{h}\right)}^{2}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=1×1\phantom{\rule{0ex}{0ex}}=1$

(RHL at x = 0) = $\underset{x\to {0}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(0+h\right)=\underset{h\to 0}{\mathrm{lim}}f\left(h\right)$$=\underset{h\to 0}{\mathrm{lim}}\left(1\right)=1$
Also, $f\left(0\right)=k$

If f(x) is continuous at x = 0, then
â€‹

$⇒1=1=k$

Hence, the required value of k is 1.

#### Question 46:

Find the relationship between 'a' and 'b' so that the function 'f' defined by

is continuous at x = 3.

Given:

We have
(LHL at x = 3) = $\underset{x\to {3}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(3-h\right)=\underset{h\to 0}{\mathrm{lim}}a\left(3-h\right)+1=3a+1$

(RHL at x = 3) = $\underset{x\to {3}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(3+h\right)=\underset{h\to 0}{\mathrm{lim}}b\left(3+h\right)+3=3b+3$

Hence, the required relationship between .

#### Question 1:

Prove that the function is everywhere continuous.

When x < 0, we have
$f\left(x\right)=\frac{\mathrm{sin}x}{x}$

We know that sin x as well as the identity function x are everywhere continuous.
So, the quotient function $\frac{\mathrm{sin}x}{x}$ is continuous at each x < 0.

When x > 0, we have $f\left(x\right)=x+1$, which is a polynomial function.
Therefore, $f\left(x\right)$ is continuous at each x > 0.

Now,
Let us consider the point x = 0.

Given:

We have
(LHL at x = 0) = $\underset{x\to {0}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(0-h\right)=\underset{h\to 0}{\mathrm{lim}}f\left(-h\right)=\underset{h\to 0}{\mathrm{lim}}\left(\frac{\mathrm{sin}\left(-h\right)}{-h}\right)=\underset{h\to 0}{\mathrm{lim}}\left(\frac{\mathrm{sin}\left(h\right)}{h}\right)=1$

(RHL at x = 0) = $\underset{x\to {0}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(0+h\right)=\underset{h\to 0}{\mathrm{lim}}f\left(h\right)=\underset{h\to 0}{\mathrm{lim}}\left(h+1\right)=1$

Also,
$f\left(0\right)=0+1=1$

$\underset{x\to {0}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{x\to {0}^{+}}{\mathrm{lim}}f\left(x\right)=f\left(0\right)$

Thus, $f\left(x\right)$ is continuous at x = 0.

Hence, $f\left(x\right)$ is everywhere continuous.

#### Question 2:

Discuss the continuity of the function

Given:

We have
(LHL at x = 0) = $\underset{x\to {0}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(0-h\right)=\underset{h\to 0}{\mathrm{lim}}f\left(-h\right)=\underset{h\to 0}{\mathrm{lim}}\left(-1\right)=-1$

(RHL at x = 0) = $\underset{x\to {0}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(0+h\right)=\underset{h\to 0}{\mathrm{lim}}f\left(h\right)=\underset{h\to 0}{\mathrm{lim}}\left(1\right)=1$

$\underset{x\to {0}^{-}}{\mathrm{lim}}f\left(x\right)\ne \underset{x\to {0}^{+}}{\mathrm{lim}}f\left(x\right)$

Thus, $f\left(x\right)$ is discontinuous at x = 0.

#### Question 3:

Find the points of discontinuity, if any, of the following functions:
(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

(ix)

(x)

(xi)

(xii)

(xiii)

(i)

When x $\ne$1, then
$f\left(x\right)={x}^{3}-{x}^{2}+2x-2$

We know that a polynomial function is everywhere continuous.
So, $f\left(x\right)={x}^{3}-{x}^{2}+2x-2$ is continuous at each x $\ne$1.

At x = 1, we have

(LHL at x = 1) = $\underset{x\to {1}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(1-h\right)=\underset{h\to 0}{\mathrm{lim}}\left({\left(1-h\right)}^{3}-{\left(1-h\right)}^{2}+2\left(1-h\right)-2\right)=1-1+2-2=0$

(RHL at x = 1) = $\underset{x\to {1}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(1+h\right)=\underset{h\to 0}{\mathrm{lim}}\left({\left(1+h\right)}^{3}-{\left(1+h\right)}^{2}+2\left(1+h\right)-2\right)=1-1+2-2=0$

Also, $f\left(1\right)=4$
$\underset{x\to {1}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{x\to {1}^{+}}{\mathrm{lim}}f\left(x\right)\ne f\left(1\right)$

Thus, $f\left(x\right)$ is discontinuous at x = 1.

Hence, the only point of discontinuity for $f\left(x\right)$ is x = 1.

(ii)
Given:

When x $\ne$2, then
$f\left(x\right)=\frac{{x}^{4}-16}{x-2}=\frac{{x}^{4}-{2}^{4}}{x-2}=\frac{\left({x}^{2}+4\right)\left(x-2\right)\left(x+2\right)}{x-2}=\left({x}^{2}+4\right)\left(x+2\right)$

We know that a polynomial function is everywhere continuous.
Therefore, the functions are everywhere continuous.
So, the product function $\left({x}^{2}+4\right)\left(x+2\right)$ is everywhere continuous.
Thus, f(x) is continuous at every x $\ne$2.

At x = 2, we have

(LHL at x = 2) = $\underset{x\to {2}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(2-h\right)=\underset{h\to 0}{\mathrm{lim}}\left[{\left(2-h\right)}^{2}+4\right]\left(2-h+2\right)=8\left(4\right)=32$

(RHL at x = 2) = $\underset{x\to {2}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(2+h\right)=\underset{h\to 0}{\mathrm{lim}}\left[{\left(2+h\right)}^{2}+4\right]\left(2+h+2\right)=8\left(4\right)=32$

Also, $f\left(2\right)=16$

$\underset{x\to {2}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{x\to {2}^{+}}{\mathrm{lim}}f\left(x\right)\ne f\left(2\right)$

Thus, $f\left(x\right)$ is discontinuous at x = 2.

Hence, the only point of discontinuity for $f\left(x\right)$ is x = 2.

(iii)

When x < 0, then
$f\left(x\right)=\frac{\mathrm{sin}x}{x}$

We know that sin x as well as the identity function x are everywhere continuous.
So, the quotient function $\frac{\mathrm{sin}x}{x}$ is continuous at each x < 0.

When x > 0, then
$f\left(x\right)=2x+3$, which is a polynomial function.
Therefore, $f\left(x\right)$ is continuous at each x > 0.

Now,
Let us consider the point x = 0.
Given:

We have
(LHL at x = 0) = $\underset{x\to {0}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(0-h\right)=\underset{h\to 0}{\mathrm{lim}}f\left(-h\right)=\underset{h\to 0}{\mathrm{lim}}\left(\frac{\mathrm{sin}\left(-h\right)}{-h}\right)=\underset{h\to 0}{\mathrm{lim}}\left(\frac{\mathrm{sin}\left(h\right)}{h}\right)=1$

(RHL at x = 0) = $\underset{x\to {0}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(0+h\right)=\underset{h\to 0}{\mathrm{lim}}f\left(h\right)=\underset{h\to 0}{\mathrm{lim}}\left(2h+3\right)=3$

$\underset{x\to {0}^{-}}{\mathrm{lim}}f\left(x\right)\ne \underset{x\to {0}^{+}}{\mathrm{lim}}f\left(x\right)$

Thus, $f\left(x\right)$ is discontinuous at x = 0.

Hence, the only point of discontinuity for $f\left(x\right)$ is x = 0.

(iv)

When x $\ne$ 0, then
$f\left(x\right)=\frac{\mathrm{sin}3x}{x}$

We know that sin 3x as well as the identity function x are everywhere continuous.
So, the quotient function $\frac{\mathrm{sin}3x}{x}$ is continuous at each x $\ne$ 0.

Let us consider the point x = 0.
Given:

We have

(LHL at x = 0) =

(RHL at x = 0) =
Also, $f\left(0\right)=4$

$\underset{x\to {0}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{x\to {0}^{+}}{\mathrm{lim}}f\left(x\right)\ne f\left(0\right)$
Thus, $f\left(x\right)$ is discontinuous at x = 0.

Hence, the only point of discontinuity for $f\left(x\right)$ is x = 0.

(v)

When x $\ne$ 0, then
$f\left(x\right)=\frac{\mathrm{sin}x}{x}+\mathrm{cos}x$

We know that sin x as well as the identity function x both are everywhere continuous.
So, the quotient function $\frac{\mathrm{sin}x}{x}$ is continuous at each x $\ne$ 0.
Also, cos x is everywhere continuous.
Therefore, is continuous at each x $\ne$ 0.

Let us consider the point x = 0.
Given:

We have
(LHL at x = 0) = $\underset{x\to {0}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(0-h\right)=\underset{h\to 0}{\mathrm{lim}}f\left(-h\right)=\underset{h\to 0}{\mathrm{lim}}\left(\frac{\mathrm{sin}\left(-h\right)}{-h}+\mathrm{cos}\left(-h\right)\right)=\underset{h\to 0}{\mathrm{lim}}\left(\frac{\mathrm{sin}\left(-h\right)}{-h}\right)+\underset{h\to 0}{\mathrm{lim}}\mathrm{cos}\left(-h\right)=1+1=2$

(RHL at x = 0) = $\underset{x\to {0}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(0+h\right)=\underset{h\to 0}{\mathrm{lim}}f\left(h\right)=\underset{h\to 0}{\mathrm{lim}}\left(\frac{\mathrm{sin}\left(h\right)}{h}+\mathrm{cos}\left(h\right)\right)=\underset{h\to 0}{\mathrm{lim}}\left(\frac{\mathrm{sin}\left(h\right)}{h}\right)+\underset{h\to 0}{\mathrm{lim}}\mathrm{cos}\left(h\right)=1+1=2$
Also, $f\left(0\right)=5$

$\underset{x\to {0}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{x\to {0}^{+}}{\mathrm{lim}}f\left(x\right)\ne f\left(0\right)$
Thus, $f\left(x\right)$ is discontinuous at x = 0.

Hence, the only point of discontinuity for  $f\left(x\right)$ is x = 0.

(vi)

When x $\ne$ 0, then
$f\left(x\right)=\frac{{x}^{4}+{x}^{3}+2{x}^{2}}{{\mathrm{tan}}^{-1}x}$

We know that ${x}^{4}+{x}^{3}+2{x}^{2}$ is a polynomial function which is everywhere continuous.
Also, ${\mathrm{tan}}^{-1}x$ is everywhere continuous.
So, the quotient function $\frac{{x}^{4}+{x}^{3}+2{x}^{2}}{{\mathrm{tan}}^{-1}x}$ is continuous at each x $\ne$ 0.

Let us consider the point x = 0.

Given:

We have
(LHL at x = 0) = $\underset{x\to {0}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(0-h\right)=\underset{h\to 0}{\mathrm{lim}}f\left(-h\right)=\underset{h\to 0}{\mathrm{lim}}\left(\frac{{\left(-h\right)}^{4}+{\left(-h\right)}^{3}+2{\left(-h\right)}^{2}}{{\mathrm{tan}}^{-1}\left(-h\right)}\right)=\underset{h\to 0}{\mathrm{lim}}\left(\frac{{\left(h\right)}^{3}-{\left(h\right)}^{2}+2\left(h\right)}{-\frac{{\mathrm{tan}}^{-1}\left(h\right)}{h}}\right)=\frac{0}{\left(-1\right)}=0$

(RHL at x = 0) = $\underset{x\to {0}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(0+h\right)=\underset{h\to 0}{\mathrm{lim}}f\left(h\right)=\underset{h\to 0}{\mathrm{lim}}\left(\frac{{\left(h\right)}^{4}+{\left(h\right)}^{3}+2{\left(h\right)}^{2}}{{\mathrm{tan}}^{-1}\left(h\right)}\right)=\underset{h\to 0}{\mathrm{lim}}\left(\frac{{\left(h\right)}^{3}+{\left(h\right)}^{2}+2\left(h\right)}{\frac{{\mathrm{tan}}^{-1}\left(h\right)}{h}}\right)=\frac{0}{1}=0$
Also, $f\left(0\right)=10$

$\underset{x\to {0}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{x\to {0}^{+}}{\mathrm{lim}}f\left(x\right)\ne f\left(0\right)$
Thus, $f\left(x\right)$ is discontinuous at x = 0.

Hence, the only point of discontinuity for  $f\left(x\right)$ is x = 0.

(vii)
Given:

We have

It is given that $f\left(0\right)=7$

$\underset{x\to 0}{\mathrm{lim}}f\left(x\right)\ne f\left(0\right)$

Hence, the given function is discontinuous at x = 0 and continuous elsewhere.

(viii)

When x > 1, then
$f\left(x\right)=\left|x-3\right|$

Since modulus function is a continuous function, $f\left(x\right)$ is continuous for each x > 1.

When x < 1, then
$f\left(x\right)=\frac{{x}^{2}}{4}-\frac{3x}{2}+\frac{13}{4}$

Since, are continuous being polynomial functions, will also be continuous.
Also, $\frac{13}{4}$ is continuous being a polynomial function.

$⇒\frac{{x}^{2}}{4}-\frac{3x}{2}+\frac{13}{4}$ is continuous for each $x<1$.

$⇒f\left(x\right)$ is continuous for each x < 1.

At x = 1, we have
(LHL at x=1) =

(RHL at x=1) =
Also, $f\left(1\right)=\left|1-3\right|=\left|-2\right|=2$

Thus,

Hence, $f\left(x\right)$ is continuous at x= 1.

Thus, the given function is nowhere discontinuous.

(ix)

At $x\le -3$, we have
$f\left(x\right)=\left|x\right|+3$

Since modulus function and constant function are continuous, $f\left(x\right)=\left|x\right|+3$ is continuous for each $x\le -3$.

At $-3, we have
$f\left(x\right)=-2x$
Since polynomial function is continuous and constant function is continuous, $f\left(x\right)=-2x$ is continuous for each$-3.

At $x>3$, we have
$f\left(x\right)=6x+2$

Since polynomial function is continuous and constant function is continuous, $f\left(x\right)=6x+2$ is continuous for each $x>3$.

Now, we check the continuity of the function at the point $x=3$.

We have
(LHL at x=3) = $\underset{x\to {3}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(3-h\right)=\underset{h\to 0}{\mathrm{lim}}-2\left(3-h\right)=-6$

(RHL at x=3) = $\underset{x\to {3}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(3+h\right)=\underset{h\to 0}{\mathrm{lim}}6\left(3+h\right)+2=20$
$⇒\underset{x\to {3}^{-}}{\mathrm{lim}}f\left(x\right)\ne \underset{x\to {3}^{+}}{\mathrm{lim}}f\left(x\right)$

Hence, the only point of discontinuity of the given function is $x=3$

(x)
Given:

The given function f is defined at all the points of the real line.

Let c be a point on the real line.

Case I:

Therefore, f is continuous at all points x, such that x < 1

Case II:

If c = 1, then the left hand limit of f at x = 1 is,

The right hand limit of f at x = 1 is,

It is observed that the left and right hand limit of f at x = 1 do not coincide.

Therefore, f is not continuous at x = 1

Case III:

Therefore, f is continuous at all points x, such that x > 1

Thus, from the above observation, it can be concluded that x = 1 is the only point of discontinuity of f.

(xi) The given function is

The given function is defined at all points of the real line.

Let c be a point on the real line.

Case I:

Therefore, f is continuous at all points x, such that x < 0

Case II:

The left hand limit of f at x = 0 is,

The right hand limit of f at x = 0 is,

Therefore, f is continuous at x = 0

Case III:

Therefore, f is continuous at all points of the interval (0, 1).

Case IV:

The left hand limit of f at x = 1 is,

The right hand limit of f at x = 1 is,

It is observed that the left and right hand limits of f at x = 1 do not coincide.

Therefore, f is not continuous at x = 1

Case V:

Therefore, f is continuous at all points x, such that x > 1

Hence, f is not continuous only at x = 1

(xii)

The given function f is

It is evident that f is defined at all points of the real line.

Let c be a real number.

Case I:

Therefore, f is continuous at all points x, such that x 0

Case II:

Therefore, f is continuous at x = 0

From the above observations, it can be concluded that f is continuous at every point of the real line.

Thus, f is a continuous function.

(xiii)
The given function
f is

The given function is defined at all points of the real line.

Let c be a point on the real line.

Case I:

Therefore, f is continuous at all points x, such that x < −1

Case II:

The left hand limit of f at x = −1 is,

The right hand limit of f at x = −1 is,

Therefore, f is continuous at x = −1

Case III:

Therefore, f is continuous at all points of the interval (−1, 1).

Case IV:

The left hand limit of f at x = 1 is,

The right hand limit of f at x = 1 is,

Therefore, f is continuous at x = 2

Case V:

Therefore, f is continuous at all points x, such that x > 1

Thus, from the above observations, it can be concluded that f is continuous at all points of the real line.

#### Question 4:

In the following, determine the value(s) of constant(s) involved in the definition so that the given function is continuous:
(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

(i) Given:

If $f\left(x\right)$ is continuous at x = 0, then
$\underset{x\to 0}{\mathrm{lim}}f\left(x\right)=f\left(0\right)$

(ii) Given:

If $f\left(x\right)$ is continuous at x = 2, then
$\underset{x\to {2}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{x\to {2}^{+}}{\mathrm{lim}}f\left(x\right)$

$⇒\underset{h\to 0}{\mathrm{lim}}f\left(2-h\right)=\underset{h\to 0}{\mathrm{lim}}f\left(2+h\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}⇒\underset{h\to 0}{\mathrm{lim}}\left(k\left(2-h\right)+5\right)=\underset{h\to 0}{\mathrm{lim}}\left(2+h-1\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}⇒2k+5=1\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}⇒2k=-4\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}⇒k=-2$

(iii) Given:

If $f\left(x\right)$ is continuous at x = 0, then
$\underset{x\to {0}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{x\to {0}^{+}}{\mathrm{lim}}f\left(x\right)$

Hence, there does not exist any value of k, which can make the given function continuous.

(iv) Given:

If $f\left(x\right)$ is continuous at x = 3 and 5, then

(v)

Given:

If $f\left(x\right)$ is continuous at x = −1 and 0, then

(vi)

Given:

If $f\left(x\right)$ is continuous at x = 0, then
$\underset{x\to {0}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{x\to {0}^{+}}{\mathrm{lim}}f\left(x\right)$

(vii)

Given:

If $f\left(x\right)$ is continuous at x = 2 and 10,  then

(viii)

Given:

If $f\left(x\right)$ is continuous at x = $\frac{\mathrm{\pi }}{2}$, then
$\underset{x\to {\frac{\mathrm{\pi }}{2}}^{-}}{\mathrm{lim}}f\left(x\right)=f\left(\frac{\mathrm{\pi }}{2}\right)$

#### Question 5:

The function
is continuous on (0, ∞), then find the most suitable values of a and b.

Given:  f is continuous on $\left(0,\infty \right)$

∴  f is continuous at x = 1 and $\sqrt{2}$

At x = 1, we have

$\underset{x\to {1}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(1+h\right)=\underset{h\to 0}{\mathrm{lim}}\left(a\right)=a$

Also,

At x = $\sqrt{2}$, we have

f is continuous at x = 1 and $\sqrt{2}$

If a = 1, then

If a = −1, then

Hence, the most suitable values of a and b are

a = −1, b = 1  or a = 1, $b=1±\sqrt{2}$

#### Question 6:

Find the values of a and b so that the function f(x) defined by

becomes continuous on [0, π].

Given: f is continuous on .

f is continuous at x = $\frac{\mathrm{\pi }}{4}$ and $\frac{\mathrm{\pi }}{2}$

At x = $\frac{\mathrm{\pi }}{4}$, we have

At x = $\frac{\mathrm{\pi }}{2}$, we have

Since f is continuous at x = $\frac{\mathrm{\pi }}{4}$ and x = $\frac{\mathrm{\pi }}{2}$, we get

#### Question 7:

The function f(x) is defined as follows:

If f is continuous on [0, 8], find the values of a and b.

Given: f is continuous on .

f is continuous at x = 2 and x = 4

At x = 2, we have

Also,
At x = 4, we have

f is continuous at x = 2 and x = 4

On simplifying eqs. (1) and (2), we get

#### Question 8:

If for x ≠ π/4, find the value which can be assigned to f(x) at x = π/4 so that the function f(x) becomes continuous every where in [0, π/2].

When $x\ne \frac{\mathrm{\pi }}{4}$,  and  are continuous in .

Thus, the quotient function  is continuous in $\left[0,\frac{\mathrm{\pi }}{2}\right]$ for each $x\ne \frac{\mathrm{\pi }}{4}$.

So, if $f\left(x\right)$ is continuous at $x=\frac{\mathrm{\pi }}{4}$, then it will be everywhere continuous in .

Now,
Let us consider the point x = $\frac{\mathrm{\pi }}{4}$.

Given:

We have
(LHL at x = $\frac{\mathrm{\pi }}{4}$) =

(RHL at x = $\frac{\mathrm{\pi }}{4}$) =

If $f\left(x\right)$ is continuous at $x=\frac{\mathrm{\pi }}{4}$, then

∴  $f\left(\frac{\mathrm{\pi }}{4}\right)=\frac{1}{2}$

Hence, for â€‹$f\left(\frac{\mathrm{\pi }}{4}\right)=\frac{1}{2}$, the function $f\left(x\right)$ will be everywhere continuous in â€‹.

#### Question 9:

Discuss the continuity of the function

When x < 2, we have

We know that a polynomial function is everywhere continuous.
So, $f\left(x\right)$ is continuous for each x < 2.

When $x>2$, we have
$f\left(x\right)=\frac{3x}{2}$

The functions 3x and 2 are continuous being the polynomial and constant function, respectively.
Thus, the quotient function $\frac{3x}{2}$ is continuous at each x > 2.

Now,
Let us consider the point x = 2.
Given:

We have
(LHL at x = 2) =

(RHL at x = 2) =

Also,
$f\left(2\right)=\frac{3\left(2\right)}{2}=3$

∴ $\underset{x\to {2}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{x\to {2}^{+}}{\mathrm{lim}}f\left(x\right)=f\left(2\right)$

Thus, $f\left(x\right)$ is continuous at x = 2.

Hence, $f\left(x\right)$ is everywhere continuous.

#### Question 10:

Discuss the continuity of f(x) = sin | x |.

This function f is defined for every real number and f can be written as the composition of two functions as,

f = h o g, where

It has to be proved first that are continuous functions.

Clearly, g is defined for all real numbers.

Let c be a real number.

Case I:

Therefore, g is continuous at all points x < 0

Case II:

Therefore, g is continuous at all points x > 0

Case III:

Therefore, g is continuous at x = 0

From the above three observations, it can be concluded that g is continuous at all points.

Now, h (x) = sin x

It is evident that h (x) = sin x is defined for every real number.

Let c be a real number.
Put
x = c + k

If x c, then k 0

h (c) = sin c

So, h is a continuous function.

is a continuous function.

#### Question 11:

Prove that

is everywhere continuous.

When x < 0, we have
$f\left(x\right)=\frac{\mathrm{sin}x}{x}$

We know that sin x as well as the identity function x are everywhere continuous.
So, the quotient function $\frac{\mathrm{sin}x}{x}$ is continuous at each x < 0.

When x > 0, we have
$f\left(x\right)=x+1$, which is a polynomial function.
Therefore, $f\left(x\right)$ is continuous at each x > 0.

Now,
Let us consider the point x = 0.
Given:

We have
(LHL at x = 0) =

(RHL at x = 0) =
Also,
$f\left(0\right)=0+1=1$

$\underset{x\to {0}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{x\to {0}^{+}}{\mathrm{lim}}f\left(x\right)=f\left(0\right)$

Thus, $f\left(x\right)$ is continuous at x = 0.

Hence, $f\left(x\right)$ is everywhere continuous.

#### Question 12:

Show that the function g (x) = x − [x] is discontinuous at all integral points. Here [x] denotes the greatest integer function.

Given:

It is evident that g is defined at all integral points.

Let $n\in Z$.

Then,

The left hand limit of f at x = n is,

The right hand limit of f at x = n is,

It is observed that the left and right hand limits of f at x = n do not coincide.
i.e.

So, f is not continuous at x = n, $n\in Z$

Hence, g is discontinuous at all integral points.

#### Question 13:

Discuss the continuity of the following functions:
(i) f(x) = sin x + cos x
(ii) f(x) = sin x − cos x
(iii) f(x) = sin x cos x

It is known that if g and h are two continuous functions, then are also continuous.

It has to proved first that g (x) = sin x and h (x) = cos x are continuous functions.

Let g (x) = sin x

It is evident that g (x) = sin x is defined for every real number.

Let c be a real number. Put x = c + h

If x c, then h 0

So, g is a continuous function.

Let h (x) = cos x

It is evident that h (x) = cos x is defined for every real number.

Let c be a real number. Put x = c + h

If x c, then h 0

h (c) = cos c

So, h is a continuous function.

Therefore, it can be concluded that

(i) f (x) = g (x) + h (x) = sin x + cos x is a continuous function.

(ii) f (x) = g (x) − h (x) = sin x − cos x is a continuous function.

(iii) f (x) = g (x$×$ h (x) = sin x cos x is a continuous function.

#### Question 14:

Show that f (x) = cos x2 is a continuous function.

Given: f (x) = cos (x2)

This function f is defined for every real number and f can be written as the composition of two functions as

f = g o h, where g (x) = cos x and h (x) = x2

It has to be first proved that g (x) = cos x and h (x) = x2 are continuous functions.

It is evident that g is defined for every real number.

Let c be a real number.

Then, g (c) = cos c

So, g (x) = cos x is a continuous function.

Now,
h
(x) = x2

Clearly, h is defined for every real number.

Let k be a real number, then h (k) = k2

So, h is a continuous function.

It is known that for real valued functions g and h, such that (g o h) is defined at x = c, if g is continuous at x = c and if f is continuous at g (c), then, (f o g) is continuous at x = c.

Therefore, is a continuous function.

#### Question 15:

Show that f (x) = | cos x | is a continuous function.

The given function is

This function f is defined for every real number and f can be written as the composition of two functions as,

f = g o h, where

It has to be first proved that are continuous functions.

Clearly, g is defined for all real numbers.

Let c be a real number.

Case I:

So, g is continuous at all points x < 0.

Case II:

So, g is continuous at all points x > 0.

Case III:

So, g is continuous at x = 0

From the above three observations, it can be concluded that g is continuous at all points.

Now, h (x) = cos x

It is evident that h (x) = cos x is defined for every real number.

Let c be a real number.
Put
x = c + h

If x c, then h 0

h (c) = cos c

So, h (x) = cos x is a continuous function.

It is known that for real valued functions g and h,such that (g o h) is defined at x = c, if g is continuous at x = c and if f is continuous at g (c), then (f o g) is continuous at x = c.

Therefore, is a continuous function.

#### Question 16:

Find all the points of discontinuity of f defined by f (x) = | x | − | x + 1 |.

Given:

The two functions, g and h, are defined as

Then, f = g h

The continuity of g and h is examined first.

Clearly, g is defined for all real numbers.

Let c be a real number.

Case I:

So, g is continuous at all points x < 0.

Case II:

So, g is continuous at all points x > 0.

Case III:

So, g is continuous at x = 0

From the above three observations, it can be concluded that g is continuous at all points.

Clearly, h is defined for every real number.

Let c be a real number.

Case I:

So, h is continuous at all points x < −1.

Case II:

So, h is continuous at all points x > −1.

Case III:

So, h is continuous at x = −1

From the above three observations, it can be concluded that h is continuous at all points of the real line.

So, g and h are continuous functions.

Thus,
f = gh is also a continuous function.

Therefore, f has no point of discontinuity.

#### Question 17:

Determine if is a continuous function?

The given function f is

It is evident that f is defined at all points of the real line.

Let c be a real number.

Case I:

So, f is continuous at all points x 0

Case II:

So, f is continuous at x = 0

From the above observations, it can be concluded that f is continuous at every point of the real line.

Thus, f is a continuous function.

#### Question 18:

Given the function $f\left(x\right)=\frac{1}{x+2}$. Find the points of discontinuity of the function f(f(x)).

$f\left[f\left(x\right)\right]=\frac{1}{\frac{1}{x+2}+2}=\frac{x+2}{2x+5}$

Hence, the function is discontinuous at

#### Question 19:

Find all point of discontinuity of the function

Hence, the function is discontinuous at

#### Question 1:

The function $f\left(x\right)=\frac{4-{x}^{2}}{4x-{x}^{3}}$
(a) discontinuous at only one point
(b) discontinuous exactly at two points
(c) discontinuous exactly at three points
(d) none of these

(C) discontinuous exactly at three points

Given: $f\left(x\right)=\frac{4-{x}^{2}}{4x-{x}^{3}}$
$⇒f\left(x\right)=\frac{4-{x}^{2}}{x\left(4-{x}^{2}\right)}$

Clearly, $f\left(x\right)$ is defined and continuous for all real numbers except .

Therefore, $f\left(x\right)$ is discontinuous exactly at three points.

#### Question 2:

If f (x) = | xa | Ï• (x), where Ï• (x) is continuous function, then
(a) f' (a+) = Ï• (a)
(b) f' (a) = −Ï• (a)
(c) f' (a+) = f' (a)
(d) none of these

(a)  $f\text{'}\left({a}^{+}\right)=\varphi \left(a\right)$
(b)  $f\text{'}\left({a}^{-}\right)=-\varphi \left(a\right)$

Here,

Also,

#### Question 3:

If , then at x = 1
(a) f (x) is continuous and f' (1+) = log10e
(b) f (x) is continuous and f' (1+) = log10e
(c) f (x) is continuous and f' (1) = log10e
(d) f (x) is continuous and f' (1) = −log10e

(a) f (x) is continuous and $f\text{'}$ (1+) = ${\mathrm{log}}_{10}e$
(d) f (x) is continuous and $f\text{'}$(1) = ${\mathrm{log}}_{10}e$

Given:

Also,

#### Question 4:

If is continuous at x = 0, then k equals
(a) $16\sqrt{2}$ log 2 log 3
(b) $16\sqrt{2}$ ln 6
(c) $16\sqrt{2}$ ln 2 ln 3
(d) none of these

Given:
If $f\left(x\right)$ is continuous at $x=0$, then
$\underset{x\to 0}{\mathrm{lim}}f\left(x\right)=f\left(0\right)$

#### Question 5:

If f (x) defined by then f (x) is continuous for all
(a) x
(b) x except at x = 0
(c) x except at x = 1
(d) x except at x = 0 and x = 1.

(d) x except at x = 0 and x = 1.

Given:

So,

$\underset{x\to {0}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(-h\right)=1$

Also,

$\underset{x\to {0}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(h\right)=-1$

$⇒\underset{x\to {0}^{+}}{\mathrm{lim}}f\left(x\right)\ne \underset{x\to {0}^{-}}{\mathrm{lim}}f\left(x\right)$

Thus, $f\left(x\right)$ is discontinuous at $x=0$.

Now,

$\underset{x\to {1}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(1-h\right)=-1$

$\underset{x\to {1}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(1+h\right)=1$

$⇒\underset{x\to {1}^{+}}{\mathrm{lim}}f\left(x\right)\ne \underset{x\to {1}^{-}}{\mathrm{lim}}f\left(x\right)$

So, $f\left(x\right)$ is discontinuous at $x=1$.

Hence, $f\left(x\right)$ is continuous for all except at $x=0$ and x = 1.

#### Question 6:

If
is continuous at x = π/2, then k =
(a) $-\frac{1}{16}$

(b) $-\frac{1}{32}$

(c) $-\frac{1}{64}$

(d) $-\frac{1}{28}$

If $f\left(x\right)$ is continuous at $x=\frac{\mathrm{\pi }}{2}$, then

#### Question 7:

If f (x) = (x + 1)cot x be continuous at x = 0, then f (0) is equal to
(a) 0
(b) 1/e
(c) e
(d) none of these

(c) e

Suppose $f\left(x\right)$ is continuous at $x=0.$

Given: $f\left(x\right)={\left(x+1\right)}^{\mathrm{cot}x}$

#### Question 8:

If
and f (x) is continuous at x = 0, then the value of k is
(a) ab
(b) a + b
(c) log a + log b
(d) none of these

Given:

If f(x) is continuous at x = 0, then
$\underset{x\to 0}{\mathrm{lim}}f\left(x\right)=f\left(0\right)$

$⇒\underset{x\to 0}{\mathrm{lim}}\left(\frac{\mathrm{log}\left(1+ax\right)-\mathrm{log}\left(1-bx\right)}{x}\right)=k$

#### Question 9:

The function
(a) is continuous at x = 0
(b) is not continuous at x = 0
(c) is not continuous at x = 0, but can be made continuous at x = 0
(d) none of these

(b) is not continuous at x = 0

Given:

We have
$\underset{x\to 0}{\mathrm{lim}}f\left(x\right)=\underset{x\to 0}{\mathrm{lim}}\left(\frac{{e}^{\frac{1}{x}}-1}{{e}^{\frac{1}{x}}+1}\right)$

If ${e}^{\frac{1}{x}}=t$, then

$\underset{x\to 0}{\mathrm{lim}}f\left(x\right)=\underset{t\to \infty }{\mathrm{lim}}\left(\frac{t-1}{t+1}\right)=\underset{t\to \infty }{\mathrm{lim}}\left(\frac{1-\frac{1}{t}}{1+\frac{1}{t}}\right)=\frac{1-0}{1+0}=1$
Also, $f\left(0\right)=0$

Hence, $f\left(x\right)$ is discontinuous at $x=0$.

#### Question 10:

Let
Then, f (x) is continuous at x = 4 when
(a) a = 0, b = 0
(b) a = 1, b = 1
(c) a = −1, b = 1
(d) a = 1, b = −1.

(d) a = 1, b = −1.

Given:

We have
(LHL at x = 4) = $\underset{x\to {4}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(4-h\right)$

$=\underset{h\to 0}{\mathrm{lim}}\left(\frac{4-h-4}{\left|4-h-4\right|}+a\right)=\underset{h\to 0}{\mathrm{lim}}\left(\frac{-h}{\left|-h\right|}+a\right)=a-1$

(RHL at x = 4) = $\underset{x\to {4}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(4+h\right)$

$=\underset{h\to 0}{\mathrm{lim}}\left(\frac{4+h-4}{\left|4+h-4\right|}+b\right)=\underset{h\to 0}{\mathrm{lim}}\left(\frac{h}{\left|h\right|}+b\right)=b+1$

Also,
$f\left(4\right)=a+b$

If f(x) is continuous at x = 4, then

$\underset{x\to {4}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{x\to {4}^{+}}{\mathrm{lim}}f\left(x\right)=f\left(4\right)$

$⇒a-1=b+1=a+b$

#### Question 11:

If the function is continuous at x = 0, then the value of k is
(a) 0
(b) 1
(c) −1
(d) e.

(b)
Given:

If$f\left(x\right)$ is continuous at $x=0$, then
$\underset{x\to 0}{\mathrm{lim}}f\left(x\right)=f\left(0\right)$

#### Question 12:

Let f (x) = | x | + | x − 1|, then
(a) f (x) is continuous at x = 0, as well as at x = 1
(b) f (x) is continuous at x = 0, but not at x = 1
(c) f (x) is continuous at x = 1, but not at x = 0
(d) none of these

(a) f (x) is continuous at x = 0, as well as at x = 1

Since modulus function is everywhere continuous ,  are also everywhere continuous.

Also,
It is known that if f and g are continuous functions, then g will also be continuous.

Thus, â€‹ â€‹$\left|x\right|+\left|x-1\right|$ is everywhere continuous.

Hence, $f\left(x\right)$ is continuous at .

#### Question 13:

Let . Then, f (x) is continuous on the set
(a) R
(b) R − {1}
(c) R − {2}
(d) R − {1, 2}

(d) R − {1, 2}

Given:

So,

$\underset{x\to {1}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(1-h\right)=\underset{h\to 0}{\mathrm{lim}}\left(1-h+1\right)\left(1-h+2\right)=2×3=6$

$\underset{x\to {1}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(1+h\right)=-\underset{h\to 0}{\mathrm{lim}}\left(1+h+1\right)\left(1+h+2\right)=-2×3=-6$

Also,

$\underset{x\to {2}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(2-h\right)=-\underset{h\to 0}{\mathrm{lim}}\left(2-h+1\right)\left(2-h+2\right)=-12$

$\underset{x\to {2}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(2+h\right)=\underset{h\to 0}{\mathrm{lim}}\left(2+h+1\right)\left(2+h+2\right)=12$

Thus,

Therefore, the only points of discontinuities of the function $f\left(x\right)$ are  .

Hence, the given function is continuous on the set  R − {1, 2}
.
.

#### Question 14:

If is continuous at x = 0, then
(a) a = $-\frac{3}{2}$, b = 0, c = $\frac{1}{2}$
(b) a = $-\frac{3}{2}$, b = 1, c = $-\frac{1}{2}$
(c) a = $-\frac{3}{2}$, bR − {0}, c = $\frac{1}{2}$
(d) none of these

(c) a = $\frac{-3}{2}$, bR − {0}, c$\frac{1}{2}$

The given function can be rewritten as

We have
(LHL at x = 0) =

(RHL at x = 0) = $\underset{x\to {0}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(0+h\right)=\underset{h\to 0}{\mathrm{lim}}f\left(h\right)$

=$\underset{h\to 0}{\mathrm{lim}}\left(\frac{\sqrt{1+bh}-1}{bh}\right)=\underset{h\to 0}{\mathrm{lim}}\left(\frac{bh}{bh\left(\sqrt{1+bh}+1\right)}\right)=\underset{h\to 0}{\mathrm{lim}}\left(\frac{1}{\left(\sqrt{1+bh}+1\right)}\right)=\frac{1}{2}$

Also, $f\left(0\right)=c$

If $f\left(x\right)$ is continuous at x = 0, then
â€‹$\underset{x\to {0}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{x\to {0}^{+}}{\mathrm{lim}}f\left(x\right)=f\left(0\right)$

$⇒a=\frac{-3}{2}$

Now, $\frac{\sqrt{1+bx}-1}{bx}$ exists only if $bx\ne 0⇒b\ne 0$.

Thus, $b\in R-\left\{0\right\}$.

#### Question 15:

If is continuous at $x=\frac{\mathrm{\pi }}{2}$, then
(a) m = 1, n = 0

(b) $m=\frac{n\mathrm{\pi }}{2}+1$

(c) $n=\frac{m\mathrm{\pi }}{2}$

(d) $m=n=\frac{\mathrm{\pi }}{2}$

(c) $n=\frac{\mathrm{m}\mathrm{\pi }}{2}$n=mπ2

Here,

$f\left(\frac{\mathrm{\pi }}{2}\right)=\frac{\mathrm{m\pi }}{2}+1$

We have
(LHL at $x=\frac{\mathrm{\pi }}{2}$) =

(RHL at $x=\frac{\mathrm{\pi }}{2}$) =

Thus,
If $f\left(x\right)$ is continuous at $x=\frac{\mathrm{\pi }}{2}$, then

$⇒\frac{\mathrm{m\pi }}{2}+1=n+1\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}⇒\frac{\mathrm{m\pi }}{2}=n$

#### Question 16:

The value of f (0), so that the function
$f\left(x\right)=\frac{\sqrt{{a}^{2}-ax+{x}^{2}}-\sqrt{{a}^{2}+ax+{x}^{2}}}{\sqrt{a+x}-\sqrt{a-x}}$ becomes continuous for all x, given by
(a) a3/2
(b) a1/2
(c) −a1/2
(d) −a3/2

(c) $-{a}^{\frac{1}{2}}$

Given: $f\left(x\right)=\frac{\sqrt{{a}^{2}-ax+{x}^{2}}-\sqrt{{a}^{2}+ax+{x}^{2}}}{\sqrt{a+x}-\sqrt{a-x}}$

If $f\left(x\right)$ is continuous for all x, then it will be continuous at x = 0 as well.

So, if $f\left(x\right)$ is continuous at x = 0, then
$\underset{x\to 0}{\mathrm{lim}}f\left(x\right)=f\left(0\right)$

#### Question 17:

The function

(a) is discontinuous at finitely many points
(b) is continuous everywhere
(c) is discontinuous only at $x=±\frac{1}{n}$, nZ − {0} and x = 0
(d) none of these

Given:

Case 1:

Here,
$f\left(x\right)=1$, which is the constant function
So, $f\left(x\right)$ is continuous for all

Case 2:

Here,
, which is also a constant function.

So, $f\left(x\right)$ is continuous for all

Case 3: Consider the points x = -1 and x = 1.

We have

Similarly,  f(x) is discontinuous at x = 1.

Case 4: Consider the point x = 0.

We have
$\underset{x\to {0}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(\frac{1}{n}-h\right)=\underset{h\to 0}{\mathrm{lim}}f\left(\frac{1}{n}-h\right)={\left(\frac{1}{n-1}\right)}^{2}$

$\underset{x\to {0}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(\frac{1}{n}+h\right)=\underset{h\to 0}{\mathrm{lim}}f\left(\frac{1}{n}+h\right)={\left(\frac{1}{n}\right)}^{2}$

$\underset{x\to {0}^{+}}{\mathrm{lim}}f\left(x\right)\ne \underset{x\to {0}^{-}}{\mathrm{lim}}f\left(x\right)$

Thus, $f\left(x\right)$ is discontinuous at $x=0$.

At = 0, we have
$\underset{x\to {0}^{-}}{\mathrm{lim}}f\left(x\right)\ne 0=f\left(0\right)$

So, $f\left(x\right)$ is discontinuous at $x=0$.

Case 5: Consider the point

We have
$\underset{x\to {\frac{1}{n}}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(\frac{1}{n}-h\right)=\underset{h\to 0}{\mathrm{lim}}f\left(\frac{1}{n}-h\right)={\left(\frac{1}{n-1}\right)}^{2}$

$\underset{x\to {\frac{1}{n}}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(\frac{1}{n}+h\right)=\underset{h\to 0}{\mathrm{lim}}f\left(\frac{1}{n}+h\right)={\left(\frac{1}{n}\right)}^{2}$

$\underset{x\to {\frac{1}{\mathrm{n}}}^{+}}{\mathrm{lim}}f\left(x\right)\ne \underset{x\to {\frac{1}{\mathrm{n}}}^{-}}{\mathrm{lim}}f\left(x\right)$

Hence, $f\left(x\right)$ is discontinuous only at  $x=±\frac{1}{n}$.

#### Question 18:

The value of f (0), so that the function
is continuous, is given by
(a) $\frac{2}{3}$
(b) 6
(c) 2
(d) 4

(c) 2

For f(x) to be continuous at x = 0, we must have

#### Question 19:

The value of f (0) so that the function
$f\left(x\right)=\frac{2-{\left(256-7x\right)}^{1/8}}{{\left(5x+32\right)}^{1/5}-2},$x ≠ 0 is continuous everywhere, is given by
(a) −1
(b) 1
(c) 26
(d) none of these

(d) none of these

Given:  $f\left(x\right)=\frac{2-{\left(256-7x\right)}^{\frac{1}{8}}}{{\left(5x+32\right)}^{\frac{1}{5}}-2}$

For $f\left(x\right)$ to be continuous at x = 0, we must have
$\underset{\mathrm{x}\to 0}{\mathrm{lim}}f\left(x\right)=f\left(0\right)$
$⇒f\left(0\right)=\underset{x\to 0}{\mathrm{lim}}f\left(x\right)=\underset{x\to 0}{\mathrm{lim}}\frac{2-{\left(256-7x\right)}^{\frac{1}{8}}}{{\left(5x+32\right)}^{\frac{1}{5}}-2}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}⇒f\left(0\right)=\underset{x\to 0}{\mathrm{lim}}\frac{{256}^{\frac{1}{8}}-{\left(256-7x\right)}^{\frac{1}{8}}}{{\left(5x+32\right)}^{\frac{1}{5}}-{32}^{\frac{1}{5}}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=-\underset{x\to 0}{\mathrm{lim}}\frac{\frac{\left[{\left(256-7x\right)}^{\frac{1}{8}}-{256}^{\frac{1}{8}}\right]}{x}}{\frac{\left[{\left(5x+32\right)}^{\frac{1}{5}}-{32}^{\frac{1}{5}}\right]}{x}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\frac{-7}{5}\underset{x\to 0}{\mathrm{lim}}\frac{\frac{\left[{\left(256-7x\right)}^{\frac{1}{8}}-{256}^{\frac{1}{8}}\right]}{7x}}{\frac{\left[{\left(5x+32\right)}^{\frac{1}{5}}-{32}^{\frac{1}{5}}\right]}{5x}}\phantom{\rule{0ex}{0ex}}=\frac{7}{5}\underset{x\to 0}{\mathrm{lim}}\frac{\frac{\left[{\left(256-7x\right)}^{\frac{1}{8}}-{256}^{\frac{1}{8}}\right]}{\left(256-7x\right)-256}}{\frac{\left[{\left(5x+32\right)}^{\frac{1}{5}}-{32}^{\frac{1}{5}}\right]}{5x+32-32}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\frac{7}{5}×\frac{\frac{1}{8}×{\left(256\right)}^{-\frac{7}{8}}}{\frac{1}{5}×{\left(32\right)}^{\frac{-4}{5}}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\frac{7}{5}×\frac{\frac{1}{8}×{2}^{4}}{\frac{1}{5}×{2}^{7}}\phantom{\rule{0ex}{0ex}}=\frac{7}{64}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

#### Question 20:

is continuous in the interval [−1, 1], then p is equal to
(a) −1
(b) −1/2
(c) 1/2
(d) 1

(b) $-\frac{1}{2}$
Given:

If $f\left(x\right)$ is continuous at x = 0,  then
$\underset{x\to {0}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{x\to {0}^{+}}{\mathrm{lim}}f\left(x\right)$

#### Question 21:

The function
is continuous for 0 ≤ x < ∞, then the most suitable values of a and b are
(a) a = 1, b = −1
(b) a = −1, b = 1 + $\sqrt{2}$
(c) a = −1, b = 1
(d) none of these

(c) a = -1, b = 1

Given: $f\left(x\right)$ is continuous for 0 ≤ x < ∞.

This means that $f\left(x\right)$ is continuous for
.

Now,

If $f\left(x\right)$ is continuous at x = 1, then
$\underset{x\to {1}^{-}}{\mathrm{lim}}f\left(x\right)=f\left(1\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}⇒\underset{h\to 0}{\mathrm{lim}}f\left(1-h\right)=a\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}⇒\frac{{\left(1-h\right)}^{2}}{a}=a\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{a}=a\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}⇒{a}^{2}=1\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}⇒a=±1$

If $f\left(x\right)$ is continuous at x = $\sqrt{2}$, thenâ€‹

$\underset{x\to {\sqrt{2}}^{-}}{\mathrm{lim}}f\left(x\right)=f\left(\sqrt{2}\right)\phantom{\rule{0ex}{0ex}}⇒\underset{h\to 0}{\mathrm{lim}}f\left(\sqrt{2}-h\right)=\frac{2{b}^{2}-4b}{2}\phantom{\rule{0ex}{0ex}}⇒\underset{h\to 0}{\mathrm{lim}}a={b}^{2}-2b\phantom{\rule{0ex}{0ex}}⇒a={b}^{2}-2b\phantom{\rule{0ex}{0ex}}⇒{b}^{2}-2b-a=0$

∴ For a = 1, we have

${b}^{2}-2b-1=0\phantom{\rule{0ex}{0ex}}⇒b=\frac{2±\sqrt{4-4\left(-1\right)}}{2}=1±\sqrt{2}$

Also,
For a = −1, we have

${b}^{2}-2b+1=0\phantom{\rule{0ex}{0ex}}⇒{\left(b-1\right)}^{2}=0\phantom{\rule{0ex}{0ex}}⇒b=1$

Thus,

#### Question 22:

If when x ≠ π/2 and f (π/2) = λ, then f (x) will be continuous function at x = π/2, where λ =
(a) 1/8
(b) 1/4
(c) 1/2
(d) none of these

(a) $\frac{1}{8}$

If $f\left(x\right)$ is continuous at $x=\frac{\mathrm{\pi }}{2}$, then
$\underset{\mathrm{x}\to \frac{\mathrm{\pi }}{2}}{\mathrm{lim}}f\left(x\right)=f\left(\frac{\mathrm{\pi }}{2}\right)$

...(1)

Suppose $\left(\frac{\mathrm{\pi }}{2}-x\right)=t$, then

#### Question 23:

The value of a for which the function
may be continuous at x = 0 is
(a) 1
(b) 2
(c) 3
(d) none of these

(d) none of these

For f(x) to be continuous at $x=0$, we must have
$\underset{\mathrm{x}\to 0}{\mathrm{lim}}f\left(x\right)=f\left(0\right)$

#### Question 24:

The function f (x) = tan x is discontinuous on the set
(a) {n π : nZ}
(b) {2n π : nZ}
(c)
(d)

When, it is not defined at the integral points. $\left[n\in Z\right]$

Hence, $f\left(x\right)$ is discontinuous on the set .

#### Question 25:

The function is continuous at x = 0, then k =
(a) 3
(b) 6
(c) 9
(d) 12

(b) 6

Given:

If $f\left(x\right)$ is continuous at $x=0$, then
$\underset{x\to 0}{\mathrm{lim}}f\left(x\right)=f\left(0\right)$

#### Question 26:

If the function

is continuous at each point of its domain, then the value of f (0) is
(a) 2

(b) $\frac{1}{3}$

(c) $-\frac{1}{3}$

(d) $\frac{2}{3}$

(b) $\frac{1}{3}$

Given:  $f\left(x\right)=\frac{2x-{\mathrm{sin}}^{-1}x}{2x+{\mathrm{tan}}^{-1}x}$

If f(x) is continuous at x = 0, then

â€‹$\underset{x\to 0}{\mathrm{lim}}f\left(x\right)=f\left(0\right)\phantom{\rule{0ex}{0ex}}⇒\underset{x\to 0}{\mathrm{lim}}\frac{2x-{\mathrm{sin}}^{-1}x}{2x+{\mathrm{tan}}^{-1}x}=f\left(0\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}⇒\underset{x\to 0}{\mathrm{lim}}\frac{x\left(2-\frac{{\mathrm{sin}}^{-1}x}{x}\right)}{x\left(2+\frac{{\mathrm{tan}}^{-1}x}{x}\right)}=f\left(0\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}⇒\underset{x\to 0}{\mathrm{lim}}\frac{\left(2-\frac{{\mathrm{sin}}^{-1}x}{x}\right)}{\left(2+\frac{{\mathrm{tan}}^{-1}x}{x}\right)}=f\left(0\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}⇒\frac{2-\underset{x\to 0}{\mathrm{lim}}\left(\frac{{\mathrm{sin}}^{-1}x}{x}\right)}{2+\underset{x\to 0}{\mathrm{lim}}\left(\frac{{\mathrm{tan}}^{-1}x}{x}\right)}=f\left(0\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}⇒\frac{2-1}{2+1}=f\left(0\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}⇒f\left(0\right)=\frac{1}{3}$

#### Question 27:

The value of b for which the function
is continuous at every point of its domain, is
(a) −1

(b) 0

(c) $\frac{13}{3}$

(d) 1

(a) −1

Given: $f\left(x\right)$ is continuous at every point of its domain. So, it is continuous at $x=1$.

$⇒\underset{x\to {1}^{+}}{\mathrm{lim}}f\left(x\right)=f\left(1\right)\phantom{\rule{0ex}{0ex}}⇒\underset{h\to 0}{\mathrm{lim}}f\left(1+h\right)=f\left(1\right)\phantom{\rule{0ex}{0ex}}⇒\underset{h\to 0}{\mathrm{lim}}\left(4{\left(1+h\right)}^{2}+3b\left(1+h\right)\right)=5\left(1\right)-4\phantom{\rule{0ex}{0ex}}⇒4+3b=1\phantom{\rule{0ex}{0ex}}⇒-3=3b\phantom{\rule{0ex}{0ex}}⇒b=-1$

#### Question 28:

If $f\left(x\right)=\frac{1}{1-x}$, then the set of points discontinuity of the function f (f(f(x))) is
(a) {1}
(b) {0, 1}
(c) {−1, 1}
(d) none of these

(b) {0, 1}

Given: $f\left(x\right)=\frac{1}{1-x}$

Clearly, $f:R-\left\{1\right\}\to R$

Now,
$f\left(f\left(x\right)\right)=f\left(\frac{1}{1-x}\right)=\left(\frac{1}{1-\left(\frac{1}{1-x}\right)}\right)=\left(\frac{1-x}{-x}\right)=\left(\frac{x-1}{x}\right)$
$fof:$

Now,

$f\left(f\left(f\left(x\right)\right)\right)=f\left(\frac{x-1}{x}\right)=\left(\frac{1}{1-\left(\frac{x-1}{x}\right)}\right)=x$

$fofof:$

Thus, â€‹$f\left(f\left(f\left(x\right)\right)\right)$ is not defined at .

Hence, â€‹$f\left(f\left(f\left(x\right)\right)\right)$ is discontinuous at {0, 1}.

#### Question 29:

Let The value which should be assigned to f (x) at $x=\frac{\mathrm{\pi }}{4},$ so that it is continuous everywhere is
(a) 1
(b) 1/2
(c) 2
(d) none of these

(b) $\frac{1}{2}$

If $f\left(x\right)$ is continuous at $x=\frac{\mathrm{\pi }}{4}$, then
$\underset{\mathrm{x}\to \frac{\mathrm{\pi }}{4}}{\mathrm{lim}}f\left(x\right)=f\left(\frac{\mathrm{\pi }}{4}\right)$

If $\frac{\mathrm{\pi }}{4}-x=y$, then .

#### Question 30:

The function $f\left(x\right)=\frac{{x}^{3}+{x}^{2}-16x+20}{x-2}$ is not defined for x = 2. In order to make f (x) continuous at x = 2, Here f (2) should be defined as
(a) 0
(b) 1
(c) 2
(d) 3

Here,

So, the given function can be rewritten as

$f\left(x\right)=\frac{{\left(x-2\right)}^{2}\left(x+5\right)}{x-2}$

$⇒f\left(x\right)=\left(x-2\right)\left(x+5\right)$

If $f\left(x\right)$ is continuous at $x=2$, then
$\underset{x\to 2}{\mathrm{lim}}f\left(x\right)=f\left(2\right)$

$⇒\underset{x\to 2}{\mathrm{lim}}\left(x-2\right)\left(x+5\right)=f\left(2\right)\phantom{\rule{0ex}{0ex}}⇒f\left(2\right)=0$

Hence, in order to make $f\left(x\right)$ continuous at  should be defined as 0.

#### Question 31:

If is continuous at x = 0, then a equals
(a) $\frac{1}{2}$

(b) $\frac{1}{3}$

(c) $\frac{1}{4}$

(d) $\frac{1}{6}$

(a) $\frac{1}{2}$

Given:
We have

(LHL at x = 0) =

(RHL at x = 0) =

#### Question 32:

If
then the value of (a, b) for which f (x) cannot be continuous at x = 1, is
(a) (2, 2)
(b) (3, 1)
(c) (4, 0)
(d) (5, 2)

(d) (5, 2)

If f(x) is continuous at x = 1, then
$\underset{\mathrm{x}\to {1}^{-}}{\mathrm{lim}}f\left(x\right)=f\left(1\right)$

Thus, the possible values of (a, b) can be . But .

Hence, for $f\left(x\right)$ cannot be continuous at = 1.

Disclaimer: The question in the book has some error. The solution here is created according to the question given in the book.

#### Question 33:

If the function f (x) defined by
is continuous at x = 0, then k =
(a) 1
(b) 5
(c) −1
(d) none of these

Given:

If f(x) is continuous at x = 0, thenâ€‹$\underset{x\to 0}{\mathrm{lim}}f\left(x\right)=f\left(0\right)$.

$⇒\underset{x\to 0}{\mathrm{lim}}\left(\frac{\mathrm{log}\left(1+3x\right)-\mathrm{log}\left(1-2x\right)}{x}\right)=k$

#### Question 34:

If
then the value of a so that f (x) may be continuous at x = 0, is
(a) 25
(b) 50
(c) −25
(d) none of these

(b) 50

If $f\left(x\right)$ is continuous at $x=0$, then
$\underset{\mathrm{x}\to {0}^{-}}{\mathrm{lim}}f\left(x\right)=f\left(0\right)$

$⇒\underset{h\to 0}{\mathrm{lim}}f\left(-h\right)=f\left(0\right)$

#### Question 35:

If then the value of the function at x = 0, so that the function is continuous at x = 0, is
(a) 0
(b) −1
(c) 1
(d) indeterminate

(a) 0

Given:

Here,

If f(x) is continuous at x = 0, then$\underset{x\to 0}{\mathrm{lim}}f\left(x\right)=f\left(0\right)$.

$⇒f\left(0\right)=0$

#### Question 36:

The value of k which makes
continuous at x = 0, is
(a) 8
(b) 1
(c) −1
(d) none of these

(d) none of these

If $f\left(x\right)$ is continuous at $x=0$, then

#### Question 37:

The values of the constants a, b and c for which the function
may be continuous at x = 0, are

(a)

(b)

(c)

(d) none of these

If $f\left(x\right)$ is continuous at $x=0$, then

Also,

#### Question 38:

The points of discontinuity of the function

(a) x = 1, $x=\frac{5}{2}$

(b) $x=\frac{5}{2}$

(c) $x=1,\frac{5}{2},4$

(d) x = 0, 4

=5

If $0\le x\le 1$, then  $f\left(x\right)=2\sqrt{x}$.

Since $f\left(x\right)=2\sqrt{x}$ is a polynomial function, it is continuous.
Thus, $f\left(x\right)$ is continuous for every $0\le x\le 1$.

If $1, then  $f\left(x\right)=4-2x$. Since $2x$ is a polynomial function and 4 is a constant function, both of them are continuous. So, their difference will also be continuous.
Thus, $f\left(x\right)$ is continuous for every $1.

If $\frac{5}{2}\le x\le 4$, then  $f\left(x\right)=2x-7$. Since $2x$ is a polynomial function and 7 is continuous function, their difference will also be continuous.
Thus, $f\left(x\right)$ is continuous for every $\frac{5}{2}\le x\le 4$.

Now,
Consider the point $x=1$. Here,

$\underset{\mathrm{x}\to {1}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(1-h\right)=\underset{h\to 0}{\mathrm{lim}}\left(2\left(\sqrt{1-h}\right)\right)=2$

$\underset{\mathrm{x}\to {1}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(1+h\right)=\underset{h\to 0}{\mathrm{lim}}\left(4-2\left(1+h\right)\right)=2$

Also, $f\left(1\right)=2\sqrt{1}=2$

$⇒\underset{\mathrm{x}\to {1}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{\mathrm{x}\to {1}^{+}}{\mathrm{lim}}f\left(x\right)=f\left(1\right)$

Thus, .

Now,
Consider the point $x=\frac{5}{2}$. Here,

$\underset{\mathrm{x}\to {\frac{5}{2}}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(\frac{5}{2}-h\right)=\underset{h\to 0}{\mathrm{lim}}\left(4-2\left(\frac{5}{2}-h\right)\right)=-1$

$\underset{\mathrm{x}\to {\frac{5}{2}}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(\frac{5}{2}+h\right)=\underset{h\to 0}{\mathrm{lim}}\left(2\left(\frac{5}{2}-h\right)-7\right)=-2$

Thus, .

#### Question 39:

If . Then, f (x) is continuous at $x=\frac{\mathrm{\pi }}{2}$, if
(a) $a=\frac{1}{3},$b = 2

(b)

(c)

(d) none of these

(b)

Given:

We have
(LHL at x = $\frac{\mathrm{\pi }}{2}$) = $\underset{x\to {\frac{\mathrm{\pi }}{2}}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(\frac{\mathrm{\pi }}{2}-h\right)$

(RHL at x = $\frac{\mathrm{\pi }}{2}$) = $\underset{x\to {\frac{\mathrm{\pi }}{2}}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(\frac{\mathrm{\pi }}{2}+h\right)$

Also, $f\left(\frac{\mathrm{\pi }}{2}\right)=a$

If f(x) is continuous at x = $\frac{\mathrm{\pi }}{2}$, then

â€‹

#### Question 40:

The points of discontinuity of the function

(a) x = 1
(b) x = 3
(c) x = 1, 3
(d) none of these

(b) x = 3

If $x\le 1$, then $f\left(x\right)=\frac{1}{5}\left(2{x}^{2}+3\right)$.
Since $2{x}^{2}+3$ is a polynomial function and $\frac{1}{5}$ is a constant function, both of them are continuous. So, their product will also be continuous.
Thus, $f\left(x\right)$ is continuous at $x\le 1$.

If $1, then $f\left(x\right)=6-5x$.

Since $5x$ is a polynomial function and $6$ is a constant function, both of them are continuous. So, their difference will also be continuous.
Thus, $f\left(x\right)$ is continuous for every $1.

If $x\ge 3$, then $f\left(x\right)=x-3$.
Since $x-3$ is a polynomial function, it is continuous. So, $f\left(x\right)$ is continuous for every $x\ge 3$.

Now,
Consider the point $x=1$. Here,

$\underset{\mathrm{x}\to {1}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(1-h\right)=\underset{h\to 0}{\mathrm{lim}}\left(\frac{1}{5}\left[2{\left(1-h\right)}^{2}+3\right]\right)=1$

$\underset{\mathrm{x}\to {1}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(1+h\right)=\underset{h\to 0}{\mathrm{lim}}\left(6-5\left(1+h\right)\right)=1$

Also,
$f\left(1\right)=\frac{1}{5}\left(2{\left(1\right)}^{2}+3\right)=1$

Thus,
$\underset{\mathrm{x}\to {1}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{\mathrm{x}\to {1}^{+}}{\mathrm{lim}}f\left(x\right)=f\left(1\right)$

Hence, $f\left(x\right)$ is continuous at $x=1$.

Now,
Consider the point $x=3$. Here,

$\underset{\mathrm{x}\to {3}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(3-h\right)=\underset{h\to 0}{\mathrm{lim}}\left(6-5\left(3-h\right)\right)=-9$

$\underset{\mathrm{x}\to {3}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(3+h\right)=\underset{h\to 0}{\mathrm{lim}}\left(\left(3+h\right)-3\right)=0$

Also,
$f\left(1\right)=\frac{1}{5}\left(2{\left(1\right)}^{2}+3\right)=1$

Thus,
$\underset{\mathrm{x}\to {3}^{-}}{\mathrm{lim}}f\left(x\right)\ne \underset{\mathrm{x}\to {3}^{+}}{\mathrm{lim}}f\left(x\right)$

Hence, $f\left(x\right)$ is discontinuous at $x=3$.

So, the only point of discontinuity of $f\left(x\right)$ is $x=3$.

#### Question 41:

The value of a for which the function
is continuous at every point of its domain, is

(a) $\frac{13}{3}$
(b) 1
(c) 0
(d) −1

(d) −1

Given:

If $f\left(x\right)$ is continuous in its domain, then it will be continuous at $x=1$.

Now,
$\underset{x\to {1}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(1-h\right)=\underset{h\to 0}{\mathrm{lim}}\left[5\left(1-h\right)-4\right]=5-4=1\phantom{\rule{0ex}{0ex}}\underset{x\to {1}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(1+h\right)=\underset{h\to 0}{\mathrm{lim}}\left[4{\left(1+h\right)}^{2}+3a\left(1+h\right)\right]=4+3a$

Since f(x) is continuous at x = 1,
$\underset{x\to {1}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{x\to {1}^{+}}{\mathrm{lim}}f\left(x\right)$

$⇒4+3a=1\phantom{\rule{0ex}{0ex}}⇒3a=-3\phantom{\rule{0ex}{0ex}}⇒a=-1$

#### Question 42:

If is continuous at x = π/2, then k is equal to
(a) 0
(b) $\frac{1}{2}$
(c) 1
(d) −1

(a) 0

Given:

If f(x) is continuous at $x=\frac{\mathrm{\pi }}{2}$, then
$\underset{x\mathit{\to }\frac{\mathrm{\pi }}{2}}{\mathrm{lim}}f\left(x\right)=f\left(\frac{\mathrm{\pi }}{2}\right)$

Now,
$\frac{\mathrm{\pi }}{2}-x=y$
$⇒\mathrm{\pi }-2x=2y$

Also,

$⇒\underset{y\mathit{\to }0}{\mathrm{lim}}\frac{\mathrm{sin}\left(\mathrm{cos}\left(\frac{\mathrm{\pi }}{2}-y\right)\right)-\mathrm{cos}\left(\frac{\mathrm{\pi }}{2}-y\right)}{4{y}^{2}}=k$

â€‹

#### Question 43:

If f(x) = 2x and $g\left(x\right)=\frac{{x}^{2}}{2}+1,$ then which of the following can be a discontinuous function

(a) f(x) + g(x)

(b) f(x) – g(x)

(c) f(x) g(x)

(d) $\frac{g\left(x\right)}{f\left(x\right)}$

f(x) = 2x and g(x) = $\frac{{x}^{2}}{2}+1$ are polynomial functions. We know polynomial functions are continuous for all values of x

So, f(x) = 2x and g(x) = $\frac{{x}^{2}}{2}+1$ are continuous functions.

Also, sum, difference and product of continuous functions is continuous functions.

∴ f(x) + g(x), f(x) − g(x) and f(x)g(x) are continuous functions.

Now, $\frac{g\left(x\right)}{f\left(x\right)}$ is continuous if f(x) ≠ 0.

But, f(x) = 2x = 0 when x = 0. So, $\frac{g\left(x\right)}{f\left(x\right)}$ is discontinuous at x = 0.

Thus, $\frac{g\left(x\right)}{f\left(x\right)}$ can be a discontinuous function.

Hence, the correct answer is option (d).

#### Question 44:

The function f(x) = cot x is discontinuous on the set

(a) {x : x = nπ, n ∈ Z}

(b) {x : x = 2nπ, n ∈ Z}

(C)

(d)

f(x) = cotx$\frac{\mathrm{cos}x}{\mathrm{sin}x}$

Now, f(x) is discontinuous when sinx = 0.

sinx = 0

⇒ xn$\mathrm{\pi }$n ∈ Z

So, f(x) = cotx is discontinuous on the set {x : x = nπ, n ∈ Z}

Hence, the correct answer is option (a).

#### Question 45:

If $f\left(x\right)={x}^{2}\mathrm{sin}\frac{1}{x},$ where x ≠ 0, then the value of the function f at x = 0, so that the function is continuous at x = 0, is

(a) 0
(b) –1
(c) 1
(d) none

The given function is $f\left(x\right)={x}^{2}\mathrm{sin}\frac{1}{x},$ where x ≠ 0.

Now, f(x) is continuous at x = 0.

$\therefore f\left(0\right)=\underset{x\to 0}{\mathrm{lim}}f\left(x\right)$

$⇒f\left(0\right)=\underset{x\to 0}{\mathrm{lim}}{x}^{2}\mathrm{sin}\frac{1}{x}$

$⇒f\left(0\right)=\underset{x\to 0}{\mathrm{lim}}{x}^{2}×\underset{x\to 0}{\mathrm{lim}}\mathrm{sin}\frac{1}{x}$

⇒ f(0) = 0 × a finite value between − 1 and 1                 $\left(-1\le \mathrm{sin}\frac{1}{x}\le 1\right)$

⇒ f(0) = 0

Thus, the value of the function f at x = 0 so that the function is continuous at x = 0 is 0.

Hence, the correct answer is option (a).

#### Question 46:

The number of points at which the function $f\left(x\right)=\frac{1}{x-\left[x\right]}$ is not continuous is
(a) 1
(b) 2
(c) 3
(d) none of these

The function $f\left(x\right)=\frac{1}{x-\left[x\right]}$ is discontinuous when x − [x] = 0.

x − [x] = 0

⇒ x = [x

⇒ x is an integer

So, f(x) is not continuous for all x ∈ Z.

Thus, the function $f\left(x\right)=\frac{1}{x-\left[x\right]}$ is not continuous at infinitely many points.

Hence, the correct answer is option (d).

#### Question 47:

The function f(x) = [x] is continuous at
(a) 4
(b) –2
(c) 1
(d) 1.5

The graph of f(x) = [x] is shown below.

It can be seen that, the function f(x) = [x] is discontinuous at all integral values of x. It is continuous at all points except the integer points.

Thus, the function f(x) = [x] is continuous at x = 1.5 and discontinuous at x = 4, x = –2 and x = 1.

Hence, the correct answer is option (d).

#### Question 48:

The function is continuous at x = 0, then the value of k is
(a) 3
(b) 2
(c) 1
(d) 1.5

It is given that, the function  is continuous at x = 0.

$\therefore f\left(0\right)=\underset{x\to 0}{\mathrm{lim}}f\left(x\right)$

$⇒k=\underset{x\to 0}{\mathrm{lim}}\left(\frac{\mathrm{sin}x}{x}+\mathrm{cos}x\right)$

$⇒k=\underset{x\to 0}{\mathrm{lim}}\frac{\mathrm{sin}x}{x}+\underset{x\to 0}{\mathrm{lim}}\mathrm{cos}x$

$⇒k=1+1=2$

Thus, the value of k is 2.

Hence, the correct answer is option (b).

#### Question 1:

If $f\left(x\right)=\left\{\begin{array}{cc}\frac{{x}^{3}-{a}^{3}}{x-a},& x\ne a\\ b,& x=a\end{array}\right\$ is continuous at x = a, then b = ______________.

It is given that, the function $f\left(x\right)=\left\{\begin{array}{cc}\frac{{x}^{3}-{a}^{3}}{x-a},& x\ne a\\ b,& x=a\end{array}\right\$is continuous at xa.

$\therefore f\left(a\right)=\underset{x\to a}{\mathrm{lim}}f\left(x\right)$

$⇒b=\underset{x\to a}{\mathrm{lim}}\frac{{x}^{3}-{a}^{3}}{x-a}$

If $f\left(x\right)=\left\{\begin{array}{cc}\frac{{x}^{3}-{a}^{3}}{x-a},& x\ne a\\ b,& x=a\end{array}\right\$ is continuous at x = a, then b = .

#### Question 2:

If the function is continuous at x = 0, then a = _____________.

The function $f\left(x\right)=\left\{\begin{array}{cc}\frac{{\mathrm{sin}}^{2}ax}{{x}^{2}},& x\ne 0\\ 1,& x=0\end{array}\right\$ is continuous at x = 0.

$\therefore \underset{x\to 0}{\mathrm{lim}}f\left(x\right)=f\left(0\right)$

$⇒\underset{x\to 0}{\mathrm{lim}}\frac{{\mathrm{sin}}^{2}ax}{{x}^{2}}=1$

$⇒{a}^{2}{\left(\underset{x\to 0}{\mathrm{lim}}\frac{\mathrm{sin}ax}{ax}\right)}^{2}=1$

$⇒{a}^{2}×{\left(1\right)}^{2}=1$             $\left(\underset{x\to 0}{\mathrm{lim}}\frac{\mathrm{sin}x}{x}=1\right)$

$⇒{a}^{2}=1$

$⇒a=±1$

Thus, the value of a is ±1.

If the function $f\left(x\right)=\left\{\begin{array}{cc}\frac{{\mathrm{sin}}^{2}ax}{{x}^{2}},& x\ne 0\\ 1,& x=0\end{array}\right\$is continuous at x = 0, then = ___±1___.

#### Question 3:

If the function $f\left(x\right)=\left\{\begin{array}{cc}a{x}^{2}-b,& 0\le x<1\\ 2,& x=1\\ x+1,& 1is continuous at x = 1, then a – b = _____________.

The function $f\left(x\right)=\left\{\begin{array}{cc}a{x}^{2}-b,& 0\le x<1\\ 2,& x=1\\ x+1,& 1 is continuous at x = 1.

$\therefore f\left(1\right)=\underset{x\to 1}{\mathrm{lim}}f\left(x\right)$

$⇒f\left(1\right)=\underset{x\to {1}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{x\to {1}^{+}}{\mathrm{lim}}f\left(x\right)$         .....(1)

Now,

f(1) = 2          ....(2)

$\underset{x\to {1}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{x\to 1}{\mathrm{lim}}\left(a{x}^{2}-b\right)=a×{\left(1\right)}^{2}-b=a-b$       .....(3)

$\underset{x\to {1}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{x\to 1}{\mathrm{lim}}\left(x+1\right)=1+1=2$          .....(4)

From (1), (2), (3) and (4), we have

2 = a − b = 2

∴ a − b = 2

Thus, the value of a − b is 2.

If the function $f\left(x\right)=\left\{\begin{array}{cc}a{x}^{2}-b,& 0\le x<1\\ 2,& x=1\\ x+1,& 1is continuous at x = 1, then a – b = ____2____.

#### Question 4:

If $f\left(x\right)=\left\{\begin{array}{cc}x+k,& x<3\\ 4,& x=3\\ 3x-5,& x>3\end{array}\right\$ is continuous at x = 3, then k = _____________.

The function $f\left(x\right)=\left\{\begin{array}{cc}x+k,& x<3\\ 4,& x=3\\ 3x-5,& x>3\end{array}\right\$ is continuous at x = 3.

$\therefore f\left(3\right)=\underset{x\to 3}{\mathrm{lim}}f\left(x\right)$

$⇒f\left(3\right)=\underset{x\to {3}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{x\to {3}^{+}}{\mathrm{lim}}f\left(x\right)$     .....(1)

Now,

f(3) = 4            .....(2)

$\underset{x\to {3}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{x\to 3}{\mathrm{lim}}\left(x+k\right)=3+k$          .....(3)

$\underset{x\to {3}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{x\to 3}{\mathrm{lim}}\left(3x-5\right)=3×3-5=9-5=4$         .....(4)

From (1), (2), (3) and (4), we have

4 = 3 + k = 4

⇒ 3 + k = 4

⇒ k = 4 − 3 = 1

Thus, the value of k is 1.

If $f\left(x\right)=\left\{\begin{array}{cc}x+k,& x<3\\ 4,& x=3\\ 3x-5,& x>3\end{array}\right\$ is continuous at x = 3, then k = ___1___.

#### Question 5:

If $f\left(x\right)=\left\{\begin{array}{cc}\frac{x-4}{\left|x-4\right|}+a,& x<4\\ a+b,& x=4\\ \frac{x-4}{\left|x-4\right|}+b,& x>4\end{array}\right\$. Then f(x) is continuous at x = 4, then a + b = _____________.

The function $f\left(x\right)=\left\{\begin{array}{cc}\frac{x-4}{\left|x-4\right|}+a,& x<4\\ a+b,& x=4\\ \frac{x-4}{\left|x-4\right|}+b,& x>4\end{array}\right\$ is continuous at x = 4.

$\therefore f\left(4\right)=\underset{x\to 4}{\mathrm{lim}}f\left(x\right)$

$⇒f\left(4\right)=\underset{x\to {4}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{x\to {4}^{+}}{\mathrm{lim}}f\left(x\right)$     .....(1)

$\left|x-4\right|=\left\{\begin{array}{ll}-\left(x-4\right),& x<4\\ x-4,& x\ge 4\end{array}\right\$

Now,

f(4) = ab       .....(2)

$\underset{x\to {4}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{x\to 4}{\mathrm{lim}}\left[\frac{x-4}{-\left(x-4\right)}+a\right]=\underset{x\to 4}{\mathrm{lim}}\frac{x-4}{-\left(x-4\right)}+\underset{x\to 4}{\mathrm{lim}}a=-1+a$        .....(3)

$\underset{x\to {4}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{x\to 4}{\mathrm{lim}}\left[\frac{x-4}{\left(x-4\right)}+b\right]=\underset{x\to 4}{\mathrm{lim}}\frac{x-4}{\left(x-4\right)}+\underset{x\to 4}{\mathrm{lim}}b=1+b$      .....(4)

From (1), (2), (3) and (4), we get

ab = −1 + a = 1 + b

So,

a + b = −1 +

⇒ b = −1

Also,

a + b = 1 + b

⇒ a = 1

∴ ab = 1 + (−1) = 0

Thus, the value of ab is 0.

If $f\left(x\right)=\left\{\begin{array}{cc}\frac{x-4}{\left|x-4\right|}+a,& x<4\\ a+b,& x=4\\ \frac{x-4}{\left|x-4\right|}+b,& x>4\end{array}\right\$. Then f(x) is continuous at x = 4, then a + b = ___0___.

#### Question 6:

If f : RR defined by is continuous at x = 0, then λ = _____________.

The function $f\left(x\right)=\left\{\begin{array}{cc}\frac{\mathrm{cos}3x-\mathrm{cos}x}{{x}^{2}},& x\ne 0\\ \lambda ,& x=0\end{array}\right\$ is continuous at x = 0.

$\therefore f\left(0\right)=\underset{x\to 0}{\mathrm{lim}}f\left(x\right)\phantom{\rule{0ex}{0ex}}$

$⇒\lambda =\underset{x\to 0}{\mathrm{lim}}\frac{\mathrm{cos}3x-\mathrm{cos}x}{{x}^{2}}\phantom{\rule{0ex}{0ex}}$

$⇒\lambda =\underset{x\to 0}{\mathrm{lim}}\frac{-2\mathrm{sin}\left(\frac{3x+x}{2}\right)\mathrm{sin}\left(\frac{3x-x}{2}\right)}{{x}^{2}}\phantom{\rule{0ex}{0ex}}$

$⇒\lambda =\underset{x\to 0}{\mathrm{lim}}\frac{-2\mathrm{sin}2x\mathrm{sin}x}{{x}^{2}}\phantom{\rule{0ex}{0ex}}$

$⇒\lambda =-2×\underset{x\to 0}{\mathrm{lim}}\frac{\mathrm{sin}2x}{x}×\underset{x\to 0}{\mathrm{lim}}\frac{\mathrm{sin}x}{x}\phantom{\rule{0ex}{0ex}}$

$⇒\lambda =-2×2\underset{x\to 0}{\mathrm{lim}}\frac{\mathrm{sin}2x}{2x}×\underset{x\to 0}{\mathrm{lim}}\frac{\mathrm{sin}x}{x}\phantom{\rule{0ex}{0ex}}$

$⇒\lambda =-4$

Thus, the value of λ is −4.

If f : R → R defined by $f\left(x\right)=\left\{\begin{array}{cc}\frac{\mathrm{cos}3x-\mathrm{cos}x}{{x}^{2}},& x\ne 0\\ \lambda ,& x=0\end{array}\right\$ is continuous at x = 0, then λ = ____−4____.

#### Question 7:

If $f\left(x\right)=\left\{\begin{array}{cc}\frac{1-\mathrm{sin}x}{\mathrm{\pi }-2x},& x\ne \frac{\mathrm{\pi }}{2}\\ k,& x=\frac{\mathrm{\pi }}{2}\end{array}\right\$is continuous at $x=\frac{\mathrm{\pi }}{2},$ then k = _______________.

The function $f\left(x\right)=\left\{\begin{array}{cc}\frac{1-\mathrm{sin}x}{\mathrm{\pi }-2x},& x\ne \frac{\mathrm{\pi }}{2}\\ k,& x=\frac{\mathrm{\pi }}{2}\end{array}\right\$ is continuous at $x=\frac{\mathrm{\pi }}{2}$.

$\therefore f\left(\frac{\mathrm{\pi }}{2}\right)=\underset{x\to \frac{\mathrm{\pi }}{2}}{\mathrm{lim}}f\left(x\right)\phantom{\rule{0ex}{0ex}}⇒k=\underset{x\to \frac{\mathrm{\pi }}{2}}{\mathrm{lim}}\frac{1-\mathrm{sin}x}{\mathrm{\pi }-2x}$

Put $x=\frac{\mathrm{\pi }}{2}-h$

When

$\therefore k=\underset{h\to 0}{\mathrm{lim}}\frac{1-\mathrm{sin}\left(\frac{\mathrm{\pi }}{2}-h\right)}{\mathrm{\pi }-2\left(\frac{\mathrm{\pi }}{2}-h\right)}$

$⇒k=\underset{h\to 0}{\mathrm{lim}}\frac{1-\mathrm{cos}h}{2h}$

$⇒k=\underset{h\to 0}{\mathrm{lim}}\frac{2{\mathrm{sin}}^{2}\frac{h}{2}}{2h}$

$⇒k=\underset{h\to 0}{\mathrm{lim}}\frac{\mathrm{sin}\frac{h}{2}}{h}×\underset{h\to 0}{\mathrm{lim}}\mathrm{sin}\frac{h}{2}$

$⇒k=\underset{h\to 0}{\mathrm{lim}}\frac{\mathrm{sin}\frac{h}{2}}{2×\frac{h}{2}}×\underset{h\to 0}{\mathrm{lim}}\mathrm{sin}\frac{h}{2}$

$⇒k=\frac{1}{2}×\underset{h\to 0}{\mathrm{lim}}\frac{\mathrm{sin}\frac{h}{2}}{\frac{h}{2}}×\underset{h\to 0}{\mathrm{lim}}\mathrm{sin}\frac{h}{2}$
$⇒k=\frac{1}{2}×1×0$

$⇒k=0$

Thus, the value of k is 0.

If $f\left(x\right)=\left\{\begin{array}{cc}\frac{1-\mathrm{sin}x}{\mathrm{\pi }-2x},& x\ne \frac{\mathrm{\pi }}{2}\\ k,& x=\frac{\mathrm{\pi }}{2}\end{array}\right\$is continuous at $x=\frac{\mathrm{\pi }}{2},$ then k = ____0____.

#### Question 8:

If is continuous at x = 0, then f(0) = ______________.

The function is continuous at x = 0.

$\therefore f\left(0\right)$

$=\underset{x\to 0}{\mathrm{lim}}f\left(x\right)$

$=\underset{x\to 0}{\mathrm{lim}}\frac{2-\sqrt{x+4}}{\mathrm{sin}2x}$

$=\underset{x\to 0}{\mathrm{lim}}\frac{2-\sqrt{x+4}}{\mathrm{sin}2x}×\frac{2+\sqrt{x+4}}{2+\sqrt{x+4}}$

$=\underset{x\to 0}{\mathrm{lim}}\frac{4-\left(x+4\right)}{\mathrm{sin}2x\left(2+\sqrt{x+4}\right)}$

$=\underset{x\to 0}{\mathrm{lim}}\frac{-x}{\mathrm{sin}2x\left(2+\sqrt{x+4}\right)}$

$=-\frac{1}{2}×\underset{x\to 0}{\mathrm{lim}}\frac{2x}{\mathrm{sin}2x\left(2+\sqrt{x+4}\right)}$

$=-\frac{1}{2}×\frac{1}{\underset{x\to 0}{\mathrm{lim}}\left(\frac{\mathrm{sin}2x}{2x}\right)}×\frac{1}{\underset{x\to 0}{\mathrm{lim}}\left(2+\sqrt{x+4}\right)}$

$=-\frac{1}{2}×\frac{1}{4}$

$=-\frac{1}{8}$

Thus, the value of f(0) is $-\frac{1}{8}$.

If is continuous at x = 0, then f(0) = .

#### Question 9:

If $f\left(x\right)=\left\{\begin{array}{cc}\frac{{x}^{2}-9}{x-3},& x\ne 3\\ 2x+k,& x=3\end{array}\right\$is continuous at x = 3, then k = ____________.

The function $f\left(x\right)=\left\{\begin{array}{cc}\frac{{x}^{2}-9}{x-3},& x\ne 3\\ 2x+k,& x=3\end{array}\right\$ is continuous at x = 3.

$\therefore f\left(3\right)=\underset{x\to 3}{\mathrm{lim}}f\left(x\right)$

$⇒2×3+k=\underset{x\to 3}{\mathrm{lim}}\frac{{x}^{2}-9}{x-3}$

$⇒6+k=\underset{x\to 3}{\mathrm{lim}}\frac{\left(x-3\right)\left(x+3\right)}{x-3}$

$⇒6+k=\underset{x\to 3}{\mathrm{lim}}\left(x+3\right)$

$⇒6+k=3+3=6$

$⇒k=0$

Thus, the value of k is 0.

If $f\left(x\right)=\left\{\begin{array}{cc}\frac{{x}^{2}-9}{x-3},& x\ne 3\\ 2x+k,& x=3\end{array}\right\$is continuous at x = 3, then k = ___0___.

#### Question 10:

If the function $f\left(x\right)=\left\{\begin{array}{cc}\frac{{x}^{2}-1}{x-1},& x\ne 1\\ k,& x=1\end{array}\right\$ is given to be continuous at x = 1, then the value of k is ____________.

The function $f\left(x\right)=\left\{\begin{array}{cc}\frac{{x}^{2}-1}{x-1},& x\ne 1\\ k,& x=1\end{array}\right\$ is continuous at x = 1.

$\therefore f\left(1\right)=\underset{x\to 1}{\mathrm{lim}}f\left(x\right)$

$⇒k=\underset{x\to 1}{\mathrm{lim}}\frac{{x}^{2}-1}{x-1}$

$⇒k=\underset{x\to 1}{\mathrm{lim}}\frac{\left(x-1\right)\left(x+1\right)}{x-1}$

$⇒k=\underset{x\to 1}{\mathrm{lim}}\left(x+1\right)$

$⇒k=1+1=2$

Thus, the value of k is 2.

If the function $f\left(x\right)=\left\{\begin{array}{cc}\frac{{x}^{2}-1}{x-1},& x\ne 1\\ k,& x=1\end{array}\right\$ is given to be continuous at x = 1, then the value of k is ___2___.

#### Question 11:

The set of points where f(x) = x – [x] is discontinuous is _____________.

The graph of f(x) = x – [x] is shown below.

It can be seen that, the function f(x) = x – [x] is discontinuous at all integral values of x. So, the set of point where where f(x) = x – [x] is discontinuous is Z i.e. the set of integers.

The set of points where f(x) = x – [x] is discontinuous is _____the set of integers i.e. Z______.

#### Question 12:

Let If f(x) is continuous in , then $f\left(\frac{\mathrm{\pi }}{4}\right)=$ _________.

The given function is

It is given that, the function f(x) is continuous in . So, the function is continuous at $x=\frac{\mathrm{\pi }}{4}$.

$\therefore f\left(\frac{\mathrm{\pi }}{4}\right)$

$=\underset{x\to \frac{\mathrm{\pi }}{4}}{\mathrm{lim}}f\left(x\right)$

$=\underset{x\to \frac{\mathrm{\pi }}{4}}{\mathrm{lim}}\frac{1-\mathrm{tan}x}{4x-\mathrm{\pi }}$

Put $x=\frac{\mathrm{\pi }}{4}+h$

When

So,

$f\left(\frac{\mathrm{\pi }}{4}\right)$
$=\underset{h\to 0}{\mathrm{lim}}\frac{1-\mathrm{tan}\left(\frac{\mathrm{\pi }}{4}+h\right)}{4\left(\frac{\mathrm{\pi }}{4}+h\right)-\mathrm{\pi }}$

$=\underset{h\to 0}{\mathrm{lim}}\frac{1-\frac{\mathrm{tan}\frac{\mathrm{\pi }}{4}+\mathrm{tan}h}{1-\mathrm{tan}\frac{\mathrm{\pi }}{4}\mathrm{tan}h}}{4h}$

$=\underset{h\to 0}{\mathrm{lim}}\frac{-2\mathrm{tan}h}{4h\left(1-\mathrm{tan}h\right)}$

$=-\frac{1}{2}×\underset{h\to 0}{\mathrm{lim}}\frac{\mathrm{tan}h}{h}×\frac{1}{\underset{h\to 0}{\mathrm{lim}}\left(1-\mathrm{tan}h\right)}$

$=-\frac{1}{2}$

Thus, the value of $f\left(\frac{\mathrm{\pi }}{4}\right)$ is $-\frac{1}{2}$.

Let If f(x) is continuous in , then $f\left(\frac{\mathrm{\pi }}{4}\right)=$.

#### Question 13:

If is everywhere continuous, then f(0) = ____________.

Thus, f(x) is a polynomial function which is continuous everywhere.

So, f(x) is continuous at x = 0.

$\therefore f\left(0\right)=\underset{x\to 0}{\mathrm{lim}}f\left(x\right)=\underset{x\to 0}{\mathrm{lim}}\frac{x}{\sqrt{2}}=\frac{1}{\sqrt{2}}×0=0$

If is everywhere continuous, then f(0) = ____0____.

#### Question 14:

The set of points at which the function $f\left(x\right)=\frac{1}{\mathrm{log}\left|x\right|}$ is not continuous, is ___________.

The given function $f\left(x\right)=\frac{1}{\mathrm{log}\left|x\right|}$ is discontinuous when $\mathrm{log}\left|x\right|=0$.

Also, x ≠ 0    (log 0 is not defined)

Now,

$\mathrm{log}\left|x\right|=0$

$⇒\left|x\right|=1$

$⇒x=±1$

Thus, the given function is not continuous at x = 0, x = −1 and x = 1.

Hence, the set of points at which the given function is not continuous is {−1, 0, 1}.

The set of points at which the function $f\left(x\right)=\frac{1}{\mathrm{log}\left|x\right|}$ is not continuous, is ___{−1, 0, 1}___.

#### Question 15:

If is continuous, then 'a' should be equal to __________.

It is given that, the function is continuous.

So, the function f(x) is continuous at x = 1.

$\therefore f\left(1\right)=\underset{x\to 1}{\mathrm{lim}}f\left(x\right)$

$⇒f\left(1\right)=\underset{x\to {1}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{x\to {1}^{+}}{\mathrm{lim}}f\left(x\right)$      .....(1)

Now,

f(1) = a × 1 + 1 = a + 1         .....(2)

$\underset{x\to {1}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{x\to 1}{\mathrm{lim}}\left(ax+1\right)=a×1+1=a+1$       .....(3)

$\underset{x\to {1}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{x\to 1}{\mathrm{lim}}\left(x+2\right)=1+2=3$      .....(4)

From (1), (2), (3) and (4), we have

a + 1 = 3 = a + 1

⇒ a = 3 − 1 = 2

Thus, the value of a is 2.

If is continuous, then 'a' should be equal to ___2___.

#### Question 16:

If f(x) is continuous at x = a and$\underset{x\to {a}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{x\to {a}^{+}}{\mathrm{lim}}f\left(x\right)=k,$ then k is equal to _____________.

It is given that, f(x) is continuous at xa.

$\therefore f\left(a\right)=\underset{x\to {a}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{x\to {a}^{+}}{\mathrm{lim}}f\left(x\right)$    .....(1)

Also,

$\underset{x\to {a}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{x\to {a}^{+}}{\mathrm{lim}}f\left(x\right)=k$            .....(2)

From (1) and (2), we have

f(a) = k

Thus, the value of k is f(a).

If f(x) is continuous at x = a and$\underset{x\to {a}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{x\to {a}^{+}}{\mathrm{lim}}f\left(x\right)=k,$ then k is equal to .

#### Question 17:

If is continuous at x = 0, then k is equal to _____________.

The given function  is continuous at x = 0.

$\therefore f\left(0\right)=\underset{x\to 0}{\mathrm{lim}}f\left(x\right)$

$⇒\frac{k}{2}=\underset{x\to 0}{\mathrm{lim}}\frac{\mathrm{sin}3x}{x}$

$⇒\frac{k}{2}=3\underset{x\to 0}{\mathrm{lim}}\frac{\mathrm{sin}3x}{3x}$

$⇒k=6$

Thus, the value of k is 6.

If  is continuous at x = 0, then k is equal to ____6____.

#### Question 18:

The set of points of discontinuity of f(x) = tan x is ___________.

The given function is $f\left(x\right)=\mathrm{tan}x$.

$f\left(x\right)=\mathrm{tan}x=\frac{\mathrm{sin}x}{\mathrm{cos}x}$

The function f(x) is discontinuous when $\mathrm{cos}x=0$.

$\mathrm{cos}x=0$

Thus, the set of point of discontinuity of $f\left(x\right)=\mathrm{tan}x$ is $\left\{\left(2n+1\right)\frac{\mathrm{\pi }}{2}:n\in \mathbf{Z}\right\}$.

The set of points of discontinuity of f(x) = tan x is .

#### Question 19:

The set of points of discontinuity of f(x) = [x] is ___________.

The graph of f(x) = [x] is shown below.

It can be seen that, the function f(x) = [x] is discontinuous at all integral values of x i.e. x ∈ Z

Thus, the set of points of discontinuity of f(x) = [x] is the set of integers i.e. Z.

The set of points of discontinuity of f(x) = [x] is __the set of integers i.e. Z___.

#### Question 20:

The set of points of discontinuity of $f\left(x\right)=\frac{1}{x-\left[x\right]}$ is ________.

The function $f\left(x\right)=\frac{1}{x-\left[x\right]}$ is discontinuous when x – [x] = 0.

x – [x] = 0

⇒ x = [x]

⇒ x is an integer

So, the function f(x) is discontinuous for all x ∈ i.e. the set of integers.

Thus, the set of points of discontinuity of $f\left(x\right)=\frac{1}{x-\left[x\right]}$ is the set of integers i.e. Z.

The set of points of discontinuity of $f\left(x\right)=\frac{1}{x-\left[x\right]}$ is  ____the set of integers i.e. Z_____.

#### Question 1:

Define continuity of a function at a point.

Continuity at a point:

A function $f\left(x\right)$ is said to be continuous at a point xa of its domain, iff $\underset{x\to \mathrm{a}}{\mathrm{lim}}f\left(x\right)=f\left(a\right)$.

#### Question 2:

What happens to a function f (x) at x = a, if $\underset{x\to a}{\mathrm{lim}}$f (x) = f (a)?

If $f\left(x\right)$ is a function defined in its domain such that $\underset{x\to \mathrm{a}}{\mathrm{lim}}f\left(x\right)=f\left(a\right)$, then $f\left(x\right)$ becomes continuous at $x=a$.

#### Question 3:

Find f (0), so that $f\left(x\right)=\frac{x}{1-\sqrt{1-x}}$ becomes continuous at x = 0.

If $f\left(x\right)$is continuous at x = 0, then $\underset{x\to 0}{\mathrm{lim}}f\left(x\right)=f\left(0\right)$          ...(1)

Given: $f\left(x\right)=\frac{x}{1-\sqrt{1-x}}$

$⇒f\left(0\right)=2$

So, for $f\left(0\right)=2$, the function f(x) becomes continuous at x = 0.

#### Question 4:

If is continuous at x = 0, then write the value of k.

If $f\left(x\right)$ is continuous at $x=0$, then
$\underset{x\to 0}{\mathrm{lim}}f\left(x\right)=f\left(0\right)$
Given:  is continuous at $x=0$.