Rd Sharma XII Vol 1 2021 Solutions for Class 12 Science Maths Chapter 17 Maxima And Minima are provided here with simple step-by-step explanations. These solutions for Maxima And Minima are extremely popular among Class 12 Science students for Maths Maxima And Minima Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma XII Vol 1 2021 Book of Class 12 Science Maths Chapter 17 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma XII Vol 1 2021 Solutions. All Rd Sharma XII Vol 1 2021 Solutions for class Class 12 Science Maths are prepared by experts and are 100% accurate.

Question 1:

f(x) = (x$-$5)4

Since f '(x) changes from negative to positive when x increases through 5, x = 5 is the point of local minima.
The local minimum value of  f (x) at x = 5 is given by
${\left(5-5\right)}^{4}=0$

Question 2:

f(x) = x3 $-$ 3x

Since f '(x) changes from negative to positive as x increases through 1, x = 1 is the point of local minima.
The local minimum value of  f (x) at x = 1 is given by
${\left(1\right)}^{3}-3\left(1\right)=-2$

Since f '(x) changes from positive to negative when x increases through -1, x = -1 is the point of local maxima.
The local maximum value of  f (x) at x = -1 is given by
${\left(-1\right)}^{3}-3\left(-1\right)=2$

.

Question 3:

f(x) = x3  (x$-$1)2

Since f '(x) changes from negative to positive when x increases through 1, x = 1 is the point of local minima.
The local minimum value of  f (x)  at x = 1 is given by
${\left(1\right)}^{3}{\left(1-1\right)}^{2}=0$

Since f '(x) changes from positive to negative when x increases through $\frac{3}{5}$, x = $\frac{3}{5}$ is the point of local maxima.
The local minimum value of  f (x) at x = $\frac{3}{5}$ is given by
${\left(\frac{3}{5}\right)}^{3}{\left(\frac{3}{5}-1\right)}^{2}=\frac{27}{125}×\frac{4}{25}=\frac{108}{3125}$

Since f '(x) does not change from positive as x increases through 0, x = 0 is a point of inflexion.

Disclaimer: The solution in the book is incorrect. The solution here is created according to the question given in the book.

Question 4:

f(x) =  (x$-$1) (x+2)2

Since  f '(x) changes from negative to positive when x increases through 0, x = 0 is the point of local minima.
The local minimum value of  f (x) at x = 0 is given by
$\left(0-1\right){\left(0+2\right)}^{2}=-4$

Since  f '(x) changes sign from positive to negative when x increases through $-2$, x = $-2$ is the point of local maxima.
The local maximum value of  f (x)  at x = $-2$ is given by
$\left(-2-1\right){\left(-2+2\right)}^{2}=0$

Question 5:

f(x) = $\frac{1}{{x}^{2}+2}$

Now, for values close to x = 0 and to the left of 0, $f\text{'}\left(x\right)>0$.
Also, for values close to x = 0 and to the right of 0, $f\text{'}\left(x\right)<0$.

Therefore, by first derivative test, x = 0 is a point of local maxima and the local maximum value of

Question 6:

f(x) =  x3 $-$ 6x2 + 9x + 15

Since f '(x) changes from negative to positive when x increases through 3, x = 3 is the point of local minima.
The local minimum value of  f (x) at x = 3 is given by
${\left(3\right)}^{3}-6{\left(3\right)}^{2}+9\left(3\right)+15=27-54+27+15=15$

Since f '(x) changes from positive to negative when x increases through 1, x = 1 is the point of local maxima.
The local maximum value of  f (x) at x = 1 is given by
${\left(1\right)}^{3}-6{\left(1\right)}^{2}+9\left(1\right)+15=1-6+9+15=19$

Question 7:

f(x) = sin 2x, 0<x<$\mathrm{\pi }$

Since f '(x) changes from positive to negative when x increases through $\frac{\mathrm{\pi }}{4}$, x = $\frac{\mathrm{\pi }}{4}$ is the point of maxima.
The local maximum value of  f (x) at x = $\frac{\mathrm{\pi }}{4}$ is given by
$\mathrm{sin}\left(\frac{\mathrm{\pi }}{2}\right)=1$

Since f '(x) changes from negative to positive when x increases through $\frac{3\mathrm{\pi }}{4}$, x = $\frac{3\mathrm{\pi }}{4}$ is the point of minima.
The local minimum value of  f (x) at x = $\frac{3\mathrm{\pi }}{4}$ is given by
$\mathrm{sin}\left(\frac{3\mathrm{\pi }}{2}\right)=-1$

Question 8:

f(x) =  sin x$-$ cos x, 0 < x<2$\mathrm{\pi }$

Since f '(x) changes from positive to negative when x increases through $\frac{3\mathrm{\pi }}{4}$, x = $\frac{3\mathrm{\pi }}{4}$ is the point of local maxima.
The local maximum value of  f (x)  at x = $\frac{3\mathrm{\pi }}{4}$ is given by
$\mathrm{sin}\left(\frac{3\mathrm{\pi }}{4}\right)-\mathrm{cos}\left(\frac{3\mathrm{\pi }}{4}\right)=\sqrt{2}$

Since f '(x) changes from negative to positive when x increases through $\frac{7\mathrm{\pi }}{4}$, x = $\frac{7\mathrm{\pi }}{4}$ is the point of local minima.
The local minimum value of  f (x)  at x = $\frac{7\mathrm{\pi }}{4}$ is given by
$\mathrm{sin}\left(\frac{7\mathrm{\pi }}{4}\right)-\mathrm{cos}\left(\frac{7\mathrm{\pi }}{4}\right)=-\sqrt{2}$

Question 9:

f(x) =  cos x, 0<x<$\mathrm{\pi }$

.

Since $0, none is in the interval .

Question 10:

f(x) = sin 2x$-$x, $-\frac{\mathrm{\pi }}{2}<\frac{<}{}x\frac{<}{}\frac{\mathrm{\pi }}{2}$

.

Since  f '(x) changes from positive to negative when x increases through $\frac{\mathrm{\pi }}{6}$, x = $\frac{\mathrm{\pi }}{6}$ is the point of local maxima.
The local maximum value of  f (x) at x = $\frac{\mathrm{\pi }}{6}$ is given by

Since f '(x) changes from negative to positive when x increases through $-\frac{\mathrm{\pi }}{6}$, x = $-\frac{\mathrm{\pi }}{6}$ is the point of local minima.
The local minimum value of  f (x)  at x = $-\frac{\mathrm{\pi }}{6}$ is given by

Question 11:

f(x) = 2sin x$-$x, $-\frac{\mathrm{\pi }}{2}<\frac{<}{}x\frac{<}{}\frac{\mathrm{\pi }}{2}$

Since f '(x) changes from positive to negative when x increases through $\frac{\mathrm{\pi }}{3}$, x = $\frac{\mathrm{\pi }}{3}$ is the point of local maxima.
The local maximum value of  f (x) at x = $\frac{\mathrm{\pi }}{3}$ is given by

Since f '(x) changes from negative to positive when x increases through $-\frac{\mathrm{\pi }}{3}$, x = $-\frac{\mathrm{\pi }}{3}$ is the point of local minima.

The local minimum value of  f (x)  at x = $-\frac{\mathrm{\pi }}{3}$ is given by

Question 12:

f(x) =

Since,  f '(x) changes from positive to negative when x increases through $\frac{2}{3}$, x = $\frac{2}{3}$ is a point of maxima.

The local maximum value of  f (x) at x = $\frac{2}{3}$ is given by
$\frac{2}{3}\sqrt{1-\frac{2}{3}}=\frac{2}{3\sqrt{3}}=\frac{2\sqrt{3}}{9}$

Question 13:

f(x) = x3 (2x$-$1)3

Since f '(x) changes from negative to positive when x increases through $\frac{1}{4}$, x = $\frac{1}{4}$ is a point of local minima.
The local minimum value of  f (x)  at x = $\frac{1}{4}$ is given by
${\left(\frac{1}{4}\right)}^{3}{\left(\frac{1}{2}-1\right)}^{3}=\frac{-1}{512}$

Question 14:

f(x) =

Since x > 0,  f '(x) changes from negative to positive when x increases through 2. So, x = 2 is a point of local minima.

The local minimum value of  f (x) at x = 2 is given by
$\frac{2}{2}+\frac{2}{2}=2$

Question 1:

(i) f(x) = x4 $-$ 62x2 + 120x + 9

(ii) f(x) = x3 $-$ 6x2 + 9x + 15

(iii) f(x) = (x$-$1) (x+2)2

(iv) f(x) = 2/x$-$2/x2 , x>0

(v) f(x) = xex

(vi) f(x) = x/2+2/x, x>0

(vii) f(x) = (x+1) (x+2)1/3, $x\frac{>}{}-2$

(viii) f(x) =

(ix) f(x) =

(x) f(x) = x ≠ 0

(xi) f(x) =

(xii) f(x) =

Question 2:

(i) f(x) = (x$-$1) (x$-$2)2

(ii) f(x) =

(iii) f(x) =

Question 3:

The function y = a log x+bx2 + x has extreme values at x=1 and x=2. Find a and b

Question 4:

Show that has a maximum value at x = e.

Question 5:

Find the maximum and minimum values of the function f(x) = $\frac{4}{x+2}+x.$

Question 6:

Find the maximum and minimum values of y = tan $x-2x$.

Question 7:

If f(x) = x3 + ax2 + bx + c has a maximum at x = $-$1 and minimum at x = 3. Determine a, b and c.

Question 8:

Prove that f(x) = sinx + $\sqrt{3}$cosx has maximum value at x = $\frac{\mathrm{\pi }}{6}$.                                                                                     [NCERT EXEMPLAR]

Question 1:

(i) f(x) = 4x $-$ $\frac{{x}^{2}}{2}$ in [$-$2,4,5]

(ii) f(x) = (x$-$1)2 + 3 in [$-$3,1]

(iii) f(x) = 3x4 $-$ 8x3 + 12x2 $-$ 48x + 25 in [0,3]

(iv) f(x) = (x$-$2)

Question 2:

Find the maximum value of 2x3 $-$ 24x + 107 in the interval [1,3]. Find the maximum value of the same function in [$-$3, $-$1].

Question 3:

Find the absolute maximum and minimum values of the function of given by

Question 4:

Find the absolute maximum and minimum values of a function f given by

Question 5:

Find the absolute maximum and minimum values of a function f given by

Question 1:

f($x$)=4${x}^{2}$-4$x$+4 on R.

Given: f(x) = 4x2 − 4x + 4
$⇒$f(x) = (4x2 − 4x + 1)+3
$⇒$f(x) = (2x − 1)2 + 3
Now,
(2x − 1)2 $\ge$ 0 for all x $\in$ R
$⇒$f(x) = (2x − 1)2 + 3 $\ge$ 3 for all x $\in$ R
$⇒$f(x) $\ge$ 3 for all x $\in$ R

The minimum value of f is attained when (x − 1) = 0.
(2x − 1) = 0
x = $\frac{1}{2}$
Thus, the minimum value of f (x) at x = $\frac{1}{2}$ is 3.

Since f(x) can be enlarged, the maximum value does not exist, which is evident in the graph also.
Hence, function f does not have a maximum value.

Question 2:

f(x)=(x-1)2+2 on R

Given: f(x) = − (x − 1)2 + 2
Now,
(x − 1)2 $\ge$ 0 for all x $\in$ R
$⇒$f(x) = − (x − 1)2 + 2 $\le$ 2 for all x $\in$ R

The maximum value of f(x) is attained when (x − 1) = 0.
(x − 1) = 0
x = 1
Therefore, the maximum value of f (x) = 2
Since f(x) can be reduced, the minimum value does not exist, which is evident in the graph also.
Hence, function f does not have a minimum value.

Question 3:

f(x)=| x+2 | on R

Given: f(x) = $\left|x+2\right|$

Now,
$\left|x+2\right|\ge 0$ for all x $\in$ R

Thus, f(x) $\ge$0 for all x $\in$ R

Therefore, the minimum value of f at x = $-$2 is 0.

Since f(x) can be enlarged, the maximum value does not exist, which is evident in the graph also.
Hence, the function f does not have a maximum value.

Question 4:

f(x)=sin 2x+5 on R

Given: f(x) = sin 2x + 5

We know that − 1 ≤ sin 2x ≤ 1.

⇒ − 1 + 5 ≤ sin 2x + 5 ≤ 1 + 5

⇒ 4 ≤ sin 2x + 5 ≤ 6

⇒ 4 ≤ f(x) ≤ 6

Hence, the maximum and minimum values of f are 6 and 4, respectively.

Question 5:

f(x) = | sin 4x+3 | on R

Given: f(x) =

We know that −1 $\le$ sin 4x $\le$ 1.

⇒ 2 $\le$ sin 4x + 3 $\le$ 4

⇒ 2 $\le$ $\le$ 4

⇒ 2 $\le$ f(x) $\le$ 4

Hence, the maximum and minimum values of f are 4 and 2, respectively.

Question 6:

f(x)=2x3 +5 on R

We can observe that f(x) increases when the values of x are increased and f(x) decreases when the values of x are decreased. Also, f(x) can be reduced by giving small values of x.
Similarly, f(x) can be enlarged by giving large values of x.
So, f(x) does not have a minimum or maximum value.

Question 7:

f(x) =$-$| x + 1 | + 3 on R

Given: f(x) =$-\left|x+1\right|$ + 3

Now,
$-\left|x+1\right|\le 0$ for all x $\in$ R

$⇒$f(x) =  $-\left|x+1\right|$ + 3$\le$ 3 for all x $\in$ R
$⇒$f(x) $\le$ 3 for all x $\in$ R

The maximum value of f is attained when $\left|x+1\right|=0.\phantom{\rule{0ex}{0ex}}⇒x=-1$

Therefore, the maximum value of f  at x = -1 is 3.

Since f(x) can be reduced, the minimum value does not exist, which is evident in the graph also.
Hence, the function f does not have a minimum value.

Question 8:

f(x) = 16x2 $-$ 16x + 28 on R

Given: f(x) = 16x2 − 16x + 28
$⇒$f(x) = 4(4x2 - 4x + 1) + 24
$⇒$f(x) = 4(2x − 1)2 + 24

Now,
4(2x − 1)2 $\ge$ 0 for all x $\in$ R

$⇒$f(x) = 4(2x − 1)2 + 24 $\ge$ 24 for all x $\in$ R
$⇒$f(x) $\ge$ 24 for all x $\in$ R

The minimum value of f is attained when (2x − 1) = 0.

(2x − 1) = 0
x = $\frac{1}{2}$

Therefore, the minimum value of f  at x = $\frac{1}{2}$ is 24.

Since f(x) can be enlarged, the maximum value does not exist, which is evident in the graph also.
Hence, the function f does not have a maximum value.

Question 9:

f(x) = x3 $-$1 on R

We can observe that f(x) increases when the values of x increase and f(x) decreases when the values of x decrease. Also, f(x) can be reduced by giving smaller values of x.
Similarly, f(x) can be enlarged by giving larger values of x.
So, f(x) does not have a minimum or maximum value.

Question 1:

Determine two positive numbers whose sum is 15 and the sum of whose squares is maximum.

Question 2:

Divide 64 into two parts such that the sum of the cubes of two parts is minimum.

.

Question 3:

How should we choose two numbers, each greater than or equal to $-$2, whose sum______________ so that the sum of the first and the cube of the second is minimum?

Question 4:

Divide 15 into two parts such that the square of one multiplied with the cube of the other is minimum.

Question 5:

Of all the closed cylindrical cans (right circular), which enclose a given volume of 100 cm3, which has the minimum surface area?

Question 6:

A beam is supported at the two ends and is uniformly loaded. The bending moment M at a distance x from one end is given by
(i)

(ii)

Find the point at which M is maximum in each case.

Question 7:

A wire of length 28 m is to be cut into two pieces. One of the pieces is to be made into a square and the other into a circle. What should be the lengths of the two pieces so that the combined area of the circle and the square is minimum?

Question 8:

A wire of length 20 m is to be cut into two pieces. One of the pieces will be bent into shape of a square and the other into shape of an equilateral triangle. Where the we should be cut so that the sum of the areas of the square and triangle is minimum?

Question 9:

Given the sum of the perimeters of a square and a circle, show that the sum of there areas is least when one side of the square is equal to diameter of the circle.

Question 10:

Find the largest possible area of a right angled triangle whose hypotenuse is 5 cm long.                                                              [CBSE 2000]

Question 11:

Two sides of a triangle have lengths 'a' and 'b' and the angle between them is $\theta$. What value of $\theta$ will maximize the area of the triangle? Find the maximum area of the triangle also.                                                                                                                                          [CBSE 2002 C]

Question 12:

A square piece of tin of side 18 cm is to be made into a box without top by cutting a square from each corner and folding up the flaps to form a box. What should be the side of the square to be cut off so that the volume of the box is maximum? Find this maximum volume.

Let the side of the square to be cut off be x cm.
Then, the length and the breadth of the box will be (18 − 2x) cm each and height of the box will be x cm.

Volume of the box, V(x) = x(18 − 2x)2

$⇒$x = 9 or x = 3

If x = 9, then length and breadth will become 0.

x ≠ 9

x = 3

Now,
$V\text{'}\text{'}\left(3\right)=-24\left(6-3\right)=-72<0$

x = 3 is the point of maxima.

Hence, if we remove a square of side 3 cm from each corner of the square tin and make a box from the remaining sheet, then the volume of the box so obtained would be the largest, i.e. 432 cm3.

Question 13:

A rectangular sheet of tin 45 cm by 24 cm is to be made into a box without top, in cutting off squares from each corners and folding up the flaps. What should be the side of the square to be cut off so that the volume of the box is maximum possible?

Question 14:

A tank with rectangular base and rectangular sides, open at the top is to the constructed so that its depth is 2 m and volume is 8 m3. If building of tank cost 70 per square metre for the base and Rs 45 per square matre for sides, what is the cost of least expensive tank?

Let l, b and h be the length, breadth and height of the tank, respectively.

Height, h = 2 m

Volume of the tank = 8 m3

Volume of the tank = l × b × h

∴  l × b × 2 = 8

$⇒lb=4\phantom{\rule{0ex}{0ex}}⇒b=\frac{4}{l}$

Area of the base = lb = 4 m2

Area of the 4 walls, A= 2h (l + b)

However, the length cannot be negative.

Thus,
l = 2 m

Thus, the area is the minimum when l = 2 m

We have
l = b = h = 2 m

Cost of building the base = Rs 70 × (lb) = Rs 70 × 4 = Rs 280

Cost of building the walls = Rs 2h (l + b) × 45 = Rs 90 (2) (2 + 2)= Rs 8 (90) = Rs 720

Total cost = Rs (280 + 720) = Rs 1000

Hence, the total cost of the tank will be Rs 1000.

Disclaimer: The solution given in the book is incorrect. The solution here is created according to the question given in the book.

Question 15:

A window in the form of a rectangle is surmounted by a semi-circular opening. The total perimeter of the window is 10 m. Find the dimension of the rectangular of the window to admit maximum light through the whole opening.

Question 16:

A large window has the shape of a rectangle surmounted by an equilateral triangle. If the perimeter of the window is 12 metres find the dimensions of the rectangle will produce the largest area of the window.

Disclaimer: The solution given in the book is incorrect. The solution here is created according to the question given in the book.

Question 17:

Show that the height of the cylinder of maximum volume that can be inscribed a sphere of radius R is $\frac{2R}{\sqrt{3}}.$

Question 18:

A rectangle is inscribed in a semi-circle of radius r with one of its sides on diameter of semi-circle. Find the dimension of the rectangle so that its area is maximum. Find also the area.

Question 19:

Prove that a conical tent of given capacity will require the least amount of  canavas when the height is times the radius of the base.

Question 20:

Show that the cone of the greatest volume which can be inscribed in a given spher has an altitude equal to $2}{3}$ of the diameter of the sphere.

Question 21:

Prove that the semi-vertical angle of the right circular cone of given volume and least curved surface is ${\mathrm{cot}}^{-1}\left(\sqrt{2}\right)$.                     [CBSE 2014]

Let:
Radius of the base = r,
Height = h,
Slant height = l,
Volume = V,
Curved surface area = C

Question 22:

An isosceles triangle of vertical angle 2$\theta$ is inscribed in a circle of radius a. Show that the area of the triangle is maximum when $\theta$ = $\frac{\mathrm{\pi }}{6}$.
[NCERT EXEMPLAR]

Let ABC be an isosceles triangle inscribed in the circle with radius a such that AB = AC.

Question 23:

Prove that the least perimeter of an isosceles triangle in which a circle of radius r can be inscribed is $6\sqrt{3}$r.

Question 24:

Find the dimensions of the rectangle of perimeter 36 cm which will sweep out a volume as large as possible when revolved about one of its sides.

Question 25:

Show that the height of the cone of maximum volume that can be inscribed in a sphere of radius 12 cm is 16 cm.

Question 26:

A closed cylinder has volume 2156 cm3. What will be the radius of its base so that its total surface area is minimum.

Question 27:

Show that the maximum volume of the cylinder which can be inscribed in a sphere of radius

Question 28:

Show that among all positive numbers x and y with x2 + y2 =r2, the sum x+y is largest when x=y=r$\sqrt{2}$.

Question 29:

Determine the points on the curve x2 = 4y which are nearest to the point (0,5).

Question 30:

Find the point on the curve y2=4x which is nearest to the point (2, $-$8).

Question 31:

Find the point on the curve x2=8y which is nearest to the point (2,4).

Question 32:

Find the point on the parabolas x2 = 2y which is closest to the point (0,5).

Question 33:

Find the coordinates of a point on the parabola y=x2+7x + 2 which is closest to the strainght line y = 3x $-$ 3.

Question 34:

Find the point on the curvey y2=2x which is at a minimum distance from the point (1,4).

Question 35:

Find the maximum slope of the curve y=$-{x}^{3}+3{x}^{2}+2x-27.$

Question 36:

The total cost of producing x radio sets per  day is Rs and the price per set  at which they may be sold is Rs. Find the daily output to maximum the total profit.

Question 37:

Manufacturer can sell x items at a price of rupees $\left(5-\frac{x}{100}\right)$ each. The cost price is Rs  $\left(\frac{x}{5}+500\right).$ Find the number of items he should sell to earn maximum profit.

Disclaimer: The solution given in the book is incorrect. The solution here is created according to the question given in the book.

Question 38:

An open tank is to be constructed with a square base and vertical sides so as to contain a given quantity of water. Show that the expenses of lining with lead with be least, if depth is made half of width.

Question 39:

A box of constant volume c is to be twice as long as it is wide. The material on the top and four sides cost three times as much per square metre as that in the bottom. What are the most economic dimensions?

Question 40:

The sum of the surface areas of a sphere and a cube is given. Show that when the sum of their volumes is least, the diameter of the sphere is equal to the edge of the cube.

Let r be the radius of the sphere, x be the side of the cube and S be the sum of the surface area of both. Then,
$S=4\mathrm{\pi }{r}^{2}+6{x}^{2}$
$⇒$$x={\left(\frac{S-4\mathrm{\pi }{r}^{2}}{6}\right)}^{\frac{1}{2}}$                                            ...(1)

Sum of volumes, V$\frac{4}{3}\mathrm{\pi }{r}^{3}+{x}^{3}$

$⇒$V$\frac{4\mathrm{\pi }{r}^{3}}{3}+{\left[\frac{\left(S-4\mathrm{\pi }{r}^{2}\right)}{6}\right]}^{\frac{3}{2}}$                        [From eq. (1)]

$⇒\frac{dV}{dr}=4\mathrm{\pi }{r}^{2}-2\mathrm{\pi }r{\left[\frac{\left(S-4\mathrm{\pi }{r}^{2}\right)}{6}\right]}^{\frac{1}{2}}$

For the minimum or maximum values of V, we must have
$\frac{dV}{dr}=0$                                                          ...(2)

Now,
$\frac{{d}^{2}V}{d{r}^{2}}=8\pi r-2\pi {\left[\frac{\left(S-4\pi {r}^{2}\right)}{6}\right]}^{\frac{1}{2}}-\frac{2\pi r}{2}{\left[\frac{\left(S-4\pi {r}^{2}\right)}{6}\right]}^{-\frac{1}{2}}\frac{\left(-8\pi r\right)}{6}\phantom{\rule{0ex}{0ex}}⇒\frac{{d}^{2}V}{d{r}^{2}}=8\pi r-2\pi {\left[\frac{\left(S-4\pi {r}^{2}\right)}{6}\right]}^{\frac{1}{2}}+\frac{4}{3}{\pi }^{2}{r}^{2}{\left[\frac{6}{\left(S-4\pi {r}^{2}\right)}\right]}^{\frac{1}{2}}\phantom{\rule{0ex}{0ex}}⇒\frac{{d}^{2}V}{d{r}^{2}}=8\pi r-2\pi x+\frac{4}{3}{\pi }^{2}{r}^{2}\frac{1}{x}=8\pi r-4\pi r+\frac{2}{3}{\pi }^{2}r\phantom{\rule{0ex}{0ex}}⇒\frac{{d}^{2}V}{d{r}^{2}}=4\pi r+\frac{2}{3}{\pi }^{2}r>0$

So, volume is minimum when x = 2r.

Question 41:

A given quantity of metal is to be cast into a half cylinder with a rectangular base and semicircular ends. Show that in order that the total surface area may be minimum the ratio of the length of the cylinder to the diameter of its semi-circular ends is .

Question 42:

The strength of a beam varies as the product of its breadth and square of its depth. Find the dimensions of the strongest beam which can be cut from a circular log of radius a.

Question 43:

A straight line is drawn through a given point P(1,4). Determine the least value of the sum of the intercepts on the coordinate axes.

Question 44:

The total area of a page is 150 cm2. The combined width of the margin at the top and bottom is 3 cm and the side 2 cm. What must be the dimensions of the page in order that the area of the printed matter may be maximum?

Question 45:

The space s described in time t by a particle moving in a straight line is given by S = Find the minimum value of acceleration.

Question 46:

A particle is moving in a straight line such that its distance at any time t is given by  S = Find when its velocity is maximum and acceleration minimum.

Question 1:

The maximum value of x1/x, x>0 is

(a) e1/e

(b) ${\left(\frac{1}{e}\right)}^{e}$

(c) 1

(d) none of these

(a)        ${e}^{\frac{1}{e}}$

Disclaimer: The answer given in the book is incorrect. The solution provided here is according to the question.

Question 2:

If for all positive x where a,b,>0, then

(a)

(b)

(c)

Question 3:

The minimum value of $\frac{x}{{\mathrm{log}}_{e}x}$ is

(a) e

(b) 1/e

(c) 1

(d) none of these

Question 4:

For the function f(x) =

(a) x = 1 is a point of maximum
(b) x = $-$1 is a point of minimum
(c) maximum value > minimum value
(d) maximum value< minimum value

Question 5:

Let f(x) = x3+3x2$-$9x+2. Then, f(x) has

(a) a maximum at x = 1
(b) a minimum at x = 1
(c) neither a maximum nor a minimum at x = $-$3
(d) none of these

Question 6:

The minimum value of f(x) = $x4-x2-2x+6$ is
(a) 6
(b) 4
(c) 8
(d) none of these

Question 7:

The number which exceeds its square by the greatest possible quantity is
(a) $\frac{1}{2}$

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

(c) $\frac{3}{4}$

(d) none of these

Question 8:

Let f(x) = (x$-$a)2 + (x$-$b)2 + (x$-$c)2. Then, f(x) has a minimum at x =

(a) $\frac{a+b+c}{3}$

(b) $\sqrt[3]{abc}$

(c) $\frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}$

(d) none of these

Question 9:

The sum of two non-zero numbers is 8, the minimum value of the sum of the reciprocals is

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

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

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

(d) none of these

Question 10:

The function f(x) = $\sum _{r=1}^{5}$ (x$-$r)2 assumes minimum value at x =
(a) 5

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

(c) 3

(d) 2

Question 11:

At x= $\frac{5\mathrm{\pi }}{6}$, f(x) = 2 sin 3x + 3 cos 3x is

(a) 0

(b) maximum

(c) minimum

(d) none of these

(d) none of these

Question 12:

If x lies in the interval [0,1], then the least value of x2 + x + 1 is
(a) 3

(b)$\frac{3}{4}$

(c) 1

(d) none of these

Question 13:

The least value of the function f(x) = $x3-18x2+96x$ in the interval [0,9] is
(a) 126
(b) 135
(c) 160
(d) 0

Question 14:

The maximum value of f(x) = $\frac{x}{4-x+{x}^{2}}$ on [$-$1, 1] is

(a) $-\frac{1}{4}$

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

(c)  $\frac{1}{6}$

(d) $\frac{1}{5}$

Disclaimer: The question in the book has some error. So, none of the options are matching with the solution. The solution here is according to the question given in the book.

Question 15:

The point on the curve y2 = 4x which is nearest to, the point (2,1) is
(a) $1,2\sqrt{2}$

(b) (1,2)

(c) (1,$-$2)

(d) ($-$2,1)

Question 16:

If x+y=8, then the maximum value of xy is

(a) 8

(b) 16

(c) 20

(d) 24

Question 17:

The least and greatest values of f(x) = x3$-$6x2+9x in [0,6], are

(a) 3,4

(b) 0,6

(c) 0,3

(d) 3,6

The least and greatest values of f(x) = x3- 6x2+9x in [0, 6] are 0 and 54, respectively.

Disclaimer: The question in the book has some error. So, none of the options are matching with the solution. The solution here is according to the question given in the book.

Question 18:

f(x) = is maximum when x =

(a) $\frac{\mathrm{\pi }}{3}$

(b) $\frac{\mathrm{\pi }}{4}$

(c) $\frac{\mathrm{\pi }}{6}$

(d) 0

Question 19:

If a cone of maximum volume is inscribed in a given sphere, then the ratio of the height of the cone to the diameter of the sphere is

(a) $\frac{3}{4}$

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

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

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

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

Question 20:

The minimum value of $\left({x}^{2}+\frac{250}{x}\right)$ is
(a) 75

(b) 50

(c) 25

(d) 55

(a) 75

Question 21:

If(x) = x+$\frac{1}{x}$, x > 0, then its greatest value is

(a) $-$2

(b) 0

(c) 3

(d) none of these

(d) none of these

Question 22:

If(x) = $\frac{1}{4x2+2x+1}$, then its maximum value is

(a) $\frac{4}{3}$

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

(c) 1

(d) $\frac{3}{4}$

(a) $\frac{4}{3}$

Question 23:

Let x, y be two variables and x>0, xy=1, then minimum value of x+y is
(a) 1

(b) 2

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

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

(b) 2

Question 24:

f(x) = 1+2 sin x+3 cos2x, $0\frac{<}{}x\frac{<}{}\frac{2\mathrm{\pi }}{3}$ is

(a) Minimum at x = $\mathrm{\pi }}{2}$

(b) Maximum at x = sin$-1$ ($1}{\sqrt{3}}$)

(c) Minimum at x = $\mathrm{\pi }}{6}$

(d) Maximum at sin$-1$($1}{6}$)

Question 25:

The function f(x) =$2{x}^{3}-15{x}^{2}+36x+4$ is maximum at x =

(a) 3

(b) 0

(c) 4

(d) 2

(d) 2

Question 26:

The maximum value of f(x) = $\frac{x}{4+x+{x}^{2}}$ on [$-$1,1] is

(a) $-\frac{1}{4}$

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

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

(d) $\frac{1}{5}$

(c) $\frac{1}{6}$

Question 27:

Let f(x) = 2x3 $-$ 3x2 $-$ 12x + 5 on [$-$2, 4]. The relative maximum occurs at x=

(a) $-$2

(b) $-$1

(c) 2

(d) 4

(c) 2

Question 28:

The minimum value of x loge x is equal to

(a) e

(b) 1/e

(c) $-$1/e

(d) 2/e

(e) $-$e

Question 29:

The minimum value of the function f(x) = 2x3$-$21x2+36x$-$20 is

(a) $-$128

(b) $-$126

(c) $-$120

(d) none of these

(a)$-$128

Question 30:

Let the function f : R → R be defined by f(x) = 2x + cos x, then f(x
(a) has a minimum at x = π                         (b) has a maximum at x = 0
(c) is a decreasing function                        (d) is an increasing function

The given function is f(x) = 2x + cosx.

f(x) = 2x + cosx

Differentiating both sides with respect to x, we get

$f\text{'}\left(x\right)=2-\mathrm{sin}x$

We know

−1 ≤ sin≤ 1

$\therefore f\text{'}\left(x\right)=2-\mathrm{sin}x>0$ ∀ x ∈ R

⇒ f(x) is an increasing function for all x ∈ R

Thus, the function f : R → R defined by f(x) = 2x + cosis an increasing function.

Hence, the correct answer is option (d).

Question 31:

If x is real, the minimum value x2 - 8x + 17 is
(a) -1          (b) 0              (c) 1              (d) 2

Let f(x) = x− 8x + 17

Differentiating both sides with respect to x, we get

$f\text{'}\left(x\right)=2x-8$

For maxima or minima,

$f\text{'}\left(x\right)=0$

$⇒2x-8=0$

$⇒x=4$

Now,

$f\text{'}\text{'}\left(x\right)=2>0$

So, x = 4 is the point of local minimum of f(x).

∴ Minimum value of f(x) = f(4) = (4)2 − 8 × 4 + 17 = 16 − 32 + 17 = 1

Thus, the minimum value of x− 8x + 17 is 1.

Hence, the correct answer is option (c).

Question 32:

The maximum value of sin x cos x is

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

$⇒f\left(x\right)=\frac{1}{2}\mathrm{sin}2x$

Differentiating both sides with respect to x, we get

$f\text{'}\left(x\right)=\frac{1}{2}\mathrm{cos}2x×2=\mathrm{cos}2x$

For maxima or minima,

$f\text{'}\left(x\right)=0$

$⇒\mathrm{cos}2x=0$

$⇒2x=\frac{\mathrm{\pi }}{2}$

$⇒x=\frac{\mathrm{\pi }}{4}$

Now,

$f\text{'}\text{'}\left(x\right)=-2\mathrm{sin}2x$

$⇒f\text{'}\text{'}\left(\frac{\mathrm{\pi }}{4}\right)=-2\mathrm{sin}\left(2×\frac{\mathrm{\pi }}{4}\right)=-2\mathrm{sin}\frac{\mathrm{\pi }}{2}=-2<0$

So, $x=\frac{\mathrm{\pi }}{4}$ is the point of local maximum.

∴ Maximum value of f(x)

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

$=\frac{1}{2}\mathrm{sin}\left(2×\frac{\mathrm{\pi }}{4}\right)$

$=\frac{1}{2}\mathrm{sin}\frac{\mathrm{\pi }}{2}$

$=\frac{1}{2}×1$

$=\frac{1}{2}$

Thus, the maximum value of sinx cosx is $\frac{1}{2}$.

Hence, the correct answer is option (b).

Question 33:

The maximum value of ${\left(\frac{1}{x}\right)}^{x}$ is
(a)              (b) ee             (c) e1/e              (d) ${\left(\frac{1}{e}\right)}^{1/e}$

Let $f\left(x\right)={\left(\frac{1}{x}\right)}^{x}$.

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

$⇒\mathrm{log}f\left(x\right)=\mathrm{log}{\left(\frac{1}{x}\right)}^{x}$

$⇒\mathrm{log}f\left(x\right)=x\mathrm{log}\left(\frac{1}{x}\right)$

$⇒\mathrm{log}f\left(x\right)=-x\mathrm{log}x$

Differentiating both sides with respect to x, we get

$\frac{1}{f\left(x\right)}×f\text{'}\left(x\right)=-\left(x×\frac{1}{x}+\mathrm{log}x×1\right)$

$⇒f\text{'}\left(x\right)=-{\left(\frac{1}{x}\right)}^{x}\left(1+\mathrm{log}x\right)$          .....(1)

For maxima or minima,

$f\text{'}\left(x\right)=0$

$⇒-{\left(\frac{1}{x}\right)}^{x}\left(1+\mathrm{log}x\right)=0$

$⇒\mathrm{log}x=-1$

$⇒x={e}^{-1}=\frac{1}{e}$

Now,

$f\text{'}\text{'}\left(x\right)=-\left[{\left(\frac{1}{x}\right)}^{x}×\frac{1}{x}+\left(1+\mathrm{log}x\right)×\left\{-{\left(\frac{1}{x}\right)}^{x}\left(1+\mathrm{log}x\right)\right\}\right]$                          [Using (1)]

$⇒f\text{'}\text{'}\left(\frac{1}{e}\right)=-\left[{\left(e\right)}^{\frac{1}{e}}×e+\left(1+\mathrm{log}\frac{1}{e}\right)×\left\{-{\left(e\right)}^{\frac{1}{e}}\left(1+\mathrm{log}\frac{1}{e}\right)\right\}\right]$                    $\left(\mathrm{log}\frac{1}{e}=\mathrm{log}{e}^{-1}=-1\right)$

$⇒f\text{'}\text{'}\left(\frac{1}{e}\right)=-{e}^{\frac{1}{e}+1}-0=-{e}^{\frac{1}{e}+1}$

$⇒f\text{'}\text{'}\left(\frac{1}{e}\right)<0$

So, $x=\frac{1}{e}$ is the point of local maximum of f(x).

∴ Maximum value of f(x) = ${\left(\frac{1}{\frac{1}{e}}\right)}^{\frac{1}{e}}={e}^{\frac{1}{e}}$

Thus, the maximum value of ${\left(\frac{1}{x}\right)}^{x}$ is ${e}^{\frac{1}{e}}$.

Hence, the correct answer is option (c).

Question 34:

The function f(x) = xx has a stationary point at
(a) x = e             (b) x$\frac{1}{e}$             (c) x = 1                   (d) x = $\sqrt{e}$

The values of x for which $f\text{'}\left(x\right)=0$ are called stationary values.

Let $f\left(x\right)={x}^{x}$.

$f\left(x\right)={x}^{x}$

$⇒\mathrm{log}f\left(x\right)=\mathrm{log}{x}^{x}$

$⇒\mathrm{log}f\left(x\right)=x\mathrm{log}x$

Differentiating both sides with respect to x, we get

$\frac{1}{f\left(x\right)}×f\text{'}\left(x\right)=x×\frac{1}{x}+\mathrm{log}x×1$

$⇒f\text{'}\left(x\right)={x}^{x}\left(1+\mathrm{log}x\right)$

For stationary point, we have

$f\text{'}\left(x\right)=0$

$⇒{x}^{x}\left(1+\mathrm{log}x\right)=0$

$⇒\mathrm{log}x=-1$

$⇒x={e}^{-1}=\frac{1}{e}$

Thus, the function f(x) = xx has a stationary point at $x=\frac{1}{e}$.

Hence, the correct answer is option (b).

Question 35:

Maximum slope of the curve y = -x3 + 3x2 + 9x - 27 is
(a) 0              (b) 12             (c) 16               (d) 32

The given curve is y = −x3 + 3x2 + 9x 27.

Slope of the curve, m = $\frac{dy}{dx}$

m = $\frac{dy}{dx}=-3{x}^{2}+6x+9$

$\frac{dm}{dx}=-6x+6$

For maxima or minima,

$\frac{dm}{dx}=0$

$⇒-6x+6=0$

$⇒x=1$

Now,

$\frac{{d}^{2}m}{d{x}^{2}}=-6$ < 0

x  = 1 is the point of local maximum

So, the slope of the given curve is maximum when x = 1.

∴ Maximum value of the slope, m

$=-3×{\left(1\right)}^{2}+6×1+9$       (m = −3x2 + 6x + 9)

$=-3+6+9$

$=12$

Thus, the maximum slope of the curve y = −x3 + 3x2 + 9x 27 is 12.

Hence, the correct answer is option (b).

Question 36:

The function f(x) = 2x3 - 3x2 - 12x + 4, has
(a) two points of local maximum              (b) two points of local minimum
(c) one maximum and one minimum        (d) no maximum no minimum

The given function is f(x) = 2x3 − 3x2 − 12x + 4.

f(x) = 2x3 − 3x2 − 12x + 4

Differentiating both sides with respect to x, we get

$f\text{'}\left(x\right)=6{x}^{2}-6x-12$

$⇒f\text{'}\left(x\right)=6\left({x}^{2}-x-2\right)$

$⇒f\text{'}\left(x\right)=6\left(x+1\right)\left(x-2\right)$

For maxima or minima,

$f\text{'}\left(x\right)=0$

⇒ 6(x + 1)(x − 2) = 0

x + 1 = 0 or x − 2 = 0

x = −1 or x = 2

Now,

$f\text{'}\text{'}\left(x\right)=12x-6$

At x = −1, we have

$f\text{'}\text{'}\left(-1\right)=12×\left(-1\right)-6=-12-6=-18$ < 0

So, x = −1 is the point of local maximum.

At x = 2, we have

$f\text{'}\text{'}\left(2\right)=12×2-6=24-6=18$ > 0

So, x = 2 is the point of local minimum.

Thus, the given function f(x) = 2x3 − 3x2 − 12x + 4 has one point of local maximum and one point of local minimum.

Hence, the correct answer is option (c).

Question 1:

If f(x) = $\frac{1}{4{x}^{2}+2x+1},$ then its maximum value is ______________.

The given function is $f\left(x\right)=\frac{1}{4{x}^{2}+2x+1}$.

The function f(x) would attain its maximum value, when the value of 4x2 + 2x + 1 is minimum.

Let g(x) = 4x2 + 2x + 1

$\therefore g\text{'}\left(x\right)=8x+2$

For maxima or minima,

$g\text{'}\left(x\right)=0$

$⇒8x+2=0$

$⇒x=-\frac{1}{4}$

Now,

$g\text{'}\text{'}\left(x\right)=8>0$

So, $x=-\frac{1}{4}$ is the point of local minimum of g(x)

Minimum value of function g(x)

$=g\left(-\frac{1}{4}\right)$

$=4×{\left(-\frac{1}{4}\right)}^{2}+2×\left(-\frac{1}{4}\right)+1$

$=\frac{1}{4}-\frac{1}{2}+1$

$=\frac{3}{4}$

∴ Maximum value of f(x) $=\frac{1}{\left(\frac{3}{4}\right)}$$=\frac{4}{3}$

Thus, the maximum value of the function $f\left(x\right)=\frac{1}{4{x}^{2}+2x+1}$ is $\frac{4}{3}$.

If f(x) = $\frac{1}{4{x}^{2}+2x+1},$ then its maximum value is .

Question 2:

The minimum value of f(x) = sin x in $\left[\frac{-\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right]$ is ____________________.

The given function is $f\left(x\right)=\mathrm{sin}x$, $x\in \left[\frac{-\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right]$.

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

Differentiating both sides with respect to x, we get

$f\text{'}\left(x\right)=\mathrm{cos}x$

For maxima or minima,

$f\text{'}\left(x\right)=0$

$⇒\mathrm{cos}x=0$

$⇒x=-\frac{\mathrm{\pi }}{2}$ or $x=\frac{\mathrm{\pi }}{2}$, for all $x\in \left[\frac{-\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right]$

Now,

$f\text{'}\text{'}\left(x\right)=-\mathrm{sin}x$

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

$f\text{'}\text{'}\left(\frac{\mathrm{\pi }}{2}\right)=-\mathrm{sin}\frac{\mathrm{\pi }}{2}=-1<0$

So, $x=\frac{\mathrm{\pi }}{2}$ is the point of local maximum of f(x).

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

$f\text{'}\text{'}\left(-\frac{\mathrm{\pi }}{2}\right)=-\mathrm{sin}\left(-\frac{\mathrm{\pi }}{2}\right)=\mathrm{sin}\frac{\mathrm{\pi }}{2}=1>0$              [sin(−θ) = −sinθ]

So, $x=-\frac{\mathrm{\pi }}{2}$ is the point of local minimum of f(x).

∴ Minimum value of f(x) = $f\left(-\frac{\mathrm{\pi }}{2}\right)=\mathrm{sin}\left(-\frac{\mathrm{\pi }}{2}\right)=-\mathrm{sin}\frac{\mathrm{\pi }}{2}=-1$

Thus, the minimum value of f(x) = sinx in $\left[\frac{-\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right]$ is −1.

The minimum value of f(x) = sin x in $\left[\frac{-\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right]$ is ___−1___.

Question 3:

The maximum value of f(x) = sin x + cos x is _______________.

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

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

Differentiating both sides with respect to x, we get

$f\text{'}\left(x\right)=\mathrm{cos}x-\mathrm{sin}x$

For maxima or minima,

$f\text{'}\left(x\right)=0$

$⇒\mathrm{cos}x-\mathrm{sin}x=0$

$⇒\mathrm{cos}x=\mathrm{sin}x$

$⇒\mathrm{tan}x=1$

$⇒x=\frac{\mathrm{\pi }}{4},\frac{5\mathrm{\pi }}{4}$          (Let us only consider the values of x ∈ [0, 2$\mathrm{\pi }$])

Now,

$f\text{'}\text{'}\left(x\right)=-\mathrm{sin}x-\mathrm{cos}x$

At $x=\frac{5\mathrm{\pi }}{4}$, we have

$f\text{'}\text{'}\left(\frac{5\mathrm{\pi }}{4}\right)$ $=-\mathrm{sin}\frac{5\mathrm{\pi }}{4}-\mathrm{cos}\frac{5\mathrm{\pi }}{4}$ $=-\left(-\frac{1}{\sqrt{2}}\right)-\left(-\frac{1}{\sqrt{2}}\right)$ $=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}$ $=\frac{2}{\sqrt{2}}$ $=\sqrt{2}$ > 0

So, $x=\frac{5\mathrm{\pi }}{4}$ is the point of local minimum of f(x).

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

$f\text{'}\text{'}\left(\frac{\mathrm{\pi }}{4}\right)$ $=-\mathrm{sin}\frac{\mathrm{\pi }}{4}-\mathrm{cos}\frac{\mathrm{\pi }}{4}$ $=-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}$ $=-\frac{2}{\sqrt{2}}$ $=-\sqrt{2}$ < 0

So, $x=\frac{\mathrm{\pi }}{4}$ is the point of local maximum of f(x).

∴ Maximum value of f(x) $=f\left(\frac{\mathrm{\pi }}{4}\right)$ $=\mathrm{sin}\frac{\mathrm{\pi }}{4}+\mathrm{cos}\frac{\mathrm{\pi }}{4}$ $=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}$ $=\frac{2}{\sqrt{2}}$ $=\sqrt{2}$

Thus, the maximum value of $f\left(x\right)=\mathrm{sin}x+\mathrm{cos}x$ is $\sqrt{2}$.

The maximum value of f(x) = sinx + cosx is .

Question 4:

If f(x) has the second order derivative at x = c such that '(c) = 0 and ''(c) > 0, then c is a point of _______________.

Second derivative test: Let f(x) be a function defined on an interval I and c ∈ I. Suppose f(x) be twice differentiable at x = c. Then, x = c is a point of local minima if '(c) = 0 and ''(c) > 0. In this case, f(c) is then the local minimum value of f(x).

So, if f(x) has the second order derivative at x = c such that '(c) = 0 and ''(c) > 0, then c is a point of local minima.

If f(x) has the second order derivative at x = c such that '(c) = 0 and ''(c) > 0, then c is a point of ___local minima___.

Question 5:

If '(x) changes its sign from positive to negative as x increases through c in the interval (c − h, c + h), then x = c is a point of ______________.

First derivative test states that if '(x) changes sign from positive to negative as x increases through c, then c is a point of local maxima, and f(c) is local maximum value.

Thus, if '(x) changes its sign from positive to negative as x increases through c in the interval (c − h, c + h), then x = c is a point of local maximum.

If '(x) changes its sign from positive to negative as x increases through c in the interval (c − h, c + h), then x = c is a point of ___local maximum___.

Question 6:

If '(x) changes its sign from negative to positive as x increases through c in the interval (c − h, c + h), then x = c is a point of ______________.

First derivative test states that if '(x) changes sign from negative to positive as x increases through c, then c is a point of local minima, and f(c) is local minimum value.

Thus, if '(x) changes its sign from negative to positive as x increases through c in the interval (c − h, c + h), then x = c is a point of local minimum.

If '(x) changes its sign from negative to positive as x increases through c in the interval (c − h, c + h), then x = c is a point of ___local minimum___.

Question 7:

The positive real number x when added to its reciprocal gives the minimum value of the sum when, x = __________________.

Let S(x) be the sum of the positive real number x and its reciprocal.

$\therefore S\left(x\right)=x+\frac{1}{x}$

Differentiating both sides with respect to x, we get

$S\text{'}\left(x\right)=1-\frac{1}{{x}^{2}}$

For maxima or minima,

$S\text{'}\left(x\right)=0$

$⇒1-\frac{1}{{x}^{2}}=0$

$⇒{x}^{2}=1$

⇒ x = −1 or x = 1

Now,

$S\text{'}\text{'}\left(x\right)=\frac{2}{{x}^{3}}$

At x = −1, we have

$S\text{'}\text{'}\left(1\right)=\frac{2}{{\left(-1\right)}^{3}}=-2<0$

So, x = −1 is the point of local maximum of S(x).

At x = 1, we have

$S\text{'}\text{'}\left(1\right)=\frac{2}{{\left(1\right)}^{3}}=2>0$

So, x = 1 is the point of local minimum of S(x).

Therefore, S(x) is minimum when x =1.

Thus, the sum of positive real number x and its reciprocal is minimum when x = 1.

The positive real number x when added to its reciprocal gives the minimum value of the sum when, x = ___1___.

Question 8:

The real number which must exceeds its cube is _______________.

Let the real number be x.

The cube of the number is x3.

Let f(x) = x − x3

Differentiating both sides with respect to x, we get

$f\text{'}\left(x\right)=1-3{x}^{2}$

For maxima or minima,

$f\text{'}\left(x\right)=0$

$⇒1-3{x}^{2}=0$

$⇒{x}^{2}=\frac{1}{3}$

$⇒x=±\frac{1}{\sqrt{3}}$

Now,

$f\text{'}\text{'}\left(x\right)=-6x$

At $x=-\frac{1}{\sqrt{3}}$, we have

$f\text{'}\text{'}\left(-\frac{1}{\sqrt{3}}\right)=-6×\left(-\frac{1}{\sqrt{3}}\right)=2\sqrt{3}>0$

So, $x=-\frac{1}{\sqrt{3}}$ is the point of local minimum of f(x).

At $x=\frac{1}{\sqrt{3}}$, we have

$f\text{'}\text{'}\left(\frac{1}{\sqrt{3}}\right)=-6×\frac{1}{\sqrt{3}}=-2\sqrt{3}<0$

So, $x=\frac{1}{\sqrt{3}}$ is the point of local maximum of f(x).

Thus, the real number which must exceeds its cube is $x=\frac{1}{\sqrt{3}}$.

The real number which must exceeds its cube is .

Question 9:

The function f(x) = ax$\frac{b}{x}$ , a, b, x > 0 takes on the least value at x equal to __________________.

The given function is $f\left(x\right)=ax+\frac{b}{x}$a, b, x > 0.

$f\left(x\right)=ax+\frac{b}{x}$

Differentiating both sides with respect to x, we get

$f\text{'}\left(x\right)=a-\frac{b}{{x}^{2}}$

For maxima or minima,

$f\text{'}\left(x\right)=0$

$⇒a-\frac{b}{{x}^{2}}=0$

$⇒{x}^{2}=\frac{b}{a}$

$x=\sqrt{\frac{b}{a}}$    (x > 0)

Now,

$f\text{'}\text{'}\left(x\right)=\frac{2b}{{x}^{3}}$

At $x=\sqrt{\frac{b}{a}}$, we have

$f\text{'}\text{'}\left(\sqrt{\frac{b}{a}}\right)=\frac{2b}{{\left(\sqrt{\frac{b}{a}}\right)}^{3}}=2a\sqrt{\frac{a}{b}}>0$

So, $x=\sqrt{\frac{b}{a}}$ is the point of local minimum of f(x).

Thus, the function takes the least value at $x=\sqrt{\frac{b}{a}}$.

The function f(x) = ax + $\frac{b}{x}$ , a, b, x > 0 takes on the least value at x equal to .

Question 10:

If y = a log x + bx2 + x has its extreme values at x = 1 and = 2, then (a, b) = ____________________.

It is given that, $y=a\mathrm{log}x+b{x}^{2}+x$ has its extreme values at x = 1 and = 2.

$\therefore \frac{dy}{dx}=0$ at x = 1 and = 2

$y=a\mathrm{log}x+b{x}^{2}+x$

Differentiating both sides with respect to x, we get

$\frac{dy}{dx}=\frac{a}{x}+2bx+1$

Now,

${\left(\frac{dy}{dx}\right)}_{x=1}=0$

$⇒a+2b+1=0$

Also,

${\left(\frac{dy}{dx}\right)}_{x=2}=0$

$⇒\frac{a}{2}+4b+1=0$

Subtracting (1) from (2), we get

6b = −1

$⇒b=-\frac{1}{6}$

Putting $b=-\frac{1}{6}$ in (1), we get

$a+2×\left(-\frac{1}{6}\right)=-1$

$⇒a=-1+\frac{1}{3}=-\frac{2}{3}$

Thus, the values of a and b are $-\frac{2}{3}$ and $-\frac{1}{6}$, respectively.

Hence, the ordered pair (a, b) is $\left(-\frac{2}{3},-\frac{1}{6}\right)$.

If y = alogx + bx2 + x has its extreme values at x = 1 and = 2, then (a, b) = .

Question 11:

If the function f(x) = a sin x has an extremum at x$\frac{\mathrm{\pi }}{3}$ then a = _________________.

It is given that, the function $f\left(x\right)=a\mathrm{sin}x+\frac{1}{3}\mathrm{sin}3x$ has an extremum at x = $\frac{\mathrm{\pi }}{3}$.

$\therefore f\text{'}\left(x\right)=0$ at x = $\frac{\mathrm{\pi }}{3}$

$f\left(x\right)=a\mathrm{sin}x+\frac{1}{3}\mathrm{sin}3x$

Differentiating both sides with respect to x, we get

$f\text{'}\left(x\right)=a\mathrm{cos}x+\frac{1}{3}×3\mathrm{cos}3x$

$⇒f\text{'}\left(x\right)=a\mathrm{cos}x+\mathrm{cos}3x$

Now,

$f\text{'}\left(\frac{\mathrm{\pi }}{3}\right)=0$

$⇒a\mathrm{cos}\left(\frac{\mathrm{\pi }}{3}\right)+\mathrm{cos}3\left(\frac{\mathrm{\pi }}{3}\right)=0$

$⇒a×\frac{1}{2}+\mathrm{cos\pi }=0$

$⇒\frac{a}{2}-1=0$

$⇒a=2$

Thus, the value of a is 2.

If the function $f\left(x\right)=a\mathrm{sin}x+\frac{1}{3}\mathrm{sin}3x$ has an extremum at x = $\frac{\mathrm{\pi }}{3}$ then a = ___2___.

Question 12:

The maximum value of f(x) = x ex is _______________.

The given function is f(x) = xex.

$f\left(x\right)=x{e}^{-x}$

Differentiating both sides with respect to x, we get

$f\text{'}\left(x\right)=x×{e}^{-x}×\left(-1\right)+{e}^{-x}×1$

$⇒f\text{'}\left(x\right)={e}^{-x}\left(-x+1\right)$

For maxima or minima,

$f\text{'}\left(x\right)=0$

$⇒{e}^{-x}\left(-x+1\right)=0$

$⇒x=1$

Now,

$f\text{'}\text{'}\left(x\right)={e}^{-x}×\left(-1\right)+\left(-x+1\right)×{e}^{-x}×\left(-1\right)$

$⇒f\text{'}\text{'}\left(x\right)={e}^{-x}\left(x-2\right)$

At x = 1, we have

$f\text{'}\text{'}\left(1\right)={e}^{-1}\left(1-2\right)=-\frac{1}{e}<0$

So, x = 1 is the point of local maximum of f(x).

∴ Maximum value of f(x) = f(1) = 1 × e−1 = $\frac{1}{e}$

Thus, the maximum value of f(x) = xex is $\frac{1}{e}$.

The maximum value of f(x) = xex is .

Question 13:

If the function f(x) = x4 - 62x2 + ax + 9 attains a local maximum at x = 1, then a = _________________.

It is given that, the function f(x) = x4 − 62x2 + ax + 9 attains a local maximum at x = 1.

$\therefore f\text{'}\left(x\right)=0$ at x = 1

f(x) = x4 − 62x2 + ax + 9

Differentiating both sides with respect to x, we get

$f\text{'}\left(x\right)=4{x}^{3}-124x+a$

Now,

$f\text{'}\left(1\right)=0$

$⇒4×{\left(1\right)}^{3}-124×1+a=0$

$⇒a=124-4=120$

Thus, the value of a is 120.

Also,

$f\text{'}\text{'}\left(x\right)=12{x}^{2}-124$

At x = 1, we have

$f\text{'}\text{'}\left(1\right)=12×{\left(1\right)}^{2}-124=12-124=-112<0$

So, x = 1 is the point of local maximum of f(x).

If the function f(x) = x4 − 62x2 + ax + 9 attains a local maximum at x = 1, then a = ___120____.

Question 14:

If the sum of two non-zero numbers is 4, then the minimum value of the sum of their reciprocals is _______________.

Let the two numbers be x and 4 − x (x ≠ 0, 4).

Suppose S be the sum of their reciprocals.

Differentiating both sides with respect to x, we get

$\frac{dS}{dx}=-\frac{1}{{x}^{2}}-\frac{1}{{\left(4-x\right)}^{2}}×\left(-1\right)$

$⇒\frac{dS}{dx}=-\frac{1}{{x}^{2}}+\frac{1}{{\left(4-x\right)}^{2}}$

For maxima or minima,

$\frac{dS}{dx}=0$

$⇒-\frac{1}{{x}^{2}}+\frac{1}{{\left(4-x\right)}^{2}}=0$

$⇒-\left(16-8x+{x}^{2}\right)+{x}^{2}=0$

$⇒8x=16$

$⇒x=2$

Now,

$\frac{{d}^{2}S}{d{x}^{2}}=\frac{2}{{x}^{3}}+\frac{2}{{\left(4-x\right)}^{2}}$

At x = 2, we have

${\left(\frac{{d}^{2}S}{d{x}^{2}}\right)}_{x=2}=$ $\frac{2}{{\left(2\right)}^{3}}+\frac{2}{{\left(4-2\right)}^{2}}=$ $\frac{1}{4}+\frac{1}{2}=$ $\frac{3}{4}>0$

So, x = 2 is the point of local minimum.

Thus, S is minimum when x = 2.

∴ Minimum value of S$\frac{1}{2}+\frac{1}{4-2}$ = $\frac{1}{2}+\frac{1}{2}$ = 1          $\left(S=\frac{1}{x}+\frac{1}{4-x}\right)$

Thus, if the sum of two non-zero numbers is 4, then the minimum value of the sum of their reciprocals is 1.

If the sum of two non-zero numbers is 4, then the minimum value of the sum of their reciprocals is ___1___.

Question 15:

If x and y are two real numbers such that x > 0 and xy = 1. The the minimum value of x + y is ________________.

It is given that, x and y are two real numbers such that x > 0 and xy = 1.

Let S = x + y.

Now, $xy=1⇒y=\frac{1}{x}$

$\therefore S=x+y=x+\frac{1}{x}$

Differentiating both sides with respect to x, we get

$\frac{dS}{dx}=1-\frac{1}{{x}^{2}}$

For maxima or minima,

$\frac{dS}{dx}=0$

$⇒1-\frac{1}{{x}^{2}}=0$

$⇒{x}^{2}=1$

x = 1    (x > 0)

Now,

$\frac{{d}^{2}S}{d{x}^{2}}=\frac{2}{{x}^{3}}$

At x = 1, we have

${\left(\frac{{d}^{2}S}{d{x}^{2}}\right)}_{x=1}=\frac{2}{{\left(1\right)}^{3}}=2>0$

So, x = 1 is the point of local minimum.

Thus, S is minimum when x = 1.

When x = 1, $y=\frac{1}{x}=1$

∴ Minimum value of S = x + y = 1 + 1 = 2

Thus, the minimum value of x + y is 2.

If x and y are two real numbers such that x > 0 and xy = 1. The the minimum value of x + y is ___2___.

Question 16:

The number that exceeds its square by the greatest amount is _______________.

Let the number be x.

The square of the number is x2.

Let f(x) = x − x2. Now, we need to find the value of x for which f(x) is maximum.

f
(x) = x − x2

Differentiating both sides with respect to x, we get

$f\text{'}\left(x\right)=1-2x$

For maxima or minima,

$f\text{'}\left(x\right)=0$

$⇒1-2x=0$

$⇒x=\frac{1}{2}$

Now,

$f\text{'}\text{'}\left(x\right)=-2<0$

So, $x=\frac{1}{2}$ is the point of local maximum of f(x). Therefore, f(x) is maximum when $x=\frac{1}{2}$.

Thus, the number that exceeds its square by the greatest amount is $x=\frac{1}{2}$.

The number that exceeds its square by the greatest amount is .

Question 17:

If m and M respectively denote the minimum and maximum values of f(x) = (x + 1)2 + 3 in the interval [−3, 1], then the ordered pair (m, M) = _________.

The given function is f(x) = (x + 1)2 + 3, x ∈ [−3, 1].

f(x) = (x + 1)2 + 3

Differentiating both sides with respect to x, we get

$f\text{'}\left(x\right)=2\left(x+1\right)$

For maxima or minima,

$f\text{'}\left(x\right)=0$

$⇒2\left(x+1\right)=0$

$⇒x+1=0$

$⇒x=-1$

Now,

$f\text{'}\text{'}\left(x\right)=2>0$

So, x = −1 is the point of local minimum of f(x).

At x = −1, we have

f(−1) = (−1 + 1)2 + 3 = 0 + 3 = 3

At x = −3, we have

f(−3) = (−3 + 1)2 + 3 = 4 + 3 = 7

At x = 1, we have

f(1) = (1 + 1)2 + 3 = 4 + 3 = 7

Thus, the minimum value of f(x) is 3 and the maximum value of f(x) is 7.

∴ m = 3 and M = 7

Thus, the ordered pair (m, M) is (3, 7).

If m and M respectively denote the minimum and maximum values of f(x) = (x + 1)2 + 3 in the interval [−3, 1], then the ordered pair (m, M) = ___(3, 7)___.

Question 18:

The minimum value of f(x) = x2$\frac{250}{x}$ is _______________.

The given function is $f\left(x\right)={x}^{2}+\frac{250}{x}$, x ≠ 0.

$f\left(x\right)={x}^{2}+\frac{250}{x}$

Differentiating both sides with respect to x, we get

$f\text{'}\left(x\right)=2x-\frac{250}{{x}^{2}}$

For maxima or minima,

$f\text{'}\left(x\right)=0$

$⇒2x-\frac{250}{{x}^{2}}=0$

$⇒{x}^{3}=\frac{250}{2}=125$

$⇒x=5$

Now,

$f\text{'}\text{'}\left(x\right)=2+\frac{500}{{x}^{3}}$

At x = 5, we have

$f\text{'}\text{'}\left(5\right)=$$2+\frac{500}{{\left(5\right)}^{3}}=$$2+\frac{500}{125}=$2 + 4 = 6 > 0

So, x = 5 is the point of local minimum.

∴ Minimum value of f(x) = $f\left(5\right)={\left(5\right)}^{2}+\frac{250}{5}$ = 25 + 50 = 75                  $\left[f\left(x\right)={x}^{2}+\frac{250}{x}\right]$

Thus, the minimum value of the given function is 75.

The minimum value of $f\left(x\right)={x}^{2}+\frac{250}{x}$ is ____75____.

Question 19:

The maximum slope of the curve y = −x3 + 3x2 + 9x 27 is _________________.

The given curve is y = −x3 + 3x2 + 9x 27.

Slope of the curve, m = $\frac{dy}{dx}$

m = $\frac{dy}{dx}=-3{x}^{2}+6x+9$

Differentiating both sides with respect to x, we get

$\frac{dm}{dx}=-6x+6$

For maxima or minima,

$\frac{dm}{dx}=0$

$⇒-6x+6=0$

$⇒x=1$

Now,

$\frac{{d}^{2}m}{d{x}^{2}}=-6<0$

So, x = 1 is the point of local maximum.

Thus, the slope of the given curve is maximum when x = 1.

∴ Maximum value of the slope

$=-3×{\left(1\right)}^{2}+6×1+9$           (Slope, m = $-3{x}^{2}+6x+9$)

= −3 + 6 + 9

= 12

Hence, the maximum slope of the curve y = −x3 + 3x2 + 9x 27 is 12.

The maximum slope of the curve y = −x3 + 3x2 + 9x 27 is ___12___.

Question 20:

The function f(x) = $\frac{x}{2}+\frac{2}{x}$ has a local minimum at x = __________________.

The given function is $f\left(x\right)=\frac{x}{2}+\frac{2}{x}$x ≠ 0.

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

Differentiating both sides with respect to x, we get

$f\text{'}\left(x\right)=\frac{1}{2}-\frac{2}{{x}^{2}}$

For maxima or minima,

$f\text{'}\left(x\right)=0$

$⇒\frac{1}{2}-\frac{2}{{x}^{2}}=0$

$⇒{x}^{2}=4$

⇒ x = −2 or x = 2

Now,

$f\text{'}\text{'}\left(x\right)=\frac{4}{{x}^{3}}$

At x = −2, we have

$f\text{'}\text{'}\left(-2\right)=\frac{4}{{\left(-2\right)}^{3}}=-\frac{1}{2}<0$

So, x = −2 is the point of local maximum of f(x).

At x = 2, we have

$f\text{'}\text{'}\left(2\right)=\frac{4}{{\left(2\right)}^{3}}=\frac{1}{2}>0$

So, x = 2 is the point of local minimum of f(x).

Thus, the given function f(x) = $\frac{x}{2}+\frac{2}{x}$ has a local minimum at x = 2.

The function f(x) = $\frac{x}{2}+\frac{2}{x}$ has a local minimum at x = ____2____.

Question 21:

The least value of the function f(x) = ax + $\frac{b}{x}$(a > 0, b > 0, x > 0) is ________________.

The given function is $f\left(x\right)=ax+\frac{b}{x}$ (a > 0, b > 0, x > 0).

$f\left(x\right)=ax+\frac{b}{x}$

Differentiating both sides with respect to x, we get

$f\text{'}\left(x\right)=a-\frac{b}{{x}^{2}}$

For maxima or minima,

$f\text{'}\left(x\right)=0$

$⇒a-\frac{b}{{x}^{2}}=0$

$⇒{x}^{2}=\frac{b}{a}$

$⇒x=\sqrt{\frac{b}{a}}$        (x > 0)

Now,

$f\text{'}\text{'}\left(x\right)=\frac{2b}{{x}^{3}}$

At $x=\sqrt{\frac{b}{a}}$, we have

$f\text{'}\text{'}\left(\sqrt{\frac{b}{a}}\right)=\frac{2b}{{\left(\sqrt{\frac{b}{a}}\right)}^{3}}=2a\sqrt{\frac{a}{b}}>0$           (a > 0, b > 0)

So, $x=\sqrt{\frac{b}{a}}$ is the point of local minimum of f(x). Thus, the function takes the least value at $x=\sqrt{\frac{b}{a}}$.

∴ Least value of the given function

$=f\left(\sqrt{\frac{b}{a}}\right)$

$=a×\sqrt{\frac{b}{a}}+\frac{b}{\sqrt{\frac{b}{a}}}$              $\left[f\left(x\right)=ax+\frac{b}{x}\right]$

$=\sqrt{ab}+\sqrt{ab}$

$=2\sqrt{ab}$

Thus, the least value of the function $f\left(x\right)=ax+\frac{b}{x}$ (a > 0, b > 0, x > 0) is $2\sqrt{ab}$.

The least value of the function $f\left(x\right)=ax+\frac{b}{x}$ (a > 0, b > 0, x > 0) is .

Question 1:

Write necessary condition for a point x = c to be an extreme point of the function f(x).

We know that at the extreme points of a function f(x), the first order derivative of the function is equal to zero, i.e.
f '(x) = 0 at x = c
$⇒$f '(c) = 0

Question 2:

Write sufficient conditions for a point x=c to be a point of local maximum.

We know that at the extreme points of a function f(x), the first order derivative of the function is equal to zero, i.e.
f '(x) = 0 at x = c
$⇒$f '(c) = 0

Also, at the point of local maximum, the second order derivative of the function at the given point must be less than zero, i.e.
f ''(c) < 0

Question 3:

If f(x) attains a local minimum at x=c, then write the values of f' (c) and f'' (c).

If f(x) attains a local minimum at x = c, then the first order derivative of the function at the given point must be equal to zero, i.e.
f '(x) = 0 at x = c
$⇒$f '(c) = 0

The second order derivative of the function at the given point must be greater than zero, i.e.
f ''(c) > 0

Question 4:

Write the minimum value of f(x) =

Question 5:

Write the maximum value of f(x) =

Question 6:

Write the point where f(x) = x log, x attains minimum value.

.

Question 7:

Find the least value of f(x) = $ax+\frac{b}{x}$, where a>0, b>0 and x>0.

Question 8:

Write the minimum value of f(x) = xx .

Question 9:

Write the maximum value of f(x) = x1/x.

Question 10:

Write the maximum value of f(x) = , if it exists.