Sir please clear my doubt in quadratic equations. i face problem in dealing with such questions. one is here:-

Find the values of m for which the equation sin^2x-(m-3)sinx+m=0 has real roots.

I can assume sinx to be t and reduce the equation to simple for. now t is binded in the interval -1,1. Now how to do sir? please clear!

sin2x-(m-3)sinx+m=0Let sinx=yy2-(m-3)y+m=0For real roots,D=b2-4ac>0So, -m-32-4(1)(m)>0m2+9-6m-4m>0m2-10m+9>0m2-m-9m+9=>0(m-9)(m-1)>0Compute the signs of the factors of (m-9)(m-1)
  m<1 m=1 1<m<9 m=9 m>9
m-9 - - - 0 +
m-1 - 0 + + +
(m-9)(m-1) + 0 - 0 +
Choosing the ranges that satisfy the required condition:>0m<1 or m>9i.e m(-,1)9,

  • -1
What are you looking for?