Question

consider the expression g(x)=sin^2x-(b+1)sinx+3(b-2) where b is parameter then: (A)No. of integral values of a for which both the roots of equation g(x)=0 has exactly one root in interval[0,pie}are? (B)If the roots of equation g(x)=0  are two distinct roots between (0,pie) then b lies in interval? (C)if g(x) is non-negative for all real x then b lies in interval

Answer 1

Answer:

For Part (A) and (B)

Given

$g(x)=\sin^2x-(b+1) \sin x+3(b-2)$

If g(x)=0

then $\sin^2x-(b+1) \sin x+3(b-2)=0$

Let sinx=t

$\t^2-(b+1) t+3(b-2)=0$

(A) For this to have one root

$(b+1)^2-4\times3(b-2)=0$

$\implies (b^2+2b+1)-12b+24=0$

$\implies b^2-10b+25=0$

$\implies (b-5)^2=0$

$\implies b=5$

Thus the number of values of b is 1

(B) If the roots are distinct

then discriminant D>0

$\implies (b-5)^2 > 0$

which is true for all b ≠ 5

Therefore b lies in the interval

$b \in (-\infty, \infty)-5$