If `f(x)=0` is a quadratic equation such that `f(-pi)=f(pi)=0` and `f(pi/2)=-(3pi^2)/4,` then `lim_(x->-pi)(f(x))/("sin"(sinx)` is equal to (a)`0` (b) `pi` (c) `2pi` (d) none of these

Options

A. 0
B. `pi`
C. `2pi`
D. none of these

Answer

C

Explanation

It is given that `f(-pi)=f(pi)=0`. Therefore ,`-pi` and `pi` are zeroes of `f(x)`.
`therefore f(x)=a(x+pi)(x-pi),a ne 0`
` rArr f((pi)/(2))=-(3api^2)/(4)rArr(-3pi^2)/(4)=-(3api^2)/(4)rArr a=1`
`therefore f(x)=(x+pi)(x-pi)`
So, `lim_(xto-pi) (f(x))/(sin(sinx))`
`=lim_(xto-pi)((x-pi)(x+pi))/((sin(sinx))/(sinx)xx sinx )=-lim_(xto-pi)((x-pi))/((sin(sinx))/(sinx))xx((x+pi))/(sin(pi+x))=-((-pi-pi))/(1)xx1=2pi`