1)

Let f(x) be a quadratic expression such that f(0)+f(1)=0.If f(-2)=0 , then 


A) $f\left(\frac{-2}{5}\right)=0$

B) $f\left(\frac{2}{5}\right)=0$

C) $f\left(\frac{-3}{5}\right)=0$

D) $f\left(\frac{3}{5}\right)=0$

Answer:

Option D

Explanation:

We have ,

 f(x) is a quadratic  equation

$\therefore$     $f(x)=ax^{2}+bx+c$

   f(0)= c, f(1)=a+b+c

$\Rightarrow$      f(0)+f(1)=0

$\Rightarrow$   c+a+b+c=0

$\Rightarrow$  a+b+2c=0        ........(i)

  $f(-2)=a(-2)^{2}+b(-2)+c$

0=4a-2b+c

$\Rightarrow$      $4a-2b+c=0$        .........(ii)

 From Eqs.(i) and (ii) , we getr

     $\frac{a}{5}=\frac{b}{7}=\frac{c}{-6}$

$\Rightarrow$      a=5k,b=7k,c=-6k

 $\therefore$   $ f(x)=k(5x^{2}+7x-6)$

$\Rightarrow$      $5x^{2}+7x-6=0$

  $\Rightarrow$     $5x^{2}+10x-3x-6=0$

$\Rightarrow$  $5x(x+2)-3(x+2)=0$

$\Rightarrow$     (x+2)(5x-3)=0

                         $x=-2,x=\frac{3}{5}$

Hence ,$f(\frac{3}{5})=0$