Question

Let $f ( x )$ be an odd function defined on $R$ such that $f (1)=2, f (3)=5$ and $f (-5)=-1$. The value of $\frac{f(f(f(-3)))+f(f(0))}{3 f(1)-2 f(3)-f(5)}$ is

• A. $\frac{-2}{5}$
• B. $\frac{-2}{3}$
• C. $ \frac{2}{5}$
• D. $\frac{2}{3}$

Answer 1

Answer: (C)

$ f ( x ) \text { is odd } \therefore f (- x )=- f ( x ) $
$f (0)=0$
$f (-1)=- f (1)=-2 $
$f (-3)=- f (3)=-5 $
$f (-5)=- f (5)=-1 $
$Nr = f ( f ( f (-3)))+ f ( f (0))= f ( f (-5))+ f (0)= f (-1)+0=-2 $
$Dr =3 f (1)-2 f (3)- f (5)=3(2)-2(5)-(1)=6-10-1=-5 $
$\frac{ Nr }{ Dr }=\frac{-2}{-5}=\frac{2}{5} .$