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 (f(f(f(-3)))+f(f(0))/3 f(1)-2 f(3)-f(5)) is

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 (f(f(f(-3)))+f(f(0))/3 f(1)-2 f(3)-f(5)) is (A) (-2/5) (B) (-2/3) (C)  (2/5) (D) (2/3)

Tardigrade Admin · Accepted Answer

f ( x ) text is odd ∴ 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 ( Nr / Dr )=(-2/-5)=(2/5) .

Q. 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

$\frac{- 2}{5}$

$\frac{- 2}{3}$

$\frac{2}{5}$

$\frac{2}{3}$

Q. 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 3f(1)−2f(3)−f(5)f(f(f(−3)))+f(f(0))​ is

5−2​

3−2​

52​

32​

f(x) is odd ∴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=3f(1)−2f(3)−f(5)=3(2)−2(5)−(1)=6−10−1=−5 DrNr​=−5−2​=52​.

Q. 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

$\frac{- 2}{5}$

$\frac{- 2}{3}$

$\frac{2}{5}$

$\frac{2}{3}$