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))÷ 3f(1) -2f(3) -f(5) is​

Answer 1

Answer:

muze easka answer pata nahi

Answer 2

The value of f(f(f(-3))) + f(f(0))÷ 3f(1) -2f(3) -f(5) is​ -0.4.

Step-by-step explanation:

Given:

• f(x) is an odd function.
• f(1) = 2
• f(3) = 5
• f(-5) = 1

To be found: f(f(f(-3))) + f(f(0))÷ 3f(1) -2f(3) -f(5)

Formula Used: f(0) = 0 when function is odd.

Solution:

Since we have to evaluate the expression: f(f(f(-3))) + f(f(0))÷ 3f(1) -2f(3) -f(5)

so let us find each term simultaneously.

$f(f(f(-3)))=f(f(-f(3)))$

Since, if f(x) is and odd function then f(-x) = -f(x).

$f(f(-5))=f(1)\\=2$

Now,

$f(f(0))= f(0)\\=0$

As, if f(x) is an odd function then f(0)=0

Now, simplified expression will be:

$=\frac{(2 + 0)}{3f(1) -2f(3) -f(5)}\\=\frac{(2 + 0)}{3\times 2 -2\times 5 -1} \\=\frac{2}{6-10-1}\\=\frac{2}{6-11}\\=\frac{2}{-5}\\=\frac{-2}{5}$

Therefore the value of  f(f(f(-3))) + f(f(0))÷ 3f(1) -2f(3) -f(5) is​ -0.4.

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