JEE MATHEMATICS MCQ: FUNCTIONS

1 Easy

1 of 1

Application +4 | -1

A polynomial function

f\left( x \right)

satisfies the condition

f\left( x \right)f\left( {\frac{1}{x}} \right) = f\left( x \right) + f\left( {\frac{1}{x}} \right)

f\left( {10} \right) = 1001,

then

f\left( {20} \right) =

View answer and explanation

Correct option: C

8001

Option C (8001) is correct because by substituting f(x) = x^n + 1 into the given condition and using f(10) = 1001, we find n = 3, thus f(20) = 20³ + 1 = 8001.

More Questions from FUNCTIONS

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f:(-\infty, 2] \longrightarrow(-\infty, 4]

, where

f(x)=x(4-x)

, then

f^{-1}(x)

is given by: If

g(f(x))=|\sin x|

and

f(g(x))=(\sin \sqrt{x})^2

, then Let

f:\left( {2,\;3} \right) \to \left( {0,\;1} \right)

be defined by

f\left( x \right) = x - \left[ x \right]

, then

{f^{ - 1}}\left( x \right)

equals Let

f: R \rightarrow R

be defined as

f(x)=x^4

, Choose the correct answer. Which of the following functions are odd functions Let

f(x)=(x+1)^2-1(x \geq-1)

. Then the set

S=\left\{x: f(x)=f^{-1}(x)\right\}

contains