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JEE MATHEMATICS MCQ: FUNCTIONS

1 Easy

1 of 1

Application +4 | -1

If

f(x)=x^2+1

, then the value of

(f \circ f)(x)

is equal to

View answer and explanation

Correct option: A

x^4+2 x^2+2

Option A (

x^4 + 2x^2 + 2

) is correct because when you compute

(f \circ f)(x)

with

f(x) = x^2 + 1

, the result simplifies to

x^4 + 2x^2 + 2

.

More Questions from FUNCTIONS

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

f(x)=x(4-x)

f^{-1}(x)

is given by: If

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

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

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

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

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

f: R \rightarrow R

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)

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