JEE MATHEMATICS MCQ: FUNCTIONS

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Analyzing +4 | -1

f\left( x \right) = \sin x + \cos x,\;\;g\left( x \right) = {x^2} - 1

then

g\left( {f\left( x \right)} \right)

is invertible in the domain

View answer and explanation

Correct option: B

\left[ { - \frac{\pi }{4},\;\frac{\pi }{4}} \right]

Option B (

[ - \frac{\pi }{4},\\;\frac{\pi }{4}]

) is correct because the composite function simplifies to sin(2x), which is invertible in the given domain.

More Questions from FUNCTIONS

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

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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

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

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