Question:

Let f(x) = sin 2x + cos 2x and g(x)=x²-1, then g(f(x)) is invertible in the domain

Updated On: Aug 18, 2024

\(x\ \epsilon\ [\frac{-\pi}{2},\frac{\pi}{2}]\)
\(x\ \epsilon\ [\frac{-\pi}{4},\frac{\pi}{4}]\)
\(x\ \epsilon\ [0,\frac{\pi}{4}]\)
\(x\ \epsilon\ [\frac{-\pi}{8},\frac{\pi}{8}]\)

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The Correct Option is D

Solution and Explanation

The correct answer is (D) : \(x\ \epsilon\ [\frac{-\pi}{8},\frac{\pi}{8}]\).

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Top Questions on composite of functions

Let \(f:\R \rightarrow \R\) and \(g:\R \rightarrow\R\) be functions defined by
\(f(x)=\left\{ \begin{array}{ll} x|x|\sin(\frac{1}{x}), & x\ne0 \\ 0, & x = 0, \end{array} \right.\text{and} \ g(x)=\left\{ \begin{array}{ll} 1-2x, & 0\leq x\leq \frac{1}{2}, \\ 0, & \text{otherwise.} \end{array} \right.\)
Let \(a,b,c,d \in \R\). Define the function \(h:\R\rightarrow\R\) by
\(h(x)=af(x)+b(g(x)+g(\frac{1}{2}-x))+c(x-g(x))+d\ g(x), x\in\R\).
Match each entry in List-I to the correct entry in List-II.

List - I		List - II
(P)	If a = 0, b = 1, c = 0 and d = 0, then	(1)	h is one-one.
(Q)	If a = 1, b = 0, c = 0 and d = 0, then	(2)	h is onto.
(R)	If a = 0, b = 0, c = 1 and d = 0, then	(3)	h is differentiable on \(\R\)
(S)	If a = 0, b = 0, c = 0 and d = 1, then	(4)	the range of h is [0, 1].
		(5)	the range of h is {0, 1}.

The correct option is

JEE Advanced - 2024
Mathematics
composite of functions

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