QJEEM24Ma0b52: JEE Main 2024 Mathematics - Class 11 JEEM24Ma0b52 Solution: Let and be defined as and . Then, gof : is :

Question

Verified Solution • 2025–26 Session moderate

Let

f: \mathbf{R} \rightarrow \mathbf{R}

and

g: \mathbf{R} \rightarrow \mathbf{R}

be defined as

f(x)=\left\{\begin{array}{ll}\log _{\mathrm{e}} x, &amp; x&gt;0 \\ \mathrm{e}^{-x}, &amp; x \leq 0\end{array}\right.

and

g(x)=\left\{\begin{array}{ll}x, &amp; x \geqslant 0 \\ \mathrm{e}^x, &amp; x&lt;0\end{array}\right.

. Then, gof :

\mathbf{R} \rightarrow \mathbf{R}

is :

Accepted Answer

Given,

f(x)=\left\{\begin{array}{ll}\log _{\mathrm{e}} x, &amp; x&gt;0 \\ \mathrm{e}^{-x}, &amp; x \leq 0\end{array}\right.

and

g(x)=\left\{\begin{array}{ll}x, &amp; x \geqslant 0 \\ \mathrm{e}^x, &amp; x&lt;0\end{array}\right.

then

g \circ f(x)

=g(f(x))

\begin{aligned} &amp; \mathrm{g}(\mathrm{f}(\mathrm{x}))=\left\{\begin{array}{l}f(x), f(x) \geq 0 \\ e^{f(x)}, f(x)&lt;0\end{array}\right. \\\\ &amp; \mathrm{g}(\mathrm{f}(\mathrm{x}))=\left\{\begin{array}{l}e^{-x},(-\infty, 0] \\ e^{\ln x},(0,1) \\ \ln x,[1, \infty)\end{array}\right.\end{aligned}

g(f(x))=\left\{\begin{array}{c}e^{-x},(-\infty, 0] \\ x,(0,1) \\ \ln x,[1, \infty)\end{array}\right.

From the graph of

g(f(x))

, we can say

\mathrm{g}(\mathrm{f}(\mathrm{x})) \Rightarrow

Many one into

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