Verified Solution • 2025–26 Session moderate

QJEEM23Ma3aa1: JEE Main 2023 Mathematics - Class 12 JEEM23Ma3aa1 Solution: Let and . Then

Let $f(x) = 2x + {\tan ^{ - 1}}x$ and $g(x) = {\log _e}(\sqrt {1 + {x^2}} + x),x \in [0,3]$ . Then

Multiple Choice Options

A there exists $\widehat x \in [0,3]$ such that $f'(\widehat x) < g'(\widehat x)$
B there exist $0 < {x_1} < {x_2} < 3$ such that $f(x) < g(x),\forall x \in ({x_1},{x_2})$
C $\min f'(x) = 1 + \max g'(x)$
D $\max f(x) > \max g(x)$

Step-by-Step Verified Solution

\begin{aligned}&amp; f^{\prime}(x)=2+\frac{1}{1+x^2}, g^{\prime}(x)=\frac{1}{\sqrt{x^2+1}} \\\\&amp; f^{\prime \prime}(x)=-\frac{2 x}{\left(1+x^2\right)^2}&lt;0 \\\\&amp; g^{\prime \prime}(x)=-\frac{1}{2}\left(x^2+1\right)^{-3 / 2} \cdot 2 x&lt;0 \\\\&amp; \left.f^{\prime}(x)\right|_{\min }=f^{\prime}(3)=2+\frac{1}{10}=\frac{21}{10} \\\\&amp; \left.g^{\prime}(x)\right|_{\max }=g^{\prime}(0)=1 \\\\&amp; \left.f^{\prime}(x)\right|_{\max }=f(3)=2+\tan ^{-1} 3 \\\\&amp; \left.g(x)\right|_{\max }=g(3)=\ln (3+\sqrt{10})&lt;\ln &lt;7&lt;2\end{aligned}

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