JEE MAIN - Mathematics (2023 - 1st February Morning Shift - No. 9)
Let $$f(x) = 2x + {\tan ^{ - 1}}x$$ and $$g(x) = {\log _e}(\sqrt {1 + {x^2}} + x),x \in [0,3]$$. Then
there exists $$\widehat x \in [0,3]$$ such that $$f'(\widehat x) < g'(\widehat x)$$
there exist $$0 < {x_1} < {x_2} < 3$$ such that $$f(x) < g(x),\forall x \in ({x_1},{x_2})$$
$$\min f'(x) = 1 + \max g'(x)$$
$$\max f(x) > \max g(x)$$
Explanation
$$
\begin{aligned}
& f^{\prime}(x)=2+\frac{1}{1+x^2}, g^{\prime}(x)=\frac{1}{\sqrt{x^2+1}} \\\\
& f^{\prime \prime}(x)=-\frac{2 x}{\left(1+x^2\right)^2}<0 \\\\
& g^{\prime \prime}(x)=-\frac{1}{2}\left(x^2+1\right)^{-3 / 2} \cdot 2 x<0 \\\\
& \left.f^{\prime}(x)\right|_{\min }=f^{\prime}(3)=2+\frac{1}{10}=\frac{21}{10} \\\\
& \left.g^{\prime}(x)\right|_{\max }=g^{\prime}(0)=1 \\\\
& \left.f^{\prime}(x)\right|_{\max }=f(3)=2+\tan ^{-1} 3 \\\\
& \left.g(x)\right|_{\max }=g(3)=\ln (3+\sqrt{10})<\ln <7<2
\end{aligned}
$$
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