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

(B) $\max \mathrm{f}(\mathrm{x})>\max \mathrm{g}(\mathrm{x})$

(A) There exists $\mathrm{x} \in[0,3]$ such that $\mathrm{f}^{\prime}(\mathrm{x})<\mathrm{g}^{\prime}(\mathrm{x})$

(C) There exist $0<\mathrm{x}_{1}<\mathrm{x}_{2}<3$ such that $\mathrm{f}(\mathrm{x})<\mathrm{g}(\mathrm{x})$, $\forall \mathrm{x} \in\left(\mathrm{x}_{1}, \mathrm{x}_{2}\right)$

(D) $\min \mathrm{f}^{\prime}(\mathrm{x})=1+\max \mathrm{g}^{\prime}(\mathrm{x})$

JEE Main · 2023 · 01 Feb 2023 · Shift 1 · Application of Derivatives

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Let

f (x) = 2 x + tan^{- 1} x

and

g (x) = lo g_{e} (1 + x^{2} + x)

x \in [0, 3]

. Then

(A)There exists $x \in [0, 3]$ such that $f^{'} (x) < g^{'} (x)$
(B) $max f (x) > max g (x)$ ✓ Correct
(C)There exist $0 < x_{1} < x_{2} < 3$ such that $f (x) < g (x)$ , $\forall x \in (x_{1}, x_{2})$
(D) $min f^{'} (x) = 1 + max g^{'} (x)$

Answer: (B)

Solution

Here

f^{'} (x) = 2 + \frac{1}{1 + x ^{2}}

and

g^{'} (x) = \frac{1}{1 + x ^{2}}

, so on

[0, 3]

we have

f^{'} (x) \geq 2

while

g^{'} (x) \leq 1

; thus

f^{'} (x) > g^{'} (x)

throughout and both functions increase. Their maxima occur at

x = 3

, where

f (3) = 6 + tan^{- 1} 3 \approx 7.25

and

g (3) = lo g_{e} (3 + 10) \approx 1.82

. Since

f (3)

far exceeds

g (3)

, the maximum of

f

is greater than the maximum of

g

JEE Main · 2023 · 01 Feb 2023 · Shift 1 · Single correct option

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JEE Main · 2023 · 01 Feb 2023 · Shift 1 · Application of Derivatives

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