Explanation: Solution Differentiate: (Value range: since max is and min is ). (Value range: ). Check Options: is always greater than for . Since and grows faster, for all . . . Clearly, . Correct Option: (2)

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JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

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

Choose the correct answer:

Explanation

Solution

Differentiate:

$f'(x) = 2 + \frac{1}{1+x^2}$ (Value range: $[2.1, 3]$ since max is $2+1$ and min is $2 + \frac{1}{10}$ ).

$g'(x) = \frac{1}{\sqrt{1+x^2}}$ (Value range: $[\frac{1}{\sqrt{10}}, 1]$ ).

Check Options:

$f'(x)$ is always greater than $g'(x)$ for $x \in [0, 3]$ . Since $f(0) = g(0) = 0$ and $f$ grows faster, $f(x) > g(x)$ for all $x > 0$ .

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

$\max g(3) = \log_e(\sqrt{10} + 3) \approx \log_e(6.16) \approx 1.82$ .

Clearly, $\max f(x) > \max g(x)$ .

Correct Option: (2)

Choose the correct answer:

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