JEE Advanced 2023 · Paper 2 · Mathematics Q8

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MathematicsInteger+4Question 8 of 51

For $x \in \mathbb{R}$ , let $\tan^{-1}(x) \in \left(-\dfrac\pi2, \dfrac\pi2\right)$ . Then the minimum value of the function $f : \mathbb{R} \to \mathbb{R}$ defined by $f(x) = \displaystyle\int_0^{x\tan^{-1}x} \frac{e^{(t-\cos t)}}{1+t^{2023}}\,dt$ is _____.

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Answer and worked solution

Official answer: 0

Upper limit. $x\tan^{-1}x\ge0$ for every real $x$ , since $x$ and $\tan^{-1}x$ have the same sign. It equals 0 only at $x=0$ .

Integrand. For $t\ge0$ , $\dfrac{e^{t-\cos t}}{1+t^{2023}}>0$ .

So $f(x)=\int_0^{x\tan^{-1}x}(\text{positive})\,dt\ge0$ , with equality at $x=0$ . The minimum value is $\mathbf0$ .

Topics:Fundamental theorem of calculus Inverse trigonometric functions

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JEE Advanced 2023 · Paper 2, Mathematics question 8