Integral Calculus & Area · JEE Advanced previous year questions

2023 2 questions

1. Q12023 · 4 Jun 2023 · Paper 2

   Let $f:[1,\infty )\to \mathbb{R}$ be a differentiable function such that $f(1)=\frac{1}{3}$ and $$3{\int }_1^{x}f(t)dt=xf(x)-\frac{x^3}{3},x\in [1,\infty ).$$ Let $e$ denote the base of the natural logarithm. Then the value of $f(e)$ is

   1. A.$\frac{e^2+4}{3}$
   2. B.$\frac{{\log }_{e}4+e}{3}$
   3. C.$\frac{4e^2}{3}$
   4. D.$\frac{e^2-4}{3}$
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   Answer: C

2. Q82023 · 4 Jun 2023 · Paper 2· Numerical answer

   For $x\in \mathbb{R}$, let ${\tan }^{-1}(x)\in \left(-\frac{\pi }{2},\frac{\pi }{2}\right)$. Then the minimum value of the function $$f:\mathbb{R}\to \mathbb{R}\text{ defined by }f(x)={\int }_0^{x{\tan }^{-1}x}\frac{e^{(t-\cos t)}}{1+t^{2023}}dt$$ is ____.

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   Answer: 0