1
JEE Advanced 2023 Paper 2 Online
Numerical
+4
-0
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 $f: \mathbb{R} \rightarrow \mathbb{R}$ defined by $f(x)=\int\limits_0^{x \tan ^{-1} x} \frac{e^{(t-\cos t)}}{1+t^{2023}} d t$ is :
2
JEE Advanced 2023 Paper 1 Online
Numerical
+4
-0
Let $n \geq 2$ be a natural number and $f:[0,1] \rightarrow \mathbb{R}$ be the function defined by

$$f(x)= \begin{cases}n(1-2 n x) & \text { if } 0 \leq x \leq \frac{1}{2 n} \\\\ 2 n(2 n x-1) & \text { if } \frac{1}{2 n} \leq x \leq \frac{3}{4 n} \\\\ 4 n(1-n x) & \text { if } \frac{3}{4 n} \leq x \leq \frac{1}{n} \\\\ \frac{n}{n-1}(n x-1) & \text { if } \frac{1}{n} \leq x \leq 1\end{cases}$$

If $n$ is such that the area of the region bounded by the curves $x=0, x=1, y=0$ and $y=f(x)$ is 4 , then the maximum value of the function $f$ is :
3
JEE Advanced 2022 Paper 2 Online
Numerical
+3
-1
The greatest integer less than or equal to

$$\int_{1}^{2} \log _{2}\left(x^{3}+1\right) d x+\int_{1}^{\log _{2} 9}\left(2^{x}-1\right)^{\frac{1}{3}} d x$$

is ___________.
4
JEE Advanced 2022 Paper 2 Online
Numerical
+3
-1
Consider the functions $f, g: \mathbb{R} \rightarrow \mathbb{R}$ defined by

$$f(x)=x^{2}+\frac{5}{12} \quad \text { and } \quad g(x)= \begin{cases}2\left(1-\frac{4|x|}{3}\right), & |x| \leq \frac{3}{4} \\ 0, & |x|>\frac{3}{4}\end{cases}$$

If $\alpha$ is the area of the region

$$\left\{(x, y) \in \mathbb{R} \times \mathbb{R}:|x| \leq \frac{3}{4}, 0 \leq y \leq \min \{f(x), g(x)\}\right\},$$

then the value of $9 \alpha$ is
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