1
JEE Main 2023 (Online) 11th April Evening Shift
MCQ (Single Correct Answer)
+4
-1

Let the function $$f:[0,2] \rightarrow \mathbb{R}$$ be defined as

$$f(x)= \begin{cases}e^{\min \left\{x^{2}, x-[x]\right\},} & x \in[0,1) \\ e^{\left[x-\log _{e} x\right]}, & x \in[1,2]\end{cases}$$

where $$[t]$$ denotes the greatest integer less than or equal to $$t$$. Then the value of the integral $$\int_\limits{0}^{2} x f(x) d x$$ is :

A
$$2 e-1$$
B
$$2 e-\frac{1}{2}$$
C
$$1+\frac{3 e}{2}$$
D
$$(e-1)\left(e^{2}+\frac{1}{2}\right)$$
2
JEE Main 2023 (Online) 11th April Morning Shift
MCQ (Single Correct Answer)
+4
-1

The value of the integral $$\int_\limits{-\log _{e} 2}^{\log _{e} 2} e^{x}\left(\log _{e}\left(e^{x}+\sqrt{1+e^{2 x}}\right)\right) d x$$ is equal to :

A
$$\log _{e}\left(\frac{(2+\sqrt{5})^{2}}{\sqrt{1+\sqrt{5}}}\right)+\frac{\sqrt{5}}{2}$$
B
$$\log _{e}\left(\frac{\sqrt{2}(2+\sqrt{5})^{2}}{\sqrt{1+\sqrt{5}}}\right)-\frac{\sqrt{5}}{2}$$
C
$$\log _{e}\left(\frac{2(2+\sqrt{5})}{\sqrt{1+\sqrt{5}}}\right)-\frac{\sqrt{5}}{2}$$
D
$$\log _{e}\left(\frac{\sqrt{2}(3-\sqrt{5})^{2}}{\sqrt{1+\sqrt{5}}}\right)+\frac{\sqrt{5}}{2}$$
3
JEE Main 2023 (Online) 10th April Evening Shift
MCQ (Single Correct Answer)
+4
-1

Let $$f$$ be a continuous function satisfying $$\int_\limits{0}^{t^{2}}\left(f(x)+x^{2}\right) d x=\frac{4}{3} t^{3}, \forall t > 0$$. Then $$f\left(\frac{\pi^{2}}{4}\right)$$ is equal to

A
$$-\pi\left(1+\frac{\pi^{3}}{16}\right)$$
B
$$\pi\left(1-\frac{\pi^{3}}{16}\right)$$
C
$$-\pi^{2}\left(1+\frac{\pi^{2}}{16}\right)$$
D
$$\pi^{2}\left(1-\frac{\pi^{2}}{16}\right)$$
4
JEE Main 2023 (Online) 6th April Evening Shift
MCQ (Single Correct Answer)
+4
-1

Let $$f(x)$$ be a function satisfying $$f(x)+f(\pi-x)=\pi^{2}, \forall x \in \mathbb{R}$$. Then $$\int_\limits{0}^{\pi} f(x) \sin x d x$$ is equal to

A
$$\pi^{2}$$
B
$$\frac{\pi^{2}}{2}$$
C
$$2 \pi^{2}$$
D
$$\frac{\pi^{2}}{4}$$
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