1
JEE Main 2023 (Online) 6th April Evening Shift
+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
$$\frac{\pi^{2}}{4}$$
2
JEE Main 2023 (Online) 6th April Evening Shift
+4
-1

$$\lim _\limits{n \rightarrow \infty}\left\{\left(2^{\frac{1}{2}}-2^{\frac{1}{3}}\right)\left(2^{\frac{1}{2}}-2^{\frac{1}{5}}\right) \ldots . .\left(2^{\frac{1}{2}}-2^{\frac{1}{2 n+1}}\right)\right\}$$ is equal to :

A
$$\sqrt{2}$$
B
1
C
$$\frac{1}{\sqrt{2}}$$
D
0
3
JEE Main 2023 (Online) 6th April Morning Shift
+4
-1

Let $$5 f(x)+4 f\left(\frac{1}{x}\right)=\frac{1}{x}+3, x > 0$$. Then $$18 \int_\limits{1}^{2} f(x) d x$$ is equal to :

A
$$10 \log _{\mathrm{e}} 2+6$$
B
$$5 \log _{e} 2-3$$
C
$$10 \log _{\mathrm{e}} 2-6$$
D
$$5 \log _{\mathrm{e}} 2+3$$
4
JEE Main 2023 (Online) 1st February Evening Shift
+4
-1

The value of the integral

$$\int\limits_{ - {\pi \over 4}}^{{\pi \over 4}} {{{x + {\pi \over 4}} \over {2 - \cos 2x}}dx}$$ is :

A
$${{{\pi ^2}} \over {6\sqrt 3 }}$$
B
$${{{\pi ^2}} \over 6}$$
C
$${{{\pi ^2}} \over {3\sqrt 3 }}$$
D
$${{{\pi ^2}} \over {12\sqrt 3 }}$$
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