JEE Main 2023 (Online) 30th January Evening Shift | Definite Integrals and Applications of Integrals Question 45 | Mathematics

JEE Main 2023 (Online) 30th January Evening Shift

MCQ (Single Correct Answer)

-1

Let $q$ be the maximum integral value of $p$ in $[0,10]$ for which the roots of the equation $x^2-p x+\frac{5}{4} p=0$ are rational. Then the area of the region $\left\{(x, y): 0 \leq y \leq(x-q)^2, 0 \leq x \leq q\right\}$ is :

$\frac{125}{3}$

243

164

JEE Main 2023 (Online) 30th January Morning Shift

MCQ (Single Correct Answer)

-1

If [t] denotes the greatest integer $$\le \mathrm{t}$$, then the value of $${{3(e - 1)} \over e}\int\limits_1^2 {{x^2}{e^{[x] + [{x^3}]}}dx} $$ is :

$$\mathrm{e^8-e}$$

$$\mathrm{e^7-1}$$

$$\mathrm{e^9-e}$$

$$\mathrm{e^8-1}$$

JEE Main 2023 (Online) 29th January Evening Shift

MCQ (Single Correct Answer)

-1

The area of the region $$A = \left\{ {(x,y):\left| {\cos x - \sin x} \right| \le y \le \sin x,0 \le x \le {\pi \over 2}} \right\}$$ is

$$\sqrt 5 + 2\sqrt 2 - 4.5$$

$$1 - {3 \over {\sqrt 2 }} + {4 \over {\sqrt 5 }}$$

$$\sqrt 5 - 2\sqrt 2 + 1$$

$${3 \over {\sqrt 5 }} - {3 \over {\sqrt 2 }} + 1$$

JEE Main 2023 (Online) 29th January Evening Shift

MCQ (Single Correct Answer)

-1

The value of the integral $$\int_1^2 {\left( {{{{t^4} + 1} \over {{t^6} + 1}}} \right)dt} $$ is

$${\tan ^{ - 1}}{1 \over 2} - {1 \over 3}{\tan ^{ - 1}}8 + {\pi \over 3}$$

$${\tan ^{ - 1}}2 - {1 \over 3}{\tan ^{ - 1}}8 + {\pi \over 3}$$

$${\tan ^{ - 1}}2 + {1 \over 3}{\tan ^{ - 1}}8 - {\pi \over 3}$$

$${\tan ^{ - 1}}{1 \over 2} + {1 \over 3}{\tan ^{ - 1}}8 - {\pi \over 3}$$

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