JEE Main 2023 (Online) 30th January Evening Shift | Area Under The Curves Question 57 | Mathematics

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) 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 Morning Shift

MCQ (Single Correct Answer)

-1

Let $$\Delta$$ be the area of the region $$\left\{ {(x,y) \in {R^2}:{x^2} + {y^2} \le 21,{y^2} \le 4x,x \ge 1} \right\}$$. Then $${1 \over 2}\left( {\Delta - 21{{\sin }^{ - 1}}{2 \over {\sqrt 7 }}} \right)$$ is equal to

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

$$2\sqrt 3 - {2 \over 3}$$

$$\sqrt 3 - {4 \over 3}$$

$$\sqrt 3 - {2 \over 3}$$

JEE Main 2023 (Online) 29th January Morning Shift

MCQ (Single Correct Answer)

-1

Let $$[x]$$ denote the greatest integer $$\le x$$. Consider the function $$f(x) = \max \left\{ {{x^2},1 + [x]} \right\}$$. Then the value of the integral $$\int\limits_0^2 {f(x)dx} $$ is

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

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

$${{8 + 4\sqrt 2 } \over 3}$$

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

PREVIOUS NEXT

Questions Asked from Area Under The Curves (MCQ (Single Correct Answer))

Number in Brackets after Paper Indicates No. of Questions