JEE Main 2020 (Online) 2nd September Morning Slot | Definite Integrals and Applications of Integrals Question 199 | Mathematics

JEE Main 2020 (Online) 2nd September Morning Slot

MCQ (Single Correct Answer)

-1

Area (in sq. units) of the region outside
$${{\left| x \right|} \over 2} + {{\left| y \right|} \over 3} = 1$$ and inside the ellipse $${{{x^2}} \over 4} + {{{y^2}} \over 9} = 1$$ is :

$$6\left( {4 - \pi } \right)$$

$$3\left( {4 - \pi } \right)$$

$$6\left( {\pi - 2} \right)$$

$$3\left( {\pi - 2} \right)$$

JEE Main 2020 (Online) 2nd September Morning Slot

MCQ (Single Correct Answer)

-1

Let y = y(x) be the solution of the differential equation,
$${{2 + \sin x} \over {y + 1}}.{{dy} \over {dx}} = - \cos x$$, y > 0,y(0) = 1.
If y($$\pi $$) = a and $${{dy} \over {dx}}$$ at x = $$\pi $$ is b, then the ordered pair (a, b) is equal to :

(2, 1)

$$\left( {2,{3 \over 2}} \right)$$

(1, -1)

(1, 1)

JEE Main 2020 (Online) 9th January Evening Slot

MCQ (Single Correct Answer)

-1

Given : $$f(x) = \left\{ {\matrix{ {x\,\,\,\,\,,} & {0 \le x < {1 \over 2}} \cr {{1 \over 2}\,\,\,\,,} & {x = {1 \over 2}} \cr {1 - x\,\,\,,} & {{1 \over 2} < x \le 1} \cr } } \right.$$

and $$g(x) = \left( {x - {1 \over 2}} \right)^2,x \in R$$

Then the area (in sq. units) of the region bounded by the curves, y = ƒ(x) and y = g(x) between the lines, 2x = 1 and 2x = $$\sqrt 3 $$, is :

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

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

$${1 \over 3} + {{\sqrt 3 } \over 4}$$

$${{\sqrt 3 } \over 4} - {1 \over 3}$$

JEE Main 2020 (Online) 9th January Morning Slot

MCQ (Single Correct Answer)

-1

The value of
$$\int\limits_0^{2\pi } {{{x{{\sin }^8}x} \over {{{\sin }^8}x + {{\cos }^8}x}}} dx$$ is equal to :

4$$\pi $$

2$$\pi $$

$$\pi $$²

2$$\pi $$²