JEE Main 2020 (Online) 7th January Evening Slot | Definite Integrals and Applications of Integrals Question 210 | Mathematics

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JEE Main 2020 (Online) 7th January Evening Slot

MCQ (Single Correct Answer)

-1

The value of $$\alpha $$ for which
$$4\alpha \int\limits_{ - 1}^2 {{e^{ - \alpha \left| x \right|}}dx} = 5$$, is:

$${\log _e}2$$

$${\log _e}\sqrt 2 $$

$${\log _e}\left( {{4 \over 3}} \right)$$

$${\log _e}\left( {{3 \over 2}} \right)$$

JEE Main 2020 (Online) 7th January Evening Slot

MCQ (Single Correct Answer)

-1

The area (in sq. units) of the region
{(x, y) $$ \in $$ R² | 4x² $$ \le $$ y $$ \le $$ 8x + 12} is :

$${{125} \over 3}$$

$${{128} \over 3}$$

$${{127} \over 3}$$

$${{124} \over 3}$$

JEE Main 2020 (Online) 7th January Evening Slot

MCQ (Single Correct Answer)

-1

If $$\theta $$₁ and $$\theta $$₂ be respectively the smallest and the largest values of $$\theta $$ in (0, 2$$\pi $$) - {$$\pi $$} which satisfy the equation,
2cot²$$\theta $$ - $${5 \over {\sin \theta }}$$ + 4 = 0, then
$$\int\limits_{{\theta _1}}^{{\theta _2}} {{{\cos }^2}3\theta d\theta } $$ is equal to :

$${\pi \over 9}$$

$${{2\pi } \over 3}$$

$${{\pi } \over 3}$$

$${\pi \over 3} + {1 \over 6}$$

JEE Main 2020 (Online) 7th January Morning Slot

MCQ (Single Correct Answer)

-1

The area of the region, enclosed by the circle x² + y² = 2 which is not common to the region bounded by the parabola y² = x and the straight line y = x, is:

$${1 \over 6}\left( {24\pi - 1} \right)$$

$${1 \over 3}\left( {12\pi - 1} \right)$$

$${1 \over 3}\left( {6\pi - 1} \right)$$

$${1 \over 6}\left( {12\pi - 1} \right)$$

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