JEE Main 2019 (Online) 10th January Evening Slot | Definite Integrals and Applications of Integrals Question 238 | Mathematics

JEE Main 2019 (Online) 10th January Evening Slot

MCQ (Single Correct Answer)

-1

The value of $$\int\limits_{ - \pi /2}^{\pi /2} {{{dx} \over {\left[ x \right] + \left[ {\sin x} \right] + 4}}} ,$$ where [t] denotes the greatest integer less than or equal to t, is

$${1 \over {12}}\left( {7\pi - 5} \right)$$

$${1 \over {12}}\left( {7\pi + 5} \right)$$

$${3 \over {10}}\left( {4\pi - 3} \right)$$

$${3 \over {20}}\left( {4\pi - 3} \right)$$

JEE Main 2019 (Online) 10th January Evening Slot

MCQ (Single Correct Answer)

-1

If $$\int\limits_0^x \, $$f(t) dt = x² + $$\int\limits_x^1 \, $$ t²f(t) dt then f '$$\left( {{1 \over 2}} \right)$$ is -

$${{18} \over {25}}$$

$${{6} \over {25}}$$

$${{24} \over {25}}$$

$${{4} \over {5}}$$

JEE Main 2019 (Online) 10th January Morning Slot

MCQ (Single Correct Answer)

-1

Let $${\rm I} = \int\limits_a^b {\left( {{x^4} - 2{x^2}} \right)} dx.$$ If I is minimum then the ordered pair (a, b) is -

$$\left( {\sqrt 2 , - \sqrt 2 } \right)$$

$$\left( {0,\sqrt 2 } \right)$$

$$\left( { - \sqrt 2 ,\sqrt 2 } \right)$$

$$\left( { - \sqrt 2 ,0} \right)$$

JEE Main 2019 (Online) 10th January Morning Slot

MCQ (Single Correct Answer)

-1

If the area enclosed between the curves y = kx² and x = ky², (k > 0), is 1 square unit. Then k is -

$$\sqrt 3 $$

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

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

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

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