JEE Main 2019 (Online) 10th January Morning Slot | Definite Integrals and Applications of Integrals Question 240 | Mathematics

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 -

$$\sqrt 3 $$

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

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

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

JEE Main 2019 (Online) 9th January Evening Slot

MCQ (Single Correct Answer)

-1

The area of the region

A = {(x, y) : 0 $$ \le $$ y $$ \le $$x |x| + 1 and $$-$$1 $$ \le $$ x $$ \le $$1} in sq. units, is :

$${2 \over 3}$$

$${4 \over 3}$$

$${1 \over 3}$$

JEE Main 2019 (Online) 9th January Evening Slot

MCQ (Single Correct Answer)

-1

If $$\int\limits_0^{{\pi \over 3}} {{{\tan \theta } \over {\sqrt {2k\,\sec \theta } }}} \,d\theta = 1 - {1 \over {\sqrt 2 }},\left( {k > 0} \right),$$ then value of k is :

$${1 \over 2}$$

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