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1

### JEE Main 2019 (Online) 11th January Evening Slot

The area (in sq. units) in the first quadrant bounded by the parabola, y = x2 + 1, the tangent to it at the point (2, 5) and the coordinate axes is :
A
$${8 \over 3}$$
B
$${{14} \over 3}$$
C
$${{187} \over {24}}$$
D
$${{37} \over {24}}$$

## Explanation

Area $$= \int\limits_0^2 {\left( {{x^2} + 1} \right)dx - {1 \over 2}} \left( {{5 \over 4}} \right)\left( 5 \right) = {{37} \over {24}}$$
2

### JEE Main 2019 (Online) 11th January Morning Slot

The area (in sq. units) of the region bounded by the curve x2 = 4y and the straight line x = 4y – 2 is :
A
$${3 \over 4}$$
B
$${5 \over 4}$$
C
$${7 \over 8}$$
D
$${9 \over 8}$$

## Explanation

x = 4y $$-$$ 2 & x2 = 4y

$$\Rightarrow$$  x2 = x + 2 $$\Rightarrow$$ x2 $$-$$ x $$-$$ 2 = 0

x = 2, $$-$$ 1

So,   $$\int\limits_{ - 1}^2 {\left( {{{x + 2} \over 4} - {{{x^2}} \over 4}} \right)\,dx = {9 \over 8}}$$
3

### JEE Main 2019 (Online) 11th January Morning Slot

If  $$\int {{{\sqrt {1 - {x^2}} } \over {{x^4}}}}$$ dx = A(x)$${\left( {\sqrt {1 - {x^2}} } \right)^m}$$ + C, for a suitable chosen integer m and a function A(x), where C is a constant of integration, then (A(x))m equals :
A
$${1 \over {27{x^6}}}$$
B
$${{ - 1} \over {27{x^9}}}$$
C
$${1 \over {9{x^4}}}$$
D
$${1 \over {3{x^3}}}$$

## Explanation

$$\int {{{\sqrt {1 - {x^2}} } \over {{x^4}}}}$$ dx = A(x)$${\left( {\sqrt {1 - {x^2}} } \right)^m}$$ + C

$$\int {{{\left| x \right|\sqrt {{1 \over {{x^2}}} - 1} } \over {{x^4}}}} \,dx,$$

Put  $${1 \over {{x^2}}} - 1 = t \Rightarrow {{dt} \over {dx}} = {{ - 2} \over {{x^3}}}$$

Case-I   $$x \ge 0$$

$$- {1 \over 2}\int {\sqrt t \,dt\, \Rightarrow {{{t^{3/2}}} \over 3}} + C$$

$$\Rightarrow - {1 \over 3}{\left( {{1 \over {{x^2}}} - 1} \right)^{3/2}}$$

$$\Rightarrow {{{{\left( {\sqrt {1 - {x^2}} } \right)}^3}} \over { - 3{x^2}}} + C$$

$$A(x) = - {1 \over {3{x^3}}}\,\,$$ and $$m = 3$$

$${(A(x))^m} = {\left( { - {1 \over {3{x^3}}}} \right)^3} = - {1 \over {27{x^9}}}$$

Case-II  $$x \le 0$$

We get  $${{{{\left( {\sqrt {1 - {x^2}} } \right)}^3}} \over { - 3{x^3}}} + C$$

$$A(x) = {1 \over { - 3{x^3}}},\,\,\,m = 3$$

$${(A(x))^m} = {{ - 1} \over {27{x^9}}}$$
4

### JEE Main 2019 (Online) 11th January Morning Slot

The value of the integral $$\int\limits_{ - 2}^2 {{{{{\sin }^2}x} \over { \left[ {{x \over \pi }} \right] + {1 \over 2}}}} \,dx$$ (where [x] denotes the greatest integer less than or equal to x) is
A
0
B
4
C
4$$-$$ sin 4
D
sin 4

## Explanation

I $$=$$ $$\int\limits_{ - 2}^2 {{{{{\sin }^2}x} \over { \left[ {{x \over \pi }} \right] + {1 \over 2}}}} \,dx$$

$${\rm I} = \int\limits_0^2 {\left( {{{{{\sin }^2}x} \over {\left[ {{x \over \pi }} \right] + {1 \over 2}}} + {{{{\sin }^2}\left( { - x} \right)} \over {\left[ { - {x \over \pi }} \right] + {1 \over 2}}}} \right)dx}$$

$$\left( {\left[ {{x \over \pi }} \right] + \left[ { - {x \over \pi }} \right] = - 1\,\,} \right.$$   as   $$\left. {\matrix{ \, \cr \, \cr } x \ne n\pi } \right)$$

$${\rm I} = \int\limits_0^2 {\left( {{{{{\sin }^2}x} \over {\left[ {{x \over \pi }} \right] + {1 \over 2}}} + {{{{\sin }^2}x} \over { - 1 - \left[ {{x \over \pi }} \right] + {1 \over 2}}}} \right)dx = 0}$$

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