1

### JEE Main 2018 (Online) 16th April Morning Slot

If $f(x) = \int\limits_0^x {t\left( {\sin x - \sin t} \right)dt\,\,\,}$ then :
A
f'''(x) + f''(x) = sinx
B
f'''(x) + f''(x) $-$ f'(x) = cosx
C
f'''(x) + f'(x) = cosx $-$ 2x sinx
D
f'''(x) $-$ f''(x) = cosx $-$ 2x sinx

## Explanation

f(x) = $\int_0^x {t(\sin x - \sin t).dt}$

= sin x$\int_0^x {t.dt - \int_0^x {t\sin t.dt} }$

= ${{{x^2}} \over 2}$ sin x +$\left[ {t\cos t} \right]_0^x$ + sin x

$\Rightarrow$f(x) = ${{{x^2}} \over 2}$ sinx + xcosx + sinx

f'(x) = ${{{x^2}} \over 2}$ cosx + 2cos x

f''(x) = x cos x $-$ ${{{x^2}} \over 2}$ sin x $-$ 2sin x

f'''(x) = cos x $-$ 2x sin x $-$ ${{{x^2}} \over 2}$ cos x $-$ 2cos x

$\therefore\,\,\,$ f'''(x) + f'(x) = cos x $-$ 2x sin x
2

### JEE Main 2019 (Online) 9th January Morning Slot

The value of $\int\limits_0^\pi {{{\left| {\cos x} \right|}^3}} \,dx$ is :
A
$4 \over 3$
B
$-$ $4 \over 3$
C
0
D
$2 \over 3$

## Explanation

$\int\limits_0^\pi {{{\left| {\cos x} \right|}^3}} \,dx$

The period of $\left| {\cos x} \right|$ = ${\pi \over 2}$

$\therefore$  I = 2 $\int\limits_0^{{\pi \over 2}} {{{\left| {\cos x} \right|}^3}} \,dx$

as in the range 0 to ${\pi \over 2}$ $\left| {\cos x} \right|$ is positive.

So,     $\left| {\cos x} \right|$ = $cosx$

$\therefore$  I = 2 $\int\limits_0^{{\pi \over 2}} {{{\cos }^3}x\,dx}$

= 2$\int\limits_0^{{\pi \over 2}} {\left( {{{\cos 3x + 3\cos x} \over 4}} \right)} dx$

I = ${1 \over 2}\left[ {{{\sin 3x} \over 3} + 3\sin x} \right]_0^{{\pi \over 2}}$

I = ${1 \over 2}\left[ {{1 \over 3}\left( {{{3\pi } \over 2}} \right) + 3.\sin {\pi \over 2}} \right]$

I = ${1 \over 2}\left[ { - {1 \over 3} + 3} \right]$

= ${1 \over 2}\left( {{8 \over 3}} \right)$

= ${4 \over 3}$
3

### JEE Main 2019 (Online) 9th January Morning Slot

The area (in sq. units) bounded by the parabolae y = x2 – 1, the tangent at the point (2, 3) to it and the y-axis is :
A
$56\over3$
B
$32\over3$
C
$8\over3$
D
$14\over3$

## Explanation

Equation of tangent at (2, 3) on the parabola y = x2 $-$ 1 is

${{y + 3} \over 2} = 2x - 1$

$\Rightarrow$  y + 3 = 4x $-$ 2

$\Rightarrow$   y = 4x $-$ 5

When x = 0 then for the tangent y = $-$ 5

$\therefore$  Tangent cuts x y axis at (0, $-$ 5) point.

$\therefore$  Area of the bounded region is

= $\int\limits_{ - 5}^3 {{{y + 5} \over 4}} \,\,\,dy - \int\limits_{ - 1}^3 {\sqrt {y + 1} } \,\,\,dy$

= ${1 \over 4}\left[ {{{{y^2}} \over 2} + 5y} \right]_{ - 5}^3 - \left[ {{2 \over 3} \times {{\left( {y + 1} \right)}^{{3 \over 2}}}} \right]_{ - 1}^3$

${1 \over 4}\left[ {\left( {{9 \over 2} + 15} \right) - \left( {{{25} \over 2} - 25} \right)} \right] - {2 \over 3}{\left( 4 \right)^{{3 \over 2}}}$

= ${1 \over 4}\left[ {{{93} \over 2} + {{25} \over 2}} \right] - {2 \over 3} \times 8$

= ${1 \over 4} \times {{64} \over 2} - {{16} \over 3}$

= $8 - {{16} \over 3}$

= ${8 \over 3}$
4

### JEE Main 2019 (Online) 9th January Evening Slot

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 :
A
${2 \over 3}$
B
2
C
${4 \over 3}$
D
${1 \over 3}$

## Explanation

Required area

$= \int\limits_{ - 1}^1 {\left( {x\left| x \right| + 1} \right)} dx$

$= 0 + \left( x \right)_{ - 1}^1 = 2$