1

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

If the area of the region bounded by the curves, $y = {x^2},y = {1 \over x}$ and the lines y = 0 and x= t (t >1) is 1 sq. unit, then t is equal to :
A
${e^{{3 \over 2}}}$
B
${4 \over 3}$
C
${3 \over 2}$
D
${e^{{2 \over 3}}}$

## Explanation

Point of intersection of y = x2 and y = ${1 \over x}$.

put y = ${1 \over x}$ in y = x2, then we get,

${1 \over x} = {x^2}$

$\Rightarrow$ $\,\,\,$ x3 $-$ 1 = 0

$\Rightarrow$ $\,\,\,$ x = 1

$\therefore\,\,\,$ y = 1

$\therefore\,\,\,$ point B = (1, 1)

Area of region ABCDA

= $\int\limits_0^1 {{x^2}}$ dx + $\int\limits_1^t {{1 \over x}}$ dx

$=$ $\left[ {{{{x^3}} \over 3}} \right]_0^1$ + $\left[ {\ell n\,x} \right]_1^t$

= ${1 \over 3}$ + $\ell n\,t$ $-$ $\ell n$ 1

= ${1 \over 3}$ + $\ell n\,t$     [ as    $\ell n$ 1 = 0]

given this Area = 1 sq unit.

$\therefore\,\,\,$ ${1 \over 3}$ + $\ell n\,t$ = 1

$\Rightarrow$    $\ell n\,t$ = ${2 \over 3}$

$\Rightarrow$    t = e${^{{2 \over 3}}}$
2

### 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
3

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

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

= ${4 \over 3}$
4

### 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

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

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

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

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

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

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

= ${8 \over 3}$