1

### JEE Main 2019 (Online) 10th January Morning Slot

The shortest distance between the point  $\left( {{3 \over 2},0} \right)$   and the curve y = $\sqrt x$, (x > 0), is -
A
${{\sqrt 3 } \over 2}$
B
${5 \over 4}$
C
${3 \over 2}$
D
${{\sqrt 5 } \over 2}$

## Explanation

Let points $\left( {{3 \over 2},0} \right),\left( {{t^2},t} \right),t > 0$

Distance = $\sqrt {{t^2} + {{\left( {{t^2} - {3 \over 2}} \right)}^2}}$

= $\sqrt {{t^4} - 2{t^2} + {9 \over 4}} = \sqrt {{{\left( {{t^2} - 1} \right)}^2} + {5 \over 4}}$

So minimum distance is $\sqrt {{5 \over 4}} = {{\sqrt 5 } \over 2}$
2

### JEE Main 2019 (Online) 10th January Evening Slot

The tangent to the curve, y = xex2 passing through the point (1, e) also passes through the point
A
$\left( {{4 \over 3},2e} \right)$
B
(3, 6e)
C
(2, 3e)
D
$\left( {{5 \over 3},2e} \right)$

## Explanation

y = xex2

${\left. {{{dy} \over {dx}}} \right|_{(1,e)}}{\left. { = \left( {e.e{x^2}.2x + {e^{{x^2}}}} \right)} \right|_{(1,e)}}$

$= 2 \cdot e + e = 3e$

T : y $-$ e = 3e (x $-$ 1)

y = 3ex $-$ 3e + e

y = $\left( {3e} \right)x - 2e$

$\left( {{4 \over 3},2e} \right)$ lies on it
3

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

The maximum value of the function f(x) = 3x3 – 18x2 + 27x – 40 on the set S = $\left\{ {x\, \in R:{x^2} + 30 \le 11x} \right\}$ is :
A
$-$ 222
B
$-$ 122
C
$122$
D
222

## Explanation

S = {x $\in$ R, x2 + 30 $-$ 11x $\le$ 0}

= {x $\in$ R, 5 $\le$ x $\le$ 6}

Now f(x) = 3x3 $-$ 18x2 + 27x $-$ 40

$\Rightarrow$  f '(x) = 9(x $-$ 1)(x $-$ 3),

which is positive in [5, 6]

$\Rightarrow$  f(x) increasing in [5, 6]

Hence maximum value = f(6) = 122
4

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

If y(x) is the solution of the differential equation ${{dy} \over {dx}} + \left( {{{2x + 1} \over x}} \right)y = {e^{ - 2x}},\,\,x > 0,\,$ where $y\left( 1 \right) = {1 \over 2}{e^{ - 2}},$ then
A
y(loge2) = loge4
B
y(x) is decreasing in (0, 1)
C
y(loge2) = ${{{{\log }_e}2} \over 4}$
D
y(x) is decreasing in $\left( {{1 \over 2},1} \right)$

## Explanation

${{dy} \over {dx}} + \left( {{{2x + 1} \over x}} \right)y = {e^{ - 2x}}$

I.F. $= {e^{\int {\left( {{{2x + 1} \over x}} \right)dx} }} = {e^{\int {\left( {2 + {1 \over x}} \right)dx} }} = {e^{2x + \ell nx}} = {e^{2x}}.x$

So,  $y\left( {x{e^{2x}}} \right) = \int {{e^{ - 2x}}.x{e^{2x}} + C}$

$\Rightarrow xy{e^{2x}} = \int {xdx + C}$

$\Rightarrow 2xy{e^{2x}} = {x^2} + 2C$

It passes through $\left( {1,{1 \over 2}{e^{ - 2}}} \right)$ we get C $=$ 0

$y = {{x{e^{ - 2x}}} \over 2}$

$\Rightarrow {{dy} \over {dx}} = {1 \over 2}{e^{ - 2x}}\left( { - 2x + 1} \right)$

$\Rightarrow f(x)$ is decreasing in $\left( {{1 \over 2},1} \right)$

$y\left( {{{\log }_e}2} \right) = {{\left( {{{\log }_e}2} \right){e^{ - 2({{\log }_e}2)}}} \over 2}$

$= {1 \over 8}{\log _e}2$