1

### 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 $\left( {{1 \over 2},1} \right)$

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

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

2

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

Let f(x) = ${x \over {\sqrt {{a^2} + {x^2}} }} - {{d - x} \over {\sqrt {{b^2} + {{\left( {d - x} \right)}^2}} }},\,\,$ x $\, \in$ R, where a, b and d are non-zero real constants. Then :
A
f is an increasing function of x
B
f is neither increasing nor decreasing function of x
C
f ' is not a continuous function of x
D
f is a decreasing function of x

## Explanation

$f\left( x \right) = {x \over {\sqrt {{a^2} + {x^2}} }} - {{d - x} \over {\sqrt {{b^2} + {{\left( {d - x} \right)}^2}} }}$

$f'\left( x \right) = {{{a^2}} \over {{{\left( {{a^2} + {x^2}} \right)}^{3/2}}}} + {{{b^2}} \over {{{\left( {{b^2} + {{\left( {d - x} \right)}^2}} \right)}^{3/2}}}} > 0\forall x \in R$

f(x) is an increasing function.
3

### JEE Main 2019 (Online) 12th January Evening Slot

The tangent to the curve y = x2 – 5x + 5, parallel to the line 2y = 4x + 1, also passes through the point :
A
$\left\{ {{1 \over 4},{7 \over 2}} \right\}$
B
$\left( { - {1 \over 8},7} \right)$
C
$\left( {{7 \over 2},{1 \over 4}} \right)$
D
$\left( {{1 \over 8}, - 7} \right)$

## Explanation

y = x2 $-$ 5x + 5

${{dy} \over {dx}} = 2x - 5 = 2 \Rightarrow x = {7 \over 2}$

at  x = ${7 \over 2}$, y = ${{ - 1} \over 4}$

Equation of tangent at

$\left( {{7 \over 2},{{ - 1} \over 4}} \right)$ is 2x $-$ y $-$ ${{29} \over 4}$ = 0

Now check options

x = ${1 \over 8}$, y = $-$7
4

### JEE Main 2019 (Online) 8th April Morning Slot

If S1 and S2 are respectively the sets of local minimum and local maximum points of the function,

ƒ(x) = 9x4 + 12x3 – 36x2 + 25, x $\in$ R, then :
A
S1 = {–1}; S2 = {0, 2}
B
S1 = {–2}; S2 = {0, 1}
C
S1 = {–2, 0}; S2 = {1}
D
S1 = {–2, 1}; S2 = {0}

## Explanation

ƒ(x) = 9x4 + 12x3 – 36x2 + 25

ƒ'(x) = 36x3 + 36x2 – 72x

ƒ'(x) = 36x(x2 + x – 2)

ƒ'(x) = 36x(x + 2)(x - 1) While moving left to right on x-axis whenever derivative changes sign from negative to positive, we get local minima, and whenever derivative changes sign from positive to negative, we get local maxima.

$\therefore$ S1 = {–2, 1}

S2 = {0}