1

### JEE Main 2016 (Online) 10th April Morning Slot

The solution of the differential equation

${{dy} \over {dx}}\, + \,{y \over 2}\,\sec x = {{\tan x} \over {2y}},\,\,$

where 0 $\le$ x < ${\pi \over 2}$, and y (0) = 1, is given by :
A
y = 1 $-$ ${x \over {\sec x + \tan x}}$
B
y2 = 1 + ${x \over {\sec x + \tan x}}$
C
y2 = 1 $-$ ${x \over {\sec x + \tan x}}$
D
y = 1 + ${x \over {\sec x + \tan x}}$

## Explanation

Given,

${{dy} \over {dx}} + {y \over 2}\sec x = {{\tan x} \over {2y}}$

$\Rightarrow$   $2y{{dy} \over {dx}} + {y^2}\sec x = \tan x$

Now, let

y2 $=$ t

$\Rightarrow$   2y${{dy} \over {dx}} = {{dt} \over {dx}}$

$\therefore$    New equation,

${{dt} \over {dx}} + t\sec x = \tan x$

$\therefore$   I.F $=$ ${e^{\int {\sec xdx} }}$

$=$   ${e^{\ln \left( {\sec x + \tan x} \right)}}$

$=$   sec x + tan x

$\therefore$   Solution is,

t(sec x + tan x) $=$ $\int {\tan x}$ (sec x + tan x) dx

$\Rightarrow$   t(sec x + tan x) $=$ sec x + tan x $-$ x + c

$\Rightarrow$    t $=$ 1 $-$ ${x \over {\sec x + \tan x}} + c$

$\Rightarrow$   y2 $=$ 1 $-$ ${x \over {\sec x + \tan x}} + c$

Given,

y(0) $=$ 1

$\therefore$   1 $=$ 1 $-$ 0 + c

$\Rightarrow$   c $=$ 0

$\therefore$   y2 $=$ 1 $-$ ${x \over {\sec x + \tan x}}$
2

### JEE Main 2017 (Offline)

Twenty meters of wire is available for fencing off a flower-bed in the form of a circular sector. Then the maximum area (in sq. m) of the flower-bed, is:
A
10
B
25
C
30
D
12.5

## Explanation

We have

Total length = r + r + r$\theta$ = 20

$\Rightarrow$ 2r + r$\theta$ = 20

$\Rightarrow$ $\theta = {{20 - 2r} \over r}$ .......(1)

A = Area = ${\theta \over {2\pi }} \times \pi {r^2}$ = ${1 \over 2}{r^2}\theta$ = ${1 \over 2}{r^2}\left( {{{20 - 2r} \over r}} \right)$

$\Rightarrow$ A = 10r – r2

For A to be maximum

${{dA} \over {dr}} = 0$

$\Rightarrow$ 10 – 2r = 0

$\Rightarrow$ r = 5

${{{d^2}A} \over {d{r^2}}} = - 2 < 0$

$\therefore$ For r = 5, A is maximum From (1)

${\theta = {{20 - 2\left( 5 \right)} \over r}}$ = 2

A = ${2 \over {2\pi }} \times \pi {\left( 5 \right)^2}$ = 25 sq. m
3

### JEE Main 2017 (Offline)

The normal to the curve y(x – 2)(x – 3) = x + 6 at the point where the curve intersects the y-axis passes through the point:
A
$\left( {{1 \over 2},{1 \over 2}} \right)$
B
$\left( {{1 \over 2}, - {1 \over 3}} \right)$
C
$\left( {{1 \over 2},{1 \over 3}} \right)$
D
$\left( { - {1 \over 2}, - {1 \over 3}} \right)$

## Explanation

Given $y = {{x + 6} \over {\left( {x - 2} \right)\left( {x - 2} \right)}}$

At y-axis, x = 0 $\Rightarrow$ y = 1

On differentiating, we get

${{dy} \over {dx}} = {{\left( {{x^2} - 5x + 6} \right)\left( 1 \right) - \left( {x + 6} \right)\left( {2x - 5} \right)} \over {{{\left( {{x^2} - 5x + 6} \right)}^2}}}$

${{dy} \over {dx}} = 1$ at point (0, 1)

$\therefore$ Slope of normal = – 1

Now equation of normal is y – 1 = –1 (x – 0)

$\Rightarrow$ y – 1 = – x

x + y = 1 ......(1)

By checking each option you can see point $\left( {{1 \over 2},{1 \over 2}} \right)$ satisfy equation (1).
4

### JEE Main 2017 (Offline)

If $\left( {2 + \sin x} \right){{dy} \over {dx}} + \left( {y + 1} \right)\cos x = 0$ and y(0) = 1,

then $y\left( {{\pi \over 2}} \right)$ is equal to
A
$- {2 \over 3}$
B
$- {1 \over 3}$
C
${4 \over 3}$
D
${1 \over 3}$

## Explanation

$\left( {2 + \sin x} \right){{dy} \over {dx}} + \left( {y + 1} \right)\cos x = 0$

$\Rightarrow$ ${d \over {dx}}\left( {2 + \sin x} \right)\left( {y + 1} \right) = 0$

On integrating, we get

(2 + sin x) (y + 1) = C

At x = 0, y = 1 we have

(2 + sin 0) (1 + 1) = C

$\Rightarrow$ C = 4

$\Rightarrow$ $\left( {y + 1} \right) = {4 \over {2 + \sin x}}$

$\Rightarrow$ y = ${4 \over {2 + \sin x}} - 1$

Now $y\left( {{\pi \over 2}} \right) = {4 \over {2 + \sin {\pi \over 2}}} - 1$

= ${4 \over 3} - 1$ = ${1 \over 3}$

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