1

### JEE Main 2017 (Online) 8th April Morning Slot

The curve satisfying the differential equation, ydx $-$(x + 3y2)dy = 0 and passing through the point (1, 1), also passes through the point :
A
$\left( {{1 \over 4}, - {1 \over 2}} \right)$
B
$\left( { - {1 \over 3},{1 \over 3}} \right)$
C
$\left( {{1 \over 3}, - {1 \over 3}} \right)$
D
$\left( {{1 \over 4}, {1 \over 2}} \right)$

## Explanation

Given,

y dx = $\left( {x + 3{y^2}} \right)dy$

$\Rightarrow $$\,\,\, y {{dx} \over {dy}} = x + 3y2 \Rightarrow$$\,\,\,$ ${{dx} \over {dy}}$ $-$ ${x \over y} = 3y$

If   =    ${e^{ - \int {{1 \over y}dy} }}$ = ${e^{ - \ln y}}$ = ${1 \over y}$

$\therefore\,\,\,$ Soluation is ,

x . ${1 \over y}$ = $\int {3y.{1 \over y}dy}$

$\Rightarrow $$\,\,\, {x \over y} = 3y + c This curve passing through (1, 1) \therefore\,\,\, 1 = 3 + c \Rightarrow$$\,\,\,$ c = $-$ 2

$\therefore\,\,\,$ Curve is, x = 3y2 $-$ 2y

Now put every point in this equation, and see which point satisfy this equation.

Following this method you can see ($-$ ${1 \over 3}$, ${1 \over 3}$) point satisfy this equation.
2

### JEE Main 2018 (Offline)

If the curves y2 = 6x, 9x2 + by2 = 16 intersect each other at right angles, then the value of b is :
A
${9 \over 2}$
B
6
C
${7 \over 2}$
D
4

## Explanation

When two curves intersect each other at right angle, then at the point of intersection the product of tangent of slopes = $-1$.

Let m1, and m2 are the tangent of the slope of the two curves respectively

$\therefore\,\,\,$ m1 m2 = $-$ 1.

Now let they intersect at point (x1, y1)

$\therefore\,\,\,$ $y_1^2 = 6x,$ and $9x_1^2 + b\,y_1^2 = 16$

y2 = 6x

$\Rightarrow \,\,\,\,2y{{dy} \over {dx}} = 6$

$\Rightarrow \,\,\,\,{{dy} \over {dx}} = {3 \over y}$

$\therefore\,\,\,$ ${\left( {{{dy} \over {dx}}} \right)_{\left( {{x_1},{y_1}} \right)}} = {3 \over {{y_1}}} = {m_1}$

9x2 + by2 = 16

$= 18x + 2by{{dy} \over {dx}} = O$

$\Rightarrow \,\,\,\,{\left( {{{dy} \over {dx}}} \right)_{({x_1},{y_1})}} = - {{9{x_1}} \over {b{y_1}}} = {m_2}$

As m1 m2 = $-$1

$\therefore\,\,\,$ ${3 \over {y{}_1}} \times - {{9{x_1}} \over {b{y_1}}} = - 1$

$\Rightarrow \,\,\,\,27{x_1} = by_1^2$

$\Rightarrow \,\,\,\,\,27{x_1} = b.6{x_1}$ $\,\,\,$ [as $y_1^2 = 6{x_1}\,]$

$\Rightarrow \,\,\,\,b = {{27} \over 6} = {9 \over 2}$
3

### JEE Main 2018 (Offline)

Let $f\left( x \right) = {x^2} + {1 \over {{x^2}}}$ and $g\left( x \right) = x - {1 \over x}$,
$x \in R - \left\{ { - 1,0,1} \right\}$.
If $h\left( x \right) = {{f\left( x \right)} \over {g\left( x \right)}}$, then the local minimum value of h(x) is
A
$2\sqrt 2$
B
3
C
-3
D
$-2\sqrt 2$

## Explanation

Given $f\left( x \right) = {x^2} + {1 \over {{x^2}}}$ and $g\left( x \right) = x - {1 \over x}$

As $h\left( x \right) = {{f\left( x \right)} \over {g\left( x \right)}}$

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

= ${{{{\left( {x - {1 \over x}} \right)}^2} + 2.x.{1 \over x}} \over {x - {1 \over x}}}$

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

Let ${x - {1 \over x} = t}$

So $f\left( t \right) = {{{t^2} + 2} \over t}$ = $t + {2 \over t}$

then $f'\left( t \right) = 1 - {2 \over {{t^2}}}$

At maximum or minimum $f'\left( t \right) = 0$.

$\therefore$ $1 - {2 \over {{t^2}}} = 0$

$\Rightarrow t = \pm \sqrt 2$

Now we need to find $f''\left( t \right)$ which will tell at which point among $+ \sqrt 2$ and $- \sqrt 2$ will be maximum and minimum.

$f''\left( t \right) = + {4 \over {{t^3}}}$

So when $t = + \sqrt 2$, then $f''\left( t \right) = + {4 \over {{{\left( {\sqrt 2 } \right)}^3}}}$ = $+ \sqrt 2$

As $f''\left( t \right) > 0$ when $t = + \sqrt 2$ then at $t = + \sqrt 2$ the function $f\left( t \right)$ will be minimum.

Min of $f\left( t \right) = {{{{\left( {\sqrt 2 } \right)}^2} + 2} \over {\sqrt 2 }}$ = ${4 \over {\sqrt 2 }}$ = $2\sqrt 2$

So local minimum of $f\left( t \right)$ = $2\sqrt 2$
4

### JEE Main 2018 (Offline)

Let S = { t $\in R:f(x) = \left| {x - \pi } \right|.\left( {{e^{\left| x \right|}} - 1} \right)$$\sin \left| x \right|$ is not differentiable at t}, then the set S is equal to
A
{0, $\pi$}
B
$\phi$ (an empty set)
C
{0}
D
{$\pi$}

## Explanation

Check differtiability at x = $\pi$ and x = 0

at x = 0 :

We have L. H. D = $\mathop {\lim }\limits_{h \to 0} {{f\left( {0 - h} \right) - f\left( 0 \right)} \over { - h}}$

= $\mathop {\lim }\limits_{h \to 0} {{\left| { - h - \pi } \right|\left( {{e^{\left| { - h} \right|}} - 1} \right)\sin \left| h \right| - 0} \over { - h}} = 0$

R. H. D = $\mathop {\lim }\limits_{h \to 0} {{f\left( {0 + h} \right) - f\left( 0 \right)} \over h}$

$= \mathop {\lim }\limits_{h \to 0} {{\left| {h - \pi } \right|\left( {{e^{\left| h \right|}} - 1} \right)\sin \left| h \right| - o} \over h}$

= 0

$\therefore,\,\,$ LHD = RHD

Therefore, function is differentiable at x = $\pi$.

at x = $\pi$ :

L. H. D = $\mathop {\lim }\limits_{h \to 0} {{f\left( {\pi - h} \right) - f\left( \pi \right)} \over { - h}}$

= $\mathop {\lim }\limits_{h \to 0} {{\left| {\pi - h - \pi } \right|\left( {{e^{\left| {\pi - h} \right|}} - 1} \right)\sin \left| {\pi - h} \right| - 0.} \over { - h}}$

= 0

RH. D = $\mathop {\lim }\limits_{h \to 0} {{f\left( {\pi + h} \right) - f\left( \pi \right)} \over h}$

= $\mathop {\lim }\limits_{h \to 0} {{\left| {\pi + h - \pi } \right|\left( {{e^{\left| {\pi + h} \right|}} - 1} \right)\sin \left| {\pi + h} \right| - 0} \over h}$

= 0

$\therefore\,\,\,$ L. H. D = R H D

Therefore, function is differentiable at x = $\pi$,

Since, the function f(x) is differentiable at all the points including 0 and $\pi$.

So, f(x) is differentiable everywhere.

Therefore, set S is empty set.

S = $\phi$