1
MCQ (Single Correct Answer)

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
MCQ (Single Correct Answer)

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
MCQ (Single Correct Answer)

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
MCQ (Single Correct Answer)

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 = $$\phi $$

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