1
MCQ (Single Correct Answer)

JEE Main 2016 (Online) 9th April Morning Slot

If   f(x) is a differentiable function in the interval (0, $$\infty $$) such that f (1) = 1 and

$$\mathop {\lim }\limits_{t \to x} $$   $${{{t^2}f\left( x \right) - {x^2}f\left( t \right)} \over {t - x}} = 1,$$ for each x > 0, then $$f\left( {{\raise0.5ex\hbox{$\scriptstyle 3$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}} \right)$$ equal to :
A
$${{13} \over 6}$$
B
$${{23} \over 18}$$
C
$${{25} \over 9}$$
D
$${{31} \over 18}$$

Explanation

$$\mathop {\lim }\limits_{t \to x} {{{t^2}f\left( x \right) - {x^2}f\left( t \right)} \over {t - x}} = 1$$

It is in $${0 \over 0}$$ form

So, applying L' Hospital rule,

$$\mathop {\lim }\limits_{t \to x} {{2tf\left( x \right) - {x^2}f'\left( x \right)} \over 1} = 1$$

$$ \Rightarrow $$   2xf(x) $$-$$ x2f '(x) = 1

$$ \Rightarrow $$   f '(x) $$-$$ $${2 \over x}$$f(x) $$=$$ $${1 \over {{x^2}}}$$

$$ \therefore $$   I.F = $${e^{\int {{{ - 2} \over x}dx} }} = {e^{ - 2\log x}} = {1 \over {{x^2}}}$$

$$ \therefore $$   Solution of equation,

f(x)$${1 \over {{x^2}}}$$ = $$\int {{1 \over x}\left( { - {1 \over {{x^2}}}} \right)} \,dx$$

$$ \Rightarrow $$   $${{f(x)} \over {{x^2}}} = {1 \over {3{x^3}}} + C$$

Given that,

f(1) = 1

$$ \therefore $$   $${1 \over 1}$$ = $${1 \over 3}$$ + C

$$ \Rightarrow $$   C $$=$$ $${2 \over 3}$$

$$ \therefore $$   f(x) $$=$$ $${2 \over 3}$$ x2 + $${1 \over {3x}}$$

$$ \therefore $$   f$$\left( {{3 \over 2}} \right) = {2 \over 3} \times {\left( {{3 \over 2}} \right)^2} + {1 \over 3} \times {2 \over 3} = {{31} \over {18}}$$
2
MCQ (Single Correct Answer)

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 $$-$$ $${x \over {\sec x + \tan x}}$$
3
MCQ (Single Correct Answer)

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
4
MCQ (Single Correct Answer)

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).

Questions Asked from Differential Equations

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