1

### JEE Main 2019 (Online) 9th January Evening Slot

Let f : [0,1] $\to$ R be such that f(xy) = f(x).f(y), for all x, y $\in$ [0, 1], and f(0) $\ne$ 0. If y = y(x) satiesfies the differential equation, ${{dy} \over {dx}}$ = f(x) with y(0) = 1, then y$\left( {{1 \over 4}} \right)$ + y$\left( {{3 \over 4}} \right)$ is equal to :
A
3
B
4
C
2
D
5

## Explanation

If f(xy) = f(x) f(y) $\forall$ x, y $\in$ R and f(0) $\ne$ 0

put x = y = 0

$\Rightarrow$  f(0) = [f(0)]2

$\Rightarrow$  f(0) = 1

put y = 0 $\Rightarrow$ f(0) = f(x) f(0)

$\Rightarrow$ f(x) = 1

given that ${{dy} \over {dx}}$ = f(x)

$\therefore$  ${{dy} \over {dx}}$ = 1 $\Rightarrow$ y = x + k

given that y(0) = 1

$\therefore$  k = 1

hence y = x + 1

y$\left( {{1 \over 4}} \right)$ + y$\left( {{3 \over 4}} \right)$ = $\left( {{1 \over 4} + 1} \right)$ + $\left( {{3 \over 4} + 1} \right)$ = 3
2

### JEE Main 2019 (Online) 10th January Evening Slot

The curve amongst the family of curves represented by the differential equation, (x2 – y2)dx + 2xy dy = 0 which passes through (1, 1) is
A
a circle with centre on the y-axis
B
an ellipse with major axis along the y-axis
C
a circle with centre on the x-axis
D
a hyperbola with transverse axis along the x-axis

## Explanation

(x2 $-$ y2) dx + 2xy dy = 0

${{dy} \over {dx}} = {{{y^2} - {x^2}} \over {2xy}}$

Put   $y = vx \Rightarrow {{dy} \over {dx}} = v + x{{dv} \over {dx}}$

Solving we get,

$\int {{{2v} \over {{v^2} + 1}}dv = \int { - {{dx} \over x}} }$

ln(v2 + 1) = $-$ ln x + C

(y2 + x2) = Cx

1 + 1 = C $\Rightarrow$ C = 2

y2 + x2 = 2x
3

### JEE Main 2019 (Online) 10th January Evening Slot

Let f be a differentiable function such that f '(x) = 7 - ${3 \over 4}{{f\left( x \right)} \over x},$ (x > 0) and f(1) $\ne$ 4. Then $\mathop {\lim }\limits_{x \to 0'} \,$ xf$\left( {{1 \over x}} \right)$
A
does not exist
B
exists and equals ${4 \over 7}$
C
exists and equals 4
D
exists and equals 0

## Explanation

$f'(x) = 7 - {3 \over 4}{{f\left( x \right)} \over x}\,\,\,\left( {x > 0} \right)$

Given f(1) $\ne$ 4    $\mathop {\lim }\limits_{x \to {0^ + }} \,xf\left( {{1 \over x}} \right)\, = ?$

${{dy} \over {dx}} + {3 \over 4}{y \over x} = 7$  (This is LDE)

IF $= {e^{\int {{3 \over {4x}}dx} }} = {e^{{3 \over 4}\ln \left| x \right|}} = {x^{{3 \over 4}}}$

$y.{x^{{3 \over 4}}} = \int {7.{x^{{3 \over 4}}}} dx$

$y.{x^{{3 \over 4}}} = 7.{{{x^{{7 \over 4}}}} \over {{7 \over 4}}} + C$

$f(x) = 4x + C.{x^{ - {3 \over 4}}}$

$f\left( {{1 \over 4}} \right) = {4 \over x} + C.{x^{{3 \over 4}}}$

$\mathop {\lim }\limits_{x \to {0^ + }} xf\left( {{1 \over x}} \right) = \mathop {\lim }\limits_{x \to {0^ + }} \left( {4 + C.{x^{{7 \over 4}}}} \right) = 4$
4

### JEE Main 2019 (Online) 10th January Evening Slot

A helicopter is flying along the curve given by y – x3/2 = 7, (x $\ge$ 0). A soldier positioned at the point $\left( {{1 \over 2},7} \right)$ wants to shoot down the helicopter when it is nearest to him. Then this nearest distance is -
A
${1 \over 6}\sqrt {{7 \over 3}}$
B
${{\sqrt 5 } \over 6}$
C
${1 \over 2}$
D
${1 \over 3}$$\sqrt {{7 \over 3}}$

## Explanation

$y - {x^{3/2}} = 7\left( {x \ge 0} \right)$

${{dy} \over {dx}} = {3 \over 2}{x^{1/2}}$

$\left( {{3 \over 2}\sqrt x } \right)\left( {{{7 - y} \over {{1 \over 2} - x}}} \right) = - 1$

$\left( {{3 \over 2}\sqrt x } \right)\left( {{{ - {x^{3/2}}} \over {{1 \over 2} - x}}} \right) = - 1$

${3 \over 2}.{x^2} = {1 \over 2} - x$

$3{x^2} = 1 - 2x$

$3{x^2} + 2x - 1 = 0$

$3{x^2} + 3x - x - 1 = 0$

$\left( {x + 1} \right)\left( {3x - 1} \right) = 0$

$\therefore$  $x = - 1$  (rejected)

$x = {1 \over 3}$

$y = 7 + {x^{3/2}} = 7 + {\left( {{1 \over 3}} \right)^{3/2}}$

${\ell _{AB}} = \sqrt {{{\left( {{1 \over 2} - {1 \over 3}} \right)}^2} + {{\left( {{1 \over 3}} \right)}^3}} = \sqrt {{1 \over {36}} + {1 \over {27}}}$

$= \sqrt {{{3 + 4} \over {9 \times 12}}}$

$= \sqrt {{7 \over {108}}} = {1 \over 6}\sqrt {{7 \over 3}}$