1

### JEE Main 2018 (Online) 16th April Morning Slot

Let M and m be respectively the absolute maximum and the absolute minimum values of the function, f(x) = 2x3 $-$ 9x2 + 12x + 5 in the interval [0, 3]. Then M $-$m is equal to :
A
5
B
9
C
4
D
1

## Explanation

Here, f(x) = 2x3 $-$ 9x2 + 12x + 5

$\Rightarrow$ f'(x) = 6x2 $-$ 18x + 12 = 0

For maxima or minima put f'(x) = 0

$\Rightarrow$ x2 $-$ 3x + 2 = 0

$\Rightarrow$ x = 1 or x = 2

Now, f''(x) = 12x $-$ 18

$\Rightarrow$ f''(1) = 12(1) $-$ 18 = $-$ 6 < 0

Hence, f(x) has maxima at x = 1

$\therefore$ maximum value = M = f(1) = 2 $-$ 9 + 12 + 5 = 10

And, f''(2) = 12(2) $-$ 18 = 6 > 0.

Hence, f(x) has minima at x = 2.

$\therefore$ minimum value = m = f(2) = 2(8) $-$ 9(4) + 12(2) + 5 = 9

$\therefore$ M $-$ m = 10 $-$ 9 = 1
2

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

If  ${{dy} \over {dx}} + {3 \over {{{\cos }^2}x}}y = {1 \over {{{\cos }^2}x}},\,\,x \in \left( {{{ - \pi } \over 3},{\pi \over 3}} \right)$  and  $y\left( {{\pi \over 4}} \right) = {4 \over 3},$  then  $y\left( { - {\pi \over 4}} \right)$   equals -
A
${1 \over 3} + {e^6}$
B
${1 \over 3}$
C
${1 \over 3}$ + e3
D
$-$ ${4 \over 3}$

## Explanation

${{dy} \over {dx}} + 3{\sec ^2}x.y = {\sec ^2}x$

I.F. = ${e^{3\int {{{\sec }^2}xdx} }} = {e^{3\tan x}}$

or   $y.e{}^{3\tan x} = \int {{{\sec }^2}x.{e^{3\tan x}}}$

or   $y.{e^{3\tan x}} = {1 \over 3}{e^{3\tan x}} + C$

Given

$y\left( {{\pi \over 4}} \right) = {4 \over 3}$

$\therefore$   ${4 \over 3}.{e^3} = {1 \over 3}{e^3} + C$

$\therefore$   C = e3
3

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

The shortest distance between the point  $\left( {{3 \over 2},0} \right)$   and the curve y = $\sqrt x$, (x > 0), is -
A
${{\sqrt 3 } \over 2}$
B
${5 \over 4}$
C
${3 \over 2}$
D
${{\sqrt 5 } \over 2}$

## Explanation

Let points $\left( {{3 \over 2},0} \right),\left( {{t^2},t} \right),t > 0$

Distance = $\sqrt {{t^2} + {{\left( {{t^2} - {3 \over 2}} \right)}^2}}$

= $\sqrt {{t^4} - 2{t^2} + {9 \over 4}} = \sqrt {{{\left( {{t^2} - 1} \right)}^2} + {5 \over 4}}$

So minimum distance is $\sqrt {{5 \over 4}} = {{\sqrt 5 } \over 2}$
4

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

The tangent to the curve, y = xex2 passing through the point (1, e) also passes through the point
A
$\left( {{4 \over 3},2e} \right)$
B
(3, 6e)
C
(2, 3e)
D
$\left( {{5 \over 3},2e} \right)$

## Explanation

y = xex2

${\left. {{{dy} \over {dx}}} \right|_{(1,e)}}{\left. { = \left( {e.e{x^2}.2x + {e^{{x^2}}}} \right)} \right|_{(1,e)}}$

$= 2 \cdot e + e = 3e$

T : y $-$ e = 3e (x $-$ 1)

y = 3ex $-$ 3e + e

y = $\left( {3e} \right)x - 2e$

$\left( {{4 \over 3},2e} \right)$ lies on it