1
MCQ (Single Correct Answer)

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
2
MCQ (Single Correct Answer)

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 $$\sqrt {{5 \over 4}} = {{\sqrt 5 } \over 2}$$
3
MCQ (Single Correct Answer)

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
4
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 11th January Morning Slot

The maximum value of the function f(x) = 3x3 – 18x2 + 27x – 40 on the set S = $$\left\{ {x\, \in R:{x^2} + 30 \le 11x} \right\}$$ is :
A
$$-$$ 222
B
$$-$$ 122
C
$$122$$
D
222

Explanation

S = {x $$ \in $$ R, x2 + 30 $$-$$ 11x $$ \le $$ 0}

= {x $$ \in $$ R, 5 $$ \le $$ x $$ \le $$ 6}

Now f(x) = 3x3 $$-$$ 18x2 + 27x $$-$$ 40

$$ \Rightarrow $$  f '(x) = 9(x $$-$$ 1)(x $$-$$ 3),

which is positive in [5, 6]

$$ \Rightarrow $$  f(x) increasing in [5, 6]

Hence maximum value = f(6) = 122

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