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Questions
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1
JEE Main 2019 (Online) 11th January Morning Slot
MCQ (Single Correct Answer)
+4
-1
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
2
JEE Main 2019 (Online) 11th January Morning Slot
MCQ (Single Correct Answer)
+4
-1
If y(x) is the solution of the differential equation $${{dy} \over {dx}} + \left( {{{2x + 1} \over x}} \right)y = {e^{ - 2x}},\,\,x > 0,\,$$ where $$y\left( 1 \right) = {1 \over 2}{e^{ - 2}},$$ then
A
y(loge2) = loge4
B
y(x) is decreasing in (0, 1)
C
y(loge2) = $${{{{\log }_e}2} \over 4}$$
D
y(x) is decreasing in $$\left( {{1 \over 2},1} \right)$$
3
JEE Main 2019 (Online) 10th January Evening Slot
MCQ (Single Correct Answer)
+4
-1
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
$$\left( {{5 \over 3},2e} \right)$$
4
JEE Main 2019 (Online) 10th January Morning Slot
MCQ (Single Correct Answer)
+4
-1
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
$$-$$ $${4 \over 3}$$
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