1
JEE Main 2019 (Online) 11th January Morning Slot
+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) 10th January Evening Slot
+4
-1
Out of Syllabus
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
$$\left( {{5 \over 3},2e} \right)$$
3
JEE Main 2019 (Online) 10th January Evening Slot
+4
-1
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
$${1 \over 3}$$$$\sqrt {{7 \over 3}}$$
4
JEE Main 2019 (Online) 10th January Morning Slot
+4
-1
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
$${{\sqrt 5 } \over 2}$$
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