1
JEE Main 2019 (Online) 11th January Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
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
Change Language
The sum of the real values of x for which the middle term in the binomial expansion of $${\left( {{{{x^3}} \over 3} + {3 \over x}} \right)^8}$$ equals 5670 is :
A
0
B
8
C
6
D
4
3
JEE Main 2019 (Online) 11th January Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
Let  $$\overrightarrow a = \widehat i + 2\widehat j + 4\widehat k,$$ $$\overrightarrow b = \widehat i + \lambda \widehat j + 4\widehat k$$ and $$\overrightarrow c = 2\widehat i + 4\widehat j + \left( {{\lambda ^2} - 1} \right)\widehat k$$ be coplanar vectors. Then the non-zero vector $$\overrightarrow a \times \overrightarrow c $$ is :
A
$$ - 10\widehat i - 5\widehat j$$
B
$$ - 10\widehat i + 5\widehat j$$
C
$$ - 14\widehat i + 5\widehat j$$
D
$$ - 14\widehat i - 5\widehat j$$
4
JEE Main 2019 (Online) 11th January Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
If  $$\int {{{\sqrt {1 - {x^2}} } \over {{x^4}}}} $$ dx = A(x)$${\left( {\sqrt {1 - {x^2}} } \right)^m}$$ + C, for a suitable chosen integer m and a function A(x), where C is a constant of integration, then (A(x))m equals :
A
$${1 \over {27{x^6}}}$$
B
$${{ - 1} \over {27{x^9}}}$$
C
$${1 \over {9{x^4}}}$$
D
$${1 \over {3{x^3}}}$$
JEE Main Papers
2023
2021
EXAM MAP