1
JEE Main 2021 (Online) 20th July Morning Shift
MCQ (Single Correct Answer)
+4
-1
Out of Syllabus
Change Language
Let $$\overrightarrow a = 2\widehat i + \widehat j - 2\widehat k$$ and $$\overrightarrow b = \widehat i + \widehat j$$. If $$\overrightarrow c $$ is a vector such that $$\overrightarrow a .\,\overrightarrow c = \left| {\overrightarrow c } \right|,\left| {\overrightarrow c - \overrightarrow a } \right| = 2\sqrt 2 $$ and the angle between $$(\overrightarrow a \times \overrightarrow b )$$ and $$\overrightarrow c $$ is $${\pi \over 6}$$, then the value of $$\left| {\left( {\overrightarrow a \times \overrightarrow b } \right) \times \overrightarrow c } \right|$$ is :
A
$${2 \over 3}$$
B
4
C
3
D
$${3 \over 2}$$
2
JEE Main 2021 (Online) 20th July Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language
The number of real roots of the equation $${\tan ^{ - 1}}\sqrt {x(x + 1)} + {\sin ^{ - 1}}\sqrt {{x^2} + x + 1} = {\pi \over 4}$$ is :
A
1
B
2
C
4
D
0
3
JEE Main 2021 (Online) 20th July Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language
Let y = y(x) be the solution of the differential equation $${e^x}\sqrt {1 - {y^2}} dx + \left( {{y \over x}} \right)dy = 0$$, y(1) = $$-$$1. Then the value of (y(3))2 is equal to :
A
1 $$-$$ 4e3
B
1 $$-$$ 4e6
C
1 + 4e3
D
1 + 4e6
4
JEE Main 2021 (Online) 20th July Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language
Let 'a' be a real number such that the function f(x) = ax2 + 6x $$-$$ 15, x $$\in$$ R is increasing in $$\left( { - \infty ,{3 \over 4}} \right)$$ and decreasing in $$\left( {{3 \over 4},\infty } \right)$$. Then the function g(x) = ax2 $$-$$ 6x + 15, x$$\in$$R has a :
A
local maximum at x = $$-$$ $${{3 \over 4}}$$
B
local minimum at x = $$-$$$${{3 \over 4}}$$
C
local maximum at x = $${{3 \over 4}}$$
D
local minimum at x = $${{3 \over 4}}$$
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