1
JEE Main 2022 (Online) 24th June Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

If $$y = {\tan ^{ - 1}}\left( {\sec {x^3} - \tan {x^3}} \right),{\pi \over 2} < {x^3} < {{3\pi } \over 2}$$, then

A
$$xy'' + 2y' = 0$$
B
$${x^2}y'' - 6y + {{3\pi } \over 2} = 0$$
C
$${x^2}y'' - 6y + 3\pi = 0$$
D
$$xy'' - 4y' = 0$$
2
JEE Main 2022 (Online) 24th June Evening Shift
MCQ (Single Correct Answer)
+4
-1
Out of Syllabus
Change Language

Consider the following statements:

A : Rishi is a judge.

B : Rishi is honest.

C : Rishi is not arrogant.

The negation of the statement "if Rishi is a judge and he is not arrogant, then he is honest" is

A
B $$\to$$ (A $$\vee$$ C)
B
($$\sim$$B) $$\wedge$$ (A $$\wedge$$ C)
C
B $$\to$$ (($$\sim$$A) $$\vee$$ ($$\sim$$C))
D
B $$\to$$ (A $$\wedge$$ C)
3
JEE Main 2022 (Online) 24th June Evening Shift
MCQ (Single Correct Answer)
+4
-1
Out of Syllabus
Change Language

The slope of normal at any point (x, y), x > 0, y > 0 on the curve y = y(x) is given by $${{{x^2}} \over {xy - {x^2}{y^2} - 1}}$$. If the curve passes through the point (1, 1), then e . y(e) is equal to

A
$${{1 - \tan (1)} \over {1 + \tan (1)}}$$
B
tan(1)
C
1
D
$${{1 + \tan (1)} \over {1 - \tan (1)}}$$
4
JEE Main 2022 (Online) 24th June Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Let $$\lambda$$$$^ * $$ be the largest value of $$\lambda$$ for which the function $${f_\lambda }(x) = 4\lambda {x^3} - 36\lambda {x^2} + 36x + 48$$ is increasing for all x $$\in$$ R. Then $${f_{{\lambda ^ * }}}(1) + {f_{{\lambda ^ * }}}( - 1)$$ is equal to :

A
36
B
48
C
64
D
72
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