1
JEE Main 2022 (Online) 28th July Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Let $$x(t)=2 \sqrt{2} \cos t \sqrt{\sin 2 t}$$ and

$$y(t)=2 \sqrt{2} \sin t \sqrt{\sin 2 t}, t \in\left(0, \frac{\pi}{2}\right)$$.

Then $$\frac{1+\left(\frac{d y}{d x}\right)^{2}}{\frac{d^{2} y}{d x^{2}}}$$ at $$t=\frac{\pi}{4}$$ is equal to :

A
$$\frac{-2 \sqrt{2}}{3}$$
B
$$\frac{2}{3}$$
C
$$\frac{1}{3}$$
D
$$ \frac{-2}{3}$$
2
JEE Main 2022 (Online) 26th July Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

The value of $$\log _{e} 2 \frac{d}{d x}\left(\log _{\cos x} \operatorname{cosec} x\right)$$ at $$x=\frac{\pi}{4}$$ is

A
$$-2 \sqrt{2}$$
B
$$2 \sqrt{2}$$
C
$$-4$$
D
4
3
JEE Main 2022 (Online) 27th June Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

If $${\cos ^{ - 1}}\left( {{y \over 2}} \right) = {\log _e}{\left( {{x \over 5}} \right)^5},\,|y| < 2$$, then :

A
$${x^2}y'' + xy' - 25y = 0$$
B
$${x^2}y'' - xy' - 25y = 0$$
C
$${x^2}y'' - xy' + 25y = 0$$
D
$${x^2}y'' + xy' + 25y = 0$$
4
JEE Main 2022 (Online) 25th June Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Let f : R $$\to$$ R be defined as $$f(x) = {x^3} + x - 5$$. If g(x) is a function such that $$f(g(x)) = x,\forall 'x' \in R$$, then g'(63) is equal to ________________.

A
$${1 \over {49}}$$
B
$${3 \over {49}}$$
C
$${43 \over {49}}$$
D
$${91 \over {49}}$$
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