$$ \text { If } x=a(\theta-\sin \theta) \text { and } y=a(1-\cos \theta) \text {, then } \frac{\left(\mathbf{1}+\boldsymbol{y}_{\mathbf{1}}{ }^{\mathbf{2}}\right)^{\mathbf{3} / \mathbf{2}}}{\boldsymbol{y}_{\mathbf{2}}}= $$
$$ -4 a \sin \left(\frac{\theta}{2}\right) $$
$$ -\frac{1}{2 \sin ^2\left(\frac{\theta}{2}\right)} $$
$$ 4 a \operatorname{cosec}^2\left(\frac{\theta}{2}\right) $$
$$ -\frac{1}{4 a} \sin \left(\frac{\theta}{2}\right) $$
The area of the region in the first quadrant enclosed by the $x$-axis, the line $x=\sqrt{3} y$ and the circle $x^2+y^2=4$ is
$\frac{\pi}{6}$ sq units
$\frac{2 \pi}{3}$ sq units
$\pi$ sq units
$\frac{\pi}{3}$ sq units
Which of the following is the simplest form of the expression $\boldsymbol{\operatorname { t a n }}^{-\mathbf{1}}\left(\frac{\sqrt{\mathbf{1 + x ^ { \mathbf { 2 } }}}-\mathbf{1}}{\boldsymbol{x}}\right)$ where $x \neq 0$
$2 \tan ^{-1} x$
$\tan ^{-1} \frac{x}{2}$
$\frac{1}{2} \tan ^{-1} x$
$\tan ^{-1} x$
$$ \mathop {\lim }\limits_{x \to 0} \frac{(1-\cos 2 x)(3+\cos x)}{x \tan 4 x} \text { is equal to: } $$
3
2
4
1
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