COMEDK 2026 Morning Shift | Differentiation Question 4 | Mathematics

$$ \text { If } x=a(\theta-\sin \theta) \text { and } y=a(1-\cos \theta) \text {, then } \frac{\left(\mathbf{1}+\boldsymbol{y}_{\mathbf{1}}{ }^{\mathbf{2}}\right)^{\mathbf{3} / \mathbf{2}}}{\boldsymbol{y}_{\mathbf{2}}}= $$

$$ -4 a \sin \left(\frac{\theta}{2}\right) $$

$$ -\frac{1}{2 \sin ^2\left(\frac{\theta}{2}\right)} $$

$$ 4 a \operatorname{cosec}^2\left(\frac{\theta}{2}\right) $$

$$ -\frac{1}{4 a} \sin \left(\frac{\theta}{2}\right) $$

COMEDK 2026 Morning Shift

MCQ (Single Correct Answer)

-0

The derivative of $y=\sin ^2\left[\cot ^{-1}\left(\sqrt{\frac{\mathbf{1}-\boldsymbol{x}}{\mathbf{1}+\boldsymbol{x}}}\right)\right]$ is

$\frac{1}{2}$

$\frac{1-x}{2}$

$$ \frac{x}{2} $$

COMEDK 2025 Evening Shift

MCQ (Single Correct Answer)

-0

If $x=a\left[\left\{\cos t+\frac{1}{2} \log \left(\tan ^2 \frac{t}{2}\right)\right\}\right]$ and $y=a \sin t$ then $\frac{d y}{d x}=$

$\frac{a^2 \cos ^3 t}{\sin t}$

$\tan t$

a $\tan t \sec t$

$\sec ^2 t$

COMEDK 2025 Evening Shift

MCQ (Single Correct Answer)

-0

If $y=x+e^x$ then $\frac{d^2 x}{d y^2}=$

$e^x$

$\frac{-e^x}{\left(1+e^x\right)^2}$

$\frac{-e^x}{\left(1+e^x\right)^3}$

$\frac{-1}{\left(1+e^x\right)^3}$

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