1
COMEDK 2026 Morning Shift
MCQ (Single Correct Answer)
+1
-0

$$ \text { The solution of } \boldsymbol{d} \boldsymbol{y}=\boldsymbol{\operatorname { c o s }} \boldsymbol{x}(\mathbf{2}-\boldsymbol{y} \boldsymbol{\operatorname { c o s e c }} \boldsymbol{x}) \boldsymbol{d} \boldsymbol{x} \quad \text { where } y=\sqrt{2} \text { when } x=\frac{\boldsymbol{\pi}}{4} \text { is } $$

A

$y=\sin x+\frac{1}{2} \operatorname{cosec} x$

B

$y=\tan \left(\frac{x}{2}\right)+\cot \left(\frac{x}{2}\right)$

C

$y \sin x=\frac{1}{2} \cos 2 x$

D

$y=\frac{1}{\sqrt{2}} \sec x+\sqrt{2} \cos \left(\frac{x}{2}\right)$

2
COMEDK 2026 Morning Shift
MCQ (Single Correct Answer)
+1
-0

$$ \text { The conjugate of } z=\frac{(\mathbf{4}+\boldsymbol{i})(\mathbf{1}-\boldsymbol{i})}{(\mathbf{1}+\boldsymbol{i})(\mathbf{2}-\boldsymbol{i})} $$

A

$\frac{6}{5}+\frac{7}{5} i$

B

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

C

$\frac{6}{5}-\frac{7}{5} i$

D

$\frac{1}{2}+\frac{1}{2} i$

3
COMEDK 2026 Morning Shift
MCQ (Single Correct Answer)
+1
-0

If A and B are two square matrices of the same order such that $\mathrm{AB}=\mathrm{A}$ and $\mathrm{BA}=\mathrm{B}$, then $(\boldsymbol{A}+\boldsymbol{B})^2$ is equal to:

A

$A^2+B^2+2 A$

B

$A+B$

C

$A^2+B^2$

D

$2(A+B)$

4
COMEDK 2026 Morning Shift
MCQ (Single Correct Answer)
+1
-0

If $f(x)=\left\{\begin{array}{l}\frac{\sqrt{1+x}-\sqrt{1-x}}{\sin x} \\ \boldsymbol{k}, x=0\end{array}, x \neq 0\right.$ is continuous at $x=0$, then $\boldsymbol{k}=$

A

1

B

$\frac{1}{2}$

C

2

D

0