$$ \text { The conjugate of } z=\frac{(\mathbf{4}+\boldsymbol{i})(\mathbf{1}-\boldsymbol{i})}{(\mathbf{1}+\boldsymbol{i})(\mathbf{2}-\boldsymbol{i})} $$
$\frac{6}{5}+\frac{7}{5} i$
$\frac{1}{2}-\frac{1}{2} i$
$\frac{6}{5}-\frac{7}{5} i$
$\frac{1}{2}+\frac{1}{2} i$
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:
$A^2+B^2+2 A$
$A+B$
$A^2+B^2$
$2(A+B)$
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 $\boldsymbol{k}=$
1
$\frac{1}{2}$
2
0
The area of the region bounded by the line $y=x+2$ and the curve $x=-y^2$ is
13.5 sq units
$\frac{7}{6}$ sq units
4.5 sq units
2.5 sq units
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