1
MHT CET 2023 11th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

If $$\mathrm{A}=\left[\begin{array}{ll}\mathrm{i} & 1 \\ 1 & 0\end{array}\right]$$ where $$\mathrm{i}=\sqrt{-1}$$ and $$\mathrm{B}=\mathrm{A}^{2029}$$, then $$\mathrm{B}^{-1}=$$

A
$$-\mathrm{A}$$
B
$$\operatorname{adj} \mathrm{A}$$
C
$$\mathrm{-I}$$
D
$$-\operatorname{adj} \mathrm{A}$$
2
MHT CET 2023 11th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

The solution of the differential equation $$\frac{\mathrm{d} y}{\mathrm{~d} x}+\frac{y}{x}=\sin x$$ is

A
$$x y+\cos x=\sin x+\mathrm{c}$$, where c is a constant of integration.
B
$$x(y+\cos x)=\sin x+\mathrm{c}$$, where c is a constant of integration.
C
$$y(x+\cos x)=\sin x+c$$, where c is a constant of integration.
D
$$x y+\sin x=\cos x+\mathrm{c}$$, where c is a constant of integration.
3
MHT CET 2023 11th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

Let $$f:[-1,2] \rightarrow[0, \infty)$$ be a continuous function such that $$\mathrm{f}(x)=\mathrm{f}(1-x), \forall x \in[-1,2]$$

Let $$\mathrm{R}_1=\int_{-1}^2 x \mathrm{f}(x) \mathrm{d} x$$ and $$\mathrm{R}_2$$ be the area of the region bounded by $$y=\mathrm{f}(x), x=-1, x=2$$ and the $$\mathrm{X}$$-axis, then $$\mathrm{R}_2$$ is

A
$$\frac{1}{2} R_1$$
B
$$2 R_1$$
C
$$3 R_1$$
D
$$\frac{1}{3} R_1$$
4
MHT CET 2023 11th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

The equation of line passing through the point $$(1,2,3)$$ and perpendicular to the lines $$\frac{x-2}{3}=\frac{y-1}{2}=\frac{z+1}{-2}$$ and $$\frac{x}{2}=\frac{y}{-3}=\frac{z}{1}$$ is

A
$$\overline{\mathrm{r}}=(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})+\lambda(4 \hat{\mathrm{i}}+7 \hat{\mathrm{j}}-13 \hat{\mathrm{k}})$$
B
$$\overline{\mathrm{r}}=(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})+\lambda(-4 \hat{\mathrm{i}}+7 \hat{\mathrm{j}}-13 \hat{\mathrm{k}})$$
C
$$\overline{\mathrm{r}}=(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})+\lambda(-4 \hat{\mathrm{i}}-7 \hat{\mathrm{j}}-13 \hat{\mathrm{k}})$$
D
$$\overline{\mathrm{r}}=(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})+\lambda(4 \hat{\mathrm{i}}-7 \hat{\mathrm{j}}-13 \hat{\mathrm{k}})$$
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