1
MHT CET 2023 14th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

The function $\mathrm{f}$ defined on $$\left(-\frac{1}{3}, \frac{1}{3}\right)$$ by $$\mathrm{f}(x)=\left\{\begin{array}{cc} \frac{1}{x} \log \left(\frac{1+3 x}{1-2 x}\right) & , \quad x \neq 0 \\ \mathrm{k} & , \quad x=0 \end{array}\right.$$ is continuous at $$x=0$$, then $$\mathrm{k}$$ is

A
6
B
1
C
5
D
$$-$$5
2
MHT CET 2023 14th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

The mirror image of $$\mathrm{P}(2,4,-1)$$ in the plane $$x-y+2 z-2=0$$ is $$(\mathrm{a}, \mathrm{b}, \mathrm{c})$$, then the value of $$a+b+c$$ is

A
4
B
5
C
7
D
9
3
MHT CET 2023 14th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

If the slope of the tangent of the curve at any point is equal to $$-y+\mathrm{e}^{-x}$$, then the equation of the curve passing through origin is

A
$$y+x \mathrm{e}^x=0$$
B
$$y \mathrm{e}^x+x=0$$
C
$$y \mathrm{e}^x-x=0$$
D
$$y-x \mathrm{e}^x=0$$
4
MHT CET 2023 14th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

If $$A=\left[\begin{array}{ll}1 & -1 \\ 2 & -1\end{array}\right], B=\left[\begin{array}{cc}1 & 1 \\ 4 & -1\end{array}\right]$$, then $$(A+B)^{-1}$$ is

A
$$\left[\begin{array}{cc}\frac{-1}{2} & 0 \\ \frac{-3}{2} & \frac{1}{2}\end{array}\right]$$
B
$$\left[\begin{array}{cc}\frac{1}{2} & 0 \\ \frac{3}{2} & \frac{-1}{2}\end{array}\right]$$
C
$$\left[\begin{array}{cc}\frac{1}{2} & 0 \\ \frac{-3}{2} & \frac{1}{2}\end{array}\right]$$
D
$$\left[\begin{array}{ll}\frac{1}{2} & 0 \\ \frac{3}{2} & \frac{1}{2}\end{array}\right]$$
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