1
MHT CET 2024 4th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

Let $\mathrm{A}=\left[\begin{array}{cc}1 & 2 \\ -5 & 1\end{array}\right]$ and $\mathrm{A}^{-1}=x \mathrm{~A}+y \mathrm{I}_2$, (where $\mathrm{I}_2$ is unit matrix of order 2), then

A
$x=\frac{-1}{11}, y=\frac{2}{11}$
B
$x=\frac{1}{11}, y=\frac{-2}{11}$
C
$x=\frac{-1}{11}, y=\frac{-2}{11}$
D
$x=\frac{1}{11}, y=\frac{2}{11}$
2
MHT CET 2024 4th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

If $\mathrm{f}(x)=\frac{x+x^2+x^3+\ldots \ldots \ldots \ldots+x^{\mathrm{n}}-\mathrm{n}}{x-1}$, for $x \neq 1$ is continuous at $x=1$, then $\mathrm{f}(1)=$

A
$\frac{\mathrm{n}(\mathrm{n}+1)(4 \mathrm{n}-1)}{6}$
B
$\frac{\mathrm{n}(\mathrm{n}+1)}{2}$
C
$\frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6}$
D
$\frac{\mathrm{n}(2 \mathrm{n}+1)}{4}$
3
MHT CET 2024 4th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

The particular solution of differential equation $\left(1+y^2\right)(1+\log x) \mathrm{d} x+x \mathrm{~d} y=0$ at $x=1, y=1$ is

A
$\log x-\frac{1}{2}(\log x)^2-\tan ^{-1} y=-\frac{\pi}{4}$
B
$\log x+\frac{1}{2}(\log x)^2+\tan ^{-1} y=\frac{\pi}{4}$
C
$\log x-\frac{1}{2}(\log x)^2+\tan ^{-1} y=\frac{\pi}{4}$
D
$\log x+\frac{1}{2}(\log x)^2-\tan ^{-1} y=\frac{\pi}{4}$
4
MHT CET 2024 4th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

If $\lim _\limits{x \rightarrow 1} \frac{x^2-a x+b}{x-1}=7$, then $a+b$ is equal to

A
$-$1
B
1
C
$-$11
D
11
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