1
MHT CET 2020 19th October Evening Shift
MCQ (Single Correct Answer)
+2
-0

$\int_\limits0^1 \tan ^{-1}\left(\frac{2 x-1}{1+x-x^2}\right) d x=$

A
2
B
1
C
0
D
4
2
MHT CET 2020 19th October Evening Shift
MCQ (Single Correct Answer)
+2
-0

If $f(x)=\frac{1-\sin x+\cos x}{1+\sin x+\cos x}$, for $x \neq \pi$ is continuous at $x=\pi$, then $f(\pi)=$

A
1
B
$-$1
C
0
D
2
3
MHT CET 2020 19th October Evening Shift
MCQ (Single Correct Answer)
+2
-0

$\mathbf{a}$ and $\mathbf{b}$ are non-collinear vectors. If $p=(2 x+1) a-b$ and $q=(x-2) a+b$ are collinear vectors, then $x=$

A
$\frac{1}{3}$
B
$-\frac{1}{3}$
C
$-3$
D
3
4
MHT CET 2020 19th October Evening Shift
MCQ (Single Correct Answer)
+2
-0

If $A=\left[\begin{array}{ll}4 & 5 \\ 2 & 1\end{array}\right]$ and $A^2-5 A-6 I=0$, then $A^{-1}=$

A
$\frac{1}{6}\left[\begin{array}{cc}-1 & 5 \\ -2 & -4\end{array}\right]$.
B
$\frac{1}{6}\left[\begin{array}{cc}-1 & 5 \\ 2 & 4\end{array}\right]$
C
$\frac{1}{6}\left[\begin{array}{cc}-1 & 5 \\ 2 & -4\end{array}\right]$
D
$\frac{1}{6}\left[\begin{array}{cc}1 & 5 \\ 2 & -4\end{array}\right]$
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