1
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
2
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
$\frac{1}{6}\left[\begin{array}{cc}1 & 5 \\ 2 & -4\end{array}\right]$
3
MHT CET 2020 19th October Evening Shift
MCQ (Single Correct Answer)
+2
-0

The quadratic equation whose roots are the numbers having arithmetic mean 34 and geometric mean 16 is

A
$x^2+68 x+256=0$
B
$x^2+68 x-256=0$
C
$x^2-68 x+256=0$
D
$x^2-68 x-256=0$
4
MHT CET 2020 19th October Evening Shift
MCQ (Single Correct Answer)
+2
-0

If $x=a \sin t-b \cos t, y=a \cos t+b \sin t$, then $y^3 \frac{d^2 y}{d x^2}+x^2+y^2=$

A
2
B
1
C
0
D
$-$1
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