1
MHT CET 2024 3rd May Evening Shift
MCQ (Single Correct Answer)
+2
-0

The vectors $\overline{\mathrm{a}}$ and $\overline{\mathrm{b}}$ are not perpendicular and $\overline{\mathrm{c}}$ and $\overline{\mathrm{d}}$ are two vectors satisfying $\overline{\mathrm{b}} \times \overline{\mathrm{c}}=\overline{\mathrm{b}} \times \overline{\mathrm{d}}$ and $\overline{\mathrm{a}} \cdot \overline{\mathrm{d}}=0$, then the vector $\overline{\mathrm{d}}$ is equal to

A
$\overline{\mathrm{b}}+\left(\frac{\overline{\mathrm{b}} \cdot \overline{\mathrm{c}}}{\overline{\mathrm{a}} \cdot \overline{\mathrm{b}}}\right) \overline{\mathrm{c}}$
B
$\overline{\mathrm{c}}-\left(\frac{\overline{\mathrm{a}} \cdot \overline{\mathrm{c}}}{\overline{\mathrm{a}} \cdot \overline{\mathrm{b}}}\right) \overline{\mathrm{b}}$
C
$\bar{b}-\left(\frac{\bar{b} \cdot \bar{c}}{\bar{a} \cdot \bar{b}}\right) \bar{c}$
D
$\overline{\mathrm{c}}+\left(\frac{\overline{\mathrm{a}} \cdot \overline{\mathrm{c}}}{\overline{\mathrm{a}} \cdot \overline{\mathrm{b}}}\right) \overline{\mathrm{b}}$
2
MHT CET 2024 3rd May Evening Shift
MCQ (Single Correct Answer)
+2
-0

If the function $\mathrm{f}(x)=\left(\frac{5 x-8}{8-3 x}\right)^{\frac{3}{2 x-4}}$ if $x \neq 2$. $=\mathrm{k}$ if $x=2$. is continuous at $x=2$, then $\mathrm{k}=$

A
$\mathrm{e}^6$
B
$\mathrm{e}^2$
C
$e^{-6}$
D
$\mathrm{e}^{-2}$
3
MHT CET 2024 3rd May Evening Shift
MCQ (Single Correct Answer)
+2
-0

The domain of definition of $\mathrm{f}(x)=\frac{\log _2(x+3)}{x^2+3 x+2}$ is

A
$\mathrm{R}-\{1,2\}$
B
$(-2, \infty)$
C
$\mathrm{R}-\{-1,-2,-3\}$
D
$(-3, \infty)-\{-1,-2\}$
4
MHT CET 2024 3rd May Evening Shift
MCQ (Single Correct Answer)
+2
-0

The equation of pair of lines $y=p x$ and $y=q x$ can be written as $(y-p x)(y-q x)=0$. Then the equation of the pair of the angle bisectors of the lines $x^2-4 x y-5 y^2=0$ is

A
$x^2-3 x y+y^2=0$
B
$x^2+4 x y-y^2=0$
C
$x^2-3 x y-y^2=0$
D
$x^2+3 x y-y^2=0$
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