1
MHT CET 2025 25th April Evening Shift
MCQ (Single Correct Answer)
+2
-0

Let $\mathrm{f}: \mathbb{R}-\{2\} \rightarrow \mathbb{R}-\{1\}$ defined by $\mathrm{f}(x)=\frac{x-3}{x-2}$ and $\mathrm{g}: \mathbb{R} \rightarrow \mathbb{R}$ defined by $\mathrm{g}(x)=3 x-2$, then sum of all values of $x$ for which $\mathrm{f}^{-1}(x)+\mathrm{g}^{-1}(x)=\frac{19}{6}$ is

A
$\frac{5}{2}$
B
$\frac{7}{2}$
C
$\frac{9}{2}$
D
$\frac{11}{2}$
2
MHT CET 2025 25th April Evening Shift
MCQ (Single Correct Answer)
+2
-0

There are 11 points in a plane of which 5 points are collinear. Then the total number of distinct quadrilaterals with vertices at these points is

A
265
B
330
C
250
D
325
3
MHT CET 2025 25th April Evening Shift
MCQ (Single Correct Answer)
+2
-0

$\int \frac{x^4 \cos \left(\tan ^{-1} x^5\right)}{1+x^{10}} d x$ equals

A

$\frac{\sin \left(\tan ^{-1} x^5\right)}{5}+\mathrm{c}$, where c is the constant of integration

B

$\quad x^4 \sin \left(\tan ^{-1} x^5\right)+\mathrm{c}$, where c is the constant of integration

C

$\frac{\sin \left(\tan ^{-1} x^5\right)}{4}+\mathrm{c}$, where c is the constant of integration

D

$\quad \cos \left(\tan ^{-1} x^5\right)+\mathrm{c}$, where c is the constant of integration

4
MHT CET 2025 25th April Evening Shift
MCQ (Single Correct Answer)
+2
-0

The length of the perpendicular drawn from the origin on the normal to the curve $x^2+2 x y-3 y^2=0$ at the point $(2,2)$ is

A
$\sqrt{2}$ units
B
$3 \sqrt{2}$ units
C
$2 \sqrt{2}$ units
D
$\frac{1}{\sqrt{2}}$ units
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