1
MHT CET 2024 9th May Morning Shift
MCQ (Single Correct Answer)
+2
-0

The area (in sq. units) bounded by the curves $y=(x+1)^2, y=(x-1)^2$ and the line $y=\frac{1}{4}$ is

A
$\frac{2}{3}$
B
$\frac{1}{6}$
C
$\frac{1}{3}$
D
$\frac{1}{4}$
2
MHT CET 2024 9th May Morning Shift
MCQ (Single Correct Answer)
+2
-0

If $\int \frac{d x}{\sqrt[3]{\sin ^{11} x \cos x}}=-\left(\frac{3}{8} f(x)+\frac{3}{2} g(x)\right)+c$ then

A
$\mathrm{f}(x)=\tan ^{\frac{-8}{3}} x, \mathrm{~g}(x)=\tan ^{\frac{-2}{3}} x$, (where c is a constant of integration)
B
$\mathrm{f}(x)=\tan ^{\frac{8}{3}} x, \mathrm{~g}(x)=\tan ^{\frac{-2}{3}} x$, (where c is a constant of integration)
C
$\mathrm{f}(x)=\tan ^{\frac{-8}{3}} x, \mathrm{~g}(x)=\tan ^{\frac{2}{3}} x$, (where c is a constant of integration)
D
$\mathrm{f}(x)=\tan ^{\frac{8}{3}} x, \mathrm{~g}(x)=\tan ^{\frac{2}{3}} x$, (where c is a constant of integration)
3
MHT CET 2024 9th May Morning Shift
MCQ (Single Correct Answer)
+2
-0

A random variable X takes values $-1,0,1,2$ with probabilities $\frac{1+3 \mathrm{p}}{4}, \frac{1-\mathrm{p}}{4}, \frac{1+2 \mathrm{p}}{4}, \frac{1-4 \mathrm{p}}{4}$ respectively, where p varies over $\mathbb{R}$. Then the minimum and maximum values of the mean of X are respectively.

A
$-\frac{7}{4}$ and $\frac{1}{2}$
B
$-\frac{1}{16}$ and $\frac{5}{16}$
C
$-\frac{7}{4}$ and $\frac{5}{16}$
D
$-\frac{1}{16}$ and $\frac{5}{4}$
4
MHT CET 2024 9th May Morning Shift
MCQ (Single Correct Answer)
+2
-0

Let C be a curve given by $y(x)=1+\sqrt{4 x-3}$, $x>\frac{3}{4}$. If P is a point on C , such that the tangent at P has slope $\frac{2}{3}$, then a point through which the normal at P passes, is

A
$(1,7)$
B
$(4,-3)$
C
$(3,-4)$
D
$(2,3)$
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