MHT CET 2025 23rd April Evening Shift | MHT CET Year Wise Previous Years Questions

MHT CET 2025 23rd April Evening Shift

MCQ (Single Correct Answer)

-0

$$ \int \sqrt{x^2+3 x} d x= $$

$\sqrt{x^2+3 x}+\log \sqrt{x^2+3 x}+c$, where c is the constant of integration.

$\frac{2 x+3}{4} \sqrt{x^2+3 x}-\frac{9}{8} \log \left(x+\sqrt{x^2+3 x}\right)+c$, where c is the constant of integration.

$x \sqrt{x^2+3 x}+\log \left(x+\sqrt{x^2+3 x}\right)+\mathrm{c}$, where c is the constant of integration.

$x+3 \sqrt{x^2+3 x}+\frac{3}{2} \log \left(x+\sqrt{x^2+3 x}\right)+\mathrm{c}$, where c is the constant of integration.

MHT CET 2025 23rd April Evening Shift

MCQ (Single Correct Answer)

-0

If the line $a x+b y+c=0$ is normal to the curve $x y=1$, then

$\mathrm{a}>0, \mathrm{~b}>0$

$\quad a>0, b<0$

a $<0$, b $\geqslant 0$

a $<0$, b $<0$

MHT CET 2025 23rd April Evening Shift

MCQ (Single Correct Answer)

-0

The sum of two nonzero numbers is 4 . The minimum value of the sum of their reciprocals is

$\frac{3}{4}$

$\frac{6}{5}$

MHT CET 2025 23rd April Evening Shift

MCQ (Single Correct Answer)

-0

The combined equation of the tangent and normal to the curve $x y=15$ at the point $(5,3)$ is________

$15 x^2-15 y^2+16 x y=480$

$15 x^2+16 x y-198 x+10 y+480-15 y^2=0$

$15 x^2-16 x y+19 x-10 y-480+15 y^2=0$

$15 x^2+16 x y+198 x-10 y-480+15 y^2=0$

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2019

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