MHT CET 2025 25th April Evening Shift | MHT CET Year Wise Previous Years Questions

MHT CET 2025 25th April Evening Shift

MCQ (Single Correct Answer)

-0

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

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

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

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

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

MHT CET 2025 25th April Evening Shift

MCQ (Single Correct Answer)

-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

$\sqrt{2}$ units

$3 \sqrt{2}$ units

$2 \sqrt{2}$ units

$\frac{1}{\sqrt{2}}$ units

MHT CET 2025 25th April Evening Shift

MCQ (Single Correct Answer)

-0

If $\mathrm{f}(x)=\log (1+x)-\frac{2 x}{2+x}$ then $\mathrm{f}(x)$ is increasing in

$(-1, \infty)$

$(-\infty, \infty)$

$(0, \infty)$

$(1, \infty)$

MHT CET 2025 25th April Evening Shift

MCQ (Single Correct Answer)

-0

The angle $\theta$, at which the curves $y=3^x$ and $y=7^x$ intersect, is given by

$\tan \theta=\frac{\log \left(\frac{3}{7}\right)}{1+(\log 3)(\log 7)}$

$\tan \theta=\frac{\log \left(\frac{7}{3}\right)}{1+(\log 3)(\log 7)}$

$\tan \theta=\frac{\log \left(\frac{3}{7}\right)}{1-(\log 3)(\log 7)}$

$\quad \tan \theta=\frac{\log \left(\frac{7}{3}\right)}{1-(\log 3)(\log 7)}$

PREVIOUS NEXT

MHT CET Papers

2022

MHT CET 2022 11th August Evening Shift

English

2020

MHT CET 2020 19th October Evening Shift

English

MHT CET 2020 16th October Evening Shift

English

MHT CET 2020 16th October Morning Shift

English

2019

MHT CET 2019 3rd May Morning Shift

English

MHT CET 2019 2nd May Evening Shift

English

MHT CET 2019 2nd May Morning Shift

English