MHT CET 2026 11th April Evening Shift | MHT CET Year Wise Previous Years Questions

MHT CET 2026 11th April Evening Shift

MCQ (Single Correct Answer)

-0

$\int \sqrt{4^x(4^x + 4)}\, dx = \cdots$

$\dfrac{1}{\log 2}\left[\dfrac{2^x}{2}\sqrt{4^x + 4} + 2\log|2^x + \sqrt{4^x + 4}|\right] + c$

$\dfrac{2^x}{2}\sqrt{4^x + 4} + 2\log|2^x + \sqrt{4^x + 4}| + c$

$\dfrac{1}{\log 2}\left[\dfrac{2^x}{2}\sqrt{4^x + 4} + \log|2^x + \sqrt{4^x + 4}|\right] + c$

$\left[\dfrac{2^x}{\log 4}\sqrt{4^x + 4} + \log|2^x + \sqrt{4^x + 4}|\right] + c$

MHT CET 2026 11th April Evening Shift

MCQ (Single Correct Answer)

-0

The value of $\int \dfrac{1}{x^2 - 5x + 8} dx$ is equal to...

$\dfrac{2}{\sqrt{7}}\tan^{-1}\left(\dfrac{2x+5}{\sqrt{7}}\right) + c$

$-\dfrac{2}{\sqrt{7}}\tan^{-1}\left(\dfrac{2x-5}{\sqrt{7}}\right) + c$

$\dfrac{2}{\sqrt{7}}\tan^{-1}\left(\dfrac{2x-5}{\sqrt{7}}\right) + c$

$-\dfrac{2}{\sqrt{7}}\tan^{-1}\left(\dfrac{2x+5}{\sqrt{7}}\right) + c$

MHT CET 2026 11th April Evening Shift

MCQ (Single Correct Answer)

-0

$\int \dfrac{2x}{4 - 3x - x^2} dx = \cdots$

$\dfrac{8}{5}\log|4 + x| - \dfrac{2}{5}\log|1 - x| + c$

$2\log|4 + x| + \dfrac{2}{5}\log|1 - x| + c$

$-\log|4 + x| + \dfrac{2}{5}\log|1 - x| + c$

$-\dfrac{8}{5}\log|4 + x| - \dfrac{2}{5}\log|1 - x| + c$

MHT CET 2026 11th April Evening Shift

MCQ (Single Correct Answer)

-0

If $\int_{2}^{e}\left[\dfrac{1}{\log x} - \dfrac{1}{(\log x)^2}\right] dx = a + \dfrac{b}{\log 2}$, then values of $a$ and $b$ are

$a = e, b = 2$

$a = -e, b = 2$

$a = e, b = -2$

$a = -e, b = -2$

PREVIOUS NEXT

MHT CET Papers

All year-wise previous year question 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