1
MHT CET 2026 18th April Evening Shift
MCQ (Single Correct Answer)
+2
-0
If $\int e^{x + \tan^{-1}x}\left(\dfrac{x^2 + 2}{\sec^2(\tan^{-1}x)}\right)dx = e^{f(x)} + c$, then $\ldots$
A
$f(x)$ is strictly decreasing on $R$.
B
$f(x)$ is strictly increasing on $R^+$ and strictly decreasing on $R^-$.
C
$f(x)$ is strictly increasing on $R$.
D
$f(x)$ is strictly decreasing on $R^+$ and strictly increasing on $R^-$.
2
MHT CET 2026 18th April Evening Shift
MCQ (Single Correct Answer)
+2
-0
Let $f(x) = x$, $f_1(x) = f(\log x)$, $f_2(x) = f_1(\log x)$, $f_3(x) = f_2(\log x)$, $\ldots$ and so on. Then $\int \dfrac{1}{f(x)\,f_1(x)\,f_2(x)\,\ldots f_{2026}(x)}\,dx = \ldots$
A
$f_{2025}(x) + c$
B
$2025 f_{2025}(x) + c$
C
$f_{2027}(x) + c$
D
$2027 f_{2027}(x) + c$
3
MHT CET 2026 18th April Morning Shift
MCQ (Single Correct Answer)
+2
-0
If $0 \leq x \leq 1$, $I_1 = \int\sin^{-1}\sqrt{1-x^2}\,dx$ and $I_2 = \int\sin^{-1}x\,dx$, then which of the following is true?
A
$I_1 = I_2$
B
$I_1 = \dfrac{\pi}{2}I_2$
C
$I_1 + I_2 = \dfrac{\pi}{2}x$
D
$I_1 + I_2 = \dfrac{\pi}{2}$
4
MHT CET 2026 18th April Morning Shift
MCQ (Single Correct Answer)
+2
-0
$\int\sin(\log x)\,dx = $
A
$\dfrac{x}{2}[\sin(\log x) - \cos(\log x)] + c$
B
$\dfrac{x}{2}[\sin(\log x) + \cos(\log x)] + c$
C
$\dfrac{x}{2}[\cos(\log x) - \sin(\log x)] + c$
D
$\dfrac{x}{4}[\cos(\log x) - \sin(\log x)] + c$

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