1
MHT CET 2025 20th April Morning Shift
MCQ (Single Correct Answer)
+2
-0

If $\mathrm{f}(x)=\left\{\begin{array}{ll}\operatorname{m} x+1, & x \leqslant \frac{\pi}{2} \\ \sin x+\mathrm{n}, & x>\frac{\pi}{2}\end{array}\right.$, is continuous at $x=\frac{\pi}{2},(\mathrm{~m}, \mathrm{n} \in \mathbb{Z})$ then

A
$\mathrm{m}=1, \mathrm{n}=0$
B
$\mathrm{m}=\frac{\mathrm{n} \pi}{2}$
C
$\mathrm{m}=\mathrm{n}=\frac{\pi}{2}$
D
$\mathrm{n}=\frac{\mathrm{m} \pi}{2}$
2
MHT CET 2025 20th April Morning Shift
MCQ (Single Correct Answer)
+2
-0

$$ \mathop {\lim }\limits_{x \to 0} \frac{\mathrm{e}^{\tan x}-\mathrm{e}^x}{\tan x-x}= $$

A
1
B
0
C
$\frac{1}{2}$
D
$\frac{1}{4}$
3
MHT CET 2025 19th April Evening Shift
MCQ (Single Correct Answer)
+2
-0

If $\mathrm{f}(x)=\frac{(27-2 x)^{\frac{1}{3}}-3}{9-3(243+5 x)^{\frac{1}{5}}}, x \neq 0$ is continuous at $x=0$, then the value of $\mathrm{f}(0)$ is

A
$\frac{2}{3}$
B
6
C
2
D
$\frac{1}{3}$
4
MHT CET 2025 19th April Evening Shift
MCQ (Single Correct Answer)
+2
-0

$$ \mathop {\lim }\limits_{x \to 0} \frac{|x|}{|x|+x^2}= $$

A
0
B
1
C
-1
D
$\frac{1}{2}$
MHT CET Subjects
EXAM MAP