1
MHT CET 2025 21st April Evening Shift
MCQ (Single Correct Answer)
+2
-0

$$ \mathop {\lim }\limits_{n \to \infty }\left[\frac{1}{1-n^4}+\frac{8}{1-n^4}+\ldots \ldots \ldots \ldots .+\frac{n^3}{1-n^4}\right]= $$

A
$\frac{1}{4}$
B
$\frac{1}{2}$
C
$-\frac{1}{2}$
D
$-\frac{1}{4}$
2
MHT CET 2025 21st April Evening Shift
MCQ (Single Correct Answer)
+2
-0

If $f(\theta)=\cos \theta_1 \cdot \cos \theta_2 \cdot \cos \theta_3$ .............. $\cos \theta_n$, then $\tan \theta_1+\tan \theta_2+\tan \theta_3+$. ............ $+\tan \theta_{\mathrm{n}}=$

A
$\frac{-\mathrm{f}^{\prime}(\theta)}{\mathrm{f}(\theta)}$
B
$\frac{\mathrm{f}^{\prime}(\theta)}{\mathrm{f}(\theta)}$
C
$\frac{-\mathrm{f}^{\prime \prime}(\theta)}{\mathrm{f}^{\prime}(\theta)}$
D
$\frac{\mathrm{f}^{\prime \prime}(\theta)}{\mathrm{f}^{\prime}(\theta)}$
3
MHT CET 2025 21st April Morning Shift
MCQ (Single Correct Answer)
+2
-0

$$ \mathop {\lim }\limits_{x \to \infty } \frac{\mathrm{e}^{x^4}-1}{\mathrm{e}^{x^4}+1}= $$

A
1
B
e
C
$\frac{1}{\mathrm{e}}$
D
not defined
4
MHT CET 2025 21st April Morning Shift
MCQ (Single Correct Answer)
+2
-0

If $f(x)= \begin{cases}\frac{8^x-4^x-2^x+1}{x^2}, & \text { if } x>0 \\ e^x \sin x+x+\lambda \log 4, & \text { if } x \leqslant 0\end{cases}$

is continuous at $x=0$ then the value of $1000 \mathrm{e}^\lambda=$

A
1000
B
3000
C
2000
D
4000 $$\mathop {\lim }\limits_{x \to {0^ - }} $$
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