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

A particular solution of $\frac{\mathrm{d} y}{\mathrm{~d} x}=(x+9 y)^2$, when $x=0, y=\frac{1}{27}$ is

A
$\quad 3 x+27 y=\tan \left[3\left(x+\frac{\pi}{12}\right)\right]$
B
$\quad 3 x+27 y=\tan \left(x+\frac{\pi}{4}\right)$
C
$3 x+27 y=\tan \left(x+\frac{\pi}{12}\right)$
D
$3 x+27 y=\tan \left[3\left(x+\frac{\pi}{4}\right)\right]$
2
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
3
MHT CET 2025 21st April Morning Shift
MCQ (Single Correct Answer)
+2
-0

The shortest distance between the lines $\bar{r}=(4 \hat{i}-\hat{j})+\lambda(\hat{i}+2 \hat{j}-3 \hat{k})$ and $\bar{r}=(\hat{i}-\hat{j}+2 \hat{k})+\mu(2 \hat{i}+4 \hat{j}-5 \hat{k})$ is

A
$\frac{1}{\sqrt{5}}$ units
B
$\frac{6}{\sqrt{5}}$ units
C
$\frac{2}{\sqrt{5}}$ units
D
$\frac{3}{\sqrt{5}}$ units
4
MHT CET 2025 21st April Morning Shift
MCQ (Single Correct Answer)
+2
-0

If $\mathrm{f}(x)=\cos (\log x)$ then $\mathrm{f}\left(x^2\right) \cdot \mathrm{f}\left(y^2\right)-\frac{1}{2}\left[\mathrm{f}\left(\frac{x^2}{y^2}\right)+\mathrm{f}\left(x^2 y^2\right)\right]$ has the value

A
-2
B
-1
C
$\frac{1}{2}$
D
0
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