1
MHT CET 2023 11th May Morning Shift
MCQ (Single Correct Answer)
+2
-0

If $$\mathrm{f}(x)=\left\{\begin{array}{ll}\frac{\sqrt{1+\mathrm{m} x}-\sqrt{1-\mathrm{m} x}}{x}, & -1 \leq x < 0 \\ \frac{2 x+1}{x-2} & , 0 \leq x \leq 1\end{array}\right.$$ is continuous in the interval $$[-1,1]$$, then $$\mathrm{m}$$ is equal to

A
$$\frac{1}{2}$$
B
$$-\frac{1}{2}$$
C
$$-1$$
D
$$-\frac{1}{4}$$
2
MHT CET 2023 10th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

Let $$\mathrm{S}=\left\{\mathrm{t} \in \mathrm{R} / \mathrm{f}(x)=|x-\pi|\left(\mathrm{e}^{|x|}-1\right) \sin |x|\right.$$ is not differentiable at $$\mathrm{t}\}$$, then $$\mathrm{S}$$ is

A
$$\phi$$ (an empty set)
B
$$\{0\}$$
C
$$\{\pi\}$$
D
$$\{0, \pi\}$$
3
MHT CET 2023 10th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

If $$\mathrm{f}(x)=\frac{4}{x^4}\left[1-\cos \frac{x}{2}-\cos \frac{x}{4}+\cos \frac{x}{2} \cdot \cos \frac{x}{4}\right]$$ is continuous at $$x=0$$, then $$\mathrm{f}(0)$$ is

A
$$\frac{1}{32}$$
B
$$\frac{1}{16}$$
C
$$\frac{1}{8}$$
D
$$\frac{1}{64}$$
4
MHT CET 2023 10th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

$$\lim _\limits{x \rightarrow a} \frac{\sqrt{a+2 x}-\sqrt{3 x}}{\sqrt{3 a+x}-2 \sqrt{x}}=$$

A
$$\frac{1}{3 \sqrt{3}}$$
B
$$\frac{2}{\sqrt{3}}$$
C
$$\frac{2}{3 \sqrt{3}}$$
D
$$\frac{-2}{3 \sqrt{3}}$$
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