1
MHT CET 2021 23rd September Evening Shift
+2
-0

\begin{aligned} & \text { If the function given by} \mathrm{f}(\mathrm{x}) \\ & =-2 \sin \mathrm{x} \quad-\pi \leq \mathrm{x}<-(\pi / 2) \\ & =a \sin x+b \quad-(\pi / 2)< x<(\pi / 2) \\ & =\cos x \quad(\pi / 2) \leq x \leq \pi \\ \end{aligned}

is continuous in $$[-\pi, \pi]$$, then the value of $$(3 a+2 b)^3$$ is

A
1
B
8
C
$$-$$1
D
$$-$$8
2
MHT CET 2021 23th September Morning Shift
+2
-0

If $$f(x)=\frac{1-\sin x+\cos x}{1+\sin x+\cos x}$$, for $$x \neq \pi$$ is continuous at $$x=\pi$$, then the value of $$f(\pi)$$ is

A
$$\frac{-1}{2}$$
B
$$-1$$
C
1
D
$$\frac{1}{2}$$
3
MHT CET 2021 23th September Morning Shift
+2
-0

$$\lim _\limits{x \rightarrow 1}\left[\frac{\sqrt{x}-1}{\log x}\right]=$$

A
$$\frac{1}{2}$$
B
2
C
$$-2$$
D
$$-\frac{1}{2}$$
4
MHT CET 2021 22th September Evening Shift
+2
-0

Let

\begin{aligned} f(x) & =x+a \sqrt{2} \sin x & & , 0 \leq x<\frac{\pi}{4} \\ & =2 x \cot x+b & & \frac{\pi}{4} \leq x<\frac{\pi}{2} \\ & =a \cos 2 x-b \sin x & & \frac{\pi}{2} \leq x \leq \pi \end{aligned}

If $$\mathrm{f}(\mathrm{x})$$ is continuous for $$0 \leq \mathrm{x} \leq \pi$$, then

A
$$a=\frac{\pi}{6}, b=\frac{\pi}{12}$$
B
$$\mathrm{a}=\frac{-\pi}{6}, \mathrm{~b}=\frac{-\pi}{12}$$
C
$$a=\frac{-\pi}{6}, b=\frac{\pi}{12}$$
D
$$a=\frac{\pi}{6}, b=\frac{-\pi}{12}$$
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