1
MHT CET 2021 21th September Evening Shift
+2
-0

If $$\lim _\limits{x \rightarrow 5} \frac{x^k-5^k}{x-5}=500$$, then the value of $$k$$, where $$k \in N$$ is

A
5
B
3
C
4
D
6
2
MHT CET 2021 21th September Morning Shift
+2
-0

\begin{aligned} & \text { If the function } \mathrm{f}(\mathrm{x})=1+\sin \frac{\pi}{2}, \quad-\infty<\mathrm{x} \leq 1 \\ & =\mathrm{ax}+\mathrm{b}, \quad 1<\mathrm{x}<3 \\ & =6 \tan \frac{x \pi}{12}, \quad 3 \leq x<6 \\ \end{aligned}

is continuous in $$(-\infty, 6)$$, then the values of $$\mathrm{a}$$ and $$\mathrm{b}$$ are respectively.

A
1, 1
B
2, 1
C
0, 2
D
2, 0
3
MHT CET 2021 21th September Morning Shift
+2
-0

$$\lim _\limits{x \rightarrow 1} \frac{(2 x-3)(\sqrt{x}-1)}{2 x^2+x-3}=$$

A
$$\frac{1}{5}$$
B
$$\frac{1}{10}$$
C
$$\frac{-1}{10}$$
D
$$\frac{-1}{5}$$
4
MHT CET 2021 20th September Evening Shift
+2
-0

If $$a=\lim _\limits{n \rightarrow \infty} \frac{1+2+3+\ldots+n}{n^2}$$ and $$b=\lim _\limits{n \rightarrow \infty} \frac{1^2+2^2+3^2+\ldots+n^2}{n^3}$$, then

A
a = b
B
2a = 3b
C
a = 2b
D
3a = 2b
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