1
AP EAPCET 2024 - 22th May Morning Shift
MCQ (Single Correct Answer)
+1
-0
$$\mathop {\lim }\limits_{n \to \infty }\left(\frac{1}{\sqrt{n^2}}+\frac{1}{\sqrt{n^2-1}}+\ldots+\frac{1}{\sqrt{n^2-(n-1)^2}}\right)= $$
A
$2 \sqrt{\pi}$
B
$\frac{2}{\sqrt{\pi}}$
C
$\frac{\pi}{2}$
D
$\frac{3 \pi}{2}$ $$\left( {{{} \over {}}} \right)$$ $$ (\because n=\infty) $$
2
AP EAPCET 2024 - 21th May Evening Shift
MCQ (Single Correct Answer)
+1
-0
$$\mathop {\lim }\limits_{x \to 0} \left( {{{\sin (\pi {{\cos }^2}x} \over {{x^2}}}} \right) = $$
A
$-\pi$
B
$\pi$
C
$\frac{\pi}{2}$
D
1
3
AP EAPCET 2024 - 21th May Evening Shift
MCQ (Single Correct Answer)
+1
-0
$$\mathop {\lim }\limits_{x \to 1} \left( {{{x + {x^2} + {x^3} + ... + {x^n} - n} \over {x - 1}}} \right) = $$
A
$\frac{n(n+1)}{2}$
B
$\frac{n+1}{2}$
C
$\frac{2}{n}$
D
$n$
4
AP EAPCET 2024 - 21th May Evening Shift
MCQ (Single Correct Answer)
+1
-0
If the function $f(x)=\frac{\sqrt{1+x}-1}{x}$ is continuous at $x=0$, then $f(0)=$
A
$-\frac{1}{2}$
B
$\frac{1}{3}$
C
$\frac{1}{2}$
D
$-\frac{1}{3}$
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