1
BITSAT 2024
MCQ (Single Correct Answer)
+3
-1

Let $ f $ be the function defined by

$ f(x)=\left\{\begin{array}{cc} \frac{x^{2}-1}{x^{2}-2|x-1|-1}, & x \neq 1 \\ \frac{1}{2}, & x=1 \end{array}\right. $

A
The function is continuous for all values of $ x $
B
The function is continuous only for $ x > 1 $
C
The function is continuous at $ x=1 $
D
The function is not continuous at $ x=1 $.
2
BITSAT 2023
MCQ (Single Correct Answer)
+3
-1

The value of $$\lim _\limits{x \rightarrow 0} \frac{8}{x^8}\left(1-\cos \frac{x^2}{2}-\cos \frac{x^2}{4}+\cos \frac{x^2}{2} \cos \frac{x^2}{4}\right)$$ is

A
$$\frac{1}{32}$$
B
$$\frac{1}{8}$$
C
$$\frac{1}{64}$$
D
$$\frac{1}{16}$$
3
BITSAT 2023
MCQ (Single Correct Answer)
+3
-1

$$ \text { The value of } \lim _\limits{x \rightarrow 0} \frac{(27+x)^{1 / 3}-3}{9-(27+x)^{2 / 3}} \text { equals to } $$

A
$$-\frac{1}{3}$$
B
$$\frac{1}{6}$$
C
$$-\frac{1}{6}$$
D
$$\frac{1}{3}$$
4
BITSAT 2022
MCQ (Single Correct Answer)
+3
-1

The value of $$\mathop {\lim }\limits_{x \to 0} {{1-\cos (1 - \cos x)} \over {{x^4}}}$$ is

A
$$\frac{1}{6}$$
B
$$\frac{1}{8}$$
C
$$\frac{1}{10}$$
D
$$\frac{1}{12}$$
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