1
AP EAPCET 2024 - 23th May Morning Shift
MCQ (Single Correct Answer)
+1
-0

Let $f(x)=\left\{\begin{array}{cl}1+\frac{2 x}{a}, & 0 \leq x \leq 1 \\ a x, & 1 < x \leq 2\end{array}\right.$.If $\lim _{x \rightarrow 1} f(x)$ exists, then the sum of the cubes of the possible values of $a$ is

A
1
B
5
C
9
D
7
2
AP EAPCET 2024 - 23th May Morning Shift
MCQ (Single Correct Answer)
+1
-0

Let $[P]$ denote the greatest integer $\leq P$. If $0 \leq a \leq 2$, then the number of integral values of ' $a$ ' such that $\lim \limits_{x \rightarrow a}\left(\left[x^2\right]-[x]^2\right)$ does not exist is

A
3
B
2
C
1
D
0
3
AP EAPCET 2024 - 23th May Morning Shift
MCQ (Single Correct Answer)
+1
-0
If $f(x)=\left\{\begin{array}{cl}\frac{\sqrt{a^2-a x+x^2}-\sqrt{x^2+a x+a^2}}{\sqrt{a+x}-\sqrt{a-x}}, & x \neq 0 \text { is } \\ K & x=0\end{array}\right.$ continuous at $x=0$, then $K$ is equal to
A
$-\sqrt{a}$
B
$\sqrt{a}$
C
-1
D
$a+\sqrt{a}$
4
AP EAPCET 2024 - 23th May Morning Shift
MCQ (Single Correct Answer)
+1
-0
If $f(x)=\left\{\begin{array}{cc}a x^2+b x-\frac{13}{8}, & x \leq 1 \\ 3 x-3, & 1 < x \leq 2 \text { is differentiable } \\ b x^3+1, & x > 2\end{array}\right.$ $\forall x \in R$, then $a-b$ is equal to
A
$\frac{9}{8}$
B
$\frac{5}{4}$
C
$\frac{11}{8}$
D
$\frac{1}{4}$
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