Limits, Continuity and Differentiability · Mathematics · TS EAMCET
MCQ (Single Correct Answer)
1
$\lim\limits_{x \rightarrow \frac{3}{2}} \frac{\left(4 x^{2}-6 x\right)\left(4 x^{2}+6 x+9\right)}{\sqrt[3]{2 x}-\sqrt[3]{3}}=$
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2
If the real valued function $f(x)=\int \frac{\left(4^{x}-1\right)^{4} \cot (x \log 4)}{\sin (x \log 4) \log \left(1+x^{2} \log 4\right)}, \quad$ if $x \neq 0$ is continuous at $x=0$, then $e^{k}=$
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3
If $0 \leq x \leq \frac{\pi}{2}$, then $\lim _{x \rightarrow a} \frac{|2 \cos x-1|}{2 \cos x-1}$
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4
The real valued function $f(x)=\frac{|x-a|}{x-a}$ is
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5
If $f(x)=3 x^{15}-5 x^{10}+7 x^{5}+50 \cos (x-1)$, then $\lim\limits_{h \rightarrow 0} \frac{f(1-h)-f(1)}{h^{3}+3 h}$
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6
If the function $f(x)=\left\{\begin{array}{cl}\frac{\left(e^{k x}-1\right) \sin k x}{4 \tan x} & x \neq 0 \\ P & x=0\end{array}\right.$ is differentiable at $x=0$, then
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7
If Rolle's Theorem is applicable for the function $f(x)=\left\{\begin{array}{cl}x^{p} \log x, & x \neq 0 \\ 0, & x=0\end{array}\right.$ on the interval $[0,1]$, then a possible value of $p$ is
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8
If $\lim \limits_{x \rightarrow 4} \frac{2 x^2+(3+2 a) x+3 a}{x^3-2 x^2-23 x+60}=\frac{11}{9}$, then $\lim \limits_{x \rightarrow a} \frac{x^2+9 x+20}{x^2-x-20}=$
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9
If the function
$$
f(x)= \begin{cases}\frac{\tan a(x-1)}{x-1}, & \text { if } 04\end{cases}
$$
domain, then $6 a+9 b^4=$
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10
$\lim _{\theta \rightarrow \frac{\pi^{-}}{2}} \frac{8 \tan ^4 \theta+4 \tan ^2 \theta+5}{(3-2 \tan \theta)^4}=$
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11
Define $ f: R \rightarrow R $ by $ f(x)=\left\{\begin{array}{cl}\frac{1-\cos 4 x}{x^{2}}, & x < 0 \\ a, & x=0 \\ \frac{\sqrt{x}}{\sqrt{16+\sqrt{x}}-4}, & x > 0\end{array}\right. $
Then, the value of $ a $ so that $ f $ is continuous at $ x=0 $ is
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12
$\lim _{x \rightarrow 0} \frac{3^{\sin x}-2^{\tan x}}{\sin x}=$
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13
If the function
$$ f(x)=\left\{\begin{array}{cc} \frac{\cos a x-\cos 9 x}{x^2} & \text {, if } x \neq 0 \\ 16 & \text {, if } x=0 \end{array}\right. $$
is continuous at $x=0$, then $a=$
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14
If $ f(x)=\left\{\begin{array}{ll}\frac{8}{x^{3}}-6 x & \text {, if } 0 < x \leq 1 \\\\ \frac{x-1}{\sqrt{x}-1} & \text {,if } x > 1\end{array}\right. $ is a real valued function, then at $ x=1, f $ is
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15
$\lim \limits_{n \rightarrow \infty}\left[\left(1+\frac{1}{n^2}\right)\left(1+\frac{4}{n^2}\right)\left(1+\frac{9}{n^2}\right) \ldots .(2)\right]^{1 / n}=$
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