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Syllabus: Limits, Continuity, and Differentiability, Maxima and minima, Mean value theorem, Integration.

$$\scriptsize{\overset{{\large{\textbf{Mark Distribution in Previous GATE}}}}{\begin{array}{|c|c|c|c|c|c|c|c|}\hline
\textbf{Year}&\textbf{2022} & \textbf{2021-1}&\textbf{2021-2}&\textbf{2020}&\textbf{2019}&\textbf{2018}&\textbf{2017-1}&\textbf{2017-2}&\textbf{2016-1}&\textbf{2016-2}&\textbf{Minimum}&\textbf{Average}&\textbf{Maximum}
\\\hline\textbf{1 Mark Count} & 1 &1&1&1&1&1&0&1&1&1&0&0.9&1
\\\hline\textbf{2 Marks Count} & 0 &0&0&0&0&0&1&0&0&0&0&0.1&1
\\\hline\textbf{Total Marks} & 1 &1&1&1&1&1&2&1&1&1&\bf{1}&\bf{1.1}&\bf{2}\\\hline
\end{array}}}$$

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Highest voted questions in Calculus

3 votes
1 answer
121
#Made Easy
Find the value of $\lim_{x \rightarrow 0 } \dfrac{x \tan2x - 2x\tan x}{(1-\cos 2x)^2} \rule{1 in}{.5 pt}.$
Find the value of $\lim_{x \rightarrow 0 } \dfrac{x \tan2x - 2x\tan x}{(1-\cos 2x)^2} \rule{1 in}{.5 pt}.$
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3 votes
1 answer
122
GO Classes Test Series 2023 | Calculus | Test 1 | Question: 2
Determine the value of following limit $ \lim _{x \rightarrow \infty}\left(\sqrt{x^{2}+4 x+1}-x\right) $ $2$ $4$ $\frac{1}{2}$ $3$
Determine the value of following limit$$\lim _{x \rightarrow \infty}\left(\sqrt{x^{2}+4 x+1}-x\right)$$$2$$4$$\frac{1}{2}$$3$
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3 votes
3 answers
123
GO Classes Test Series 2023 | Calculus | Test 1 | Question: 3
The function $f(x)=x^{4}-6 x^{2}$ is increasing on the intervals $(0, \sqrt{3})$ only $(\sqrt{3}, \infty)$ only $(-\infty,-\sqrt{3})$ and $(0, \sqrt{3})$ only $(-\sqrt{3}, 0)$ and $(\sqrt{3}, \infty)$ only
The function $f(x)=x^{4}-6 x^{2}$ is increasing on the intervals$(0, \sqrt{3})$ only$(\sqrt{3}, \infty)$ only$(-\infty,-\sqrt{3})$ and $(0, \sqrt{3})$ only$(-\sqrt{3}, 0)...
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3 votes
1 answer
126
GATE Overflow Test Series | Mock GATE | Test 6 | Question: 31
The value of $\displaystyle{} \int_{0}^{\frac{\pi}{2}} \sin ^{4}x \cos^{4}x dx$ is _____ $\left(\dfrac{3\pi}{256}\right)$ $\left(\dfrac{5\pi}{768}\right)$ $\left(\dfrac{7\pi}{768}\right)$ $\left(\dfrac{3\pi}{384}\right)$
The value of $\displaystyle{} \int_{0}^{\frac{\pi}{2}} \sin ^{4}x \cos^{4}x dx$ is _____$\left(\dfrac{3\pi}{256}\right)$$\left(\dfrac{5\pi}{768}\right)$$\left(\dfrac{7\pi...
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3 votes
1 answer
127
GATE Overflow Test Series | Mock GATE | Test 4 | Question: 21
The minimum value of the function $f(x) = \frac{x^4}{4} - x^2 -3$ occurs at $x = 1$ $x =\sqrt 2$ $x = 0$ $ x = \frac{1}{\sqrt{4}}$
The minimum value of the function $$f(x) = \frac{x^4}{4} - x^2 -3$$ occurs at$x = 1$$x =\sqrt 2$$x = 0$$ x = \frac{1}{\sqrt{4}}$
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3 votes
2 answers
128
GATE Overflow Test Series | Mock GATE | Test 3 | Question: 59
If the function $f(x) =\left\{ \begin{array}{rcl} \alpha \sqrt{x+1} &;0\leq x \leq 3 \\\beta x + 2&;3 < x\leq 5\end{array}\right.$ is differentiable, then the value of $\alpha - \beta$ is _________
If the function $f(x) =\left\{ \begin{array}{rcl} \alpha \sqrt{x+1} &;0\leq x \leq 3 \\\beta x + 2&;3 < x\leq 5\end{array}\right.$ is differentiable, then the value of $\...
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3 votes
1 answer
129
GATE Overflow Test Series | Calculus | Test 1 | Question: 30
The value of the integral $I =\int_{-\infty }^{\infty}\frac{dx}{1+x^2}$ $-\pi$ $-\frac{\pi}{2}$ $\frac{\pi}{2}$ $\pi$
The value of the integral $$I =\int_{-\infty }^{\infty}\frac{dx}{1+x^2}$$$-\pi$$-\frac{\pi}{2}$$\frac{\pi}{2}$$\pi$
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3 votes
2 answers
130
NIELIT 2016 MAR Scientist C - Section B: 2
The function $f(x)=x^{5}-5x^{4}+5x^{3}-1$ has one minima and two maxima two minima and one maxima two minima and two maxima one minima and one maxima
The function $f(x)=x^{5}-5x^{4}+5x^{3}-1$ hasone minima and two maximatwo minima and one maximatwo minima and two maximaone minima and one maxima
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3 votes
2 answers
131
NIELIT 2016 MAR Scientist B - Section B: 5
The greatest and the least value of $f(x)=x^4-8x^3+22x^2-24x+1$ in $[0,2]$ are $0,8$ $0,-8$ $1,8$ $1,-8$
The greatest and the least value of $f(x)=x^4-8x^3+22x^2-24x+1$ in $[0,2]$ are$0,8$$0,-8$$1,8$$1,-8$
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3 votes
1 answer
132
ISI2014-DCG-2
Let $a_n=\bigg( 1 – \frac{1}{\sqrt{2}} \bigg) \cdots \bigg( 1 – \frac{1}{\sqrt{n+1}} \bigg), \: n \geq 1$. Then $\underset{n \to \infty}{\lim} a_n$ equals $1$ does not exist equals $\frac{1}{\sqrt{\pi}}$ equals $0$
Let $a_n=\bigg( 1 – \frac{1}{\sqrt{2}} \bigg) \cdots \bigg( 1 – \frac{1}{\sqrt{n+1}} \bigg), \: n \geq 1$. Then $\underset{n \to \infty}{\lim} a_n$equals $1$does not...
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3 votes
4 answers
133
ISI2014-DCG-4
$\underset{n \to \infty}{\lim} \dfrac{1}{n} \bigg( \dfrac{n}{n+1} + \dfrac{n}{n+2} + \cdots + \dfrac{n}{2n} \bigg)$ is equal to $\infty$ $0$ $\log_e 2$ $1$
$\underset{n \to \infty}{\lim} \dfrac{1}{n} \bigg( \dfrac{n}{n+1} + \dfrac{n}{n+2} + \cdots + \dfrac{n}{2n} \bigg)$ is equal to$\infty$$0$$\log_e 2$$1$
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3 votes
1 answer
134
ISI2014-DCG-19
It is given that $e^a+e^b=10$ where $a$ and $b$ are real. Then the maximum value of $(e^a+e^b+e^{a+b}+1)$ is $36$ $\infty$ $25$ $21$
It is given that $e^a+e^b=10$ where $a$ and $b$ are real. Then the maximum value of $(e^a+e^b+e^{a+b}+1)$ is$36$$\infty$$25$$21$
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3 votes
1 answer
135
ISI2014-DCG-31
For real $\alpha$, the value of $\int_{\alpha}^{\alpha+1} [x]dx$, where $[x]$ denotes the largest integer less than or equal to $x$, is $\alpha$ $[\alpha]$ $1$ $\dfrac{[\alpha] + [\alpha +1]}{2}$
For real $\alpha$, the value of $\int_{\alpha}^{\alpha+1} [x]dx$, where $[x]$ denotes the largest integer less than or equal to $x$, is$\alpha$$[\alpha]$$1$$\dfrac{[\alph...
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3 votes
2 answers
136
ISI2015-MMA-10
The value of the infinite product $P=\frac{7}{9} \times \frac{26}{28} \times \frac{63}{65} \times \cdots \times \frac{n^3-1}{n^3+1} \times \cdots \text{ is }$ $1$ $2/3$ $7/3$ none of the above
The value of the infinite product$$P=\frac{7}{9} \times \frac{26}{28} \times \frac{63}{65} \times \cdots \times \frac{n^3-1}{n^3+1} \times \cdots \text{ is }$$$1$$2/3$$7/...
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3 votes
3 answers
137
ISRO2018-15
The domain of the function $\log (\log \sin(x))$ is: $0<x<$\pi$ $2n$\pi$<$x$<$(2n+1)$\pi$, for $n$ in $N$ Empty set None of the above
The domain of the function $\log (\log \sin(x))$ is:$0<x<$$\pi$$2n$$\pi$$<$$x$$<$$(2n+1)$$\pi$, for $n$ in $N$Empty setNone of the above
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3 votes
2 answers
138
ISI2017-MMA-14
Let $(v_n)$ be a sequence defined by $v_1 = 1$ and $v_{n+1} = \sqrt{v_n^2 +\left(\dfrac{1}{5}\right)^n}$ for $n\geq1$. Then $\displaystyle{\lim_{n \rightarrow \infty}v_n}$ is $\sqrt{5/3}$ $\sqrt{5/4}$ $1$ $\text{non existent}$
Let $(v_n)$ be a sequence defined by $v_1 = 1$ and $v_{n+1} = \sqrt{v_n^2 +\left(\dfrac{1}{5}\right)^n}$ for $n\geq1$. Then $\displaystyle{\lim_{n \rightarrow \infty}v_n}...
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