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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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Most answered questions in Calculus

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541
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For f(x)=√x x€[0,b] the number c satisfying mean value therom is c=1
For f(x)=√x x€[0,b] the number c satisfying mean value therom is c=1
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1 votes
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543
Engineering Mathematics
Please verify the solution(the differentiation part ),
Please verify the solution(the differentiation part ),
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544
Mathematics for Natural Science
Sketch the graph of $f(x) = \frac{x^{3}-1}{x^{2}-1}$
Sketch the graph of $f(x) = \frac{x^{3}-1}{x^{2}-1}$
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545
iit kanpur Calculas practice question
Does there exist a differentiable function f : [0, 2] → R satisfying f(0) = −1, f(2) = 4 and f’(x) ≤ 2 for all x ∈ [0, 2]?
Does there exist a differentiable function f : [0, 2] → R satisfying f(0) = −1, f(2) = 4 and f’(x) ≤ 2 for all x ∈ [0, 2]?
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1 votes
0 answers
546
#ISI2021
Let f(x-y) = $\frac{f(x)}{f(y)}$ for all x,y $\epsilon$ R and f’(0) = p, f’(5) = q. Then the value of f’(-5) is q -q $\frac{p}{q}$ $\frac{p^2}{q}$
Let f(x-y) = $\frac{f(x)}{f(y)}$ for all x,y $\epsilon$ R and f’(0) = p, f’(5) = q. Then the value of f’(-5) is q-q$\frac{p}{q}$$\frac{p^2}{q}$
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548
Applied Test Series
What is the correct procedure to solve this limit ?
What is the correct procedure to solve this limit ?
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549
math book
$\int_{0}^{1}\tan^{-1} (1-\frac{1}{x})$ d(x) find
$\int_{0}^{1}\tan^{-1} (1-\frac{1}{x})$ d(x) find
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2 votes
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550
#calculus #limits
$\lim_{x\rightarrow a} f(x)^{g(x)} = e^{\lim_{x\rightarrow a}g(x)[f(x)-1]}$ Solve the below limit without using the above formula, $\lim_{x \rightarrow 0} ({\frac{sin x}{x}})^{\frac{sin x}{x - sin x}}$
$\lim_{x\rightarrow a} f(x)^{g(x)} = e^{\lim_{x\rightarrow a}g(x)[f(x)-1]}$ Solve the below limit without using the above formula, $\lim_{x \rightarrow 0} ({\frac{sin x...
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0 votes
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552
TIFR-2018-Maths-A: 5
True/False Question : The function $f\left ( x \right )=cos\left ( e^{x} \right )$ is not uniformly continuous on $\mathbb{R}$.
True/False Question :The function $f\left ( x \right )=cos\left ( e^{x} \right )$ is not uniformly continuous on $\mathbb{R}$.
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553
TIFR-2019-Maths-A: 6
The limit $\underset{n\rightarrow \infty }{\lim}\:n^{2}\int_{0}^{1}\:\frac{1}{\left ( 1+x^{2} \right )^{n}}\:dx$ is equal to $1$ $0$ $+\infty$ $1/2$
The limit$$\underset{n\rightarrow \infty }{\lim}\:n^{2}\int_{0}^{1}\:\frac{1}{\left ( 1+x^{2} \right )^{n}}\:dx$$is equal to$1$$0$$+\infty$$1/2$
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554
NIELIT 2017 OCT Scientific Assistant A (IT) - Section D: 6
A solution for the differential equation $x’(t) + 2x(t) = \delta(t)$ with initial condition $x(\overline{0}) = 0$ $e^{-2t}u(t)$ $e^{2t}u(t)$ $e^{-t}u(t)$ $e^{t}u(t)$
A solution for the differential equation $x’(t) + 2x(t) = \delta(t)$ with initial condition $x(\overline{0}) = 0$$e^{-2t}u(t)$$e^{2t}u(t)$$e^{-t}u(t)$$e^{t}u(t)$
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1 votes
0 answers
555
NIELIT 2016 MAR Scientist C - Section B: 12
$\displaystyle \lim_{x \rightarrow a}\frac{1}{x^{2}-a^{2}} \displaystyle \int_{a}^{x}\sin (t^{2})dt=$? $2a \sin (a^{2})$ $2a$ $\sin (a^{2})$ None of the above
$\displaystyle \lim_{x \rightarrow a}\frac{1}{x^{2}-a^{2}} \displaystyle \int_{a}^{x}\sin (t^{2})dt=$?$2a \sin (a^{2})$$2a$$\sin (a^{2})$None of the above
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556
NIELIT 2016 MAR Scientist B - Section B: 15
Differential equation, $\dfrac{d^2x}{dt^2}+10\dfrac{dx}{dt}+25x=0$ will have a solution of the form $(C_1+C_2t)e^{-5t}$ $C_1e^{-2t}$ $C_1e^{-5t}+C_2e^{5t}$ $C_1e^{-5t}+C_2e^{2t}$ where $C_1$ and $C_2$ are constants.
Differential equation, $\dfrac{d^2x}{dt^2}+10\dfrac{dx}{dt}+25x=0$ will have a solution of the form $(C_1+C_2t)e^{-5t}$$C_1e^{-2t}$$C_1e^{-5t}+C_2e^{5t}$$C_1e^{-5t}+C_2e^...
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1 votes
0 answers
557
ISI2014-DCG-33
Let $f(x)$ be a continuous function from $[0,1]$ to $[0,1]$ satisfying the following properties. $f(0)=0$, $f(1)=1$, and $f(x_1)<f(x_2)$ for $x_1 < x_2$ with $0 < x_1, \: x_2<1$. Then the number of such functions is $0$ $1$ $2$ $\infty$
Let $f(x)$ be a continuous function from $[0,1]$ to $[0,1]$ satisfying the following properties.$f(0)=0$,$f(1)=1$, and$f(x_1)<f(x_2)$ for $x_1 < x_2$ with $0 < x_1, \: x_...
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0 votes
0 answers
558
ISI2014-DCG-42
Let $f(x)=\sin x^2, \: x \in \mathbb{R}$. Then $f$ has no local minima $f$ has no local maxima $f$ has local minima at $x=0$ and $x=\pm\sqrt{(k+\frac{1}{2} ) \pi}$ for odd integers $k$ and local maxima at $x=\pm\sqrt{(k+\frac{1}{2} ) \pi}$ for even integers $k$ None of the above
Let $f(x)=\sin x^2, \: x \in \mathbb{R}$. Then$f$ has no local minima$f$ has no local maxima$f$ has local minima at $x=0$ and $x=\pm\sqrt{(k+\frac{1}{2} ) \pi}$ for odd i...
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0 votes
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559
ISI2014-DCG-43
Let $f(x) = \begin{cases}\mid \:x \mid +1, & \text{ if } x<0 \\ 0, & \text{ if } x=0 \\ \mid \:x \mid -1, & \text{ if } x>0. \end{cases}$ Then $\underset{x \to a}{\lim} f(x)$ exists if $a=0$ for all $a \in R$ for all $a \neq 0$ only if $a=1$
Let $$f(x) = \begin{cases}\mid \:x \mid +1, & \text{ if } x<0 \\ 0, & \text{ if } x=0 \\ \mid \:x \mid -1, & \text{ if } x>0. \end{cases}$$ Then $\underset{x \to a}{\lim}...
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