Difficulty distribution
How the classified questions are distributed by difficulty.
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Practice Calculus - Mathematics previous year questions organised from real papers, with year-wise coverage and clear topic navigation.
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Year-wise coverage for Calculus. Each bar uses a separate theme-derived color.
How the classified questions are distributed by difficulty.
MCQ, numerical, multiple-select and other formats found in these papers.
Top subjects by unique question coverage.
Top topics across the included previous year papers.
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Question coverage for the most populated papers. Every active PYP paper remains listed below.
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| Paper | Year / session | Questions in this view | Open |
|---|---|---|---|
| TS EAMCET 2020 (Online) 10th September Evening Shift | 2020 | 2 | View paper |
| TS EAMCET 2020 (Online) 11th September Evening Shift | 2020 | 21 | View paper |
| TS EAMCET 2020 (Online) 11th September Morning Shift | 2020 | 22 | View paper |
| TS EAMCET 2020 (Online) 14th September Evening Shift | 2020 | 22 | View paper |
| TS EAMCET 2020 (Online) 14th September Morning Shift | 2020 | 21 | View paper |
| TS EAMCET 2020 (Online) 14th September Morning Shift | 2020 | 21 | View paper |
A varied preview from the papers represented in this selection, with every available option.
Let $f:[2,5] \rightarrow \mathbf{R}$ be a differentiatiable function and $\frac{f(5)}{f(2)}=1$. If there is a $c \in(2,5)$ such that $c f^{\prime}(c)=2 f(c)-2 c^3$, then $f(x)=$
Match the functions of List-I with derivates given in List-II
| $$ \[\text { List-I }\] $$ |
$$ \[\text { List-II }\] $$ |
||
|---|---|---|---|
| A. | $$ \sec ^{-1} x $$ |
I. | $$ \[\frac{1}{1-x^2}, x \in(-1,1)\] $$ |
| B. | $$ \tanh ^{-1} x $$ |
II. | $$ \[\frac{-1}{|x| \sqrt{x^2+1}}, x \neq 0\] $$ |
| C. | $$ \[\operatorname{coth}^{-1} x\] $$ |
III. | $$ \[\frac{1}{|x| \sqrt{x^2-1}},|x|>1\] $$ |
| D. | $$ \[\operatorname{cosech}^{-1} x\] $$ |
IV. | $$ \[\frac{1}{1-x^2}, x \in \mathbf{R}-[-1,1]\] $$ |
| V. | $$ \[\frac{-1}{|x| \sqrt{1-x^2}},|x|<1, x \neq 0\] $$ |
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If $x \sqrt{1+y}+y \sqrt{1+x}=0$, then $\frac{d y}{d x}=$
If $\operatorname{Lt}_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}=e^x(x+1)$ and $f(0)=0$, then $\frac{d}{d x}\left(f(x) e^{-x}\right)+\frac{d}{d x}\left(\frac{f(x)}{x}\right)=$
If $y(x)=\tan ^{-1}\left(\frac{\sqrt{1+a^2 x^2}-1}{a x}\right)$ and $\left(1+a^2 x^2\right) y^{\prime \prime}+g(x) y^{\prime}=0$ then, the sum of the roots of the equation $1+a^2 x^2+g(x)=0$ is
$$\frac{d}{d x}\left[\operatorname{cosech}^{-1}(\tan 2 x)\right]=$$