JEE Main Syllabus 2025 PDF (Out) for Physics, Chemistry, Maths

Mathematical Tools - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

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35 Questions around this concept.

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A particle motion along an ellipse $\mathrm{x^{2}+4 y^{2}=0}$ when the particle is at point At $(2,1)$ its $\mathrm{x}$ - component of velocity is 6, then the $\mathrm{y}$ -component of velocity is.

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Concepts Covered - 1

Mathematical tool used in Kinematics

Differentiation

Differentiation is very useful when we have to find rates of change of one quantity compared to another.

If y is one quantity and we have to find the rate of change of y with respect to x which is another quantity

Then the differentiation of y w.r.t x is given as $\frac{dy}{dx}$

For a y V/s x graph

We can find the slope of graph using differentiation

I.e Slope of y V/s x graph = $\frac{dy}{dx}$

Some important Formulas of differentiation

$\frac{d}{dx}\left ( x^{n} \right )={n} x^{n-1}$

Example-

$\frac{d}{dx}\left ( x^{5} \right )=\left ( n=5 \right )$

$\because \frac{\mathrm{d} x^{n}}{\mathrm{d} x} ={n}x^{n-1}$

$\therefore \frac{\mathrm{d} x^{5}}{\mathrm{d} x} ={5}x^{5-1}$

$\Rightarrow \frac{\mathrm{d} x^{5}}{\mathrm{d} x} ={5}x^{4}$

Similarly

$\begin{array}{l}{\frac{d}{d x} \sin x=\cos x} \\ \\ {\frac{d}{d x} \cos x=-\sin x} \\ \\ {\frac{d}{d x} \tan x=\sec ^{2} x}\end{array}$

$\begin{array}{l}{\frac{d}{d x} \cot x=-\csc ^{2} x} \\\\ {\frac{d}{d x} \sec x=\sec x \tan x} \\\\ {\frac{d}{d x} \csc x=-\csc x \cot x} \\\\ {\frac{d}{d x} e^{x}=e^{x}} \\ \\ {\frac{d}{d x} a^{x}=a^{x} \ln a} \\ \\ {\frac{d}{d x} \ln |x|=\frac{1}{x}}\end{array}$

Integration

Opposite process of differentiation is known as integration.
Let x, y are two quantities

Using differentiation we can find the rate of change of y with respect to x

Which is given by $\frac{dy}{dx}$

But using integration we can get direct relationship between quantities x and y

So let $\frac{dy}{dx}=K$ where K is constant

Or we can write $dy=Kdx$

Now integrating on both sides we get direct relationship between x and y

I.e $\int dy=\int Kdx$

$y=Kx+C$

Where C is some constant

For a y V/s x graph

We can find the area of graph using integration

Some important Formulas of integration

$\int x^{n}dx= \frac{x^{n+1}}{n+1}+C$ where (C = constant)

E.g- $\int x^{n}dx=,\, \, \, \, \, n=3$

$\Rightarrow \frac{x^{n+1}}{n+1}+C$

$\Rightarrow \frac{x^{3+1}}{3+1}+C$

$\Rightarrow \frac{x^{4}}{4}+C$

$\begin{array}{l}{\int \frac{d x}{x}=\ln |x|+C} \\\\ {\int e^{x} d x=e^{x}+C} \\\\ {\int a^{x} d x=\frac{1}{\ln a} a^{x}+C} \\ \\ {\int \ln x d x=x \ln x-x+C}\end{array}$
$\begin{array}{l}{\int \sin x d x=-\cos x+C} \\ {\int \cos x d x=\sin x+C} \\ {\int \tan x d x=-\ln |\cos x|+C} \\ {\int \cot x d x=\ln |\sin x|+C} \\ {\int \sec x d x=\ln |\sec x+\tan x|+C} \\ {\int \csc x d x=-\ln |\csc x+\cot x|+C}\end{array}$
$\begin{array}{l}{\int \sec ^{2} x d x=\tan x+C} \\ {\int \csc ^{2} x d x=-\cot x+C} \\ {\int \sec x \tan x d x=\sec x+C} \\ {\int \csc x \cot x d x=-\csc x+C}\end{array}$
$\begin{array}{l}{\int \frac{d x}{\sqrt{a^{2}-x^{2}}}=\sin ^{-1} \frac{x}{a}+C} \\\\ {\int \frac{d x}{a^{2}+x^{2}}=\frac{1}{a} \tan ^{-1} \frac{x}{a}+C} \\\\ {\int \frac{d x}{x \sqrt{x^{2}-a^{2}}}=\frac{1}{a} \sec ^{-1} \frac{|x|}{a}+C}\end{array}$