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    NEET 2026 Preparation Tips for Chemistry, Biology and Physics

    Mathematical Tools MCQ - Practice Questions with Answers

    Edited By admin | Updated on Sep 25, 2023 25:23 PM | #NEET

    Quick Facts

    • 58 Questions around this concept.

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    The motion of a particle along a straight line is described by the equation: x = 8 + 12 t - t3 where x is in metres and t in seconds. The retardation of the particle when its velocity becomes zero, is:

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    Mathematical tool used in Kinematics
    1. 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 )

    \Rightarrow {n}x^{n-1}

    \Rightarrow {5}x^{5-1}

    \Rightarrow {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 \operatorname{sc} 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}

    1. 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}

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