Class 12 revision of cbse math part-II 2022-2023

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class 12 revision of cbse math part-I 2022-2023

Class 12 revision of cbse math part-II 2022-2023

Class 12 revision of cbse math part-II 2022-2023

Integration: Chapter 7

Exercise 7.1

(1) \int 1dx = x+C

(2) \int x dx=\frac{x^2}{2}+C

(3) \int x^n dx = \dfrac{x^{n+1}}{n+1}+C

(4) \int e^x dx=e^x+C

(5) \int a^x dx = \dfrac{a^x}{\log_e a}+C

(6) \int \frac{1}{x}dx=\log_e x+C

(7) \int \sin x dx = -\cos x+C

(8) \int \cos x dx =\sin x +C

(9) \int \sec^2 xdx=\tan x+C

(10) \int \operatorname{cosec}^2x dx = -\cot x+C

(11) \int \sec x\tan x dx=\sec x+C

(12) \int \operatorname{cosec}x\cot x dx = -\operatorname{cosec}x+C

(13) \int \dfrac{1}{\sqrt{1-x^2}}dx=\sin^{-1}x+C

(14) \int \dfrac{1}{\sqrt{1-x^2}}dx=-\cos^{-1}x+C

(15) \int\dfrac{1}{1+x^2}dx=\tan^{-1}x+C

(16) \int\dfrac{1}{1+x^2}dx=-\cot^{-1}x+C

(17) \int\dfrac{1}{x\sqrt{x^2-1}}dx=\sec^{-1}x+C

(18) \int\dfrac{1}{x\sqrt{x^2-1}}dx=-\operatorname{cosec}^{-1}x+C

Methods of Integration:-
1. Integration by Substitution
2. Integration using Partial Fractions
3. Integration by Parts

Exercise 7.2

Integration by substitution

⇒ Some other formulae:

(1) \int\tan x dx =\log_e|\sec x|+C

(2) \int\cot xdx =\log_e|\sin x|+C

(3) \int \sec x dx=\log_e|\sec x+\tan x|+C

(4) \int \operatorname{cosec}x dx=\log_e|\operatorname{cosec}x-\cot x|+C

Exercise 7.3

Integration using trigonometric identities

(1) \sin 2x = 2\sin x\cos x

(2) \sin 2x = \dfrac{2\tan x}{1+\tan^2x}

(3) \cos 2x = \cos^2x-\sin^2x

(4) \cos 2x =1-2\sin^2x

(5) \cos 2x = 2\cos^2x-1

(6) 1-\cos 2x =2\sin^2 x

(7) 1+\cos 2x = 2\cos^2x

(8) \sin 3x = 3\sin x-4\sin^3x

(9) \cos 3x = 4\cos^3x-3\cos x

(10)(a) 2\sin x\cos y =\sin(x+y)+\sin(x-y)

(b) 2\cos x\sin y = \sin(x+y)-\sin(x-y)

(c) 2\cos x\cos y=\cos(x+y)+\cos(x-y)

(d) 2\sin x\sin y=\cos(x-y)-\cos(x+y)

(11) (a) \sin x+\sin y=2\sin\dfrac{x+y}{2}\cos\dfrac{x-y}{2}

(b) \sin x-\sin y=2\cos\dfrac{x+y}{2}\sin\dfrac{x-y}{2}

(c) \cos x+\cos y = 2\cos\dfrac{x+y}{2}\cos\dfrac{x-y}{2}

(d) \cos x-\cos y=-2\sin\dfrac{x+y}{2}\sin\dfrac{x-y}{2}

Exercise 7.4

Integrals of Some Particular Functions

(1) \displaystyle \int\dfrac{1}{x^2-a^2}dx=\dfrac{1}{2a}\log\left|\dfrac{x-a}{x+a}\right|+C

(2) \displaystyle\int\dfrac{1}{a^2-x^2}dx=\dfrac{1}{2a}\log\left|\dfrac{a+x}{a-x}\right|+C

(3) \displaystyle \int\dfrac{1}{x^2+a^2}dx=\dfrac{1}{a}\tan^{-1}\frac{x}{a}+C

(4) \displaystyle \int\dfrac{1}{\sqrt{x^2-a^2}}dx=\log\left|x+\sqrt{x^2-a^2}\right|+C

(5) \displaystyle\int \dfrac{1}{\sqrt{a^2-x^2}}dx=\sin^{-1}\frac{x}{a}+C

(6) \displaystyle \int\dfrac{1}{\sqrt{x^2+a^2}}dx=\log\left|x+\sqrt{x^2+a^2}\right|+C

(7) \displaystyle \int\dfrac{dx}{ax^2+bx+c}

(8) \displaystyle \int\dfrac{dx}{\sqrt{ax^2+bx+c}}

(9) \displaystyle \int\dfrac{px+q}{ax^2+bx+c}dx


\Rightarrow px+q=A(2ax+b)+B

(10) \displaystyle \int\dfrac{px+q}{\sqrt{ax^2+bx+c}}dx

From (7) to (10) types form using completing the square

x^2+bx+c for completing the square add and subtract by \left(\dfrac{\text{cofficient of x}}{2}\right)^2

Exercise 7.5

Integration by Partial Fractions:

⇒ Partial fraction expression:

(1) \dfrac{px+q}{(x+a)(x+b)}=\dfrac{A}{(x+a)}+\dfrac{B}{(x+b)}

(2) \dfrac{px+q}{(x-a)^2}=\dfrac{A}{(x-a)}+\dfrac{B}{(x-a)^2}

(3) \dfrac{px^2+qx+r}{(x-a)(x-b)(x-c)}=\dfrac{A}{(x-a)}+\dfrac{B}{(x-b)}+\dfrac{C}{(x-c)}

(4) \dfrac{px^2+qx+r}{(x-a)^2(x-b)}=\dfrac{A}{(x-a)}+\dfrac{B}{(x-a)^2}+\dfrac{C}{(x-b)}

(5) \dfrac{px^2+qx+r}{(x-a)(x^2+b^2)}=\dfrac{A}{(x-a)}+\dfrac{Bx+C}{(x^2+b^2)}

Exercise 7.6

Integration by Parts


(1) \displaystyle\int u.vdx=u\int vdx-\int\left[\frac{d}{dx}u\int vdx\right]dx+C


I= inverse tigonometry(\sin^{-1}x,\cos^{-1}x...)

L= logrithmic function(\log x..)

A= Algebraic function

T = trigonometric function

E= Exponential function

(2) \int\left[f(x)+f'(x)\right]dx=e^xf(x)+C

Exercise 7.7

Integrals of some more types

(1) \displaystyle\int \sqrt{x^2-a^2}dx=\frac{x}{2}\sqrt{x^2-a^2}-\frac{a^2}{2}\log\left|x+\sqrt{x^2-a^2}\right|+C

(2) \displaystyle\int \sqrt{x^2+a^2}dx=\frac{x}{2}\sqrt{x^2-a^2}+\frac{a^2}{2}\log\left|x+\sqrt{x^2+a^2}\right|+C

(3) \displaystyle\int \sqrt{a^2-a^2}dx=\frac{x}{2}\sqrt{x^2-a^2}+\frac{a^2}{2}\sin^{-1}\frac{x}{a}+C

Exercise 7.9 & 7.10

\displaystyle \int_a^bf(x) dx=\left[F(x)\right]_a^b= F(b)-F(a)

Evaluation of Definite Integrals by Substitution

Exercise 7.11

Some Properties of Definite Integrals

(1) P_0:\int_a^bf(x)dx = \int_a^bf(t)dt

(2) P_1:\int_a^bf(x)dx =-\int_b^af(x)dx\text{ and }\int_a^af(x)dx =0

(3) P_2:\int_a^bf(x)dx =\int_a^cf(x)dx +\int_c^bf(x)dx

(4) P_3:\int_a^bf(x)dx =\int_a^bf(a+b-x)dx

(5) P_4:\int_0^af(x)dx =\int_0^af(a-x)dx

(6) P_5:\int_0^{2a}f(x)dx =\int_0^af(x)dx+\int_a^bf(2a-x)dx

(7) P_6:(i)\int_0^{2a}f(x)dx =2\int_0^af(x)dx \text{ if } f(2a-x)=f(x)

(ii) \int_0^{2a}f(x)dx =0 \text{ if } f(2a-x)=-f(x)

(8) P_7:(i) \int_{-a}^{a}f(x) dx=2\int_0^af(x)dx,\text{If f is an even function,i.e. if} f(-x)=f(x)

(ii) \int_{-a}^{a}f(x) dx=0,\text{If f is an odd function,i.e. if} f(-x)=-f(x)

Chapter 8

Area under Simple Curves

Area of region when strip is perpendicular to x-axis =\int ydx

Area of region when strip is perpendicular to x-axis =\int x dy

Exercise 8.2

Area between Two Curves = \int_a^b(f(x)-g(x))dx when f(x)\geq g(x)

Chapter 9(Differential equation): Exercise 9.1:

Order of a differential equation:-Order of a differential equation is defined as the order of the highest order derivative of the dependent variable with respect to the independent variable involved in the given differential equation.

Ex:-\dfrac{d^2y}{dx^2}+\dfrac{dy}{dx}+x =0

Order of d.e. = 2

Degree of a differential equation:-By the degree of a differential equation, when it is a polynomial equation in
derivatives, we mean the highest power (positive integral index) of the highest order derivative involved in the given differential equation.

EX:- \left(\dfrac{d^2y}{dx^2}\right)^3+\dfrac{dy}{dx}+x =0

Degree of d.e. = 3

NOTE:- Order and degree (if defined) of a differential equation are always positive integers.

Exercise 9.2

General solution:The solution which contains arbitrary constants is called the general solution(primitive) of the differential equation.

particular solution:-The solution free from arbitrary constants i.e., the solution obtained from the general solution by giving particular values to the arbitrary constants is called a particular solution of the differential equation.

Methods of Solving First Order, First Degree Differential Equations

Exercise 9.4

Differential equations with variables separable

Exercise: 9.5

Homogeneous differential equations:- In homogeneous differential equation substitute y = vx or x=vy

Exercise 9.6

Linear differential equations:-

(1) (i) Write the given differential equation in the form \dfrac{dy}{dx}+Py=Q where P, Q are constants or functions of x only.
(ii) Find the Integrating Factor (I.F) = e^{\int Pdx}

(iii) Solution of differential equation of this form is
y.(I.F) = \int Q(I.F)dx+C

(2) (i) Write the given differential equation in the form \dfrac{dx}{dy}+Px=Q where P, Q are constants or functions of x only.
(ii) Find the Integrating Factor (I.F) = e^{\int Pdy}

(iii) Solution of differential equation of this form is
x.(I.F) = \int Q(I.F)dy+C

Vector algebra:

Exercise 10.1

Let The point P(x, y, z)

Position vector of P


dr’s of line OP = (x, y, z)

dc’s of the line OP


Relation between dc’s


Types of Vectors:

(1) Zero Vector: A vector whose initial and terminal points coincide, is called a zero vector (or null vector)

Ex:- \overrightarrow{AA}=0

(2) Unit Vector: A vector whose magnitude is unity (i.e., 1 unit) is called a unit vector. The unit vector in the direction of \vec{a} given vector is denoted by \hat{a}.

(3) Coinitial Vectors: Two or more vectors having the same initial point are called coinitial vectors.

(4) Collinear Vectors :Two or more vectors are said to be collinear if they are parallel to the same line, irrespective of their magnitudes and directions.

(5) Equal Vectors:  Two vectors \vec{a} and \vec{b} are said to be equal, if they have the same magnitude and direction regardless of the positions of their initial points, and written
as \vec{a}=\vec{b}.

(6) Negative of a Vector :-A vector whose magnitude is the same as that of a given vector(say,\overrightarrow{AB} ), but direction is opposite to that of it, is called negative of the given vector.
For example, \overrightarrow{BA} vector is negative of the vector\overrightarrow{AB} , and written as \overrightarrow{BA}= -\overrightarrow{AB} .

Exercise 10.2

Let \vec{a}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k}

(i) Modulus of \vec{a}=|\vec{a}|


(ii) Unit vector:- \hat{a}=\dfrac{\vec{a}}{|\vec{a}|}


Components of a vector:

Let \overrightarrow{OP}=x\hat{i}+y\hat{j}+\hat{k}

Scalar component of \overrightarrow{OP}=(x, y, z)

Vector component of \overrightarrow{OP}=(x\hat{i}, y\hat{j}, z\hat{k})

\vec{a}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k} and \vec{b}=b_1\hat{i}+b_2\hat{j}+b_3\hat{k}

(i) Let two vectors \vec{a} and \vec{b} are collinear vector


(ii)  If \vec{a}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k}, then a_1, a_2, a_3 are also called direction ratios of .

(iii) In case if it is given that l, m, n are direction cosines of a vector, then l\hat{i} + m\hat{j} + n\hat{k} = (cos\alpha)\hat{i} + (cos\beta) \hat{j} + (cos \gamma)\hat{k} is the unit vector in the direction of that vector, where \alpha, \beta and \gamma are the angles which the vector makes with x, y and z axes respectively.

Vector joining two points:

If P_1(x_1, y_1, z_1) and P_2(x_2, y_2, z_2) are any two points, then the vector joining P_1 and P_2 is the
vector \overrightarrow {P_1P_2}



Then, \overrightarrow {P_1P_2}=\overrightarrow{OP_2}-\overrightarrow{OP_1}

= (x_2\hat{i}+y_2\hat{j}+z_2\hat{k})-(x_1\hat{i}+y_1\hat{j}+z_1\hat{k})


Magnitude of \overrightarrow {P_1P_2} is |\overrightarrow {P_1P_2}|=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}

Section formula: P and q are two points on line and point R divide PQ in m:n Then

\overrightarrow{OR} =\dfrac{m\overrightarrow{OQ}+n\overrightarrow{OP}}{m+n}

And the mid point

\overrightarrow{OR} =\dfrac{\overrightarrow{OQ}+\overrightarrow{OP}}{2}

Product of Two Vectors

Scalar (or dot) product of two vectors: The scalar product of two nonzero vectors\vec{a} and \vec{b} , denoted by \vec{a}.\vec{b}


Observation: (1) Let \vec[a] and \vec{b} be two nonzero vectors, then \vec{a}.\vec{b}=0 if and only if \vec[a] and \vec{b} are perpendicular to each other. i.e.

\vec{a}.\vec{b}=0\Leftrightarrow \vec{a}\perp\vec{b}

(2) If \theta=0 then, \vec{a}.\vec{b}=|\vec{a}||\vec{b}|

In particular, \vec{a}.\vec{a}=|\vec{a}|^2 in this case \theta =0

(3) If \theta=\pi then, \vec{a}.\vec{b}=-|\vec{a}||\vec{b}|

(4) \hat{i}.\hat{i}=\hat{j}.\hat{j}=\hat{k}.\hat{k}=1


(5) \cos\theta =\left(\dfrac{\vec{a}.\vec{b}}{|\vec{a}||\vec{b}|}\right)

(6) The scalar product is commutative, i.e.


(7) If \vec{a}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k} and \vec{b}=b_1\hat{i}+b_2\hat{j}+b_3\hat{k}

Then \vec{a}.\vec{b}=a_1.b_1+a_2.b_2+a_3.b_3

Projection of a vector on a line

Projection of a vector \vec{a} on other vector \vec{b}, is given by

\vec{a}.\hat{b} or \vec{a}\left(\dfrac{\vec{b}}{|\vec{b}}\right) or \dfrac{(\vec{a},\vec{b})}{|\vec{b}|}

Remarks: \vec{a}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k}

\cos\alpha =\dfrac{a_1}{|\vec{a}|},\cos\beta =\dfrac{a_2}{|\vec{a}|},\cos\gamma=\dfrac{a_3}{|\vec{a}|}

Vector (or cross) product of two vectors

(1) \vec{a}\times \vec{b} is vector.

(2) Let \vec{a} and \vec{b} be two non zero vectors, then \vec{a}\times \vec{b}=0 if and only if \vec{a} and \vec{b} are parallel(or collinear)to each other i.e.

\vec{a}\times \vec{b}=0\Leftrightarrow \vec{a}||\vec{b}

In Particular \vec{a}\times \vec{a}=0

(3). If \theta =\dfrac{\pi}{2} and \vec{a}\times \vec{b}=|\vec{a}||\vec{b}|.

(4) \hat{i}\times \hat{i}=\hat{j}\times \hat{j}=\hat{k}\times \hat{k}=0

(5) \hat{i}\times \hat{j}=\hat{k}, \hat{j}\times \hat{k}=\hat{i}, \hat{k}\times \hat{i}=\hat{j}

(6) \sin\theta =\dfrac{|\vec{a}\times \vec{b}|}{|\vec{a}||\vec{b}|}

(7) \vec{a}\times \vec{b}=-\vec{b}\times \vec{a}

(8) \hat(j)\times \hat{i}=-\hat{k}, \hat{k}\times \hat{j}=-\hat{i},\hat{i}\times \hat{k}=-\hat{j}

(9) If \vec{a} and \vec{b|} are the adjacent sides of a triangle then its area is given as =\dfrac{1}{2}|\vec{a}\times \vec{b}|

(10) If \vec{a} and \vec{b|} are the adjacent sides of a parallelogram then its area is given as =|\vec{a}\times \vec{b}|

(11) If \vec{a}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k} and \vec{b}=b_1\hat{i}+b_2\hat{j}+b_3\hat{k}

\vec{a}\times \vec{b}=\begin{vmatrix} \hat{i} &\hat{j} &\hat{k}\\ a_1 & a_2 & a_3 \\b_1 & b_2 & b_3 \end{vmatrix}

Three dimension geometry

Direction cosine: If \vec{a} = a_1\hat{i}+a_2\hat{j}+a_3\hat{k} makes angles \alpha,\beta,\gamma with x, y, and z-axis respectively

l=\cos\alpha=\dfrac{a_1}{\sqrt{a_1^2+a_2^2+a_3^2}}, m=\cos\beta=\dfrac{a_2}{\sqrt{a_1^2+a_2^2+a_3^2}}


Relation between direction cosine of a line


Direction cosines of a line passing through two points:

Let P(x_1, y_1, z_1) and Q(x_2, y_2, z_2)

Then \vec{PQ}=(x_2-x_1)\hat{i}+(y_2-y_1)\hat{j}+(z_2-z_1)\hat{k}

PQ= \sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}


Equation of a Line in Space:

(i) it passes through a given point and has given direction,

(ii) it passes through two given points.

Equation of a line through a given point and parallel to a given vector \vec{b}

vector form:


Cartesian form:

If point P(x_1,y_1,z_1) and direction ratio = (a,b,c)


Note: –If l, m, n are the direction cosines of the line, the equation of the line is


Equation of a line passing through two given points:

Two points are \vec{a} and \vec{b} then equation of line


In cartesian form:

Two points P(x_1,y_1,z_1) and Q(x_2,y_2,z_2)

Equation of line


Angle between Two Lines:

\vec{a} = a_1\hat{i}+a_2\hat{j}+a_3\hat{k} and \vec{b} = b_1\hat{i}+b_2\hat{j}+b_3\hat{k} angle between them is \theta



\cos \theta =\left|\dfrac{a_1b_1+a_2b_2+a_3b_3}{\sqrt{a_1^2+a_2^2+a_3^2}\sqrt{b_1^2+b_2^2+b_3^2}}\right|

(i) If the lines are perpendicular,\theta =0

\vec{a}.\vec{b}=0 or a_1b_1+a_2b_2+a_3b_3=0

(ii) If two lines are parallel \theta =0 or \pi


Shortest Distance between Two Lines

(i) Distance between two skew lines



Shortest distance=\left|\dfrac{(\vec{b_1}\times \vec{b_2})(a_2-a_1)}{|\vec{b_1}\times \vec{b_2}|}\right|

In cartesian form:


S.D. =\left|\dfrac{\begin{vmatrix}x_2-x_1 & y_2-y_1 & z_2-z_1\\a-1 & b_1 &c_1 \\a_2 & b_2 & c_2\end{vmatrix}}{\sqrt{(b_1c_2-b_2c_1)^2+(c_1a_2-c_2a_1)^2+(a_1b_2-a_2b_1)^2}}\right|

(ii) Distance between parallel lines

\vec{r}=a_1+\lambda \vec{b}


Distance d = \left|\dfrac{\vec{b}\times(\vec{a_2}-\vec{a_1})}{|\vec{b}|}\right|

Linear Programming Problems


x+2y\geq 10

2x+y\geq 12

Objective function;


x, y are called decision variables

Max or min value of Z is called optimal value


(i) P(A)+P(A')=1

(ii) P(A'\cap B)=P(B)-P(A\cap B)

(iii) P(A\cap B')=P(A)-P(A\cap B)

(iv) P(A'\cup B')=P(A\cap B)'=1-P(A\cap B)

(v) P(A'\cap B')=P(A\cup B)'=1-P(A\cup B)

(vi) P(A\cup B)=P(A)+P(B)-P(A\cap B)

Exercise 13.1

Conditional Probability:

P(A/B)=\dfrac{P(A\cap B)}{P(B)}

Properties of conditional probability:- Let E and F be events of a sample space S of an experiment, then we have

(a) P(S/F)=P(F/F)=1

(b) P((A \cup B)/F) = P(A/F) + P(B/F) - P((A \cap B)|F)

(c) P(E'/F)=1-P(E/F)

Exercise 13.2

Multiplication Theorem on Probability:-

(i) P(E \cap F) = P(E) P(F/E)

(ii) P(E \cap F \cap G) = P(E) P(F/E) P(G|(E \cap F))

= P(E) P(F/E) P(G/EF)

Independent Events: P(E \cap F) = P(E) . P (F)

Bayes’ Theorem:- E_1,E_2 and E_3 are three events and A be a non zero events


Theorem of total probability:-


Mean =\Sigma x_iP_i

class 12 revision of cbse math part-I 2022-2023

Ex 7.11 integration ncert maths solution class 12


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