### Dictionary Rank of a Word | Permutations & Combinations

PERMUTATIONS & COMBINATIONS Rank of the word or Dictionary order of the English words like COMPUTER, COLLEGE, SUCCESS, SOCCER, RAIN, FATHER, etc. Dictionary Rank of a Word Method of finding the Rank (Dictionary Order) of the word  “R A I N” Given word: R A I N Total letters = 4 Letters in alphabetical order: A, I, N, R No. of words formed starting with A = 3! = 6 No. of words formed starting with I = 3! = 6 No. of words formed starting with N = 3! = 6 After N there is R which is required R ----- Required A ---- Required I ---- Required N ---- Required RAIN ----- 1 word   RANK OF THE WORD “R A I N” A….. = 3! = 6 I……. = 3! = 6 N….. = 3! = 6 R…A…I…N = 1 word 6 6 6 1 TOTAL 19 Rank of “R A I N” is 19 Method of finding the Rank (Dictionary Order) of the word  “F A T H E R” Given word is :  "F A T H E R" In alphabetical order: A, E, F, H, R, T Words beginni

## COMMON ERRORS DONE BY STUDENTS

Class 12, Mathematics: Common errors or common mistakes done by students while solving the questions in the examination, Common mistakes done by the students of class XII while solving the mathematical problems.

### Let A = { 1, 2, 3}. Check whether R = { (1, 2), (2, 1), (1, 1), (1, 3)} is symmetric or not.Common Error : Here, (1, 2) ∈ R , (2,1) ∈ R. So, it is symmetric.

Correction : The student thinks that only an ordered pair is to check for Symmetric. So, he
forgets to check for (1, 3).

### If A = {1, 2, 3}, check whether R = {(1, 1), (1, 2), (2, 1)} is transitive or not.Common Error : (1,2) ∈ R , (2,1) ∈ R ⇒ that (1,1) ∈ R. So it is transitive.

Correction : Here the student forgets to see for (2, 1) ∈ R , (1, 2) ∈ R implies (2, 2) ∈ R or not

### Show that relation R in the set A ={1, 2, 3, 4, 5 } given by R = {(a, b) : |a - b| is even }, is an equivalence relation.Common Error : Student forget to write ± while proving transitive

Correction :
Transitive: Let (a, b) ∈ R and (b, c) ∈ R, where a, b, c ∈ A
⇒ |a - b| is even and |b - c| is even
⇒ a - b = ± 2k1 and  b - c = ± 2k2

a - c = ±2k1 ± 2k= ±2 (k+ 2k2)
⇒ |a - c| = 2(k1+2k2)
⇒ |a - c| is even
⇒ (a, c) ∈ R, so R is transitive.

## Chapter - 2INVERSE TRIGONOMETRIC FUNCTIONS

$Error : cos^{-1}\left ( -\frac{1}{2} \right )=-\frac{\pi }{3}$

$Remedy : cos^{-1}\left ( -\frac{1}{2} \right )=\pi -\frac{\pi }{3}=\frac{2\pi }{3}$
Students must know the principal value branch of the Inverse Trigonometric Functions
 Functions Domain (Value of x) Range (Principal Value Branch) (Value of y) $y=sin^{-1}x$ $[-1,1]$ $\left [ -\frac{\pi }{2},\frac{\pi }{2} \right ]$ $y=cos^{-1}x$ $[-1,1]$ $[0,\pi ]$ $y=tan^{-1}x$ $R$ $\left (-\frac{\pi }{2},\frac{\pi }{2} \right )$ $y=cot^{-1}x$ $R$ $(0,\pi )$ $y=sec^{-1}x$ $R-(-1,1)$ $[0,\pi ]-\left \{ \frac{\pi }{2} \right \}$ $y=cosec^{-1}x$ $R-(-1,1)$ $\left [ -\frac{\pi }{2},\frac{\pi }{2} \right ]-\left \{ 0 \right \}$

GENERAL DISCUSSION:-

$cot^{-1}\left [ \frac{\sqrt{1+sinx}+\sqrt{1-sinx}}{\sqrt{1+sinx}-\sqrt{1-sinx}} \right ]=\frac{x}{2},\: \: x\: \epsilon \left ( 0,\: \frac{\pi }{4} \right )$
Note

$For \; x\: \varepsilon \: \left [ 0,\: \frac{\pi }{2} \right ],\: \: \sqrt{\left ( cos\frac{x}{2}-sin\frac{x}{2} \right )^{2}}=cos\frac{x}{2}-sin\frac{x}{2}$

$For \; x\: \varepsilon \: \left [ \frac{\pi }{2},\: \pi \right ],\: \: \sqrt{\left ( cos\frac{x}{2}-sin\frac{x}{2} \right )^{2}}=-\left ( cos\frac{x}{2}-sin\frac{x}{2} \right )$

Trigonometric Substitution in Algebraic Expressions

 S. No. Expression Substitution 1 $\sqrt{1+x^{2}}$ x = tan Ө  or  x = cot Ө 2 $\sqrt{a^{2}+x^{2}}$ x = a tan Ө  or x = a cot Ө 3 $\sqrt{1-x^{2}}$ x = sin Ө  or x = cos Ө 4 $\sqrt{a^{2}-x^{2}}$ x = a sin Ө  or  x = a cos Ө 5 $\sqrt{x^{2}-a^{2}}$ x = a sec Ө or x = a cosec Ө 6 $\sqrt{1+x^{2}}\pm \sqrt{1-x^{2}},\: \sqrt{\frac{1+x^{2}}{1-x^{2}}},\: \sqrt{\frac{1-x^{2}}{1+x^{2}}}$ x2 = cos 2Ө 7 $\sqrt{a^{2}+x^{2}}\pm \sqrt{a^{2}-x^{2}},\: \sqrt{\frac{a^{2}+x^{2}}{a^{2}-x^{2}}},\: \sqrt{\frac{a^{2}-x^{2}}{a^{2}+x^{2}}}$ x2 = a2cos 2Ө 8 $\sqrt{1+x}\pm \sqrt{1-x},\: \sqrt{\frac{1+x}{1-x}},\: \sqrt{\frac{1-x}{1+x}}$ x = cos 2Ө 9 $\sqrt{a+x}\pm \sqrt{a-x},\: \sqrt{\frac{a+x}{a-x}},\: \sqrt{\frac{a-x}{a+x}}$ x = a cos 2Ө

## Chapter - 4Determinants

### Common error : Finding the adjoint of a matrix. Students find co-factors without taking proper signStudents don't take the transpose of a matrix made from the co-factors.

Remedy:
Teachers must tell students to
Find co-factors by using proper sign and insist them to take transpose of a matrix made from the co-factors by giving more questions for practice

### Common Error:$\left|\begin{matrix}2 & -1 & 4 \\0 & 1 & 6 \\1 & -3 & 2 \\\end{matrix} \right|=2(2+18)-1(0-6)+4(0-1)=42$ It is wrong answer

Remedy
For expanding a determinant we must use sign as follows

$\left|\begin{matrix}+ & - & + \\- & + & - \\+ & - & + \\\end{matrix} \right|$
Correct Solution is = 2(2+18)+1(0-6)+4(0-1)= 30

### Common Error :  While using the matrix method of finding the area of triangle

Find the value of k, such that area of triangle with vertices (3, k), (2, 2), (-4, 1) is 3 square units.
Common Error
$Area \: of\: Triangle = \frac{1}{2}\left| \begin{matrix}3 & k & 1 \\2 & 2 & 1 \\-4 & 1 & 1 \\\end{matrix}\right|=3$

$\Rightarrow \left| \begin{matrix}3 & k & 1 \\2 & 2 & 1 \\-4 & 1 & 1 \\\end{matrix}\right|=6$
3(2-1) - k(2+4)+1(2+8) = 6
3-6k+10 = 6
-6k = 6-13
-6k = -7
k=7/6  (Wrong Value )
Remedy
Here area of triangle = 3 square unit
So the values of determinant can be 3 or -3
So we must take both the cases

$Area\: of\: triangle=\frac{1}{2} \left| \begin{matrix}3 & k & 1 \\2 & 2 & 1 \\-4 & 1 & 1 \\\end{matrix}\right|=\pm 3$

3(2-1) - k(2+4)+1(2+8) = ± 6
3-6k+10 = ±6
Either  -6k = 6-13  or   -6k = -6 - 13
Either -6k = -7       or   -6k = -19
Either k=7/6          or       k = 19/6
So  k = { 7/6, 19/6 }   (Correct Values)

$Given \: matrix \: A =\left [ \begin{matrix}3 & 1 \\-1 & 2 \\\end{matrix} \right ]$

### Prove that A2 - 5A + 7I = O, and hence find A-1Common Error: For finding the value of A-1 , some students apply the formula

$A^{-1}=\frac{1}{|A|}\times Adjoint A$

In this question using above formula is a wrong working

Remedy: Here first of all we prove that A2 - 5A + 7I = O

Now multiply on both side by A-1 we get

A-1 ( A2 - 5A + 7I) = A-1.O

(A-1A)A -5(A-1A) + 7 A-1I = O

IA - 5I + 7A-1 = O

$A^{-1}=\frac{1}{7}(5I - A)$

Now we substitute the value of A and I to get A-1

$A^{-1}=\frac{1}{7}(5I - A)$

### Common Mistakes while solving the linear equations

$Given\: A=\left|\begin{matrix}1 & -1 & 0 \\2 & 3 & 4 \\0 & 1 & 2 \\\end{matrix} \right|\: and\: B=\left| \begin{matrix}2 & 2 & -4 \\-4 & 2 & -4 \\2 & -1 & -5 \\\end{matrix}\right|$
Find the product of matrices A and B and hence solve the following system of equations
x - y = 3,     2x + 3y + 4z = 17, y + 2z = 7

### Common Error: In this problem some of the students opts the following procedure

Step -1: Find AB
Step -2: Find A-1 by applying the following formula
$A^{-1}=\frac{1}{|A|}\times Adjoint A$
Step -3: Find the value of x, y and z

Remedy: Correct procedure is
Step -1: Find the product AB

$AB=\left|\begin{matrix}1 & -1 & 0 \\2 & 3 & 4 \\0 & 1 & 2 \\\end{matrix} \right|\left| \begin{matrix}2 & 2 & -4 \\-4 & 2 & -4 \\2 & -1 & -5 \\\end{matrix}\right|=\left| \begin{matrix} 6& 0 & 0 \\0 & 6 & 0 \\0 & 0 & 6 \\\end{matrix}\right|=6I$
Step -2: Find A-1 by using  AB = 6I
Multiply on both side by A-1 we get
(A-1A)B  = 6(A-1I)
IB = 6 A-1  or  B = 6 A-1
$A^{-1}=\frac{1}{6}B=\frac{1}{6}\left|\begin{matrix}2 & 2 & -4 \\-4 & 2 & -4 \\ 2& -1 & 5 \\\end{matrix} \right|$
Now given equations can be written in matrix form as  AX = C where

$A=\left [ \begin{matrix}1 & -1 & 0 \\2 & 3 & 4 \\0 & 1 & 2 \\\end{matrix} \right ],\: X=\left [ \begin{matrix}x \\y \\z\end{matrix} \right ],\: C=\left [ \begin{matrix}3 \\17 \\7\end{matrix} \right ]$

Step -3: Find the value of X by using the formula
X = A-1C
$X=\frac{1}{6}\left [ \begin{matrix}2 & 2 & -4 \\-4 & 2 & -4 \\2 & -1 & 5 \\\end{matrix} \right ]\left [ \begin{matrix}3 \\17 \\7\end{matrix} \right ]$

$\left [ \begin{matrix}x \\y \\z\end{matrix} \right ]=\frac{1}{6}\left [ \begin{matrix}12 \\-6 \\24\end{matrix} \right ]=\left [ \begin{matrix}2 \\-1 \\4\end{matrix} \right ]$
⇒ x = 2,  y = -1,  z = 4

## Chapter 5Continuity and Differentiability

### Common Error: Solving Implicit functions when y is a explicit function of x

For Example:  y5 = x2
When students differentiating this function w. r. t. x, on both side, then sometimes students by mistake or in rush or unknowingly forget to write the term dy/dx to the L.H.S.
Students simply write here
y5 = x2
Differentiating both side w.r.t x we get
5y4 = 2x  (Which is wrong)

Remedy:    y5 = x2
Differentiating both side w.r.t x we get

$5y^{4}\frac{dy}{dx}=2x\frac{d}{dx}(x)=2x$
Students should use chain rule here.

### Common Error: Finding second derivative when parametric function is givenNCERT Book Page No.  192 Q. No. 17

If x = a(cost + t sint) and y = a(sint - t cost) then find d2y/dx2
x = a(cost + t sint)
Differentiating w.r.t x we get
dx/dt = at cost  ........... (1)
y = a(sin t - t cost)
Differentiating w.r.t x we get
dy/dt = at sint

$\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{at\: sint}{at\: cost}=\frac{sint}{cost}=tant$
$Now\: we\: have\: \: \frac{dy}{dx}=tant$
Again differentiating w.r.t x we get

$\frac{d^{2}y}{dx^{2}}=\frac{d}{dx}(tant)=sec^{2}t \: \: \: (Error)$

Remedy

$\frac{d^{2}y}{dx^{2}}=\frac{d}{dx}(tant)=sec^{2}t\times \frac{dt}{dx}$
From Eqn. (1) dx/dt = at cost

$\Rightarrow \: \frac{dt}{dx}=\frac{1}{at\: cost}=\frac{sect}{at}$

Putting this value above we get

$\frac{d^{2}y}{dx^{2}}=sec^{2}t\times \frac{sect}{at}=\frac{sec^{3}t}{at}$

### Common Error: Improper use of log during logarithmic differentiation. Students generally use wrong log properties.

For Example : Log (a + b) = Log a + Log b
(This is not a property and is a wrong perception made by the students)

Remedy
Students should know the correct formula for logarithmic differentiation
Correct formula is   Log (ab) = Log a + Log b
For Example:
y = sin x+ x
Taking log on both side we get
Log y = Log (sin x+ xx )
Log y = Log ((sin x) + Log (xx(Error)

Remedy
y = sin x+ xx
Here teacher should insist the students to take  y = u + v
Where  u = sin x  and   v = xx

## Chapter 6Application of Derivatives

### Common Error: Incorrect identifications of intervals after obtaining the critical points.

Remedy: After obtaining the critical points it is better to draw the table for identifying the sign of f '(x) in different intervals or students may use Baby Curve Method.

### Common Error: Incorrect sign of f ' (x) to identify the increasing / decreasing functions.

Remedy: Correctly observing the sign of f '(x) in the interval before deciding the nature of the function.

### Common Error: Incorrect factorization of an algebraic function to obtain the critical point.

Remedy: Students should Revise the question after solving.

### Common Error: Finding intervals for Trigonometric Functions involving multiples and sub-multiples of angles.

Remedy: Here students should recapitulate and make use of the general solutions of Trigonometric Functions.

## Chapter 7Integration

### Common Error: Solving Definite Integral Problems by using Substitution Method.While making the substitution students forget to change the limit and committed mistake

For Example:

$I=\int_{0}^{1}2x(x^{2}+3)^{1/2}dx$
Putting  x+ 3 = t  so 2xdx = dt

$I=\int_{0}^{1}(t)^{1/2}dx = \left [ \frac{2}{3}(t)^{3/2} \right ]_{0}^{1}=\frac{2}{3}\:\: \: [Error]$

Remedy:
While making the substitution students should remember to change the limit also
$I=\int_{0}^{1}2x(x^{2}+3)^{1/2}dx$
Putting  x+ 3 = t  so 2xdx = dt
If x = 0 then t = 3 and if x = 1, then t = 4

$I=\int_{3}^{4}(t)^{1/2}dx = \left [ \frac{2}{3}(t)^{3/2} \right ]_{3}^{4}$

$I=\frac{2}{3}(8-3\sqrt{3})\: \: \: [Correct Ans.]$

### Common Error: Dropping the absolute value when integrating   $I=\int \frac{1}{x}dx$

Remedy: Recall that the following formula

$I=\int \frac{1}{x}dx=log|x|+c$
Here absolute value bars on the argument are required. It is certainly true that on occasion they can be dropped after the integration is done. But in most of the cases they are required.
Let us take the following two examples

$1)\: \: I=\int \frac{2x}{x^{2}+10}dx=log|x^{2}+10|+c$

$2)\: \: I=\int \frac{2x}{x^{2}-10}dx=log|x^{2}-10|+c$
In first case x+ 10 is always > zero. So here absolute bars are not required
But in second case x- 10 may be + ve, - ve or 0. Its value depends upon the value of x. As the domain of log function is (0, ∞) so x- 10 should always be positive. So it is compulsory to apply here absolute bars.

### Common Error: In indefinite integral students forget to write constant.

Remedy:- Student must learn to write  constant in indefinite integral.

### Common Error: Some time students write more than one constants when they do integration of terms in one problem separately.

Remedy:- This is not wrong but it may create problem in finding particular result. So students should write constant once at last of the final answer. Otherwise they should combined all constants.

## Chapter 9 Differential Equations

### Common Error: In finding the degree of a differential equation when it is not defined.

Remedy: Students should learn, " when a differential equation is not a polynomial in derivatives its degree is not defined".
Students should have more practice and understanding of questions in which degree is not defined.

### Common Error: Difference between General solution and Particular solution .

Remedy: Here students should learn General solution contains an arbitrary constant where as Particular solution contains particular value in place of arbitrary constant.

## Chapter 10 Vector Algebra

Colinear Vectors
Show that the points   $A(-2\hat{i}+3\hat{j}+5\hat{k}),\: \: B(\hat{i}+2\hat{j}+3\hat{k})\: and\: \: C(7\hat{i}-\hat{k})$  are collinear.

### Common Error : Here students commonly use the following concept

Three points are collinear if   $\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}$

Remedy: Correct form is
Three points are collinear if   $|\overrightarrow{AB}|+|\overrightarrow{BC}|=|\overrightarrow{AC}|$

Triangle inequality : For any two vectors  $\vec{a}$  and  $\vec{b}$

### Common Mistake :  $|\vec{a}+\vec{b}| = |\vec{a}|+|\vec{b}|$

Correct Form :         $|\vec{a}+\vec{b}| \leq |\vec{a}|+|\vec{b}|$

$|\vec{a}+\vec{b}| = |\vec{a}|+|\vec{b}|$  is true only when points are collinear.

Equality of two vectors: Find the value of x, y and z so that  $\vec{a}$  and  $\vec{b}$  are equal vectors. Where

$\vec{a}=x\vec{i}+2\vec{j}+z\vec{k}\: \: and\: \: \vec{b}=2\vec{i}+y\vec{j}+\vec{k}$

### Common Error: Here students may use the concept  $|\vec{a}|=|\vec{b}|$

Correct Form: If two vectors are equal then their components are equal
$x\vec{i}+2\vec{j}+z\vec{k}=2\vec{i}+y\vec{j}+\vec{k}$
⇒ x = 2, y = 2, z = 1

## Chapter 11Three Dimensional Geometry

### Common Mistake : Students start using the cartesian form of equation of line without converting them into standard form.

$\frac{2x-1}{2}=\frac{4-y}{7}=\frac{z+1}{2}$
Direction ratios of the line are  2, 7, 2  ( wrong)

Remedy:

$\frac{x-1/2}{1}=\frac{y-4}{-7}=\frac{z-(-1)}{2}$
Direction ratios of the line are  1, -7, 2  (correct)
So students should learn and understand the standard form of equation of line

$\frac{x-x_{1}}{a}=\frac{y-y_{1}}{b}=\frac{z-z_{1}}{c}$