Law	Symbolic Representation	Definition
Commutative Law	$a \oplus b = b \oplus a$	The order of applying the operation does not affect the result.
Associative Law	$(a \oplus b) \oplus c = a \oplus (b \oplus c)$	The grouping of operands does not affect the result.
Closure Law	$a, b \in S, a \oplus b \in S$ (where $S$ is the set)	The result of applying the operation to two elements from a set is still an element of that set.

If A and B are real numbers then A+B is a real number This law is called

a)Closureb)Commutativec)Associatived)Distributive

Supplementary angles: whose sum is 180(degree)

$\sqrt{n} = \frac{n}{\sqrt{n}}; \frac{1}{\sqrt{n}} = \frac{\sqrt{n}}{n}$

- ratio of area of ABC to DEF= 1/4
- perimeter of ABC to DEF = 1/2
- measure of angles of DEF= 60º
- all triangles are congruent
- ratio of length of AC to DE = 1/2
- CDEF forms a rhombus

Part 1

1: Number System -+

$a + b i = (a, b)$
- $i^{2 n} = - 1$ ;
- $i^{3 n} = - i$ ;
- $i^{4 n} = 1$

(- i)^{31}

a) $i$ b) $- i$ c)1d)-1

$(- i)^{31} = (- 1)^{31} \times (i)^{30} \times i = (- 1) \times (- 1)^{15} \times i = i$ -

when solving $(a + b i)^{n}$ you can make it simpler by doing $[(a + b i)^{2}]^{n / 2}$

(\frac{1 + i}{\sqrt{2}})^{8} =

a)b) $t b$ c) $t b$ d) $t b$
m

$\frac{[(1 + i)^{2}]^{4}}{{\sqrt{2}}^{8}} = \frac{(1 + (- 1) + 2 i)^{4}}{16} = \frac{16 (- 1)^{2}}{16}$

Modulus or Absolute value= $| Z |$ = $r = \sqrt{a^{2} + b^{2}}$
argument = $θ = \tan^{- 1} (\frac{b}{a})$
theorem for complex numbers also called polar forms are:
- De-moivre's: $(a + b i)^{n} = r^{n} (C o s (n θ) + i S i n (n θ))$
- Euler formula: $r e^{i θ} = r (C o s θ + i S i n θ)$
polar co-ordinates of a point $(r, θ)$
$c i s θ = C o s θ + i S i n θ$
$\frac{c i s (x)}{c i s (y)} = c i s (x - y)$
$(c i s θ)^{x} = c i s (θ x)$

(\cos 20 + i \sin 20)^{5} \div (\cos 30 + i \sin 30)^{3}

a) $\cos 20 + i \sin 20$ b) $\cos 30 + i \sin 30$ c) $\cos 10 + i \sin 10$ d) $\cos 20 - i \sin 20$

multiplication of complex numbers

(a + b i) (c + d i) = (a c - b d) + (a d + b c) i

$Q$ field of rational numbers
$C$ is field of complex numbers
$\frac{1}{i} = - i; ∵ \frac{i}{i} \times \frac{1}{i} = \frac{i}{i^{2}}$
Inverse of $(a, b)$ is: $(\frac{a}{a^{2} + b^{2}}, \frac{- b}{a^{2} + b^{2}})$
MCQ
Multiplicative Inverse: when multiplied turns the number into 1
Additive Inverse: When added turns the number into 0

a \in R

then multiplicative inverse of a is

a)0b)ac)-ad) $\frac{1}{a}$

Additive inverse of (3, 3) in C is

a)(3, 0)b)(0, 3)c)(-3, 3)d)(-3, 3)

2: Sets and Sub-Sets -+

Power of a Set = No. of subsets = $2^{n}; n = no. of elements in set$
Absorption Law: $A \cup (A \cap B) = A \cap (A \cup B) = A$

\begin{array}{ccccccc} AND & OR & NOT & 2nd should not be false while first is true & X N O R \\ P & Q & P \land Q (Conjunction) & P \lor Q (Disjunction) & \neg P (Negation) & P \to Q (Implication) & P \leftrightarrow Q (Biconditional) \\ True & True & True & True & False & True & True \\ True & False & False & True & False & False & False \\ False & True & False & True & True & True & False \\ False & False & False & False & True & True & True \end{array}

$A^{'} = U - A$ ;all the elements that are not in A
A\B = $A \cap B^{'} = A - B; B^{'} = U - B;$ where A\B means all that elements that are in A but not in B
$(A \cup B)^{'} = A^{'} \cap B^{'}$
$(A \cap B)^{'} = A^{'} \cup B^{'}$
there is only one inverse for each element in a set/group


Converse	$p \to q$	$q \to p$
Inverse	$p \to q$	$\neg p \to \neg q$
Contra positive	$p \to q$	$\neg q \to \neg p$

if function's elements are in sets of ordered pairs swapping their position is the function's inverse
Groupoid: if elements are closed with respect to an operator(closure property)
Semi Group: groupoid but order when applying operator doesn't affect the result(associative property)
- order of a group is its number of elements

Name	Notation	Definition
Subset	A ⊆ B	all elements of A are in B
Superset	B ⊃ A	B has every element of A(basically A is subset of B)

if $A$ is a set, $| A |$ represents the number of elements in $A$
$| A \cup B | = | A | + | B | - | A \cap B |$
$Probability = P (A) = \frac{Number of favorable outcomes for event A}{Total number of possible outcomes}$
$P (A ∣ B) = P (B) P (B ∣ A) P (A)$

3: Matrices -+

Minor $= M_{i j}$ : determinant of matrix after removing $i^{t h}$ row and $j^{t h}$ column
Co factor $= A_{i j} = (- 1)^{i + j} M_{i j}$
Symmetric: $A^{t} = A$
Skew-Symmetric $A^{t} = - A$
Order: Row x Column
$| 4 A | = 4^{n} A;$ for all square matrices where $n$ is the order of the matrix
if a whole row/column is zero, determinant is zero
if a two whole rows or columns are equal then determinant is zero
$A \times B$ if $A_{j} = B_{i}$
if A is diagonal matrix, then its inverse is reciprocal of its elements i.e
$A = (\begin{matrix} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{matrix}); A^{- 1} = (\begin{matrix} \frac{1}{a} & 0 & 0 \\ 0 & \frac{1}{b} & 0 \\ 0 & 0 & \frac{1}{c} \end{matrix})$
if A is diagonal then its determinant |A| is simple product of diagonal elements.i.e
$| A | = | \begin{matrix} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{matrix} | = (a b c)$
solve determinant of $4 \times 4$ or higher using determinant properties

| \begin{array}{cc} x & 2 \\ 18 & x \end{array} | = | \begin{array}{cc} 6 & 2 \\ 18 & 6 \end{array} |

, then

x =

a)6b)16c)-6d)0

4: Quadratic Equations

Quadratic Formula : x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}

Fourth roots of 16 are: $2 i, - 2 i, 2, - 2$
sum of fourth roots is zero
A quadratic has a degree of 2, always
if $a$ is root of polynomial equation $P (x)$ then $P (a)$ is zero

x^{4} - 9 x^{3} + 6 x^{2} + 2

a)1b)2c)-2d)-1

\begin{align} \\ \omega^3 = \omega^2 \times \omega \times 1= 1 \\ \\ \omega^2 + \omega + 1 = 0 \\ \\ \omega = \dfrac{-1 + i\sqrt{3}}{2} \\ \\ \omega^2 =\dfrac{-1 - i\sqrt{3}}{2} \\ \\ \omega = \dfrac{1}{\omega^2};\qquad \omega {#2} = \dfrac{1}{\omega} \end{align}

at the point of intersection of two rays their functions are equal. $a x + b y = c x + d y$
for $a x^{2} + b x + c$ whose roots are $\begin{array}{r} α and β \end{array}$ then $$\begin{align*} S = \alpha + \beta &= -\dfrac{b}{a} \ P = \alpha \times \beta &= \dfrac{c}{a} \ \ ax^2 + bx + c &= x^2 - Sx + P \end{align*}$$

α

β

are roots of

a x^{2} + b x + c = 0

, then the equation with roots

\frac{1}{α}, \frac{1}{β}

is:

a) $a x^{2} - b x - c = 0$ b) $c x^{2} + a x + b = 0$ c) $b x^{2} + a x + c = 0$ d) $c x^{2} + b x + a = 0$

since $α + β = - b$ and $α β = c$ do the same for the other roots and put values

In a homogeneous system, if the determinant of the coefficient matrix is non-zero, the system only has trivial solution (i.e. $x = 0, y = 0$

System of equations:

{\begin{cases} 3 x + 2 y = 0 \\ x + 2 y = 0 \end{cases}

has

a)Trivial Solutionb)Non-Trivial Solutionc)infinite solutionsd)none

since

\begin{aligned} | \begin{array}{c} 3 & 2 \\ 1 & 2 \end{array} | = & 6 - 4 = 4 \\ 4 & \neq 0 \end{aligned}

6: Sequences and Series+

\hline
\textbf{Type} & \textbf{General Term} & \textbf{Sum} & \textbf{Mean} & \textbf{Example} \
\hline
\text{A.P.} & a_n = a + (n-1)d & S_n = \dfrac{n}{2}[2a_{1} + (n-1)d] = \dfrac{n}{2}(a_{1} + a_n) & \text{A.M} = \dfrac{a + b}{2} & 2, 5, 8 \
\hline
\text{G.P.} & a_n = ar^{n-1} & S_n = \dfrac{a_{1}(1 - r^n)}{1 - r} , \text{for} , r < 1, , \text{or} , S_n = \dfrac{a_{1}(r^n - 1)}{r - 1} , \text{for} , r> 1 & \text{G.M} = \sqrt{ab} & 3, 6, 12 \
\hline
\text{H.P.} & a_n = \dfrac{1}{a + (n-1)d} & S_n = \dfrac{n}{\dfrac{1}{a_1} + \dfrac{1}{a_2} + \dots + \dfrac{1}{a_n}} & \text{H.M} = \dfrac{2ab}{a+B} & 1, \dfrac{1}{2}, \dfrac{1}{3} \
\hline
\end

- x/9 = 0.xxxxx... like if you divide x digit number by 9 with x 9s this repeats that number in decimal e.g. 34/99 = 0.343434... or 234/999= 0.234234.... * $A>G>H$ if a & b are +ve $A<G<H$ if a & b are -ve $G^2 = A \times H$ - to find d in AP from $a_{x}=m \text{ to } a_{y}=n$ $\dfrac{n-m}{y-x} =d$ - Sum of natural numbers starting from one to n is: $\dfrac{n(n+1)}{2}$ >[!question]- $\sum ^{100}_{n=2}:?$ simply do $\sum ^{100}_{n=1}=\dfrac{100(100+1)}{2}=50\times 101=5050$ now just -1 from it since the orignal series starts from 2 ANS: 5049 ### A.P - to find d from $a_{n} = x$ and $a_{m} = y$ - $d =\dfrac{(x-y)}{(n-m)}$ - e.g. $a_{2}​=5$ and $a_{6}=17$ - $d=\dfrac{17-5}{6-2}=\dfrac{12}{4}=3$ - find any $a_{n}$ if from any $a_{m}$ with $d$ - $a_{n}= a_{m}+d(n-m)$ e.g. $a_{20} =a_5 +d(20-5) = a_5 +15d$ > [!question]- if $a_{3}=4\text{ and } a_{5}=8$ find $a_{50}$ > a)100$\qquad$b)102$\qquad$**c)98**$\qquad$d) none - $a_{n} = a_{1}+(n-1)d$ $n=\dfrac{a_{n}-a_{1}}{d}+1$ > [!question]- which term is 148, in A.P -2, 4, 10... > **a)26th**$\qquad$b)25th$\qquad$c)31th$\qquad$d) 28th - $A_{m} = a+\dfrac{m(b-a)}{(n+1)}$ - m = AM to be found - n = Total number of AMs > [!question]- if there are 6 AM b/w 2 and 5 then $A_{5}$ is: > a)$\dfrac{26}{7}$th$\qquad$b)$\dfrac{6}{5}$th$\qquad$**c)**$\dfrac{29}{7}$th$\qquad$d) $\dfrac{31}{7}$th - $a_{n} =S_{n}-S_{n-1}$ for all A.P - $a_{5} =S_{5}-S_{4}$ > [!question]- If $S_{n} = 2n^{2} -n$ then $a_{8} =?$ a)120$\qquad$b)53$\qquad$***c)29***$\qquad$d)43 - Sum of interior angles of a polygon = $(n-2)\times 180^{o}$ > [!question]- Find the sum of interior angles of 16 sided polygon: > a)2880$\qquad$***b)2520***$\qquad$c)1800$\qquad$d)360 - if n is *odd* then Sum of n terms = n x middle term if n is *even* then Sum of n terms = $n\times \dfrac{(\text{middle term}+2)}{2}$. > [!question]- If $5^{th}$ term of an A.P. is 5, then sum of 9 terms of A.P. is: ***a) 45***$\qquad$b) 40$\qquad$c) 50$\qquad$d) 35 ### G.P - for an infinite G.P series$$S_{\infty}=\sum_{n=1}^{\infty} a^n = \dfrac{a}{1 - r} , |r| < 1

0.1 + 0.01 + 0.0001 + \dots

sum to infinity

$\frac{a_{2}}{a_{1}} = r = 0.1, S_{n} = \frac{0.1}{1 - 0.1} = \frac{0.1}{0.9} = \frac{1}{9}$

for two terms in GP $a_{n} = x$ and $a_{m} = y$
$r^{n - m} = (\frac{x}{y})$
e.g. $a_{4} = 24 and a_{7} = 192$
$r^{7 - 4} = r^{3} = \frac{192}{24} = 8$
$r = 2$
find any $a_{n}$ from any $a_{m}$ with $r$
- $a_{n} = a_{m} r^{n - m}$
  e.g. $a_{10} = a_{7} r^{10 - 7} = a_{7} r^{3}$

In GP, if

a_{4}

= 24, and

a_{7}

= 192 then

a_{10}

a) 1536 b) 1236 c) 1436d)1336

term of GP remain in GP even after reciprocal

7: Permutation and Combination +

$n! = (n) \times (n - 1)!$
permutation: different arrangements of n numbers taken r at a time
Combination: selection of r objects from a total of n objects, where the order of selection does not matter.

\begin{aligned} Permutation: & ^{n} P_{r} = \frac{n!}{(n - r)!} = (n) \times (n - 1) \times (n - 2) \dots (n - (r - 1)) \\ e . g . & ^{7} P_{3} = 7 \times 6 \times \underset{7 - (3 - 1)}{\underset{⏟}{5}} simply write r digits starting from n 5 \\ Combination: & ^{n} C_{r} = (\binom{n}{r}) = \frac{n!}{r! (n - r)!} = \frac{(n) \times (n - 1) \times (n - 2) \dots (n - (r - 1))}{r!} \end{aligned}

if in combination or permutation q* objects are always included: $^{n - p} C_{r - p}$ or $^{n - q} P_{r - q}$ if never included $^{n - p} C_{r}$ or $^{n - q} P_{r}$

$^{n} C_{0} =^{n} C_{n} = 1$	$^{n} C_{r} =^{n} C_{r - 1}$	$^{n} C_{r} . r! =^{n} P_{r}$	$^{n} C_{1} = n$
$^{n} P_{n} = n!$	$^{n} C_{r} =^{n} C_{n - r}$

if some object in permutation is being repeated say $p$ times e.g. A,B,B,C,C,C,D (B is repeated 2 times and C 3 times so) then its permutation taking 5 at a time is

\frac{^{7} P_{5}}{2! \times 3!} = \frac{\frac{7!}{(7 - 5)!}}{2! \times 3!} = \frac{7!}{(2! \times 3!) \times (7 - 5)!}

In how many ways can the letters of the word 'LEADER' be arranged?

a)72b)44c)360d)72065e)None of these

and

In how many different ways can the letters of the word 'LEADING' be arranged in such a way that the vowels always come together?

a)360b)480c)720d)5040e) None of these

also try for corporation , note that $r$ comes twice and o thrice

How many 3-digit numbers can be formed from the digits 2, 3, 5, 6, 7 and 9 which are divisible by 5 and none of the digits is repeated?

a)5b)10c)15d)20

since divisible by 5, 5 is fixed at the end, no digits repeat so $^{5} p_{2}$ = $5 \times 4$ since only two places remain,

How many positive integers 'n' can be form using the digits 3, 4, 4, 5, 6, 6, 7, we want 'n' to exceed 60,00,000?

a)320b)360c)540d.720

how many natural numbers less than a lakh can be formed with the digits 0, 6 and 9?

a)242b)243c)728d)none

less than a lakh has 5 digits, each digit can be replaced by 3 possible numbers so $3 \times 3 \times 3 \times 3 \times 3 = 3^{5}$
excluding 00000, so 243-1=242

Diagonals in an $n$ sided polygon: $^{n} C_{2} - n$
circular permutation = $(n - 1)!$
circular permutation on a flipable ring = $\frac{(n - 1)!}{2}$
common factorials

0!	1!	2!	3!	4!	5!	6!	7!
1	1	2	6	24	120	720	5040

in permutation order of elements matters while in combination only the presence of an elements matters
how to determine whether the question requires permutation or combinations

Combination:

\begin{aligned} Combination: & ^{n} C_{r} = (\binom{n}{r}) = \frac{n!}{r! (n - r)!} \end{aligned}

$^{n} C_{x} =^{n} C_{y}; if x + y = n$

^{n} C_{5} =^{n} C_{4}; find n

a)5 b)9c)6d)8

for consecutive $r$ 's with same n: $^{n} C_{r} +^{n} C_{r - 1} =^{n + 1} C_{r}$ keeping the larger $r$

^{13} C_{5} +^{13} C_{7} = ?

a)Both B and C b) $^{14} C_{6}$ c) $^{14} C_{8}$ d) $^{14} C_{7}$

solution: $=^{13} C_{8} +^{13} C_{7} ∵^{n} C_{r} +^{n} C_{n - r}$
and using above given rule, 8 is higher so we $^{14} C_{8}$ or $^{14} C_{14 - 8}$ = $^{14} C_{6}$

$^{n} C_{r} \times r! =^{n} P_{r}$

^{n} C_{r} = 28 +^{n} P_{r} = 112

then

r!

a)noneb)4 c)12d)2

$^{n} C_{x} =^{m} C_{x}$ are equal for all n and m, if $x = 0$ as $^{n} C_{0} = 1$

Geometrical problems

if n is the number of non-collinear points

\begin{aligned} ^{n} C_{2} = & Number of line segments \\ ^{n} C_{2} - n = & Number of diagonals \\ ^{n} C_{3} = & Number of triangles \\ ^{n} C_{4} = & Number of quadrilaterals \\ ^{n} C_{5} = & Number of Pentagonals \end{aligned}

Diagonals in 8 sided polygon?

a)28b)52 c)20d)56

Triangles in pentagon?

a)40b)50 c)10d)70

$^{5} C_{3}$

Pecrmutation	Combinations
Arrangements standing or sitting in row or in a circle problem regarding digits letters(A,B,C...) Formation of words, number,etc	Selection, Choice, Draw,etc Distributation, formation of a group, commitee, team,etc problem regarding geometry

Probability:

Rules of probability:
- Probability of an event = $P (E) = \frac{n (E)}{n (S)} = \frac{no. favourible outcomes}{total outcomes}$

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

Tickets numbered 1 to 20 are mixed up and then a ticket is drawn at random. What is the probability that the ticket drawn has a number which is a multiple of 3 or 5?

a) $\frac{1}{2}$ b) $\frac{2}{5}$ c) $\frac{8}{15}$ d) $\frac{9}{20}$

$\frac{20}{5} = 4; \frac{20}{3} \approx 6$ and $15$ which comes with both so $P = \frac{4}{20} + \frac{6}{20} - \frac{1}{20}$

blueeeeh:
- $P (E) + P \overset{―}{(E)} = 1; P \overset{―}{(E)} = 1 - P (E)$
- probability can never be greater than 1
- Improbability of an event = $P \overset{―}{(E)}$

In a box, there are 2 red, 2 blue and 3 green balls. Two balls are picked up randomly. What is the probability that neither is is blue?

a) $\frac{10}{21}$ b) $\frac{11}{21}$ c) $\frac{2}{7}$ d) $\frac{5}{7}$

Total outcomes: $^{7} C_{2} = 21$ ,probability for NOT BLUE(not including its two balls): $\frac{^{7 - 2} C_{2}}{21} = \frac{10}{21}$

Die	1	2	3	4	5	6
1	2	3	4	5	6	7
2	3	4	5	6	7	8
3	4	5	6	7	8	9
4	5	6	7	8	9	10
5	6	7	8	9	10	11
6	7	8	9	10	11	12

What is the probability of getting a sum 9 from two throws of a dice?

a) $\frac{1}{6}$ b) $\frac{1}{8}$ c) $\frac{1}{9}$ d) $\frac{1}{12}$

Two dice are tossed. The probability that the total score is a prime number is:

a) $\frac{1}{6}$ b) $\frac{5}{12}$ c) $\frac{1}{2}$ d) $\frac{7}{9}$

Card in a deck: $52$
- Spades, Hearts, Diamonds, Clubs: $(13 \times 4) = 52$
- Black Cards: 26, Red Cards: 26
- Face Cards: $12$
- All Hearts and Diamonds are Red
- All Spades and Clubs are Black

One card is drawn from a pack of 52 cards. What is the probability that the card drawn is either a red card or a king?

a) $\frac{1}{2}$ b) $\frac{6}{13}$ c) $\frac{7}{13}$ d) $\frac{27}{52}$

Red card Prob: $\frac{26}{52}$ , King Prob: $\frac{4}{52}$ , Red King Porb: $\frac{2}{52}$ ; Total Prob: $\frac{26}{52} + \frac{4}{52} - \frac{2}{52} = \frac{28}{52} = \frac{14}{26} = \frac{7}{13}$

1/2 Probability of even and odd sum from one or multiple dice rolls $; ∵ one die \frac{3}{6}, two die \frac{18}{36}$

$P (A \cap B)$ is the probability of an event where both A and B occur, $P (A \cup B)$ is the probability where in a given set either A or B may occur

possible outcomes of an experiment with $x$ outcomes performed $n$ times: $x^{n}$ , then probability is $\frac{favourable outcomes from x^{n}}{x^{n}}$
- Possible outcomes of a coin tossed $n$ times: $2^{n}$
- Possible outcomes of a dice rolled $n$ times: $6^{n}$

In a simultaneous throw of two coins, the probability of getting at least one head is:

a) $\frac{1}{2}$ b) $\frac{1}{3}$ c) $\frac{2}{3}$ d) $\frac{3}{4}$

Four dice are thrown simultaneously. Find the probability that all of them show the same face:

a) $\frac{1}{216}$ b) $\frac{1}{36}$ c) $\frac{4}{216}$ d) $\frac{3}{216}$

Total outcomes: $6^{4}$ , same face: $\underset{since they have only 6 faces which should be same}{\underset{⏟}{6}}$ , $P (E) = \frac{6}{6^{4}} = \frac{1}{6^{3}} = \frac{1}{216}$

$S = {H H, H T, T H, T T}$

probability of an
dependent events:
a work is to be and involves 3 steps, step 1 has probability of completion 0.7, step 2 has probability of completion 0.8, step 3 has probability of completion 0.9, total probability of work being done is?

simply multiply them $0.7 \times 0.8 \times 0.9$

8: Mathematical Induction and Binomial Theorem

Mathematical Induction

Use back solving method

1 + 2 + 3 + 4 + \dots + (2 \times 3^{n - 1}) = ?; n \in N

a) $3^{n - 1}$ b) $3^{n - 1} + 1$ c) $3^{n - 1} - 1$ d) $3^{n + 1}$

since n is a natural number, using 2 and adding only first two numbers (1+2) = 3, now put n=2 in each option and whichever equal 3 is correct

for positive integral values of n,

5^{n} - 2^{n}

is divisible by:

a)7b)3c)2d)5

n^{2} > n + 3

is true for integral values:

a) $n \geq 2$ b) $n > 8$ c) $n \geq 3$ d) $n \leq 6$

since $n \geq 3$ includes more numbers than $n > 8$ so we use it

Binomial Theorem

(a + x)^{n} =^{n} C_{0} \cdot a^{n} x^{0} +^{n} C_{1} \cdot a^{n - 1} x^{1} +^{n} C_{2} \cdot a^{n - 2} x^{2} + \dots +^{n} C_{n - 1} \cdot a^{1} x^{n - 1} +^{n} C_{n} \cdot a^{0} x^{n}

where $^{n} C_{0},^{n} C_{1},^{n} C_{2} \dots$ are binomial co-efficient

binomial expression of A is:

1 - 8 x + 24 x^{2} - 32^{3} + 16 x^{4}

then A is?

a) $(1 - 2 x)^{4}$ b) $(1 - 4 x)^{4}$ c) $(1 + 2 x)^{4}$ d) $(1 + 4 x)^{4}$

since there are alternating minus sign, A contain a minus, and 2nd term is 8x, which is n( $^{n} C_{1} = n = 4$ ) times $2 x$

number of terms is $n + 1$

Number of terms in the expansion of

(1 + 2 x + x^{2})^{4}

are:

a)5b)8c)9d)none

first make it binomial and it becomes $[(1 + x)^{2}]^{4}$

Middle term if n is even: $(\frac{n + 2}{2})^{th}$ position
if n is odd: $(\frac{n + 1}{1})^{th} and (\frac{n + 3}{1})^{th}$ position
Sum of binomial co-efficient: $2^{n}$
Sum of even or odd binomial co-efficient: $2^{n - 1}$

In a binomial expansion, if the sum of binomial coefficients is 128, then the number of terms in expansion

a)7b)9c)8d)6

since $a^{7} = 128$ , $n = 7$ and number of terms is $n + 1 = 8$

Find Sum of co-efficient $_{not binomial coefficient}$ of $(a + b x)^{n}$ by replacing $x$ by 1 and solving

Sum of co-efficients in the expansion of

(1 - 3 x)^{5}

is:

a)32b)16c)-32d)-16

Sum of exponent of $a$ and $x$ in each term is equal to $n$

Which of the following is the term in the expansion of

(x + 3 y)^{5}

a) $80 x y^{2}$ b) $90 y^{2} x^{3}$ c) $80 x^{5} y^{2}$ d) $5 x y^{5}$

$n$ is 5 so the degree of the correct option should be 5

$x^{th}$ term or $r^{th}$ power of b: $T_{r + 1} = (\binom{n}{r}) \times a^{n - r} \times b^{r} $ $ w h e r e $ r + 1 $ i s f o r 4 t h t e r m s o w e p u t r = 3$

a) $198 x^{- 2}$ b) $198 x^{2}$ c) $- 198 x^{- 2}$ d) $- 198 x^{2}$

8th term so r=7, since all coefficient same in option, don't need to calculate $^{12} C_{7}$ , since all even terms in are negatives when a negative is present in equation, its negative, now $a^{n - r} \times b^{r}$ $= 2 x^{12 - 7} \times \frac{1}{(2 x)^{7}}$

if asked to find Xth terms from last, simply exchange $a$ and $b$ : $T_{r + 1} = (\binom{n}{r}) \times b^{n - r} \times a^{r}$

the 8th term from last in the expansion of

(2 x - \frac{1}{2 x})^{12}

a) $1526 x^{2}$ b) $- 1498 x^{2}$ c) $- 3168 x^{2}$ d)none

this time we calculate the combination, $^{12} C_{7} \times (- \frac{1}{2 x})^{12 - 7} \times 2 x^{7} = 729 \times - \frac{1}{32} \times 128 = 3168$

Term involving Cth power of x as $x^{c}$ in an equation of form
- $(a x + \frac{1}{b x})^{n}$ : we get value of r by doing $n - t r = c; r = \frac{n - c}{t}$ where t is the combined degree of $a x$ and $\frac{1}{b x}$ here its 2
- $(a - b x)^{n} :$ it is simply nth position, if $b x$ has a degree of 1
- term independent of $x$ means $x^{0}$ , where c=0

Term involving

x^{2}

in the expansion of

(2 x - \frac{1}{2 x})^{12}

a)b)c) $t b$ d) $t b$

here $n = 12$ , and $t = 2$ so $r = \frac{12 - 2}{2} = 5$ , that means 6th term involves $x^{2}$

9-14: Trigonometry

9: Fundamentals of Trigonometry

System	units	conversion	definition
sexagesimal	$θ^{\circ} M_{i n u t e}^{'} S_{e c o n d s}^{″}$	$\frac{M^{'}}{60} = θ^{\circ}, \frac{S^{″}}{60} = M$	1 degree is the angle, the 360th part of a circle subtends on the centre
circular	rad.	$2 π r a d . = 1 r e v = 360^{\circ}$	the angle subtended by an arc equal in length to radius

basic unit of angle is second (60th part of a minute and 3600th part of a degree
Calculate angle between clock hands: $30^{o} \times Hrs - (\frac{11}{2})^{o} \times Mins$

angle between hands of watch at 11:10am, in degree

a) 275b)225c)A and Dd)85

$(30 \times 11) - (10 \times \frac{11}{2}) = 275$
275 is in 4th quad so (360-275) = 85 so both

Polygon	Area	Perimeter	arc
Circle	$A = π r^{2}$	$L = 2 π r$	$l = θ r$
Sector	$A = \frac{1}{2} r^{2} θ = \frac{1}{2} r l$	$l = θ r$
Triangle	$A = \frac{1}{2} a b$
![sign of trigonometric functions.png	center	300](/img/user/Notes/entry-test-notes/Entry%20Test/attachments/sign%20of%20trigonometric%20functions.png)

	co-ratios	Identity	addition	double angle	triple angle	half angle
Sin $θ$	Cos $90 - θ$	$\sin^{2} θ + \cos^{2} θ = 1$	$\sin (α + β) = \sin (α) \cos (β) + \cos (α) \sin (β)$ $\sin (α - β) = \sin (α) \cos (β) - \cos (α) \sin (β)$	$\sin 2 θ = 2 \cos θ \sin θ$ $= \frac{2 \tan θ}{1 + \tan^{2} θ}$	$\sin (3 θ) = 3 \sin (θ) - 4 \sin^{3} (θ)$	$\sin (\frac{θ}{2}) = \pm \sqrt{\frac{1 - \cos (θ)}{2}}$
Cos $θ$	Sin $90 - θ$	$\sec^{2} θ = \tan^{2} θ + 1$	$\cos (α + β) = \cos (α) \cos (β) - \sin (α) \sin (β)$ $\cos (α - β) = \cos (α) \cos (β) + \sin (α) \sin (β)$	$\cos 2 θ = 2 \cos^{2} θ - 1$ $= 1 - 2 \sin^{2} θ$ $= \cos^{2} θ - \sin^{2} θ$ $= \frac{1 - \tan^{2} θ}{1 + \tan^{2} θ}$	$\cos (3 θ) = 4 \cos^{3} (θ) - 3 \cos (θ)$	$\cos (\frac{θ}{2}) = \pm \sqrt{\frac{1 + \cos (θ)}{2}}$
Tan $θ$	Cot $90 - θ$	$c o s e c^{2} θ = \cot^{2} θ + 1$	irrelevant	$\tan 2 θ = \frac{2 \tan θ}{1 - \tan^{2} θ}$	irrelevant	$\tan (\frac{θ}{2}) = \pm \sqrt{\frac{1 - \cos (θ)}{1 + \cos (θ)}}$
Sec $θ$	Cosec $90 - θ$

\cos 75

$\cos (45 + 30) = \cos (30) \times (\cos 45) - \sin (30) \times \sin (45) = (\frac{\sqrt{3}}{2} \times \frac{1}{\sqrt{2}}) - (\frac{1}{2} \times \frac{1}{\sqrt{2}}) = (\frac{\sqrt{3}}{2} \times \frac{\sqrt{2}}{2}) - (\frac{1}{2} \times \frac{\sqrt{2}}{2})$

if sum of two angles is 90 then those co ratios are equal

\frac{\tan 71}{\cot 19}

a) $\cot 42$ b)Both C and Dc) $\cot 45$ d) $\tan 45$

\tan 9 \times \tan 18 \times \tan 72 \times \tan 81

a)0b)1c)9d)none

$= \cot 81 \times \cot 72 \times \tan 72 \times \tan 81$

\frac{1 - \cos t}{\sin t}

a) $\cot \frac{t}{2}$ b) $\tan \frac{3}{t}$ c) $\tan \frac{t}{2}$ d) $\cos 3 t$

simplify put t=90 and compare answers with each option

8 \cos 15 \cos \frac{15}{2} \sin \frac{15}{4} \cos \frac{15}{4} = ?

a)0.2b)0.5c)0.1d)1

$= 4 \cos 15 \cos \frac{15}{2} \sin \frac{15}{2} ∵ 2 \sin θ \cos θ = \sin 2 θ$
$= 2 \cos 15 \sin 15 = \sin 30 = 0.5$

6 \sin 10 - 8 \sin 10

a)2b)1c) $\frac{1}{2}$ d) $\sqrt{3}$

$2 (3 \sin 10 - 4 \sin^{3} 10)$

\sin θ = \frac{1}{2} \cos e c θ

θ

$\sin θ = \frac{1}{2} \times \frac{1}{\sin θ}; \sin θ = \frac{1}{\sqrt{2}}$

for a right angled triangle
$C o s θ = \frac{B a s e}{H y p o t e n u s e}$
$S i n θ = \frac{p e r p e n d i c u l a r}{H y p o t e n u s e}$
$\tan θ = \frac{\sin θ}{c o s θ} = \frac{p e r p e n d i c u l a r}{B a s e}$

since 71+19= 90, so tan 71 = Cot(90 - 71) = Cot 19
and tan 45 and cot 45 both equal to one

Allied angle concept: if $θ = n \frac{π}{2} \pm ϕ_{r a d} = ((n \times 90) \pm ϕ^{o})$
if $n$ is odd, then the function applied on $θ$ is converted to it co-ratio with angle $ϕ$ , otherwise remains same
if the function at $θ$ is negative, then has a negative sign
e.g. $\sin (90 + 20) = - \cos (20)$ since $\sin$ is negative in 2nd quad
for a right angled triangle

\begin{array}{r} C o s θ = \frac{B a s e}{H y p o t e n u s e} \\ S i n θ = \frac{p e r p e n d i c u l a r}{H y p o t e n u s e} \\ \tan θ = \frac{\sin θ}{c o s θ} = \frac{p e r p e n d i c u l a r}{B a s e} \end{array}

	product to sum/diff	Sum/diff to product
sin	$\cos (A) \cos (B) = \frac{1}{2} (\cos (A + B) + \cos (A - B))$ $\cos (A) \sin (B) = \frac{1}{2} (\sin (A + B) - \sin (A - B))$	$\sin (A) + \sin (B) = 2 \sin (\frac{A + B}{2}) \cos (\frac{A - B}{2})$ $\sin (A) - \sin (B) = 2 \cos (\frac{A + B}{2}) \sin (\frac{A - B}{2})$
cos	$\sin (A) \sin (B) = \frac{1}{2} (\cos (A - B) - \cos (A + B))$ $\sin (A) \cos (B) = \frac{1}{2} (\sin (A + B) + \sin (A - B))$	$\cos (A) + \cos (B) = 2 \cos (\frac{A + B}{2}) \cos (\frac{A - B}{2})$ $\cos (A) - \cos (B) = - 2 \sin (\frac{A + B}{2}) \sin (\frac{A - B}{2})$

$s i n (- θ) = - s i n (θ)$	$\cos e c (- θ) = - \cos e c (θ)$
$\cos (- θ) = \cos (θ)$	$\sec (- θ) = \sec (θ)$
$\tan (- θ) = - \tan (θ)$	$\cot (- θ) = - \cot (θ)$

function	$θ = 0$	$θ = 30$	$θ = 45$	$θ = 60$	$θ = 90$
$\sin$	$0$	$\frac{1}{2}$	$\frac{1}{\sqrt{2}}$	$\frac{\sqrt{3}}{2}$	1
$\cos$	$1$	$\frac{\sqrt{3}}{2}$	$\frac{1}{\sqrt{2}}$	$\frac{1}{2}$	0
$\tan$	0	$\frac{1}{\sqrt{3}}$	1	$\sqrt{3}$	undefined

10: Trigonometric Identities

reference angle.png|center|600

11:

func	period formula
sin(ax)	period = $\frac{2 π}{a}$
cos(ax)	period = $\frac{2 π}{a}$
tan(ax)	period = $\frac{π}{a}$

period of

\cos b n

ANS: $\frac{2 π}{b}$

$y = a \sin (b x - c)$
a: determines amplitude
b: affects the period
c: and b affects the phase shift
phase shift simply refers to the angle $θ$ or any equation in a trigonometric function()

Phase shift of

y = 2 + 3 \cos (4 x + 9)

a) $- \frac{9}{2}$ b)2c) $- \frac{4}{9}$ d) $- \frac{27}{4}$

$4 x + 9 = 0; x = - \frac{9}{4}$

any number added or subtracted from the variable is ignored when finding period
any number being multiplied to the variable is divides it in period e.g. $\tan (\frac{3}{2} x + 50)$ has a period of $\frac{2}{3} x$ , $\sin (8 x + 2)$ has period of $\frac{x}{4}$ (since $4 (2 x)$ ) has $2 x$ is period of normal $\sin$
if $S i n, C o s, S e c, C o \sec$ trigonometric function have even $_{2, 4, 6, \dots}$ exponential power, the period becomes half, $\sin^{2} x, \cos^{6} x$ both have period of $\frac{x}{2}$

period of

y = 3 \sin^{2} 5 x

a) $\frac{2 π}{5}$ b) $\frac{π}{5}$ c) $\frac{5}{π}$ d) $2 π$

since $\sin$ has square, period becomes $2 π \to π$ and since $θ = 5 x$ , period becomes $π \to \frac{π}{5}$

Frequency of a trigonometric function is simply the reciprocal of its period

frequency of

y = 3 + 2 \tan (x - 9)

a) $\frac{π}{9}$ b) $\frac{1}{π}$ c) $\frac{9}{π}$ d) $\frac{2}{π}$

Amplitude is the constant multiplied with the trigonometric function and is not negative
$\tan, \cot$ have no amplitude, their amplitude does not exist

amplitude of

y = - 3 \sin x

a)-3b)1c)3d) $\pm 3$

amplitude of

y = 4 \tan x

a)4b)1c) $π$ d)Does not Exist

for:

0, π

0, π, \frac{π}{3}

0, π, \frac{π}{6}

0, \frac{π}{8}

12: Applications of trigonometry

Law of Sines	Law of Cosines	Law of tangents
$\frac{a}{\sin α} = \frac{b}{\sin β} = \frac{c}{\sin γ}$	$\cos α = \frac{b^{2} + c^{2} - a^{2}}{2 b c}$	$\frac{a - b}{a + b} = \frac{\tan (\frac{α - β}{2})}{\tan (\frac{α + β}{2})}$

Sine	Cosine	Tangent
$\sin \frac{α}{2} = \sqrt{\frac{(s - b) (s - c)}{b c}}$	$\cos \frac{α}{2} = \sqrt{\frac{s (s - a)}{b c}}$	$\tan \frac{α}{2} = \sqrt{\frac{(s - b) (s - c)}{s (s - a)}}$
where
$s = \frac{a + b + c}{2} = \frac{perimeter}{2}$

Two sides and included angle	one side and two angles	three sides are given
$Δ = \frac{1}{2} b c \sin α$	$Δ = \frac{a^{2} \sin β \sin γ}{2 \sin α}$	$Δ = \sqrt{s (s - a) (s - b) (s - c)}$

	Circum Cricle	In Circle	Ex Circle
relation	$R = \frac{a b c}{4 Δ}$ $R = \frac{a}{2 \sin α}$	$r = \frac{Δ}{s}$
	!center	!center

Δ A B C

b = 2

β

=30° then area of circum-circle is:

a) $π$ b) $2 π$ c) $4 π$ d) $8 π$

The area of

Δ A B C

72 c m

and its perimeter is

36 c m

, the radius of its in-circle is:

a)1cmb)2cmc) $\frac{1}{2} c m$ d)4cm

Which of the triangle is oblique with sides:

a)3, 4, 5b)5, 12, 13c)6, 7 ,16d)none

Type	Angle Condition	Description	Side Relation
Acute	All angles < 90°	Every angle is sharp or narrow	All sides follow a² + b²> c²
Obtuse	One angle> 90°	Contains one wide angle	a² + b² < c²
Right	One angle = 90°	Contains one perfect right angle	a² + b² = c² (Pythagorean)
Oblique	No 90° angle (either acute or obtuse)	A general term for non-right triangles	Covers acute and obtuse
where $c$ is the greatest side

13: Inverse Trigonometric Functions

\begin{aligned} \sin^{- 1} x + \cos^{- 1} x = & \frac{π}{2}; - 1 \leq x \leq 1 \\ \cos e c^{- 1} x + \sec^{- 1} x = & \frac{π}{2}; x \leq - 1 or x \geq 1 \\ \tan^{- 1} x + \cot^{- 1} x = & \frac{π}{2}; x \in R \end{aligned}

$\arcsin = \sin^{- 1}; \arccos = \cos^{- 1}$

\cos e c^{- 1} (\frac{1}{2})

a) $\frac{π}{2}$ b) $\frac{π}{3}$ c) $\frac{π}{6}$ d)doesn't exist

since $\cos e c^{- 1} (\frac{1}{2}) = \sin^{- 1} (2)$ and $\sin$ never gives 2, and range of $\cos e c$ is $x \leq - 1$ or $x \geq 1$

\tan^{- 1} (\frac{1}{6}) + \cot^{- 1} (\frac{1}{6}) = ?

a)cannot be calculatedb) $\frac{π}{2}$ c) $\frac{π}{6}$ d) $\frac{π}{4}$

\cos e c^{- 1} (2 x) + \sec^{- 1} (5) = \frac{π}{2}

, then

x

is:

a)5b) $\frac{2}{5}$ c) $\frac{5}{2}$ d)none

\arcsin (\frac{3}{2}) + \arccos (\frac{3}{2}) = ?

a) $\frac{π}{2}$ b) $\frac{π}{3}$ c) $\frac{π}{4}$ d) Does not exist

since $\frac{3}{2} = 1.5$ and that is out of the rage of $\cos$ and $\sin$

if asked to express one trigonometric function in terms of other, draw a triangle use that,

sin	cos	tan
$\sin = \frac{perp.}{h y p o .}$	$\cos = \frac{base}{hypo.}$	$\tan = \frac{base}{perp.}$

base= $\sqrt{a - x^{2}}$	perp. = $\sqrt{a + x^{2}}$	hypo. = $\sqrt{1 + x^{2}}$

\tan^{- 1} x

in terms of

\arccos :

a) $\cos^{- 1} \sqrt{1 + x^{2}}$ b) $\cos^{- 1} \frac{1}{\sqrt{1 + x^{2}}}$ c) $\cos^{- 1} \frac{1}{\sqrt{1 - x^{2}}}$ d) $\cos^{- 1} \sqrt{1 - x^{2}}$

\cos^{- 1} (\sin^{- 1} x) = ?

a) $\frac{1}{x}$ b) $1 - x$ c) $\sqrt{1 - x^{2}}$ d) $\sqrt{1 + x^{2}}$

Part 2

1: Function & Limits +


Injective(one to one)	every elements of A maps to only one element of B
surjective (onto)	all elements of B are mapped by A, multiple
bijective	both

vertical line test determine if a graph is a function or not
- if intersected at two points it is not a function
Functions:
- Even: $f (x) = f (- x);$ graph is symmetric with respect to y-axis
- Odd: $f (- x) = - f (x);$ graph is symmetric with respect to origin
if $f (x) = f^{'} (x)$ its called involution and give same results for same value of all values of x e.g. $e^{x}$
simplify a fractional limit by taking the derivative of numerator and denominator separately, called rule of De l $^{'}$ Hopital $$\begin{align*}&\text{if f(a) and g(a) are equal to zero \quad } \ & \lim_{x \to a} \dfrac{f(x)}{g(x)} =\lim_{x \to a} \dfrac{f'(x)}{g'(x)}\end{align*}$$

Evaluate the limit: $l i m_{x \to 3} \frac{ x^{2} - 9}{x - 3} $ .

a)0b)undefinedc)6d)3

Important: $lim_{n \to \infty} {(1 + \frac{1}{n})}^{n} = e = lim_{x \to 0} {(1 + x)}^{\frac{1}{x}}$
$lim_{x \to 0} \frac{\sin x}{x} = 1$
$lim_{x \to a} \frac{x^{n} - a^{n}}{n - a} = n a^{n - 1}$
Vertical asymptotes: in a rational functions $\frac{P (x)}{Q (x)}$ when the $Q (x)$ equals zero while the $P (x)$ does not, if both are zero its a hole
find inverse by isolating the $x$ and then replacing $x$ with $x^{- 1}$ and $y$ with $x$

What is the inverse function of

f (x) = 2 x - 5

a) $\frac{x - 5}{2}$ b) $\frac{x + 5}{2}$ c) $\frac{x}{2} - 5$ d) $\frac{x}{2} + 5$

In functions with limit of $x \to \infty$ , compare the degrees of the numerator and denominator
- if degree of denominator > numerator (rational function): limit is 0
- if degree of denominator = numerator : limit is ratio of leading coefficients
- if degree of denominator < numerator (irrational function): limit is $\infty$ or undefined

l i m_{x \to \infty} ​ \frac{2 x^{2} + 3 x - 1 ​}{5 x^{2} - 4 x + 7}

a) $\frac{2}{5}$ b)0c)undefinedd) $\infty$

l i m_{x \to \infty} ​ \frac{2 x^{2} + 3 x - 1 ​}{5 x^{3} - 4 x + 7}

$a) \frac{2}{5}$ b)0c)undefinedd) $\infty$

function	value
$\sinh (x)$	= $\frac{e^{x} - e^{- x}}{2}$
$\cosh (x)$	= $\frac{e^{x} + e^{- x}}{2}$
$\tanh (x)$	= $\frac{e^{x} - e^{- x}}{e^{x} + e^{- x}}$

domain and range

typically domain is represented by $x$ and range is represented by $y$
linear functions: $a x + b$ = 0 have domain and range of R
- functions with $\sqrt{x}, | x |, (x)^{2}$ here $x$ can never be negative so, range is $[0, + \infty]$ and domain (except $\sqrt{x}$ ) is all real numbers R
- $\sqrt{x}$ only has domain of positive integers
What is the range of the function $f (x) = x^{2}$ ?

a) $y \geq 0$ b) $x \geq 0$ c)all real numbers d) $y \leq 0$

Domain and Range of $f (x)$ are Range And Domain of its inverse $f^{'} (x)$ respectively
in linear and rational functions $\frac{1}{x}$ , domain is excluding number that makes the function irrational
and range has a shortcut if numerator and denominator have linear polynomial functions $\frac{P (x)}{Q (x)}$ then exclude ratio of coefficients of x from range e.g. $\frac{a x + b}{c x + d}$ or $\frac{x + 1}{x - 5}$ then Range = $R - \frac{a}{c}$ or $R - {1}$ or $(- \infty, 1) U (1, \infty)$

2: Differentiation and Derivatives

if a third variable is present, don't use chain rule, simply do $\frac{d y}{d x} = \frac{y equation}{x equation}$ , now differentiate separately the numerator and denominator with respect to the third variable and divide

y = 2 a t

and

x = a t^{2}

then

\frac{d y}{d x} = ?

a) $\frac{1}{t}$ b) $\frac{2 a}{y}$ c) $t$ **d)**Both a) and b)

derivative of

\cos^{4} x

with respect to

\sin^{2} x

is:

For implicit function: $\frac{d y}{d x} = - (\frac{Derivative w.r.t x, treating y as constant}{Derivative w.r.t y, treating x as constant})$
important $\begin{array}{cc} Mathematician & Notation \\ Cauchy & D f (x) \\ Newton & \dot{f (x)} \\ Leibniz & \frac{d y}{d x} o r \frac{d f}{d x} \\ Lagrange & f^{'} (x) \end{array}$
$f (x) = a S i n (x) + b C o s (x)$ has
- max. value = $\sqrt{a^{2} + b^{2}}$
- min. value = $- \sqrt{a^{2} + b^{2}}$
order of derivative is highest derivative in the equation
$\frac{d^{3} y}{d x^{3}} + {(\frac{x d^{2} y}{d x^{2}})}^{2} + \frac{d y}{d x}$
has highest order of 3,
degree is the highest power of x, which in this case is 2
a function is:
- increasing if $f^{'} (x) > 0$
- decreasing if $f^{'} (x) < 0$

For which of the following intervals is the function

y = x^{2} - 6 x + 5

Increasing?

a) $1 < x < 5$ b) $x > 5$ c) $x > 3$ d) $x < 1$

name
product rule	$\frac{d}{d x} (f (x) . g (x)) = f^{'} (x) . g (x) + f (x) . g^{'} (x)$
quotience rule	$\frac{d}{d x} [\frac{f (x)}{g (x)}] = \frac{g (x) f^{'} (x) . - f (x) . g^{'} (x)}{(g (x))^{2}}$
reciprocal rule	$\frac{d}{d x} (\frac{1}{f (x)}) = - \frac{f^{'} (x)}{[f (x)]^{2}}$
power rule	$\frac{d}{d x} (x^{n}) = n x^{n - 1}$

	simple	inverse	hyperbolic
Sin	$\frac{d}{d x} \sin x = \cos x$	$\frac{d}{d x} \sin^{- 1} = \frac{1}{\sqrt{1 - x^{2}}}$	$\frac{d}{d x} \sinh x = \cosh x$
Cos	$\frac{d}{d x} \cos x = - \sin x$	$\frac{d}{d x} \cos^{- 1} = \frac{- 1}{\sqrt{1 - x^{2}}}$	$\frac{d}{d x} \cosh x = \sinh x$
Tan	$\frac{d}{d x} \tan x = \sec^{2} x$	$\frac{d}{d x} \tan^{1} = \frac{1}{1 + x^{2}}$	$\frac{d}{d x} \tanh x = \sec h^{2} x$
Cot	$\frac{d}{d x} \cot x = - \cos e c^{2} x$	$\frac{d}{d x} \cot^{- 1} = \frac{- 1}{1 + x^{2}}$	$\frac{d}{d x} \coth x = - c o s e c h^{2} x$
Cosec	$\frac{d}{d x} \cos e c x = - \cos e c (x) . \cot x$	$\frac{d}{d x} \cos e c^{- 1} = \frac{- 1}{x \sqrt{1 - x^{2}}}$	$\frac{d}{d x} \cosh e c x = - \cosh e c (x) . \coth x$
Sec	$\frac{d}{d x} \sec x = \sec x \tan x$	$\frac{d}{d x} \sec^{- 1} = \frac{1}{x \sqrt{1 - x^{2}}}$	$\frac{d}{d x} \sec h x = - \sec h x \tan h x$

$\log_{a} a = 1; \log_{a} 1 = 0$

\log_{7}^{1} = ?

ANS = 0


$\frac{d}{d x} e^{x} = e^{x}$
$\frac{d}{d x} e^{f (x)} = e^{f (x)} \times f^{'} (x)$	$\frac{d}{d x} a^{f (x)} = a^{f (x)} \times \ln a \times f^{'} (x)$
$\frac{d}{d x} (\log_{a} x) = \frac{1}{x} \times \frac{1}{\ln a}$

derivative of

e^{x^{- 2}} :

a) $- 2 e^{x^{- 2}}$ b) $- \frac{1}{x^{2}} e^{x^{- 1}}$ c) $- \frac{2}{x^{3}} e^{x^{- 2}}$ d) $e^{x^{- 2}}$

Ripples in shape of circles are produced when a stone is thrown in water. If the rate of change of circumference of these circles is

12 π

, what is the rate of change of Area of the circle when theradius is 3cm?

a) $28 π$ b) $36 π$ c) $54 π$ d) $60 π$

Rate of change of circumference of circle = $\frac{d c}{d t} = 12 π$
Circumference = $C = 2 π r$
$\frac{d C}{d t} = 2 π \frac{d r}{d t}; \frac{d r}{d t} = \frac{1}{2 π} (\frac{d C}{d t}) = \frac{1}{2 π} \times_{1} 2 π = 6 =$ rate of change of radius
Area of circle = $A = π r^{2}; \frac{d A}{d t} = 2 π \frac{d r}{d t} = 2 π (3) (6) = 36 π$

LOOK FOR RELATOON BETWEEN MACLAURAN AND GEOMETRIC SERIES AAAAAAAAAAAAAAAAAAH

3: Integration

for any integral limit of form $\int_{- a}^{a} f^{'} (x) d x$

\begin{aligned} if f^{'} (x) is odd: & = 0 \\ if f^{'} (x) is even: & = \int_{0}^{a} f^{'} (x) d x \end{aligned}

in $\int_{a}^{b} f^{'} (x) d x$ a and b are range of $f (x)$
Integration cancels out der uivative: $\int \frac{d [f (x)]}{d x} d x = f (x)$

\int_{0}^{1} \frac{d}{d x} (\sin^{- 1} (\frac{x^{2}}{1 + 2 x})) d x

simply put upper and lower limit: $\sin^{- 1} (\frac{(1)^{2}}{1 + 2 (1)}) - \sin^{- 1} (\frac{(0)^{2}}{1 + 2 (0)}) = \sin^{- 1} (\frac{1}{3})$

if we have

\int \ln [f (x)] f^{'} (x) d x = f (x) \times [\ln (f (x)) - 1] + c

if upper limit is a variable $t$ and lower limit is a constant $a$

\begin{array}{r} if: \int_{a}^{t} f^{'} (x) d x = f (t) then \\ \int_{0}^{t} x d x = {[\frac{x^{2}}{2}]}_{0}^{t} = \frac{t^{2}}{2} \end{array}

order: highest derivative in an equation
degree: power of the term with the highest order
$\int e^{a \times g (x)} (a \times f (x) + f^{'} (x)) = e^{a \times g (x)} f (x) + c$

\int 0 d x =

a) $x + C$ b) $C$ c) $1 + C$ d)none

$(x \times 0) + C = C$

$e^{x}$ has same derivative and integral
$\int \frac{f^{'} (x)}{f (x)} = \ln (f (x)) + C$

integral of

\frac{e^{x}}{e^{x} + 1}

a) $\ln | e^{x} + 1 | + C$ b) $\ln | e^{x} | + C$ c) $e^{x} + C$ d)none

\frac{6 x^{2} + 8 x + 16}{2 x^{3} + 4 x^{2} + 16 x}

$\ln (2 x^{3} + 4 x^{2} + 16 x)$

4: Analytical Geometry

To determine the unknown point in a parallelogram ABCD, D is equal to sum of its adjacent sides A and C, subtracted by B:
$D = A + C - B$
a linear equation of two variables represents a line i.e.

\begin{aligned} general form of eq. of a line: & a x + b y + c = 0 \\ slope = m = - \frac{a}{b} \\ x-intercept = - \frac{c}{a} \\ y-intercept = - \frac{c}{b} \end{aligned}

two lines are coinciding if their slopes and $x$ or $y$ intercept are equal

\begin{aligned} m_{1} = m_{2}; & only parallel \\ x or y intercepts are equal; & also coinciding \end{aligned}

thing	formula	explainations
distance	$d = \| A B \| = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}$	distance between two points
midpoint	$(\frac{x_{2} + x_{1}}{2}, \frac{y_{2} + y_{1}}{2})$	the point in the middle of 2 points
division of line internally	$P = (\frac{k_{2} x_{1} + k_{1} x_{2}}{k_{1} + k_{2}}, \frac{k_{2} y_{1} + k_{1} y_{2}}{k_{1} + k_{2}})$ $m = k_{1}$ , $n = k_{2}$	P is the position vector of the point	!center
division of line externally	$P = (\frac{k_{2} x_{1} - k_{1} x_{2}}{k_{1} - k_{2}}, \frac{k_{2} y_{1} - k_{1} y_{2}}{k_{1} - k_{2}})$	P is the position vector where of the point
distance between point and line	$d = \frac{\| a x_{1} + b y_{1} + c \|}{\sqrt{a^{2} + b^{2}}}$	where $a x + b y + c$ is the line $P (x_{1}, y_{1})$ is the point
slope	$\tan θ = m = (\frac{y_{2} - y_{1}}{x_{2} - x_{1}})$	the tangent of inclination(θ)
Centroid of triangle	$(\frac{x_{1} + x_{2} + x_{3}}{3}, \frac{y_{1} + y_{2} + y_{3}}{3})$		!center
in-center of a triangle	$(\frac{a x_{1} + b x_{2} + c x_{3}}{a + b + c}, \frac{a y_{1} + b y_{2} + c y_{3}}{a + b + c})$ where $x_{1}, x_{2}, x_{3} \dots$ are coordinates of the corner points and a,b,c the sides	from bisector of the angles	!center

\begin{aligned} Standard: & A x + B y = C \\ General: & A x + B y + C = 0 \\ Slope-Intercept: & y = m x + c \\ Point-Slope: & y - y_{1} = m (x - x_{1}) \\ Two-Slope: & y - y_{1} = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} (x - x_{1}) \\ Two-Intercept: & \frac{x}{a} + \frac{y}{b} = 1 \end{aligned}

slope of horizontal line is $0$ or $t a n (0 or 180)$ and of vertical line $\infty$ or undefined
if two lines are perpendicular then $m_{1} \times m_{2} = - 1$
if two lines are parallel then $m_{1} = m_{2}$
for three collinear points

\begin{array}{r} | \begin{array}{c} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{array} | = 0 \end{array}

if slope of two lines AB and BC are equal, then A, B and C are co-linear
in $(x, y)$ , x is the distance from y-axis and vice versa

Area of triangle = Δ = \frac{1}{2} | \begin{matrix} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{matrix} |

if in the equation of two lines coefficients of $x$ , and coefficients of $y$ are equal then they are parallel a₁=a₂ and b₁=b₂
Position of a point $P (x_{1}, y_{1})$ relative to a line $a x + b y + c = 0$

\begin{aligned} a x_{1} + b y_{1} + c > 0; & lies above line \\ a x_{1} + b y_{1} + c < 0; & lies below line \\ a x_{1} + b y_{1} + c = 0; & lies on the line \end{aligned}

similarly position of a line $a x + b y + c = 0$ relative to $x and y$ axes

\begin{aligned} a > 0; & above x-axis | & b > 0; right y-axis \\ a < 0; & below x-axis | & b < 0; left y-axis \\ a = 0; & on x-axis | & b = 0; on y-axis \end{aligned}

if axis $(x, y)$ is shifted through $(h, k)$ then new axis $$\begin{align}
(X,Y) = (x-h,y-k) \
\
\text{or} \quad (x,y) = (X+h,y+k)
\end{align}$$
in other words, a point $(x, y)$ shifted through point $(h, k)$ , becomes $(X, Y)$
three lines $a x_{n} + b y_{n} + c_{n} = 0$ are concurrent if

| \begin{matrix} a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3} \end{matrix} | = 0

angle between two lines having selopes $m_{1} and m_{2}$ , from $l_{1}$ to $l_{2}$ going counter clockwise(arrow strikes $l_{2}$ )

\tan θ = \frac{m_{2} - m_{1}}{1 + m_{1} m_{2}}

$a x^{2} + 2 h x y + b y^{2} = 0$ represents two lines through origin, they are

\begin{aligned} real and distinct: & h^{2} > a b \\ real and coincident: & h^{2} = a b \\ imaginary: & h^{2} < a b \end{aligned}

they are perpendicular if $a + b = 0$

the angle between them is:

\tan θ = \frac{2 \sqrt{h^{2} - a b}}{a + b}

pair of lines perpendicular to $a x^{2} + 2 h x y + b y^{2} = 0$ are given by $b x^{2} - 2 h x y + a y^{2} = 0$

5: Linear Inequalities and Programming

if an equality $a x + b y >$ or $< 0$ its associated equation $a x + b x = 0$ which represents the line, if the equations includes $\leq or \geq$ then this line is included otherwise its shown as dotted
a function being maximized or minimized is an objective function
the solution which maximizes or minimizes a function is called an optimal solution
feasible regions, region restricted to first quadrant
optimal solution only exists in feasible region
intersection of two boundary lines is a corner point if its is in feasible region
four symbols of inequality $<, >, \geq, \leq$
the corner point satisfies both inequalities, and can be obtained by solving the associated equation of both lines

6: Conic Section

Notes/Entry Test/attachments/conic section gif.gif|center|400

Circle: x^{2} + y^{2} = 1 Ellipse: \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 Hyperbola: \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 Parabola: y^{2} = 4 a x

Most General Equation: A x^{2} + B y^{2} + 2 h x y + 2 g x + 2 f y + c = 0

For MCQs i would recommend these 4 four videos by fakhar STEM sphere
Degenerated conic:
- a point is a degenerated circle or ellipse (plane parallel or at angle not equal to angle of cone)
- a straight line is a degenerated parabola (plane at angle equal to cone)
- two intersecting lines are a degenerated hyperbola (plane is perpendicular to the cone)
eccentricity; $e = \frac{| P F |}{| P L |}$ is a positive constant, and if

\begin{aligned} e = 1 \to & parabola \\ e = 0 \to & circle \\ e > 1 \to & hyperbola \\ 0 < e < 1 \to & ellipse \end{aligned}

at maximum, two conic can intersect each other at 4 points, min at 0 points
determine the conic

\begin{aligned} A = B; & same sign & \to & circle \\ A \neq B; & same sign & \to & ellipse \\ A \neq B; & opposite sign & \to & hyperbola \\ A = 0 & or B = 0 & \to & parabola \end{aligned}

if $x y$ term is present use the following method:

\begin{aligned} h^{2} = A B : & \to Parabola \\ h^{2} > A B : & \to Hyperbola \\ h^{2} < A B : & \to Circle or Ellipse \\ h = 0; A = B : & \to Circle \end{aligned}

any point that lies on a line or conic, satisfies its equation, this core concept is used to solve many MCQs b y back solving

circle:

&\text{General Equation: \quad }Ax^2 + By^2+ 2gx + 2fy + c = 0 \
&\text{Standard Equation: \quad}r^2=(x-h)^2 + (y-k)^2 \
\end

i f i n g e n e r a l e q u a t i o n $ A and B $ a r e e q u a l t o 1 t h e n $ $ \begin{aligned} center = & (- g, - f) or (h, k) \\ radius = r = & \sqrt{g^{2} + f^{2} - c} \end{aligned}

if values $c (h, k)$ and $r$ are given, use standard equation to form equation of circle.

while solving any MCQ, make sure $A and B$ are equal to $1$
if radius = 0 then its a point circle
no $x y$ term (product of x and y) is present in equation of circle
if $g^{2} = c$ then circle touches x-axis
if $f^{2} = c$ then circle touches y-axis
if $f^{2} = g^{2} = c$ then circle touches both axes

The circle

x^{2} + y^{2} + 4 x - 2 y + 4 = 0

touches___

a)X-axisb)y-axisc)Both X and Y Axisd)None of the Axis

length of a chord $= 2 \sqrt{r^{2} - d^{2}}$ , where d is distance of midpoint of chord from center
circle is a special form of ellipse when minor and major axes are equal
Position of a point $P (x_{1}, y_{1})$ relative to a circle is done by putting point in equation of circle(standard or general)

\begin{aligned} r^{2} = (x_{1} - h)^{2} + (y_{1} - k)^{2} > 0; & lies outside circle \\ r^{2} = (x_{1} - h)^{2} + (y_{1} - k)^{2} < 0; & lies inside circle \\ r^{2} = (x_{1} - h)^{2} + (y_{1} - k)^{2} = 0; & lies on circle \end{aligned}

to find the quadrants the circle passes though
1.) find center, this will give one quadrant
2.) find radius, compare with the distance of origin from axes, the second quadrant will be the one, which is less than radius

The circle

x^{2} + y^{2} - 8 x + 4 y + 4 = 0

passes through:

a)I and IIb)II and IIIc)III and IVd)I and IV

Circle passes through
- all 4 quadrants if $c$ is negative
- 3 quadrants and origin if $c$ is zero

tangents

to find tangent to a curve at any point $(x_{1}, y_{1})$ , replace

\begin{array}{r} x^{2} \to x . x_{1} \\ 2 x \to x + x_{1} \\ y^{2} \to y . y 1 \\ 2 y \to y + y 1 \end{array}

or take derivative and separate $\frac{d y}{d x}$ and then put $(x_{1}, y_{1}) = (x, y)$
length of tangent of circle from $(x_{1}, y_{1})$ is done by putting $x = x_{1} and y = y_{1}$ in equation of circle and taking square root
CO-EFFICIENT OF $X^{2}$ AND $Y^{2}$ MUST BE ONE

Length of tangent = \sqrt{(x_{1} - h)^{2} + (y_{1} - k)^{2} - r^{2}} = \sqrt{x_{1}^{2} + y_{1}^{2} + 2 g x_{1} + 2 f y_{1} + c}

$x^{2} + y^{2} + 4 x + 2 y = 0$ from the point P(-1,2)

Parabola:

\begin{array}{r} y^{2} = 4 a x y^{2} = - 4 a x are symmetric about x-axis \\ x^{2} = 4 a y x^{2} = - 4 a y are symmetric about y-axis \end{array}

Notes/Entry Test/attachments/table of parabola.jpg|center|800

the parabola opens where the term with degree 1 points; x-right, -x-left, y-up, -y-down
$a$ is called the focal length
for form
line $y = m x + c$ is tangent on parabola if: $c = \frac{a}{m}$
in any equation: $(y - k)^{2} = 4 a (x - h)$ , the term $(y - k)$ is the axis
for any equation of form $A x + B y^{2} + h x y + 2 g x + 2 f y + c = 0$ solve it by completing square

Ellipse:

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1

$2 a$ : length of major axis
$2 b$ : length of minor axis
$c = \sqrt{a^{2} - b^{2}};$ distance between center and foci:
$2 c$ distance between foci
- $c^{2} = a^{2} - b^{2}$
semi major axis means only $a$ , as in half of major axis
- eq. directrix: x or y = ± a²/c
distance b/w directrices: d= 2a²/c
Area: $π a b$
Perimeter≈ $π (a + b)$
focal parameter: distance between focus and directrix: $\frac{b^{2}}{c}$
in standard form of ellipse $A x + B y^{2} + h x y + 2 g x + 2 f y + c = 0$ if certain conditions are met, of A and B, the bigger can be a² and smaller can be b²

largest circle inscribed in ellipse

16 x^{2} + 4 y^{2} = 64

$= \frac{x^{2}}{4} + \frac{y^{2}}{16} = 1; a = 4, b = 2$
the largest circle inside will have radius equal to semi minor axis = b, $π r^{2} = π (b)^{2} = π 2^{2}$

Hyperbola:

\frac{(x - h)^{2}}{a^{2}} - \frac{(y - k)^{2}}{b^{2}} = 1

product of distances of foci to any perpendicular is $b^{2}$
hyperbola is along the axis which is positive in the equation
symmetric on both axes
Foci: $c^{2} = a^{2} + b^{2}$
distance between foci: 2ck
$e = \frac{c}{a}$
$c^{²} = a (1 - e) = a - a e$
Directrices = $x = y = \pm \frac{c}{e^{2}}$
Length of latus-rectum: = $\frac{2 b^{2}}{a}$
Length of traverse axis(where vertices lie): 2a
Length of conjugate axis(where co-vertices lie): 2b

\frac{(x - 2)^{2}}{9} - \frac{(y + 1)^{2}}{16} = 1

Center = (2,-1)
Vertices(∓(a+h), k) = (∓(3+2),-1)
Co Vertices(∓(b+h), k) = (∓(4+2),-1)
Foci (∓(c+h),k) = (∓(5+2),-1) ; $c = \sqrt{a^{2} + b^{2}}$

for asymptotes simply replace the 1 with 0 and solve for y

asymptote of

4 x^{2} - y^{2} = 1

a) $y = \pm x$ b) $y = \pm 2 x$ c) $y = 1$ d)none

$4 x^{2} - y^{2} = 0$
$4 x^{2} = y^{2}; y = \pm 2 x$

In rectangular hyperbola
- $a^{2} = b^{2}$
- $x y = c; c \neq 0;$ where c is a constant
- its eccentricity is always $\sqrt{2}$
- their asymptotes are perpendicular

7: Vectors

$P Q = O Q - O P$
$P Q = - Q P$
sum of vectors forming a closed loop is zero

A(1,1,2), B(2,2,3), C(3,3,4) and D(4,4,5) then

A B + B C + C D + D A = ?

a)(10,10,10)b)0c)(1,1,1)d)none

for a triangle sum of its sides as vectors is zero $$\underline{u} + \underline{v} + \underline{w} = 0$$
or sum of its any two sides, is equal to the third$$\underline{u} + \underline{v} = \underline{w} $$

i + j + k, 2 i + 3 j + 4 k, and 3 i + λ j + 5 k

is a triangle, value of

λ

is:

a)4b)-4c)2d)-2

line segment from origin to point is a position vector
from point to point is a vector

\begin{array}{cccccc} Operation & Trigonometric Formula & Component Formula & Perpendicular & Parallel & Angle \\ Dot Product & A \cdot B = | A | | B | \cos θ & A \cdot B = A_{x} B_{x} + A_{y} B_{y} + A_{z} B_{z} & 0 if θ = 90^{\circ} & Max. if θ = 0^{\circ} & \cos θ = \frac{\vec{A} \vec{B}}{| A | | B |} \\ Cross Product & A \times B = | A | | B | \sin θ \hat{n} & A \times B = | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ A_{x} & A_{y} & A_{z} \\ B_{x} & B_{y} & B_{z} \end{matrix} | & Max. if θ = 0^{\circ} & 0 if θ = 180^{\circ} & \sin θ = \frac{\vec{A} \vec{B}}{| A | | B |} \end{array}

if two vectors are perpendicular and you need to find one of their coordinates, set their dot product equal to zero and solve

a = 2 α i + j - k

and

b = i + α j + 4 k

, are perpendicular then

α = ?

a) $\frac{4}{3}$ b)4c)3d) $\frac{5}{3 +}$

projection of B on A where $θ$ is the angle b/w them is: = component of B along A = $B \cos θ = \frac{\vec{A} \vec{B}}{A} = \vec{B} \hat{A}$

if the projection of

a = 3 i + j - k

along

b = λ i - j + k

- \frac{8}{\sqrt{6}}

then

λ

is?

a)-1b)-2c)0d)1

$\frac{3 λ - 2}{\sqrt{λ^{2}} + 2} = - \frac{8}{\sqrt{6}}$ ; now back solve and chose the one that satisfies the equation

orthogonal: if two curves are perpendicular
for any vector, where $α, β and γ$ are its angles with corresponding axes
- $\cos^{2} α + \cos^{2} β + \cos^{2} γ = 1$
- $\sin^{2} α + \sin^{2} β + \sin^{2} γ = 2$
For any two parallel vectors $u = a i + b j$ and $v = c i + d j$ , the ratios of coefficient of corresponding components are equal
i.e. $\frac{a}{c} = \frac{b}{d}$

u = 4 i - 6 j

and

v = α i - 18 j

, are parallel then

α

is:

a)4b)6c)8d)12

a vector $u = a i + b j + c k$ is
- is $\sqrt{a^{2} + b^{2}}$ units away from z-axis
- is $\sqrt{c^{2} + a^{2}}$ units away from y-axis
- is $\sqrt{c^{2} + b^{2}}$ units away from x-axis

A(3, 4, 9) then distance of A from z-axis is:

a)5b)6c)3d)8

Direction cosines of a vector are the components of its unit vector
- e.g. $u = 4 i - 6 j$ has direction cosines of $\cos α = \frac{4}{\sqrt{52}}$ and $\cos \frac{6}{\sqrt{52}}$
Scalar Triple Product:
- its zero if the vectors are co planar
- it gives the volume of the parallelepiped formed by the three vectors.

\vec{a} \cdot (\vec{b} \times \vec{c}) = | \begin{matrix} a_{x} & a_{y} & a_{z} \\ b_{x} & b_{y} & b_{z} \\ c_{x} & c_{y} & c_{z} \end{matrix} |

4 i . (2 j \times 5 k)

a)30b)20c)11d)40

Volume of	formula	diagram
Parallelogram	$A \times B$	!center
Tetrahedron	$\frac{1}{6} [A . (B \times C)]$	!center
Parallelepiped	$A . (B \times C)$	!center

» Volume of tetrahedron whose vertices are A(0, 0, 0), B(1,1, 0), C(O, 1, 1) and D(1, 0, 1)

a) $\frac{1}{3}$ b) $\frac{1}{2}$ c) $\frac{1}{8}$ d) $\frac{1}{6}$

$A B = i + j; A C = j + k; A D = i + k$

\frac{1}{6} A B . (A C \times A D) = \frac{1}{6} | \begin{matrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \end{matrix} | = \frac{1}{6} \times 2 = \frac{1}{3}

Part 1

1: Number System -+

2: Sets and Sub-Sets -+

3: Matrices -+

4: Quadratic Equations

6: Sequences and Series+

7: Permutation and Combination +

Combination:

Geometrical problems

Probability:

8: Mathematical Induction and Binomial Theorem

Mathematical Induction

Binomial Theorem

here n=12, and t=2 so r=12−22=5, that means 6th term involves x2

9-14: Trigonometry

9: Fundamentals of Trigonometry

10: Trigonometric Identities

11:

12: Applications of trigonometry

13: Inverse Trigonometric Functions

Part 2

1: Function & Limits +

domain and range

2: Differentiation and Derivatives

3: Integration

4: Analytical Geometry

5: Linear Inequalities and Programming

6: Conic Section

circle:

tangents

Parabola:

Ellipse:

Hyperbola:

7: Vectors

here $n = 12$ , and $t = 2$ so $r = \frac{12 - 2}{2} = 5$ , that means 6th term involves $x^{2}$