Algebra is one of the most important topics tested under the JEE Main Mathematics section. It is one of the four major topics from which several questions are asked. A minimum of 11-12 questions are asked from Algebra which bear a total of around 44-46 marks. Check JEE Main 2020 Mathematics Syllabus

• The weightage of this topic is about 15-16% in JEE Main.
• Some of the most important topics tested under Algebra include - Matrices and Determinants, Complex Numbers, Quadratic Equations, Statistics and Probability, Binomial Theorem and Its Application, Permutations and Combinations, and Mathematical Reasoning. 

Algebra is a topic that must not be skipped while preparing for the examination. It is a very scoring and easy topic under the Mathematics section. Read the article to find the study notes for Algebra for JEE Main Preparation. 

Matrices and Determinants 

Matrix: It is a set of numbers, objects or symbols expressed in the form of the rectangular array. The order of the matrix is described by the number of rows and number of columns held in the rectangular array of representation.

For example, 

The matrix consists 2 rows and 3 columns. Thus, its order is said to be 2 × 3.

A general element of the matrix is expressed as , where displays the elements of the ith row and jth column.

Operations on matrices: The algebraic operations on matrices are addition, multiplication, subtraction, and division. 

Transpose of the matrix: Suppose A is a matrix, the matrix that is found by changing the columns of a matrix with rows or rows with columns is known as the transpose of the matrix. 

For example,

Conjugate of the matrix: Suppose a matrix A has a complex number because it is an element, the matrix found by replacing that complex number with its conjugate is known as the conjugate of the matrix A and it is expressed as . 

The determinant of a matrix: This is a number that is found from the matrix. For a determinant to be present, the matrix A must be a square matrix. The determinant of a matrix is represented as det A or |A|.

Minor and cofactor of an element in a matrix/determinant: The minor of any element , where j is the number of columns, i is the number of rows, is the det of the matrix that remains after eliminating the ith row and jth column.

Adjoint of the matrix: The transpose of the cofactor of the component of the matrix is known as the adjoint of the matrix.

The inverse of a matrix: A non-singular square matrix “A” is considered to be invertible when there exists a non-singular square matrix B such that AB = I=BA and the matrix B is known as the inverse of the matrix A.

Read in detail JEE Main Study Notes for Matrices & Determinants

Complex Numbers

Complex number: It is the sum of an imaginary number and real number. For example, 5 + 2i is a complex number, where 5 is the real number and 2 is the imaginary number.

Algebra: The algebra of complex numbers involves subtraction, addition, multiplication, and division of complex numbers. For example, the addition of 3 + 4i and 9 + 8i is (3+4i) + (9+4i) = 12 + 12i.

Conjugate of complex numbers: The basis of conjugates is that they differ by the sign of the imaginary number. When the imaginary part of the number is +ve, the conjugate imaginary part turns -ve.

For example,

z = 2 + 2i

z= 2 - 2i

Modulus and argument of a complex number: The modulus of a complex number is the distance of the point from the origin. If z = a+bi is the complex number, then it |z| = . The argument is the angle that the complex number z makes with the +ve x-axis.

Read in detail JEE Main Study Notes for Complex Numbers

Quadratic Equation

Suppose the degree of a polynomial equation is 2, it is considered to be a quadratic equation. The value of x for which a quadratic equation is satisfied is known as its roots. The standard form of a quadratic equation is ax2 + b x + c = 0, where a, b, and c are constant. This is known as the coefficient of the equation.

The discriminant of quadratic equations: The discriminant of quadratic equations is expressed as D = b2 - 4ac. The discriminant of a quadratic equation is utilized in the Sridharacharya formula of quadratic equations to calculate the roots. This is -

By placing + and - one after another, two necessary solutions of the equation are found. Suppose D>0, then both the roots are distinct. When D = 0, then the roots are the same. Iff D<0, then the roots are complex and occur in pairs of conjugate numbers.

Relation of coefficient and roots:

If are the two roots, then the sum of the roots are linked to the coefficient as and the product of roots is linked with the coefficient as .

Read in detail JEE Main Study Notes for Quadratic Equation

Statistics

• The arithmetic mean of individual series (ungrouped data): Suppose the series in this scenario is x1, x2, … xn, then, the arithmetic mean is expressed as, 

• Arithmetic mean for discrete frequency distribution: Suppose the terms of the series are x1, x2, … xn and the frequencies are f1, f2, … , fn, then the arithmetic mean is expressed as,

• Arithmetic mean for grouped or continuous frequency distribution: Suppose A is the assumed mean, f the frequency, and x – A = deviation of every item from the assumed mean, then the arithmetic mean is represented as, 

• Combined Arithmetic Mean: When (i = 1, 2, …... , k) are the means of k- component series of sizes ni (i = 1, 2, … ,k) respectively, the mean of the composite series found on combining the component series is expressed as,

• Weighted Arithmetic Mean: The weighted arithmetic mean points to the arithmetic mean that is calculated after assigning weights to distinct values of variables. This is appropriate when the relative importance of distinct items of variable is not the same. It is represented as,

?

• Individual Series: When n is odd, Median = value of (n+1/2)th item. When n is even, Median = ½ [ value of (n/2)th item + value of (n/2 + 1)th item].
• Discrete Series: Median = (n/2 + 1)th item, where n is the cumulative frequency.
• Grouped or continuous distribution:

where, l = Lower limit of the median class

f = frequency of the median class

N = the sum of all frequencies

i = The width of the median class 

C = The cumulative frequency of the class preceding the median class

• Quartile: As a median divides a distribution into two parts, similarly the quartiles, quantiles, deciles, and percentiles also divide the distribution into 4, 5, 10, and 100 equal parts. The jth quartile is expressed as, 

Read in detail JEE Main Study Notes for Statistics

Probability 

• 1 is the sum of all the probabilities in the sample space.
• 0 is the probability of an event that cannot occur. 
• 0 is the probability of any event that is not in the sample space. 
• 1 is the probability of an event that must occur.
• 1 is the probability of the sample space.
• The complement of an event E is expressed as E' and is represented by P (E') = 1 - P (E).
• P (A∪B) is represented as P (A + B) and P (A ∩ B) is expressed as P (AB).
• When A and B are mutually exclusive events, P(A or B) = P (A) + P (B).
• When the events A and B are not independent, the probability of the intersection of A and B is represented by P (A and B) = P (A) P (B|A).
• A and B are independent if P (B/A) = P(B) and P(A/B) = P(A).
• If E1, E2, ......... En are n independent events then P (E1 ∩ E2 ∩ ... ∩ En) = P (E1) P (E2) P (E3)...P (En).
• Events E1, E2, E3, ......... En will be pair-wise independent if P(Ai ∩ Aj) = P(Ai) P(Aj) i ≠ j.
• P(Hi | A) = P(A | Hi) P(Hi) / ∑i P(A | Hi) P(Hi).
• If A1, A2, ……An are exhaustive events and S is the sample space, then A1 U A2 U A3 U ............... U An = S
• If E1, E2,….., En are mutually exclusive events, then P(E1 U E2 U ...... U En) = ∑P(Ei)
• When the events are not mutually exclusive, then P (A or B) = P (A) +P (B) – P (A and B)
• The three events A, B, and C are considered mutually independent if P(A∩B) = P(A).P(B), P(B∩C) = P(B).P(C), P(A∩C) = P(A).P(C), P(A∩B∩C) = P(A).P(B).P(C)
• If two events A and B are mutually exclusive,

P (A ∩ B) = 0 but P(A) P(B) ≠ 0 (In general)

⇒ P(A ∩ B) ≠ P(A) P(B)

⇒ Mutually exclusive events will not be independent.

• The probability distribution of a count variable X is considered as the binomial distribution with parameters n and abbreviated B (n,p) if it fulfils the conditions given below -
  • The total number of observations remains fixed.
  • The observations are independent.
  • Every outcome shows either a success or a failure.
  • The probability of success, that is, p is the same for every outcome.

Read in detail JEE Main Study Notes for Probability

Binomial Theorem and its Application 

Binomial Expression: It is an algebraic expression that contains two dissimilar terms. For example, a + b, a3 + b3.

Binomial Theorem: Let n ∈ N,x,y,∈ R, then,

nΣr=0 nCr xn – r · yr + nCr xn – r · yr + …………. + nCn-1 x · yn – 1 + nCn · yn

i.e.(x + y)n = nΣr=0 nCr xn – r · yr where,

Binomial Expansion:

• The total number of terms in the expansion of (x+y)n are (n+1).
• The sum of exponents of x and y is typically n.
• nC0, nC1, nC2, … .., nCn are known as binomial coefficients and are expressed as C0, C1, C2, ….., Cn
• The binomial coefficients that are equidistant from the beginning and from the ending are equal, that is, nC0 = nCn, nC1 = nCn-1 , nC2 = nCn-2 ,….. etc.
• To find the binomial coefficients, Pascal’s Triangle can also be used. 
• (x + y)n + (x−y)n = 2[C0 xn + C2 xn-1 y2 + C4 xn-4 y4 + …]
• (x + y)n – (x−y)n = 2[C1 xn-1 y + C3 xn-3 y3 + C5 xn-5 y5 + …]
• (1 + x)n = nΣr-0 nCr . xr = [C0 + C1 x + C2 x2 + … Cn xn]
• (1+x)n + (1 − x)n = 2[C0 + C2 x2+C4 x4 + …]
• (1+x)n − (1−x)n = 2[C1 x + C3 x3 + C5 x5 + …]
• Number of terms in expansion of (x + a)n + (x−a)n are (n+2)/2 when “n” is even or (n+1)/2 when “n” is odd.
• Number of terms in expansion of (x + a)n − (x−a)n are (n/2) when “n” is even or (n+1)/2 when “n” is odd.

Binomial Coefficients: These are the integers that are the coefficients in the binomial theorem. The most significant properties of binomial coefficients are - 

• C0 + C1 + C2 + … + Cn = 2n
• C0 + C2 + C4 + … = C1 + C3 + C5 + … = 2n-1
• C0 – C1 + C2 – C3 + … +(−1)n . nCn = 0
• nC1 + 2.nC2 + 3.nC3 + … + n.nCn = n.2n-1
• C1 − 2C2 + 3C3 − 4C4 + … +(−1)n-1 Cn = 0 for n > 1
• C02 + C12 + C22 + …Cn2 = [(2n)!/ (n!)2]

Read in detail JEE Main Study Notes for Binomial Theorem

Permutations and Combinations

Permutation

• The idea of permutation is utilized for the arrangement of objects in a particular order, that is, in areas where the order is crucial, permutation is used.
• The total number of permutations on a set of n distinct objects is represented as n! and is given by nPn = n!
• The total number of permutations on a set of n objects considered r at a time is expressed as nPr = n!/ (n-r)!
• The number of ways to allocate n objects of which r are the same is represented as n!/ r!
• To arrange a total of n objects, out of which ‘p’ are of one kind, q of the second kind are alike, and r of a third kind are the same, then such a computation is given by n!/p!q!r!
• The number of permutations of n different objects when a particular object is not taken in the arrangement is expressed as n-1Pr.
• The number of permutations of n different objects when a particular object has to be always included in the arrangement is given as r.n-1Pr-1.
• Circular permutation is utilized when some arrangement has to be done in the form of a ring or circle.
• When ‘n’ different or unlike objects have to be arranged in a ring in a way that the clockwise and anticlockwise arrangements are distinct, the number of such arrangements is represented as (n – 1)!
• When r things are considered at a time out of n different things and arranged along a circle, the number of ways to do this is represented as nCr(r-1)!.
• When the clockwise and anti-clockwise are taken to be the same, the total number of circular permutations is represented as (n-1)!/2.
• Suppose n persons have to be seated around a round table in a way that no person has a similar neighbor, it is represented as ½ (n – 1)!
• nP0 =1
• nP1 = n
• nPn = n!/(n-n)! = n! /0! = n!/1 = n!

Combination

• When certain objects have to be arranged in a way that the order of objects is not crucial, the idea of combinations is used.
• The number of combinations of n things considered as r (0 < r < n) at a time is represented as nCr = n!/r!(n-r)!
• The relationship among combinations and permutations is nCr = nPr/r!
• The number of ways of choosing r objects from n different objects with respect to certain condition are - 

1. k particular objects that are always included = n-kCr-k
2. k particular objects that are never included = n-kCr

• The arrangement of n different objects that are taken r at a time such that k particular objects are, 

(i) Always included = n-kCr-k.r!,

(ii) Never included = n-kCr.r!.

• To be able to compute the combination of n distinct items taken r at a time, the chances of occurrence of any item are not established and may be one, twice, thrice, …. up to r times is represented as n+r-1Cr
• When there are m men and n women (m > n) and they should be seated or accommodated in a row in a way that no two women sit together, the total number of such arrangements = m+1Cn. m! This is also known as the Gap Method. 
• If there are n different things taken r at a time, then the required number of arrangements is given as nCr(r-1)!/2.
• When there is an issue that needs n number of persons to be accommodated in a way that a fixed number ‘p’ are always together, then the particular set of p persons must be treated as one person. Here, the total number of people becomes (n-m+1). Thus, the total number of possible arrangements is (n-m+1)! m! This is also known as the String Method.
• The number of selections from n different objects, taking a minimum of one = nC1 + nC2 + nC3 + ... + nCn = 2n - 1.
• The total number of selections of zero or more objects from n identical objects is n+1.

Read in detail JEE Main Study Notes for Permutation & Combination

Mathematical Reasoning 

• An open statement is a declarative sentence consisting of variables that turns into a statement when the variables get replaced by a few definite values.
• A compound statement is a statement that has two or more statements. Every statement is known to be a compound statement.
• The compound statements are linked by the word “and” (^). The resulting statement is known as a conjunction that is represented by p ∧ q.
• The compound statement with “And” holds true if each of its component statements are true.

The many truth tables are given below - 

• Idempotent Laws: When p is any statement, then, p ∨ p = p and p ∧ p = p.
• Associative Laws: When p, q, and r are any three statements, then, p ∨ (q ∨ r) = (p ∨ q) ∨ r and p ∧ (q ∧ r) = (p ∧ q) ∧ r.
• Commutative Laws: When p and q are any two statements, then, p ∨ q = q ∨ p and p ∧ q = q ∧ p.
• Distributive Laws: When p, q, and r are any three statements, then, p ∧ (q ∨ r) = (p ∧ q) ∨ (p ∧ r) and p ∨ (q ∧ r) = (p ∨ q) ∧ (p ∨ r).
• Identity Laws: When p is any statement, t is a tautology, and c is a contradiction, then, p ∨ t = t, p ∧ t = p, p ∨ c = p and p ∧ c = c.
• Complement Laws: When t is a tautology, c is a contradiction, and p is any statement, then, p ∨ (~p) = t, p ∧ (~p) = c, ~t = c and ~c = t.
• Involution Law: When p is any statement, then, ~(~p) = p.
• De-Morgan’s Law: When p and q are two statements, then, ~(p ∨ q) = (~p) ∧ (~q) and ~(p ∧ q) = (~p) ∨ (~q).
• Two compound statements, S1 and S2, are considered as duals of one another if one can be found from the other by replacing ∧ with ∨, or ∨ with ∧, that is, S∗ (p,q)=p∨q.

Read in detail JEE Main Study Notes for Mathematical Reasoning

Vector Algebra 

• a.a = |a|2 = a2
  a.b = b.a
  a.0 = 0
  a.b = (a cos q)b = (projection of a on b)b = (projection of b on a) a
  a.(b + c) = a.b + a.c (This is also termed as the distributive law)
  (la).(mb) = lm (a.b)
  (a ± b)2 = (a ± b) . (a ± b) = a2 + b2 ± 2a.b
  If a and b are non-zero, then the angle between them is given by cos θ = a.b/|a||b|
  a x a = 0
  a x b = - (b x a)
  a x (b + c) = a x b + a x c
• Any vector that is perpendicular to the plane of a and b is l(a x b), where l is a real number.
• A unit vector that is perpendicular to a and b is ± (a x b)/ |a x b|
• The position of dot and cross could be interchanged without any alteration of the product. Therefore, it is also represented as [a b c]
• [a b c] = [b c a] = [c a b]
  [a b c] = - [b a c]
  [ka b c] = k[a b c]
• [a+b c d] = [a c d] + [b c d]
  a x (b x c) = (a x b) x c, if some or all of a, b and c are zero vectors or a and c are collinear.

Solved Examples on Algebra

Question 1:

Solution: 

Question 2: Suppose the number of ways of choosing n cards from an unlimited number of cards carrying the numbers 0, 2, 3, such that it cannot be used to write the number 203 is 93. Then, n is equal to?

a) 3 

b) 4

c) 5 

d) 6

Solution: 203 cannot be written when in the selection of n cards, either (2 or 3), (3 or 0), ( 0 or 2), (only 0), (only 2) or (only 3) is found. 

Question 3: The set of all real numbers x for which x2 - |x + 2| + x > 0 is? 

a) (-∞, -2) ∪ (2, ∞)

b) (-∞, -√2) ∪ (√2, ∞)

c) (-∞, -1) ∪ (1, ∞)

d) (√2, ∞)

Solution: Given that, x2 - |x + 2| + x > 0

Case 1: When (x+2) ≥ 0.

x2 - x - 2 + x > 0

Hence, x2 – 2 > 0

Either x < - √2 or x > √2.

Hence, x ∈ [-2, -√2) ∪ (√2, ∞) ……. (1)

Case 2: When (x+2) < 0

x2 + x + 2 + x > 0

Thus, x2 + 2x + 2 > 0

This provides (x+1)2 + 1 > 0 and this is true for every x

Hence, x ≤ -2 or x ∈ (-∞, -2) ...…… (2)

From equations (2) and (3), x ∈ (-∞, -√2) ∪ (√2, ∞).

Question 4: When a = (i+j+k), a.b = 1 and a x b = j-k, then b is?

a) (i-j+k)

b) 2j-k) 

c) i

d) 2i

Solution: a x (a x b) = (a.b) a – (a.a) b

Hence, (i+j+k) x (j-k) = (i + j + k) – (√3)2 b

This means, -2i + j + k = i + j + k – 3b

Hence, 3b = 3i

Therefore, b = i.

Question 5: What is the contrapositive of the inverse of p ⇒ ~q? 

Solution: The inverse of p ⇒ ∼q is ∼ p ⇒ q

The contrapositive of ∼ p ⇒ q is ∼ q ⇒ p as the contrapositive of p ⇒ q is ∼ q ⇒ p.

Tricks to Solve Questions from Algebra 

1. Use graphical analysis to attempt algebraic questions. Roughly plot graphs for questions and analyse the properties to arrive at a solution. 

2. For more complex algebraic questions, begin by entering values to the variables and then analyze which options can give you an answer. 

3. Use the technique of reverse engineering. Once you look at an algebraic question, instead of moving forward with it, take a step back. Attempt to build the question using the options provided. 

Study Tips for Mathematics in JEE Main 

1. Lay more emphasis on Algebra and Calculus which are the most scoring and highest weightage topics. 

2. Solve questions from Arihant, NCERT, and R. D Sharma books.

3. Solve as many previous years’ question papers as you can. 

We hope with proper practice of mathematics assisted by the study notes, candidates will be able to accurately solve questions in JEE Main with a record time.