NCERT Solutions for Class 11 Maths

NCERT Solutions for Class 11 Maths in PDF format are available to download updated for new academic session 2020-2021. CBSE and UP Board NCERT books as well as Offline Apps and NCERT Solutions of Mathematics for class XI (11th).

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NCERT Solutions for Class 11 Maths

Contents:NCERT Solutions

CBSE NCERT Solutions for Class 11 Maths

NCERT Solutions for class 11 Maths all chapters are given below updated for new academic session 2020-2021. Download NCERT Solutions 2020-21 for other subjects also. If you are having any suggestion for the improvement, your are welcome. The improvement of the website and its contents are based on your suggestion and feedback.

Point to be covered in Chapter 1, 2, 3

Chapter 1: Sets
Sets and their representations.Empty set.Finite and Infinite sets.Equal sets.Subsets.Subsets of a set of real numbers especially intervals (with notations). Power set. Universal set. Venn diagrams. Union and Intersection of sets.Difference of sets. Complement of a set. Properties of Complement Sets.

Chapter 2: Relations & Functions
Ordered pairs, Cartesian product of sets.Number of elements in the cartesian product of two finite sets. Cartesian product of the set of reals with itself (upto R x R x R). Definition of relation, pictorial diagrams, domain, co-domain and range of a relation. Function as a special type of relation. Pictorial representation of a function, domain, co-domain and range of a function. Real valued functions, domain and range of these functions, constant, identity, polynomial, rational, modulus, signum, exponential, logarithmic and greatest integer functions, with their graphs. Sum, difference, product and quotient of functions.

Chapter 3: Trigonometric Functions
Positive and negative angles. Measuring angles in radians and in degrees and conversion from one measure to another.Definition of trigonometric functions with the help of unit circle. Truth of the identity sin2x+cos2x=1, for all x. Signs of trigonometric functions. Domain and range of trignometric functions and their graphs. Expressing sin (x±y) and cos (x±y) in terms of sinx, siny, cosx & cosy and their simple applications. Deducing the identities. Identities related to sin 2x, cos2x, tan 2x, sin3x, cos3x and tan3x. General solution of trigonometric equations.

Point to be covered in Chapter 4, 5, 6

Chapter 4: Principle of Mathematical Induction
Process of the proof by induction, motivating the application of the method by looking at natural numbers as the least inductive subset of real numbers. The principle of mathematical induction and simple applications.

Chapter 5: Complex Numbers and Quadratic Equations
Need for complex numbers, especially √−1, to be motivated by inability to solve some of the quardratic equations. Algebraic properties of complex numbers. Argand plane and polar representation of complex numbers. Statement of Fundamental Theorem of Algebra, solution of quadratic equations (with real coefficients) in the complex number system. Square root of a complex number.

Chapter 6: Linear Inequalities
Linear inequalities. Algebraic solutions of linear inequalities in one variable and their representation on the number line.Graphical representation of linear inequalities in two variables.Graphical method of finding a solution of system of linear inequalities in two variables.

Main points of Chapter 7, 8, 9

Chapter 7: Permutations and Combinations
Fundamental principle of counting. Factorial n. (n!) Permutations and combinations, derivation of formulae and their connections, simple applications.

Chapter 8: Binomial Theorem
History, statement and proof of the binomial theorem for positive integral indices.Pascal’s triangle, General and middle term in binomial expansion, simple applications.

Chapter 9: Sequence and Series
Sequence and Series. Arithmetic Progression (A. P.). Arithmetic Mean (A.M.) Geometric Progression (G.P.), general term of a G.P., sum of first n terms of a G.P., infinite G.P. and its sum, geometric mean (G.M.), relation between A.M. and G.M. Formulae for the special series sums.

Main points of Chapter 10, 11, 12

Chapter 10: Straight Lines
Brief recall of two dimensional geometry from earlier classes. Shifting of origin. Slope of a line and angle between two lines. Various forms of equations of a line: parallel to axis, point-slope form, slope-intercept form, two-point form, intercept form and normal form. General equation of a line.Equation of family of lines passing through the point of intersection of two lines.Distance of a point from a line.

Chapter 11: Conic Sections
Sections of a cone: circle, ellipse, parabola, hyperbola, a point, a straight line and a pair of intersecting lines as a degenerated case of a conic section. Standard equations and simple properties of parabola, ellipse and hyperbola.Standard equation of a circle.

Chapter 12: Introduction to Three-dimensional Geometry
Coordinate axes and coordinate planes in three dimensions. Coordinates of a point. Distance between two points and section formula.

Point to be covered in Chapter 13, 14

Chapter 13: Limits and Derivatives
Derivative introduced as rate of change both as that of distance function and geometrically. Intutive idea of limit.Limits of polynomials and rational functions trigonometric, exponential and logarithmic functions. Definition of derivative relate it to scope of tangent of the curve, Derivative of sum, difference, product and quotient of functions. Derivatives of polynomial and trigonometric functions.

Chapter 14: Mathematical Reasoning
Mathematically acceptable statements. Connecting words/ phrases – consolidating the understanding of “if and only if (necessary and sufficient) condition”, “implies”, “and/or”, “implied by”, “and”, “or”, “there exists” and their use through variety of examples related to real life and Mathematics. Validating the statements involving the connecting words, Difference between contradiction, converse and contrapositive.

Point to be covered in Chapter 15, 16

Chapter 15: Statistics
Measures of dispersion: Range, mean deviation, variance and standard deviation of ungrouped/grouped data. Analysis of frequency distributions with equal means but different variances.

Chapter 16: Probability
Random experiments; outcomes, sample spaces (set representation). Events; occurrence of events, ‘not’, ‘and’ and ‘or’ events, exhaustive events, mutually exclusive events, Axiomatic (set theoretic) probability, connections with other theories studied in earlier classes. Probability of an event, probability of ‘not’, ‘and’ and ‘or’ events.

Important Questions on 11th Maths

Write down all the subsets of the following set: {1, 2, 3}.
The subsets of {1, 2, 3} are

{1, 2},
{2, 3},
{1, 3}
{1, 2, 3}

In a group of students 100 students know Hindi, 50 know English and 25 know both. Each of the students knows either Hindi or English. How many students are there in the group?
Let U be the set of all students in the group.
Let E be the set of all students who know English.
Let H be the set of all students who know Hindi.
∴ H ∪ E = U
Accordingly, n(H) = 100 and n(E) = 50
n( H U E ) = n(H) + n(E) – n(H ∩ E)
= 100 + 50 – 25
= 125
Hence, there are 125 students in the group.
If A = {-1, 1}, find A × A × A.
A = {-1,1},
={(-1,-1),(-1,1),(1,-1),(1,1)} × {-1,1}
Write the relation R = {(x, x^3): x is a prime number less than 10} in roster form.
R = {(x, x^3): x is a prime number less than 10}
The prime numbers less than 10 are 2, 3, 5, and 7.
∴ The roster form R = {(2, 8), (3, 27), (5, 125), (7, 343)}
मान ज्ञात कीजिए: sin⁡〖765°〗
=sin⁡(2×360° + 45°)
〖=sin 45°〗
[∵ पहले चतुर्थांश में sin धनात्मक होता है]
Prove the following by using the principle of mathematical induction for all n ∈ N: 〖10〗^(2n-1)+1 is divisible by 11.
Let the given statement be P(n), therefore,
P(n):〖10〗^(2n-1)+1 is divisible by 11.
For n = 1, we have
〖10〗^(2-1)+1=11,which is divisible by 11.
So, P(1) is true.
Let P(k) be true for some positive integer k, such that
P(k):〖10〗^(2k-1)+1 is divisible by 11.
Let 〖10〗^(2n-1)+1 =11m … (1)
Where, m is any natural number.
Now, to prove that P(k + 1) is true. i.e.
P(k+1):〖10〗^(2k+1)+1 is divisible by 11.
Consider 〖10〗^(2k+1)+1
[From the equation (1),〖10〗^(2n-1)=11m-1]
=1100m – 100 + 1
=1100m – 99
=11[100m – 9],which is divisible by 11.
Thus, P(k + 1) is true whenever P(k) is true.
Hence, by the Principle of Mathematical Induction, statement P(n) is true for all natural numbers.
Solve the following equation: 2x^2 + x + 1 = 0.
The given quadratic equation is 2x^2 + x + 1 = 0.
On comparing the given equation with 〖ax〗^2+bx+c=0,
we obtain a = 2, b = 1, and c = 1
Therefore, the discriminant of the given equation is given by
D = b^2-4ac
= 1^2-4×2×1
Therefore, the required solutions are
x = (-b±√D)/2a=(-1±√(-7))/(2×2)
= (-1±√7.√(-1))/4
= (-1±√7 i)/4 [∵ √(-1)=i]
Find all pairs of consecutive odd positive integers both of which are smaller than 10 such that their sum is more than 11.
Let x be the smaller of the two consecutive odd positive integers. Then, the other integer is x + 2.
Since both the integers are smaller than 10, therefore
x + 2 < 10 ⇒ x < 10 – 2 ⇒ x < 8 … (i) Also, the sum of the two integers is more than 11. ∴ x + (x + 2) > 11
⇒ 2x + 2 > 11
⇒ 2x > 11 – 2
⇒ 2x > 9
⇒x>4.5 …(ii)
From (i) and (ii), we obtain that the value of x can be 4,5,6 or 7. .
Since x is an odd number, x can take the values, 5 and 7.
Hence, the required possible pairs of numbers are (5, 7) and (7, 9).
Given 5 flags of different colours, how many different signals can be generated if each signal requires the use of 2 flags, one below the other?
Each signal requires the use of 2 flags.
There will be as many flags as there are ways of filling in 2 vacant places _ _ in succession by the given 5 flags of different colours.
The upper vacant place can be filled in 5 different ways by any one of the 5 flags following which, the lower vacant place can be filled in 4 different ways by any one of the remaining 4 different flags.
Thus, by multiplication principle, the number of different signals that can be generated is 5 × 4 = 20
Find the sum of all natural numbers lying between 100 and 1000, which are multiples of 5.
The natural numbers lying between 100 and 1000, which are multiples of 5, are 105, 110, 115 … 995.
Here, first term, a = 105 and common difference, d = 5
⇒ 995 = 105+(n-1)×5
⇒ 890=(n-1)×5
⇒ 178=(n-1)
⇒ n=179
S_n = n/2 [2a+(n-1)d]
⇒ S_179 = 179/2 [2×105+(179-1)×5]
⇒ S_179 = (179)[550]
= 98450
Hence, the sum of all natural numbers lying between 100 and 1000, which are multiples of 5, is 98450.
150 workers were engaged to finish a job in a certain number of days. 4 workers dropped out on second day, 4 more workers dropped out on third day and so on. It took 8 more days to finish the work. Find the number of days in which the work was completed.
Let x be the number of days in which 150 workers finish the work.
According to the given information,
150x = 150 + 146 + 142 + ….(x + 8) terms
The series 150 + 146 + 142 + …. (x + 8) terms is an A.P. with first term 146, common difference –4 and number of terms as (x + 8).
⇒150x=(x+8)/2 [2×150+(x+8-1)×(-4)]
⇒x=-32 or 17
However, x cannot be negative.
∴ x = 17
Therefore, originally, the number of days in which the work was completed is 17.
Hence, required number of days = (17 + 8) = 25
Find the centre and radius of the circle (x + 5)^2 + (y – 3)^2 = 36
The equation of the given circle is (x + 5)^2 + (y – 3)^2 = 36.
(x + 5)^2 + (y – 3)^2 = 36
⇒ {x – (–5)}^2 + (y – 3)^2 = 62,
which is of the form (x – h)^2 + (y – k)^2 = r^2,
where h = – 5, k = 3, and r = 6.
Thus, the centre of the given circle is (–5, 3), while its radius is 6.
Find the derivative of x^n+ax^(n-1)+a^2 x^(n-2)+⋯+a^(n-1) x+a^n for some fixed real number a.
Let f(x)=x^n+ax^(n-1)+a^2 x^(n-2)+⋯+a^(n-1) x+a^n
∴f^’ (x)=d/dx(x^n+ax^(n-1)+a^2 x^(n-2)+⋯+a^(n-1) x+a^n)
=d/dx (x^n )+a d/dx (x^(n-1) )+a^2 d/dx (x^(n-2) )+⋯+(a^(n-1 ) d)/dx (x)+a^n d/dx (1)
On using theorem d/dx x^n=nx^(n-1) ,we obtain
f^’ (x)=nx^(n-1)+a(n-1) x^(n-2)+a^2 (n-2) x^(n-3)+⋯+a^(n-1)+a^n (0)
=nx^(n-1)+a(n-1) x^(n-2)+a^2 (n-2) x^(n-3)+⋯+a^(n-1)
Find the component statements of the following compound statements and check whether they are true or false: Number 3 is prime or it is odd.
The component statements are as follows.
p: Number 3 is prime.
q: Number 3 is odd.
Both the statements are true.
What is mean Deviation?
The mean deviation about a central value ‘a’ is the mean of the absolute values of the deviations of the observations from ‘a’. The mean deviation from ‘a’ is denoted as M.D. (a).
One die of red colour, one of white colour and one of blue colour are placed in a bag. One die is selected at random and rolled, its colour and the number on its uppermost face is noted. Describe the sample space.
A die has six faces that are numbered from 1 to 6, with one number on each face. Let us denote the red, white, and blue dices as R, W, and B respectively.
Accordingly, when a die is selected and then rolled, the sample space is given by
S = {R1, R2, R3, R4, R5, R6, W1, W2, W3, W4, W5, W6, B1, B2, B3, B4, B5, B6}