NCERT Solutions for Class 11 Maths PDF

NCERT Solutions for Class 11 Maths in PDF format are available to download. CBSE and UP Board NCERT books as well as Offline Apps and NCERT Solutions of maths for class XI ( 11th ). You can Buy NCERT Books Online or download in PDF form. Solutions are strictly based on Latest CBSE NCERT Syllabus for 2019 – 20 for UP Board as well as CBSE Board the current academic year.

 Class 11: Maths – गणित Medium: English – हिंदी

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Points to be covered

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.

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.

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.

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.

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.

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.

In a survey of 600 students in a school, 150 students were found to be taking tea and 225 taking coffee, 100 were taking both tea and coffee. Find how many students were taking neither tea nor coffee?

Let U be the set of all students who took part in the survey.
Let T be the set of students taking tea.
Let C be the set of students taking coffee.
Accordingly, n(U) = 600, n(T) = 150, n(C) = 225, n(T ∩ C) = 100
To find: Number of student taking neither tea nor coffee i.e.,
we have to find n(T’ ∩ C’).
n(T’ ∩ C’) = n(T ∪ C)’
= n(U) – n(T ∪ C)
= n(U) – [n(T) + n(C) – n(T ∩ C)] = 600 – [150 + 225 – 100] = 600 – 275
= 325
Hence, 325 students were taking neither tea nor coffee.

यदि G = {7, 8} और H = {5, 4, 2} तो G × H और H × G ज्ञात कीजिए।

G = {7, 8} और
H = {5, 4, 2},
इसलिए,

G × H
= {(7,5),(7,4),(7,2),(8,5),(8,4),(8,2)}

तथा
H × G
= {(5,7),(5,8),(4,7),(4,8),(2,7),(2,8)}

मान लीजिए कि A = {9, 10, 11, 12, 13} तथा f : A→N, f (n) = n का महत्तम अभाज्य गुणक द्वारा, परिभाषित है। f का परिसर ज्ञात कीजिए।

दिया है: A = {9,10,11,12,13} तथा f : A→N, f (n) = n का महत्तम अभाज्य गुणक द्वारा, परिभाषित है।
9 का अभाज्य गुणक = 3
10 का महत्तम अभाज्य गुणक = 2, 5
11 का अभाज्य गुणक = 11
12 का अभाज्य गुणक = 2, 3
13 का अभाज्य गुणक = 13

∴ f(9) = 9 का महत्तम अभाज्य गुणक = 3
f(10) = 10 का महत्तम अभाज्य गुणक = 5
f(11) = 11 का महत्तम अभाज्य गुणक = 11
f(12) = 12 का महत्तम अभाज्य गुणक = 3
f(13) = 13 का महत्तम अभाज्य गुणक = 13
∴ f का परिसर = {3, 5, 11, 13}

Prove that: sin^2⁡〖π/6〗+cos^2⁡〖π/3〗-tan^2⁡〖π/4〗= -1/2

LHS
= sin^2⁡〖π/6〗+cos^2⁡〖π/3〗-tan^2⁡〖π/4〗
= (1/2)^2 + (1/2)^2 – (1)^2
= 1/4 + 1/4 – 1
= 1/2 – 1
= – 1/2
= RHS

Prove the following by using the principle of mathematical induction for all n ∈ N: (2n+7)<(n+3)^2.

Let the given statement be P(n),
therefore,
P(n):(2n+7)<(n+3)^2. For n = 1, we have (2×1+7)<(1+3)^2 ⇒9<16 ⇒ P(1) is true. Let P(k) be true for some positive integer k, such that P(k):(2k+7)<(k+3)^2. Now, to prove that P(k + 1) is true. i.e. P(k+1):(2k+9)<(k+4)^2 Considering the statement P(k), we have (2k+7)<(k+3)^2 [Difference between (k+4)^2 and (k+3)^2=(k+4)^2-(k+3)^2=k^2+8k+16-(k^2+6k+9)=2k+7] Adding both sides (2k+7), we have (2k+7)+(2k+7)<(k+3)^2+(2k+7) ⇒(2k+9)+(2k+5)<(k^2+6k+9)+(2k+7) ⇒(2k+9)+(2k+5)<(k^2+6k+9+2k+7) ⇒(2k+9)+(2k+5)<(k^2+8k+16) ⇒(2k+9)+(2k+5)<(k+4)^2 ⇒(2k+9)<(k+4)^2 [As k>0,so 2k+1>0] 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.

Express the following complex number in the form a + ib: i^9+i^19

i^9+i^19
= (i^4 )^2.i + (i^4 )^4.i^2.i
=(1)^2.i + (1)^4.(-1).i
[∵i^4 = 1 and i^2 = – 1] = i – i
= 0
= 0 + i0

Solve the following inequality: 2 ≤ 3x – 4 ≤ 5.

2 ≤ 3x – 4 ≤ 5
⇒ 2 + 4 ≤ 3x – 4 + 4 ≤ 5 + 4
⇒ 6 ≤ 3x ≤ 9
⇒ 2 ≤ x ≤ 3
Therefore, all the real numbers, x, which are greater than or equal to 2 but less than or equal to 3, are the solutions of the given inequality. The solution set for the given inequality is [2, 3].

It is required to seat 5 men and 4 women in a row so that the women occupy the even places. How many such arrangements are possible?

5 men and 4 women are to be seated in a row such that the women occupy the even places.
The 5 men can be seated in 5! ways.
The 4 women can be seated only at the cross marked places (so that women occupy the even places).
M×M×M×M×M
Therefore, the women can be seated in 4! ways.
Hence, possible number of arrangements = 4! × 5! = 24 × 120 = 2880

Using Binomial theorem,indicate which is larger (1.1)^10000 or 1000.

By splitting 1.1 and then applying Binomial Theorem, the first few terms of (1.1)10000 can be obtained as
(1.1)^10000 = (1 + 0.1)^10000
= C(10000, 0) + C(10000, 1) (10000)^1 (0.1)^1 + Other positive terms.
= 1 + 10000×0.1 + Other positive terms.
= 1001 + Other positive terms.
> 1000
Hence,(1.1)^10000 > 1000.

Insert five numbers between 8 and 26 such that the resulting sequence is an A.P.

Let A1, A2, A3, A4 and A5 be five numbers between 8 and 26 such that 8, A1, A2, A3, A4, A5, 26 is an A.P.
Here, a = 8, b = 26, n = 7
Therefore, 26 = 8 + (7 – 1) d
⇒ 6d = 26 – 8 = 18
⇒ d = 3
A1 = a + d = 8 + 3 = 11
A2 = a + 2d = 8 + 2 × 3 = 8 + 6 = 14,
A3 = a + 3d = 8 + 3 × 3 = 8 + 9 = 17
A4 = a + 4d = 8 + 4 × 3 = 8 + 12 = 20,
A5 = a + 5d = 8 + 5 × 3 = 8 + 15 = 23
Hence, the required five numbers between 8 and 26 are 11, 14, 17, 20 and 23.

If the sum of three numbers in A.P., is 24 and their product is 440, find the numbers.

Let the three numbers in A.P. be a – d, a, and a + d.
According to question,
(a – d)+ (a)+ (a + d)= 24 … (1)
⇒ 3a = 24 ⇒ a = 8
Now, product of the numbers:
(a – d)a (a + d)= 440 … (2)
⇒ (8 – d)(8)(8 + d)= 440
⇒ (8 – d)(8 + d)= 55
⇒ 64 – d2 = 55
⇒ d2 = 64 – 55 = 9
⇒ d = ± 3
Therefore, when d = 3, the numbers are 5, 8, and 11 and when d = –3, the numbers are 11, 8, and 5.
Hence, the three numbers are 5, 8 and 11.

Write the equations for the x and y-axes.

The y-coordinate of every point on the x-axis is 0.
Therefore, the equation of the x-axis is y = 0.
The x-coordinate of every point on the y-axis is 0.
Therefore, the equation of the y-axis is y = 0.

Find the equation of the parabola that satisfies the following conditions: Focus (6, 0); directrix x = –6.

Focus (6, 0); directrix, x = –6
Since the focus lies on the x-axis, the x-axis is the axis of the parabola.
Therefore, the equation of the parabola is either of the form y2 = 4ax or
y^2 = – 4ax.
It is also seen that the directrix, x = –6 is to the left of the y-axis, while the focus (6, 0) is to the right of the y-axis.
Hence, the parabola is of the form y^2 = 4ax.
Here, a = 6
Thus, the equation of the parabola is y^2 = 24x.

A point is on the x-axis. What are its y-coordinates and z-coordinates?

If a point is on the x-axis, then its y-coordinates and z-coordinates are zero.

Give three examples of sentences which are not statements. Give reasons for the answers.

The three examples of sentences, which are not statements, are as follows.
(i) He is a doctor.
It is not evident from the sentence as to whom ‘he’ is referred to. Therefore, it is not a statement.
(ii) Geometry is difficult.
This is not a statement because for some people, geometry can be easy and for some others, it can be difficult.
(iii) Where is she going?
This is a question, which also contains ‘she’, and it is not evident as to who ‘she’ is. Hence, it is not a statement.

5 thoughts on “NCERT Solutions for Class 11 Maths PDF”

1. Ravi says:

हिन्दी माध्यम की कक्षा 11-12 की salutions pdf file s कब आएगी श्री मान

2. Viresh kumar says:

Sir its so nice

3. Lokesh Malakar says:

sir ,
11 th maths sub. hai mera and meko sare notes ap hi ki class ke