# 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).

NCERT Books 2020-21 are available Online or download in PDF form to use it offline. Solutions are strictly based on Latest CBSE NCERT Syllabus for 2020–21 for UP Board as well as CBSE Board the current academic year 2020-2021.## NCERT Solutions for Class 11 Maths

Class: | 11 |

Subject: | Mathematics |

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.

### Select the Chapter from the following

- Chapter 1: Sets
- Chapter 2: Relations and Functions
- Chapter 3: Trigonometric Functions
- Chapter 4: Principle of Mathematical Induction
- Chapter 5: Complex Numbers and Quadratic Equations
- Chapter 6: Linear Inequalities
- Chapter 7: Permutations and Combinations
- Chapter 8: Binomial Theorem
- Chapter 9: Sequences and Series
- Chapter 10: Straight Lines
- Chapter 11: Conic Sections
- Chapter 12: Introduction to Three Dimensional Geometry
- Chapter 13: Limits and Derivatives
- Chapter 14: Mathematical Reasoning
- Chapter 15: Statistics
- Chapter 16: Probability

#### 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

Φ,

{1},

{2},

{3},

{1, 2},

{2, 3},

{1, 3}

{1, 2, 3}

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.

Therefore,

A×A

={-1,1}×{-1,1}={(-1,-1),(-1,1),(1,-1),(1,1)}

and

A×A×A

={(-1,-1),(-1,1),(1,-1),(1,1)} × {-1,1}

={(-1,-1,-1),(-1,-1,1),(-1,1,-1),(-1,1,1),(1,-1,-1),(1,-1,1),(1,1,-1),(1,1,1)}

The prime numbers less than 10 are 2, 3, 5, and 7.

∴ The roster form R = {(2, 8), (3, 27), (5, 125), (7, 343)}

=sin(2×360° + 45°)

〖=sin 45°〗

[∵ पहले चतुर्थांश में sin धनात्मक होता है]

=1/√2

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

=〖10〗^(2k-1+2)+1

=〖10〗^2.〖10〗^(2k-1)+1

=〖10〗^2.(11m-1)+1

[From the equation (1),〖10〗^(2n-1)=11m-1]

=100.(11m-1)+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.

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

=-7

Therefore, the required solutions are

x = (-b±√D)/2a=(-1±√(-7))/(2×2)

= (-1±√7.√(-1))/4

= (-1±√7 i)/4 [∵ √(-1)=i]

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>9/2

⇒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).

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

Here, first term, a = 105 and common difference, d = 5

Now,a_n=a+(n-1)d

⇒ 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.

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)]

⇒300x=(x+8)(300-4x-28)

⇒300x=(x+8)(272-4x)

⇒300x=272x-4x^2+2176-32x

⇒4x^2+60x-2176=0

⇒x^2+15x-544=0

⇒x^2+32x-17x-544=0

⇒x(x+32)-17(x+32)=0

⇒(x+32)(x-17)=0

⇒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

(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.

∴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)

p: Number 3 is prime.

q: Number 3 is odd.

Both the statements are true.

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}