# NCERT Solution

## NCERT solution class 8 SCIENCE

S.no Chapters Links 1 Crop Production and Management Exercise 1.1 Exercise 1.2 Exercise 1.3 Exercise 1.4 Exercise 1.5 2 Microorganisms: Friend and Foe Exercise 2.1 Exercise 2.2 Exercise 2.3 3 Synthetic Fibres and Plastics Exercise 3.1 Exercise 3.2 Exercise 3.3 Exercise 3.4 Exercise 3.5 4 Materials: Metals and Non-Metals Exercise 4.1 Exercise 4.2 Exercise …

## NCERT solution class 12 chapter 7 Probability exercise 7.6 mathematics part 2

EXERCISE 7.6 Page No 582: Question 1: A and B are two events such that P (A) ≠ 0. Find P (B|A), if (i) A is a subset of B (ii) A ∩ B = Φ Answer: It is given that, P (A) ≠ 0 (i) A is a subset of B. (ii)  Question 2: …

## NCERT solution class 12 chapter 7 Probability exercise 7.5 mathematics part 2

EXERCISE 7.5 Page No 576: Question 1: A die is thrown 6 times. If ‘getting an odd number’ is a success, what is the probability of (i) 5 successes? (ii) at least 5 successes? (iii) at most 5 successes? Answer: The repeated tosses of a die are Bernoulli trials. Let X denote the number of …

## NCERT solution class 12 chapter 7 Probability exercise 7.4 mathematics part 2

EXERCISE 7.4 Page No 569: Question 1: State which of the following are not the probability distributions of a random variable. Give reasons for your answer. (i) X 0 1 2 P (X) 0.4 0.4 0.2 (ii) X 0 1 2 3 4 P (X) 0.1 0.5 0.2 − 0.1 0.3 (iii) Y −1 0 1 P …

## NCERT solution class 12 chapter 7 Probability exercise 7.3 mathematics part 2

EXERCISE 7.3 Page No 555: Question 1: An urn contains 5 red and 5 black balls. A ball is drawn at random, its colour is noted and is returned to the urn. Moreover, 2 additional balls of the colour drawn are put in the urn and then a ball is drawn at random. What is …

## NCERT solution class 12 chapter 7 Probability exercise 7.2 mathematics part 2

EXERCISE 7.2 Page No 546: Question 1: If, find P (A ∩ B) if A and B are independent events. Answer: It is given that A and B are independent events. Therefore, Question 2: Two cards are drawn at random and without replacement from a pack of 52 playing cards. Find the probability that both …

## NCERT solution class 12 chapter 7 Probability exercise 7.1 mathematics part 2

EXERCISE 7.1 Page No 538: Question 1: Given that E and F are events such that P(E) = 0.6, P(F) = 0.3 and P(E ∩ F) = 0.2, find P (E|F) and P(F|E). Answer: It is given that P(E) = 0.6, P(F) = 0.3, and P(E ∩ F) = 0.2 Question 2: Compute P(A|B), if …

## NCERT solution class 12 chapter 6 Linear Programming exercise 6.3 mathematics part 2

EXERCISE 6.3 Page No 525: Question 1: Refer to Example 9. How many packets of each food should be used to maximize the amount of vitamin A in the diet? What is the maximum amount of vitamin A in the diet? Answer: Let the diet contain x and y packets of foods P and Q respectively. Therefore, x ≥ 0 …

## NCERT solution class 12 chapter 6 Linear Programming exercise 6.2 mathematics part 2

EXERCISE 6.2 Page No 519: Question 1: Reshma wishes to mix two types of food P and Q in such a way that the vitamin contents of the mixture contain at least 8 units of vitamin A and 11 units of vitamin B. Food P costs Rs 60/kg and Food Q costs Rs 80/kg. Food …

## NCERT solution class 12 chapter 6 Linear Programming exercise 6.1 mathematics part 2

EXERCISE 6.1 Page No 513: Question 1: Maximise Z = 3x + 4y Subject to the constraints: Answer: The feasible region determined by the constraints, x + y ≤ 4, x ≥ 0, y ≥ 0, is as follows. The corner points of the feasible region are O (0, 0), A (4, 0), and B (0, 4). The values of Z at these points …

## NCERT solution class 12 chapter 5 Three Dimensional Geometry exercise 5.4 mathematics part 2

EXERCISE 5.4 Page No 497: Question 1: Show that the line joining the origin to the point (2, 1, 1) is perpendicular to the line determined by the points (3, 5, −1), (4, 3, −1). Answer: Let OA be the line joining the origin, O (0, 0, 0), and the point, A (2, 1, 1). …

## NCERT solution class 12 chapter 5 Three Dimensional Geometry exercise 5.3 mathematics part 2

EXERCISE 5.3 Page No 493: Question 1: In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin. (a)z = 2 (b)  (c)  (d)5y + 8 = 0 Answer: (a) The equation of the plane is z = 2 or 0x + 0y + z = 2 … (1) The direction ratios of …

## NCERT solution class 12 chapter 5 Three Dimensional Geometry exercise 5.2 mathematics part 2

EXERCISE 5.2 Page No 477: Question 1: Show that the three lines with direction cosines  are mutually perpendicular. Answer: Two lines with direction cosines, l1, m1, n1 and l2, m2, n2, are perpendicular to each other, if l1l2 + m1m2 + n1n2 = 0 (i) For the lines with direction cosines,  and , we obtain Therefore, the lines are perpendicular. (ii) For the lines with direction cosines,  and , we obtain Therefore, the lines …

## NCERT solution class 12 chapter 5 Three Dimensional Geometry exercise 5.1 mathematics part 2

EXERCISE 5.1 Page No 467: Question 1: If a line makes angles 90°, 135°, 45° with x, y and z-axes respectively, find its direction cosines. Answer: Let direction cosines of the line be l, m, and n. Therefore, the direction cosines of the line are Question 2: Find the direction cosines of a line which makes equal angles with the coordinate axes. …

## NCERT solution class 12 chapter 4 Differential Equations exercise 4.5 mathematics part 2

EXERCISE 4.5 Page No 458: Question 1: Write down a unit vector in XY-plane, making an angle of 30° with the positive direction of x-axis. Answer: If is a unit vector in the XY-plane, then  Here, θ is the angle made by the unit vector with the positive direction of the x-axis. Therefore, for θ = 30°: Hence, the required unit vector …

## NCERT solution class 12 chapter 4 Differential Equations exercise 4.4 mathematics part 2

EXERCISE 4.4 Page No 454: Question 1: Find, if and. Answer: We have, and Question 2: Find a unit vector perpendicular to each of the vector and, where and. Answer: We have, and Hence, the unit vector perpendicular to each of the vectors and is given by the relation, Question 3: If a unit vector  makes an angleswith with and an acute angle θ with, …

## NCERT solution class 12 chapter 4 Differential Equations exercise 4.3 mathematics part 2

EXERCISE 4.3 Page No 447: Question 1: Find the angle between two vectorsandwith magnitudesand 2, respectively having. Answer: It is given that, Now, we know that. Hence, the angle between the given vectors andis. Question 2: Find the angle between the vectors Answer: The given vectors are. Also, we know that. Question 3: Find the projection …

## NCERT solution class 12 chapter 4 Differential Equations exercise 4.2 mathematics part 2

EXERCISE 4.2 Page No 440: Question 1: Compute the magnitude of the following vectors: Answer: The given vectors are: Question 2: Write two different vectors having same magnitude. Answer: Hence, are two different vectors having the same magnitude. The vectors are different because they have different directions. Question 3: Write two different vectors having same direction. …

## NCERT solution class 12 chapter 4 Differential Equations exercise 4.1 mathematics part 2

EXERCISE 4.1 Page No 428: Question 1: Represent graphically a displacement of 40 km, 30° east of north. Answer: Here, vector represents the displacement of 40 km, 30° East of North. Question 2: Classify the following measures as scalars and vectors. (i) 10 kg (ii) 2 metres north-west (iii) 40° (iv) 40 watt (v) 10–19 coulomb (vi) …

## NCERT solution class 12 chapter 3 Differential Equations exercise 3.7 mathematics part 2

EXERCISE 3.7 Page No 419: Question 1: For each of the differential equations given below, indicate its order and degree (if defined). (i)  (ii)  (iii)  Answer: (i) The differential equation is given as: The highest order derivative present in the differential equation is. Thus, its order is two. The highest power raised to is one. Hence, its …

## NCERT solution class 12 chapter 3 Differential Equations exercise 3.5 mathematics part 2

EXERCISE 3.5 Page No 406: Question 1: Answer: The given differential equation i.e., (x2 + xy) dy = (x2 + y2) dx can be written as: This shows that equation (1) is a homogeneous equation. To solve it, we make the substitution as: y = vx Differentiating both sides with respect to x, we get: Substituting the values of v and in equation (1), we get: Integrating both sides, …

## NCERT solution class 12 chapter 3 Differential Equations exercise 3.4 mathematics part 2

EXERCISE 3.4 Page No 395: Question 1: Answer: The given differential equation is: Now, integrating both sides of this equation, we get: This is the required general solution of the given differential equation. Question 2: Answer: The given differential equation is: Now, integrating both sides of this equation, we get: This is the required general …

## NCERT solution class 12 chapter 3 Differential Equations exercise 3.3 mathematics part 2

EXERCISE 3.3 Page No 391: Question 1: Answer: Differentiating both sides of the given equation with respect to x, we get: Again, differentiating both sides with respect to x, we get: Hence, the required differential equation of the given curve is Question 2: Answer: Differentiating both sides with respect to x, we get: Again, differentiating both sides with …

## NCERT solution class 12 chapter 3 Differential Equations exercise 3.2 mathematics part 2

EXERCISE 3.2 Page No 385: Question 1: Answer: Differentiating both sides of this equation with respect to x, we get: Now, differentiating equation (1) with respect to x, we get: Substituting the values ofin the given differential equation, we get the L.H.S. as: Thus, the given function is the solution of the corresponding differential equation. Question 2: …

## NCERT solution class 12 chapter 3 Differential Equations exercise 3.1 mathematics part 2

EXERCISE 3.1 Page No 382: Question 1: Determine order and degree(if defined) of differential equation  Answer: The highest order derivative present in the differential equation is. Therefore, its order is four. The given differential equation is not a polynomial equation in its derivatives. Hence, its degree is not defined. Question 2: Determine order and degree(if …

## NCERT solution class 12 chapter 2 Application of Integrals exercise 2.3 mathematics part 2

EXERCISE 2.3 Page No 375: Question 1: Find the area under the given curves and given lines: (i) y = x2, x = 1, x = 2 and x-axis (ii) y = x4, x = 1, x = 5 and x –axis Answer: The required area is represented by the shaded area ADCBA as The required area is represented by the shaded area ADCBA as Question 2: Find the area between the …

## NCERT solution class 12 chapter 2 Application of Integrals exercise 2.2 mathematics part 2

EXERCISE 2.2 Page No 371: Question 1: Find the area of the circle 4×2 + 4y2 = 9 which is interior to the parabola x2 = 4y Answer: The required area is represented by the shaded area OBCDO. Solving the given equation of circle, 4×2 + 4y2 = 9, and parabola, x2 = 4y, we obtain the point of intersection as. It can …

## NCERT solution class 12 chapter 2 Application of Integrals exercise 2.1 mathematics part 2

EXERCISE 2.1 Page No 365: Question 1: Find the area of the region bounded by the curve y2 = x and the lines x = 1, x = 4 and the x-axis. Answer: The area of the region bounded by the curve, y2 = x, the lines, x = 1 and x = 4, and the x-axis is the area ABCD. Question 2: Find the area of the region bounded by y2 = 9x, x = …

## NCERT solution class 12 chapter 1 Integrals exercise 1.12 mathematics part 2

EXERCISE 1.12 Page No 352: Question 1: Answer: Equating the coefficients of x2, x, and constant term, we obtain −A + B − C = 0 B + C = 0 A = 1 On solving these equations, we obtain From equation (1), we obtain Question 2: Answer: Question 3:  [Hint: Put] Answer: Question 4: Answer: Question 5:   Answer: On dividing, we obtain Question 6: …

## NCERT solution class 12 chapter 1 Integrals exercise 1.11 mathematics part 2

EXERCISE 1.11 Page No 347: Question 1: Answer: Adding (1) and (2), we obtain Question 2: Answer: Adding (1) and (2), we obtain Question 3: Answer: Adding (1) and (2), we obtain Question 4: Answer: Adding (1) and (2), we obtain Question 5: Answer: It can be seen that (x + 2) ≤ 0 on [−5, …

## NCERT solution class 12 chapter 1 Integrals exercise 1.10 mathematics part 2

EXERCISE 1.10 Page No 340: Question 1: Answer: When x = 0, t = 1 and when x = 1, t = 2 Question 2: Answer: Also, let  Question 3: Answer: Also, let x = tanθ ⇒ dx = sec2θ dθ When x = 0, θ = 0 and when x = 1,  Takingθas first function and sec2θ as second function and integrating by parts, we obtain Question 4: Answer: Let x + 2 = t2 ⇒ dx = 2tdt When x = 0,  and …

## NCERT solution class 12 chapter 1 Integrals exercise 1.9 mathematics part 2

EXERCISE 1.9 Page No 338: Question 1: Answer: By second fundamental theorem of calculus, we obtain Question 2: Answer: By second fundamental theorem of calculus, we obtain Question 3: Answer: By second fundamental theorem of calculus, we obtain Question 4: Answer: By second fundamental theorem of calculus, we obtain Question 5: Answer: By second fundamental …

## NCERT solution class 12 chapter 1 Integrals exercise 1.8 mathematics part 2

EXERCISE 1.8 Page No 334: Question 1: Answer: It is known that, Question 2: Answer: It is known that, Question 3: Answer: It is known that, Question 4: Answer: It is known that, From equations (2) and (3), we obtain Question 5: Answer: It is known that, Question 6: Answer: It is known that,

## NCERT solution class 12 chapter 1 Integrals exercise 1.6 mathematics part 2

EXERCISE 1.6 Page No 327: Question 1: x sin x Answer: Let I =  Taking x as first function and sin x as second function and integrating by parts, we obtain Question 2: Answer: Let I =  Taking x as first function and sin 3x as second function and integrating by parts, we obtain Question 3: Answer: Let  Taking x2 as first function and ex as second function and integrating by parts, …

## NCERT solution class 12 chapter 1 Integrals exercise 1.5 mathematics part 2

EXERCISE 1.5 Page No 322: Question 1: Answer: Let  Equating the coefficients of x and constant term, we obtain A + B = 1 2A + B = 0 On solving, we obtain A = −1 and B = 2 Question 2: Answer: Let  Equating the coefficients of x and constant term, we obtain A + B = 0 −3A + 3B = 1 On solving, we obtain Question 3: Answer: Let  Substituting x = …

## NCERT solution class 12 chapter 1 Integrals exercise 1.4 mathematics part 2

EXERCISE 1.4 Page No 315: Question 1: Answer: Let x3 = t ∴ 3×2 dx = dt Question 2: Answer: Let 2x = t ∴ 2dx = dt Question 3: Answer: Let 2 − x = t ⇒ −dx = dt Question 4: Answer: Let 5x = t ∴ 5dx = dt Question 5: Answer: Question 6: Answer: Let x3 = t ∴ 3×2 dx = dt Question 7: Answer: From (1), we obtain Question 8: Answer: Let x3 = t ⇒ 3×2 dx = dt …

## NCERT solution class 12 chapter 1 Integrals exercise 1.3 mathematics part 2

EXERCISE 1.3 Page No 307: Question 1: Answer: Question 2: Answer: It is known that,  Question 3: cos 2x cos 4x cos 6x Answer: It is known that, Question 4: sin3 (2x + 1) Answer: Let  Question 5: sin3 x cos3 x Answer: Question 6: sin x sin 2x sin 3x Answer: It is known that,  Question 7: sin 4x sin 8x Answer: It is known …

## NCERT solution class 12 chapter 1 Integrals exercise 1.2 mathematics part 2

EXERCISE 1.2 Page No 304: Question 1: Answer: Let = t ∴2x dx = dt Question 2: Answer: Let log |x| = t ∴  Question 3: Answer: Let 1 + log x = t ∴  Question 4: sin x ⋅ sin (cos x) Answer: sin x ⋅ sin (cos x) Let cos x = t ∴ −sin x dx = dt Question 5: Answer: Let  ∴ 2adx = dt Question 6: Answer: Let ax …

## NCERT solution class 12 chapter 1 Integrals exercise 1.1 mathematics part 2

EXERCISE 1.1 Page No 299: Question 1: sin 2x Answer: The anti derivative of sin 2x is a function of x whose derivative is sin 2x. It is known that, Therefore, the anti derivative of Question 2: Cos 3x Answer: The anti derivative of cos 3x is a function of x whose derivative is cos 3x. It is known that, Therefore, …

## NCERT solution class 12 chapter 6 Application of Derivatives exercise 6.6 mathematics part 1

EXERCISE 6.6 Page No 242: Question 1: Using differentials, find the approximate value of each of the following. (a)  (b)  Answer: (a) Consider Then, Now, dy is approximately equal to Δy and is given by, Hence, the approximate value of = 0.667 + 0.010 = 0.677. (b) Consider. Let x = 32 and Δx = 1. Then, Now, dy is approximately equal to Δy and is …

## NCERT solution class 12 chapter 6 Application of Derivatives exercise 6.5 mathematics part 1

EXERCISE 6.5 Page No 231: Question 1: Find the maximum and minimum values, if any, of the following functions given by (i) f(x) = (2x − 1)2 + 3        (ii) f(x) = 9×2 + 12x + 2 (iii) f(x) = −(x − 1)2 + 10    (iv) g(x) = x3 + 1 Answer: (i) The given function is f(x) = (2x − 1)2 + 3. It can be observed that (2x − …

## NCERT solution class 12 chapter 6 Application of Derivatives exercise 6.4 mathematics part 1

EXERCISE 6.4 Page No 216: Question 1: 1. Using differentials, find the approximate value of each of the following up to 3 places of decimal (i)  (ii)  (iii)  (iv)  (v)  (vi)  (vii)  (viii)  (ix)  (x)  (xi)  (xii)  (xiii)  (xiv)  (xv)  Answer: (i)  Consider. Let x = 25 and Δx = 0.3. Then, Now, dy is approximately equal to Δy and is given by, Hence, the approximate value ofis 0.03 + 5 = …

## NCERT solution class 12 chapter 6 Application of Derivatives exercise 6.3 mathematics part 1

EXERCISE 6.3 Page No 211: Question 1: Find the slope of the tangent to the curve y = 3×4 − 4x at x = 4. Answer: The given curve is y = 3×4 − 4x. Then, the slope of the tangent to the given curve at x = 4 is given by, Question 2: Find the slope of the tangent to the curve, x ≠ 2 at x = 10. …

## NCERT solution class 12 chapter 6 Application of Derivatives exercise 6.2 mathematics part 1

EXERCISE 6.2 Page No 205: Question 1: Show that the function given by f(x) = 3x + 17 is strictly increasing on R. Answer: Letbe any two numbers in R. Then, we have: Hence, f is strictly increasing on R. Question 2: Show that the function given by f(x) = e2x is strictly increasing on R. Answer: Letbe any two numbers in R. Then, we have: Hence, f is …

## NCERT solution class 12 chapter 6 Application of Derivatives exercise 6.1 mathematics part 1

EXERCISE 6.1 Page No 197: Question 1: Find the rate of change of the area of a circle with respect to its radius r when (a) r = 3 cm (b) r = 4 cm Answer: The area of a circle (A)with radius (r) is given by, Now, the rate of change of the area with respect to its radius is …

## NCERT solution class 12 chapter 5 Continuity and Differentiability exercise 5.9 mathematics part 1

EXERCISE 5.9 Page No 191: Question 1: Answer: Using chain rule, we obtain Question 2: Answer: Question 3: Answer: Taking logarithm on both the sides, we obtain Differentiating both sides with respect to x, we obtain Question 4: Answer: Using chain rule, we obtain Question 5: Answer: Question 6: Answer: Therefore, equation (1) becomes Question 7: …

## NCERT solution class 12 chapter 5 Continuity and Differentiability exercise 5.8 mathematics part 1

EXERCISE 5.8 Page No 186: Question 1: Verify Rolle’s Theorem for the function Answer: The given function,, being a polynomial function, is continuous in [−4, 2] and is differentiable in (−4, 2). ∴ f (−4) = f (2) = 0 ⇒ The value of f (x) at −4 and 2 coincides. Rolle’s Theorem states that there is a point c ∈ (−4, 2) …

## NCERT solution class 12 chapter 5 Continuity and Differentiability exercise 5.7 mathematics part 1

EXERCISE 5.7 Page No 183: Question 1: Find the second order derivatives of the function. Answer: Let Then, Question 2: Find the second order derivatives of the function. Answer: Let Then, Question 3: Find the second order derivatives of the function. Answer: Let Then, Question 4: Find the second order derivatives of the function. Answer: …

## NCERT solution class 12 chapter 5 Continuity and Differentiability exercise 5.6 mathematics part 1

EXERCISE 5.6 Page No 181: Question 1: If x and y are connected parametrically by the equation, without eliminating the parameter, find. Answer: The given equations are Question 2: If x and y are connected parametrically by the equation, without eliminating the parameter, find. x = a cos θ, y = b cos θ Answer: The given equations are x = a cos θ and y = b cos θ Question 3: If x and y are connected parametrically by the equation, without eliminating the parameter, find. …

## NCERT solution class 12 chapter 5 Continuity and Differentiability exercise 5.5 mathematics part 1

EXERCISE 5.5 Page No 178: Question 1: Differentiate the function with respect to x. Answer: Taking logarithm on both the sides, we obtain Differentiating both sides with respect to x, we obtain Question 2: Differentiate the function with respect to x. Answer: Taking logarithm on both the sides, we obtain Differentiating both sides with respect to x, we obtain …

## NCERT solution class 12 chapter 5 Continuity and Differentiability exercise 5.4 mathematics part 1

EXERCISE 5.4 Page No 174: Question 1: Differentiate the following w.r.t. x: Answer: Let By using the quotient rule, we obtain Question 2: Differentiate the following w.r.t. x: Answer: Let By using the chain rule, we obtain Question 3: Differentiate the following w.r.t. x: Answer: Let  By using the chain rule, we obtain Question 4: Differentiate the following …

## NCERT solution class 12 chapter 5 Continuity and Differentiability exercise 5.3 mathematics part 1

EXERCISE 5.3 Page No 169: Question 1: Find : Answer: The given relationship is Differentiating this relationship with respect to x, we obtain Question 2: Find  : Answer: The given relationship is Differentiating this relationship with respect to x, we obtain Question 3: Find  : Answer: The given relationship is Differentiating this relationship with respect to x, we …

## NCERT solution class 12 chapter 5 Continuity and Differentiability exercise 5.2 mathematics part 1

EXERCISE 5.2 Page No 166: Question 1: Differentiate the functions with respect to x. Answer: Let f(x)=sinx2+5, ux=x2+5, and v(t)=sint   Then, vou=vux=vx2+5=tanx2+5=f(x) Thus, f is a composite of two functions. Alternate method Question 2: Differentiate the functions with respect to x. Answer: Thus, f is a composite function of two functions. Put t = u (x) = sin x By chain rule, Alternate method Question 3: Differentiate the functions with …

## NCERT solution class 12 chapter 5 Continuity and Differentiability exercise 5.1 mathematics part 1

EXERCISE 5.1 Page No 159: Question 1: Prove that the functionis continuous at Answer: Therefore, f is continuous at x = 0 Therefore, f is continuous at x = −3 Therefore, f is continuous at x = 5 Question 2: Examine the continuity of the function. Answer: Thus, f is continuous at x = 3 Question 3: Examine the following functions for continuity. (a)  (b) (c)  (d)  Answer: (a) The given function …

## NCERT solution class 12 chapter 4 Determinants exercise 4.7 mathematics part 1

EXERCISE 4.7 Page No 141: Question 1: Prove that the determinant is independent of θ. Answer: Hence, Î” is independent of Î¸. Question 2: Without expanding the determinant, prove that Answer: Hence, the given result is proved. Question 3: Evaluate  Answer: Expanding along C3, we have: Question 4: If a, b and c are real numbers, and, Show that either a + b + c = 0 or a = b = c. Answer: …

## NCERT solution class 12 chapter 4 Determinants exercise 4.6 mathematics part 1

EXERCISE 4.6 Page No 136: Question 1: Examine the consistency of the system of equations. x + 2y = 2 2x + 3y = 3 Answer: The given system of equations is: x + 2y = 2 2x + 3y = 3 The given system of equations can be written in the form of AX = B, where ∴ A is non-singular. Therefore, A−1 exists. Hence, the given system of …

## NCERT solution class 12 chapter 4 Determinants exercise 4.5 mathematics part 1

EXERCISE 4.5 Page No 131: Question 1: Find adjoint of each of the matrices. Answer: Question 2: Find adjoint of each of the matrices. Answer: Question 3: Verify A (adj A) = (adj A) A = I . Answer: Question 4: Verify A (adj A) = (adj A) A = I . Answer: Page No 132: Question 5: Find the inverse of each of the matrices (if …

## NCERT solution class 12 chapter 4 Determinants exercise 4.4 mathematics part 1

EXERCISE 4.4 Page No 126: Question 1: Write Minors and Cofactors of the elements of following determinants: (i)  (ii)  Answer: (i) The given determinant is. Minor of element aij is Mij. ∴M11 = minor of element a11 = 3 M12 = minor of element a12 = 0 M21 = minor of element a21 = −4 M22 = minor of element a22 = 2 Cofactor of aij is Aij = (−1)i + j Mij. ∴A11 = …

## NCERT solution class 12 chapter 4 Determinants exercise 4.3 mathematics part 1

EXERCISE 4.3 Page No 122: Question 1: Find area of the triangle with vertices at the point given in each of the following: (i) (1, 0), (6, 0), (4, 3) (ii) (2, 7), (1, 1), (10, 8) (iii) (−2, −3), (3, 2), (−1, −8) Answer: (i) The area of the triangle with vertices (1, 0), …

## NCERT solution class 12 chapter 4 Determinants exercise 4.2 mathematics part 1

EXERCISE 4.2 Page No 119: Question 1: Using the property of determinants and without expanding, prove that: Answer: Question 2: Using the property of determinants and without expanding, prove that: Answer: Here, the two rows R1 and R3 are identical. Δ = 0. Question 3: Using the property of determinants and without expanding, prove that: Answer: Question …

## NCERT solution class 12 chapter 4 Determinants exercise 4.1 mathematics part 1

EXERCISE 4.1 Page No 108: Question 1: Evaluate the determinants in Exercises 1 and 2. Answer:  = 2(−1) − 4(−5) = − 2 + 20 = 18 Question 2: Evaluate the determinants in Exercises 1 and 2. (i)  (ii)  Answer: (i)  = (cos θ)(cos θ) − (−sin θ)(sin θ) = cos2 θ+ sin2 θ = 1 (ii)  = (x2 − x + 1)(x + 1) − (x − 1)(x + …

## NCERT solution class 12 chapter 3 Matrices exercise 3.5 mathematics part 1

EXERCISE 3.5 Page No 100: Question 1: Let, show that, where I is the identity matrix of order 2 and n ∈ N Answer: It is given that  We shall prove the result by using the principle of mathematical induction. For n = 1, we have: Therefore, the result is true for n = 1. Let the result be true for n = k. That is, Now, …

## NCERT solution class 12 chapter 3 Matrices exercise 3.4 mathematics part 1

EXERCISE 3.4 Page No 97: Question 1: Find the inverse of each of the matrices, if it exists. Answer: We know that A = IA Question 2: Find the inverse of each of the matrices, if it exists. Answer: We know that A = IA Question 3: Find the inverse of each of the matrices, if it exists. Answer: We know …

## NCERT solution class 12 chapter 3 Matrices exercise 3.3 mathematics part 1

EXERCISE 3.3 Page No 88: Question 1: Find the transpose of each of the following matrices: (i)  (ii)  (iii)  Answer: (i)  (ii)  (iii)  Question 2: If and, then verify that (i)  (ii)  Answer: We have: (i) (ii) Question 3: If and, then verify that (i)  (ii)  Answer: (i) It is known that Therefore, we have: (ii) Question 4: …

## NCERT solution class 12 chapter 3 Matrices exercise 3.2 mathematics part 1

EXERCISE 3.2 Page No 80: Question 1: Let Find each of the following (i)  (ii)  (iii)  (iv)  (v)  Answer: (i) (ii) (iii) (iv) Matrix A has 2 columns. This number is equal to the number of rows in matrix B. Therefore, AB is defined as: (v) Matrix B has 2 columns. This number is equal to the number of rows in matrix A. Therefore, BA is defined as: Question 2: …

## NCERT solution class 12 chapter 3 Matrices exercise 3.1 mathematics part 1

EXERCISE 3.1 Page No 64: Question 1: In the matrix, write: (i) The order of the matrix (ii) The number of elements, (iii) Write the elements a13, a21, a33, a24, a23 Answer: (i) In the given matrix, the number of rows is 3 and the number of columns is 4. Therefore, the order of the matrix is 3 × 4. (ii) Since …

## NCERT solution class 12 chapter 2 Inverse Trigonometric Functions exercise 2.3 mathematics part 1

EXERCISE 2.3 Page No 51: Question 1: Find the value of  Answer: We know that cos−1 (cos x) = x if, which is the principal value branch of cos −1x. Here, Now, can be written as: Question 2: Find the value of  Answer: We know that tan−1 (tan x) = x if, which is the principal value branch of tan −1x. Here, Now, can be written …

## NCERT solution class 12 chapter 2 Inverse Trigonometric Functions exercise 2.2 mathematics part 1

EXERCISE 2.2 Page No 47: Question 1: Prove  Answer: To prove:  Let x = sinθ. Then,  We have, R.H.S. = = 3θ = L.H.S. Question 2: Prove  Answer: To prove: Let x = cosθ. Then, cos−1 x =θ. We have, Question 3: Prove  Answer: To prove: Question 4: Prove  Answer: To prove:  Question 5: Write the function in the simplest form: …

## NCERT solution class 12 chapter 2 Inverse Trigonometric Functions exercise 2.1 mathematics part 1

EXERCISE 2.1 Page No 41: Question 1: Find the principal value of  Answer: Let sin-1  Then sin y =  We know that the range of the principal value branch of sin−1 is  and sin Therefore, the principal value of  Question 2: Find the principal value of  Answer: We know that the range of the principal value branch of cos−1 is …

## NCERT solution class 12 chapter 1 Relations and Functions exercise 1.5 mathematics part 1

EXERCISE 1.5 Page No 29: Question 1: Let f: R → R be defined as f(x) = 10x + 7. Find the function g: R → R such that g o f = f o g = 1R. Answer: It is given that f: R → R is defined as f(x) = 10x + 7. One-one: Let f(x) = f(y), where x, y ∈R. ⇒ 10x + 7 = 10y + 7 ⇒ x = y ∴ f is a one-one function. Onto: For y ∈ R, let y = 10x + 7. Therefore, for any y ∈ R, there exists such that  ∴ f is …

## NCERT solution class 12 chapter 1 Relations and Functions exercise 1.4 mathematics part 1

EXERCISE 1.4 Page No 24: Question 1: Determine whether or not each of the definition of given below gives a binary operation. In the event that * is not a binary operation, give justification for this. (i) On Z+, define * by a * b = a − b (ii) On Z+, define * by a * b = ab (iii) On R, define * by a * b = ab2 (iv) On Z+, define * …

## NCERT solution class 12 chapter 1 Relations and Functions exercise 1.3 mathematics part 1

EXERCISE 1.3 Page No 18: Question 1: Let f: {1, 3, 4} → {1, 2, 5} and g: {1, 2, 5} → {1, 3} be given by f = {(1, 2), (3, 5), (4, 1)} and g = {(1, 3), (2, 3), (5, 1)}. Write down gof. Answer: The functions f: {1, 3, 4} → {1, 2, 5} and g: {1, 2, 5} → …

## NCERT solution class 12 chapter 1 Relations and Functions exercise 1.2 mathematics part 1

EXERCISE 1.2 Page No 10: Question 1: Show that the function f: R* → R* defined byis one-one and onto, where R* is the set of all non-zero real numbers. Is the result true, if the domain R* is replaced by N with co-domain being same as R*? Answer: It is given that f: R* → R* is defined by One-one: ∴f is one-one. Onto: It is clear that for y∈ R*, there existssuch that …

## NCERT solution class 12 chapter 1 Relations and Functions exercise 1.1 mathematics part 1

EXERCISE 1.1 Page No 5: Question 1: Determine whether each of the following relations are reflexive, symmetric and transitive: (i)Relation R in the set A = {1, 2, 3…13, 14} defined as R = {(x, y): 3x − y = 0} (ii) Relation R in the set N of natural numbers defined as R = {(x, y): y = x + 5 and x < 4} (iii) Relation R in …

## NCERT solution class 11 chapter 16 Probability exercise 16.4 mathematics

EXERCISE 16.4 Page No 408: Question 1: A box contains 10 red marbles, 20 blue marbles and 30 green marbles. 5 marbles are drawn from the box, what is the probability that (i) all will be blue? (ii) atleast one will be green? Answer: Total number of marbles = 10 + 20 + 30 = …

## NCERT solution class 11 chapter 16 Probability exercise 16.3 mathematics

EXERCISE 16.3 Page No 403: Question 1: Which of the following can not be valid assignment of probabilities for outcomes of sample space S =  Assignment ω1 ω2 ω3 ω4 ω5 ω6 ω7 (a) 0.1 0.01 0.05 0.03 0.01 0.2 0.6 (b) (c) 0.1 0.2 0.3 0.4 0.5 0.6 0.7 (d) –0.1 0.2 0.3 0.4 …

## NCERT solution class 11 chapter 16 Probability exercise 16.2 mathematics

EXERCISE 16.2 Page No 393: Question 1: A die is rolled. Let E be the event “die shows 4” and F be the event “die shows even number”. Are E and F mutually exclusive? Answer: When a die is rolled, the sample space is given by S = {1, 2, 3, 4, 5, 6} Accordingly, …

## NCERT solution class 11 chapter 16 Probability exercise 16.1 mathematics

EXERCISE 16.1 Page No 386: Question 1: Describe the sample space for the indicated experiment: A coin is tossed three times. Answer: A coin has two faces: head (H) and tail (T). When a coin is tossed three times, the total number of possible outcomes is 23 = 8 Thus, when a coin is tossed three …

## NCERT solution class 11 chapter 15 Statistics exercise 15.4 mathematics

EXERCISE 15.4 Page No 380: Question 1: The mean and variance of eight observations are 9 and 9.25, respectively. If six of the observations are 6, 7, 10, 12, 12 and 13, find the remaining two observations. Answer: Let the remaining two observations be x and y. Therefore, the observations are 6, 7, 10, 12, 12, 13, x, y. From …

## NCERT solution class 11 chapter 15 Statistics exercise 15.3 mathematics

EXERCISE 15.3 Page No 375: Question 1: From the data given below state which group is more variable, A or B? Marks 10-20 20-30 30-40 40-50 50-60 60-70 70-80 Group A 9 17 32 33 40 10 9 Group B 10 20 30 25 43 15 7 Answer: Firstly, the standard deviation of group A …

## NCERT solution class 11 chapter 15 Statistics exercise 15.2 mathematics

EXERCISE 15.2 Page No 371: Question 1: Find the mean and variance for the data 6, 7, 10, 12, 13, 4, 8, 12 Answer: 6, 7, 10, 12, 13, 4, 8, 12 Mean,  The following table is obtained. xi 6 –3 9 7 –2 4 10 –1 1 12 3 9 13 4 16 4 …

## NCERT solution class 11 chapter 15 Statistics exercise 15.1 mathematics

EXERCISE 15.1 Page No 360: Question 1: Find the mean deviation about the mean for the data 4, 7, 8, 9, 10, 12, 13, 17 Answer: The given data is 4, 7, 8, 9, 10, 12, 13, 17 Mean of the data,  The deviations of the respective observations from the mean are –6, – 3, –2, …

## NCERT solution class 11 chapter 14 Mathematical Reasoning exercise 14.6 mathematics

EXERCISE 14.6 Page No 345: Question 1: Write the negation of the following statements: (i) p: For every positive real number x, the number x – 1 is also positive. (ii) q: All cats scratch. (iii) r: For every real number x, either x > 1 or x < 1. (iv) s: There exists a number x such that 0 < x < 1. Answer: (i) The negation of statement p is as …

## NCERT solution class 11 chapter 14 Mathematical Reasoning exercise 14.5 mathematics

EXERCISE 14.5 Page No 342: Question 1: Show that the statement p: “If x is a real number such that x3 + 4x = 0, then x is 0” is true by (i) direct method (ii) method of contradiction (iii) method of contrapositive Answer: p: “If x is a real number such that x3 + 4x = 0, then x is 0”. Let q: x is a real number such that x3 + 4x = …

## NCERT solution class 11 chapter 14 Mathematical Reasoning exercise 14.4 mathematics

EXERCISE 14.4 Page No 338: Question 1: Rewrite the following statement with “if-then” in five different ways conveying the same meaning. If a natural number is odd, then its square is also odd. Answer: The given statement can be written in five different ways as follows. (i) A natural number is odd implies that its …

## NCERT solution class 11 chapter 14 Mathematical Reasoning exercise 14.3 mathematics

EXERCISE 14.3 Page No 334: Question 1: For each of the following compound statements first identify the connecting words and then break it into component statements. (i) All rational numbers are real and all real numbers are not complex. (ii) Square of an integer is positive or negative. (iii) The sand heats up quickly in …

## NCERT solution class 11 chapter 14 Mathematical Reasoning exercise 14.2 mathematics

EXERCISE 14.2 Page No 329: Question 1: Write the negation of the following statements: (i) Chennai is the capital of Tamil Nadu. (ii) is not a complex number. (iii) All triangles are not equilateral triangle. (iv) The number 2 is greater than 7. (v) Every natural number is an integer. Answer: (i) Chennai is not the …

## NCERT solution class 11 chapter 14 Mathematical Reasoning exercise 14.1 mathematics

EXERCISE 14.1 Page No 324: Question 1: Which of the following sentences are statements? Give reasons for your answer. (i) There are 35 days in a month. (ii) Mathematics is difficult. (iii) The sum of 5 and 7 is greater than 10. (iv) The square of a number is an even number. (v) The sides …

## NCERT solution class 11 chapter 13 Limits and Derivatives exercise 13.3 mathematics

EXERCISE 13.3 Page No 317: Question 1: Find the derivative of the following functions from first principle: (i) –x (ii) (–x)–1 (iii) sin (x + 1) (iv)  Answer: (i) Let f(x) = –x. Accordingly, By first principle, (ii) Let. Accordingly, By first principle, (iii) Let f(x) = sin (x + 1). Accordingly, By first principle, (iv) Let. Accordingly, By first principle, …

## NCERT solution class 11 chapter 13 Limits and Derivatives exercise 13.2 mathematics

EXERCISE 13.2 Page No 312: Question 1: Find the derivative of x2 – 2 at x = 10. Answer: Let f(x) = x2 – 2. Accordingly, Thus, the derivative of x2 – 2 at x = 10 is 20. Question 2: Find the derivative of 99x at x = 100. Answer: Let f(x) = 99x. Accordingly, Thus, the derivative of 99x at x = 100 is 99. Question 3: Find the derivative of x at x = …

## NCERT solution class 11 chapter 13 Limits and Derivatives exercise 13.1 mathematics

EXERCISE 13.1 Page No 301: Question 1: Evaluate the Given limit: Answer: Question 2: Evaluate the Given limit: Answer: Question 3: Evaluate the Given limit: Answer: Question 4: Evaluate the Given limit: Answer: Question 5: Evaluate the Given limit: Answer: Question 6: Evaluate the Given limit: Answer: Put x + 1 = y so that y → 1 as x → 0. Question …

## NCERT solution class 11 chapter 12 Introduction to Three Dimensional Geometry exercise 12.4 mathematics

EXERCISE 12.4 Page No 278: Question 1: Three vertices of a parallelogram ABCD are A (3, –1, 2), B (1, 2, –4) andC (–1, 1, 2). Find the coordinates of the fourth vertex. Answer: The three vertices of a parallelogram ABCD are given as A (3, –1, 2), B (1, 2, –4), and C (–1, …

## NCERT solution class 11 chapter 12 Introduction to Three Dimensional Geometry exercise 12.3 mathematics

EXERCISE 12.3 Page No 277: Question 1: Find the coordinates of the point which divides the line segment joining the points (–2, 3, 5) and (1, –4, 6) in the ratio (i) 2:3 internally, (ii) 2:3 externally. Answer: (i) The coordinates of point R that divides the line segment joining points P (x1, y1, z1) and Q …

## NCERT solution class 11 chapter 12 Introduction to Three Dimensional Geometry exercise 12.2 mathematics

EXERCISE 12.2 Page No 273: Question 1: Find the distance between the following pairs of points: (i) (2, 3, 5) and (4, 3, 1) (ii) (–3, 7, 2) and (2, 4, –1) (iii) (–1, 3, –4) and (1, –3, 4) (iv) (2, –1, 3) and (–2, 1, 3) Answer: The distance between points P(x1, y1, z1) and …

## NCERT solution class 11 chapter 12 Introduction to Three Dimensional Geometry exercise 12.1 mathematics

EXERCISE 12.1 Page No 271: Question 1: A point is on the x-axis. What are its y-coordinates and z-coordinates? Answer: If a point is on the x-axis, then its y-coordinates and z-coordinates are zero. Question 2: A point is in the XZ-plane. What can you say about its y-coordinate? Answer: If a point is in the XZ plane, then its y-coordinate is zero. …

## NCERT solution class 11 chapter 11 Conic Sections exercise 11.5 mathematics

EXERCISE 11.5 Question 1: If a parabolic reflector is 20 cm in diameter and 5 cm deep, find the focus. Answer: The origin of the coordinate plane is taken at the vertex of the parabolic reflector in such a way that the axis of the reflector is along the positive x-axis. This can be diagrammatically represented …

## NCERT solution class 11 chapter 11 Conic Sections exercise 11.4 mathematics

EXERCISE 11.4 Page No 262: Question 1: Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola Answer: The given equation is. On comparing this equation with the standard equation of hyperbola i.e.,, we obtain a = 4 and b = 3. We know that a2 + b2 = c2. Therefore, The coordinates …

## NCERT solution class 11 chapter 11 Conic Sections exercise 11.3 mathematics

EXERCISE 11.3 Page No 255: Question 1: Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse  Answer: The given equation is. Here, the denominator of is greater than the denominator of. Therefore, the major axis is along …

## NCERT solution class 11 chapter 11 Conic Sections exercise 11.2 mathematics

EXERCISE 11.2 Page No 246: Question 1: Find the coordinates of the focus, axis of the parabola, the equation of directrix and the length of the latus rectum for y2 = 12x Answer: The given equation is y2 = 12x. Here, the coefficient of x is positive. Hence, the parabola opens towards the right. On comparing this equation with y2 = 4ax, we …

## NCERT solution class 11 chapter 11 Conic Sections exercise 11.1 mathematics

EXERCISE 11.1 Page No 241: Question 1: Find the equation of the circle with centre (0, 2) and radius 2 Answer: The equation of a circle with centre (h, k) and radius r is given as (x­ – h)2 + (y ­– k)2 = r2 It is given that centre (h, k) = (0, 2) and radius (r) = 2. Therefore, the equation of the circle …

## NCERT solution class 11 chapter 10 Straight Lines exercise 10.4 mathematics

EXERCISE 10.4 Page No 233: Question 1: Find the values of k for which the lineis (a) Parallel to the x-axis, (b) Parallel to the y-axis, (c) Passing through the origin. Answer: The given equation of line is (k – 3) x – (4 – k2) y + k2 – 7k + 6 = 0 … (1) (a) If the given line is parallel to the x-axis, then Slope …

## NCERT solution class 11 chapter 10 Straight Lines exercise 10.3 mathematics

EXERCISE 10.3 Page No 227: Question 1: Reduce the following equations into slope-intercept form and find their slopes and the y-intercepts. (i) x + 7y = 0 (ii) 6x + 3y – 5 = 0 (iii) y = 0 Answer: (i) The given equation is x + 7y = 0. It can be written as This equation is of the form y = mx + c, where. Therefore, equation (1) is …

## NCERT solution class 11 chapter 10 Straight Lines exercise 10.2 mathematics

EXERCISE 10.2 Page No 219: Question 1: Write the equations for the x and y-axes. Answer: 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 x = 0. Question 2: Find the equation of the line which passes through the …

## NCERT solution class 11 chapter 10 Straight Lines exercise 10.1 mathematics

EXERCISE 10.1 Page No 211: Question 1: Draw a quadrilateral in the Cartesian plane, whose vertices are (–4, 5), (0, 7), (5, –5) and (–4, –2). Also, find its area. Answer: Let ABCD be the given quadrilateral with vertices A (–4, 5), B (0, 7), C (5, –5), and D (–4, –2). Then, by plotting …

## NCERT solution class 11 chapter 9 Sequences and Series exercise 9.5 mathematics

EXERCISE 9.5 Page No 199: Question 1: Show that the sum of (m + n)th and (m – n)th terms of an A.P. is equal to twice the mth term. Answer: Let a and d be the first term and the common difference of the A.P. respectively. It is known that the kth term of an A. P. is given by ak = a + (k –1) d ∴ am + n = a + (m + n –1) d am – n = a + (m – n –1) d am = a + (m –1) d …

## NCERT solution class 11 chapter 9 Sequences and Series exercise 9.4 mathematics

EXERCISE 9.4 Page No 196: Question 1: Find the sum to n terms of the series 1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 + … Answer: The given series is 1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 + … nth term, an = n ( n + 1) …

## NCERT solution class 11 chapter 9 Sequences and Series exercise 9.3 mathematics

EXERCISE 9.3 Page No 192: Question 1: Find the 20th and nthterms of the G.P. Answer: The given G.P. is  Here, a = First term =  r = Common ratio =  Question 2: Find the 12th term of a G.P. whose 8th term is 192 and the common ratio is 2. Answer: Common ratio, r = 2 Let a be the first term of the G.P. …

## NCERT solution class 11 chapter 9 Sequences and Series exercise 9.2 mathematics

EXERCISE 9.2 Page No 185: Question 1: Find the sum of odd integers from 1 to 2001. Answer: The odd integers from 1 to 2001 are 1, 3, 5, …1999, 2001. This sequence forms an A.P. Here, first term, a = 1 Common difference, d = 2 Thus, the sum of odd numbers from 1 to 2001 is 1002001. …

## NCERT solution class 11 chapter 9 Sequences and Series exercise 9.1 mathematics

EXERCISE 9.1 Page No 180: Question 1: Write the first five terms of the sequences whose nth term is  Answer: Substituting n = 1, 2, 3, 4, and 5, we obtain Therefore, the required terms are 3, 8, 15, 24, and 35. Question 2: Write the first five terms of the sequences whose nth term is  Answer: Substituting n = 1, …

## NCERT solution class 11 chapter 8 Binomial Theorem exercise 8.3 mathematics

EXERCISE 8.3 Page No 175: Question 1: Find a, b and n in the expansion of (a + b)n if the first three terms of the expansion are 729, 7290 and 30375, respectively. Answer: It is known that (r + 1)th term, (Tr+1), in the binomial expansion of (a + b)n is given by . The first three terms of the expansion are given as 729, 7290, and 30375 …

## NCERT solution class 11 chapter 8 Binomial Theorem exercise 8.2 mathematics

EXERCISE 8.2 Page No 171: Question 1: Find the coefficient of x5 in (x + 3)8 Answer: It is known that (r + 1)th term, (Tr+1), in the binomial expansion of (a + b)n is given by . Assuming that x5 occurs in the (r + 1)th term of the expansion (x + 3)8, we obtain Comparing the indices of x in x5 and in Tr +1, we obtain r = 3 Thus, the coefficient of x5 is …

## NCERT solution class 11 chapter 8 Binomial Theorem exercise 8.1 mathematics

EXERCISE 8.1 Page No 166: Question 1: Expand the expression (1– 2x)5 Answer: By using Binomial Theorem, the expression (1– 2x)5 can be expanded as Question 2: Expand the expression Answer: By using Binomial Theorem, the expression  can be expanded as Question 3: Expand the expression (2x – 3)6 Answer: By using Binomial Theorem, the expression (2x – 3)6 can …

## NCERT solution class 11 chapter 7 Permutations and Combinations exercise 7.5 mathematics

EXERCISE 7.5 Page No 156: Question 1: How many words, with or without meaning, each of 2 vowels and 3 consonants can be formed from the letters of the word DAUGHTER? Answer: In the word DAUGHTER, there are 3 vowels namely, A, U, and E, and 5 consonants namely, D, G, H, T, and R. …

## NCERT solution class 11 chapter 7 Permutations and Combinations exercise 7.4 mathematics

EXERCISE 7.4 Page No 153: Question 1: If, find. Answer: It is known that, Therefore, Question 2: Determine n if (i)  (ii)  Answer: (i) (ii) Question 3: How many chords can be drawn through 21 points on a circle? Answer: For drawing one chord on a circle, only 2 points are required. To know the number of chords …

## NCERT solution class 11 chapter 7 Permutations and Combinations exercise 7.3 mathematics

EXERCISE 7.3 Page No 148: Question 1: How many 3-digit numbers can be formed by using the digits 1 to 9 if no digit is repeated? Answer: 3-digit numbers have to be formed using the digits 1 to 9. Here, the order of the digits matters. Therefore, there will be as many 3-digit numbers as …

## NCERT solution class 11 chapter 7 Permutations and Combinations exercise 7.2 mathematics

EXERCISE 7.2 Page No 140: Question 1: Evaluate (i) 8! (ii) 4! – 3! Answer: (i) 8! = 1 × 2 × 3 × 4 × 5 × 6 × 7 × 8 = 40320 (ii) 4! = 1 × 2 × 3 × 4 = 24 3! = 1 × 2 × 3 = …

## NCERT solution class 11 chapter 7 Permutations and Combinations exercise 7.1 mathematics

EXERCISE 7.1 Page No 138: Question 1: How many 3-digit numbers can be formed from the digits 1, 2, 3, 4 and 5 assuming that (i) repetition of the digits is allowed? (ii) repetition of the digits is not allowed? Answer: (i) There will be as many ways as there are ways of filling 3 …

## NCERT solution class 11 chapter 6 Linear Inequalities exercise 6.4 mathematics

EXERCISE 6.4 Page No 132: Question 1: Solve the inequality 2 ≤ 3x – 4 ≤ 5 Answer: 2 ≤ 3x – 4 ≤ 5 ⇒ 2 + 4 ≤ 3x – 4 + 4 ≤ 5 + 4 ⇒ 6 ≤ 3x ≤ 9 ⇒ 2 ≤ x ≤ 3 Thus, all the real numbers, x, which are greater than or …

## NCERT solution class 11 chapter 6 Linear Inequalities exercise 6.3 mathematics

EXERCISE 6.3 Page No 129: Question 1: Solve the following system of inequalities graphically: x ≥ 3, y ≥ 2 Answer: x ≥ 3 … (1) y ≥ 2 … (2) The graph of the lines, x = 3 and y = 2, are drawn in the figure below. Inequality (1) represents the region on the right hand side of the line, x = 3 (including the …

## NCERT solution class 11 chapter 6 Linear Inequalities exercise 6.2 mathematics

EXERCISE 6.2 Page No 127: Question 1: Solve the given inequality graphically in two-dimensional plane: x + y < 5 Answer: The graphical representation of x + y = 5 is given as dotted line in the figure below. This line divides the xy-plane in two half planes, I and II. Select a point (not on the line), which lies in one of the half planes, to …

## NCERT solution class 11 chapter 6 Linear Inequalities exercise 6.1 mathematics

EXERCISE 6.1 Page No 122: Question 1: Solve 24x < 100, when (i) x is a natural number (ii) x is an integer Answer: The given inequality is 24x < 100. (i) It is evident that 1, 2, 3, and 4 are the only natural numbers less than. Thus, when x is a natural number, the solutions of the given inequality are 1, …

## NCERT solution class 11 chapter 5 Complex Numbers and Quadratic Equations exercise 5.4 mathematics

EXERCISE 5.4 Page No 112: Question 1: Evaluate:  Answer: Question 2: For any two complex numbers z1 and z2, prove that Re (z1z2) = Re z1 Re z2 – Im z1 Im z2 Answer: Question 3: Reduce to the standard form. Answer: Question 4: If x – iy =prove that. Answer: Question 5: Convert the following in the polar form: (i) , (ii)  Answer: (i) …

## NCERT solution class 11 chapter 5 Complex Numbers and Quadratic Equations exercise 5.3 mathematics

EXERCISE 5.3 Page No 109: Question 1: Solve the equation x2 + 3 = 0 Answer: The given quadratic equation is x2 + 3 = 0 On comparing the given equation with ax2 + bx + c = 0, we obtain a = 1, b = 0, and c = 3 Therefore, the discriminant of the given equation is D = b2 – 4ac = 02 – 4 × 1 × 3 = –12 …

## NCERT solution class 11 chapter 5 Complex Numbers and Quadratic Equations exercise 5.2 mathematics

EXERCISE 5.2 Page No 108: Question 1: Find the modulus and the argument of the complex number Answer: On squaring and adding, we obtain Since both the values of sin θ and cos θ are negative and sinθ and cosθ are negative in III quadrant, Thus, the modulus and argument of the complex number are 2 and respectively. Question 2: Find the modulus …

## NCERT solution class 11 chapter 5 Complex Numbers and Quadratic Equations exercise 5.1 mathematics

EXERCISE 5.1 Page No 103: Question 1: Express the given complex number in the form a + ib: Answer: Question 2: Express the given complex number in the form a + ib: i9 + i19 Answer: Question 3: Express the given complex number in the form a + ib: i–39 Answer: Page No 104: Question 4: Express the given complex number in the form a + ib: 3(7 + i7) + i(7 + i7) …

## NCERT solution class 11 chapter 4 Principle of Mathematical Induction exercise 4.1 mathematics

EXERCISE 4.1 Page No 94: Question 1: Prove the following by using the principle of mathematical induction for all n ∈ N:    Answer: Let the given statement be P(n), i.e., P(n): 1 + 3 + 32 + …+ 3n–1 = For n = 1, we have P(1): 1 =, which is true. Let P(k) be true for some positive integer k, i.e., …

## NCERT solution class 11 chapter 3 Trigonometric Functions exercise 3.5 mathematics

EXERCISE 3.5 Page No 81: Question 1: Prove that:  Answer: L.H.S. = 0 = R.H.S Question 2: Prove that: (sin 3x + sin x) sin x + (cos 3x – cos x) cos x = 0 Answer: L.H.S. = (sin 3x + sin x) sin x + (cos 3x – cos x) cos x = RH.S. Page No 82: Question 3: Prove that:  Answer: L.H.S. =  Question 4: Prove that:  …

## NCERT solution class 11 chapter 3 Trigonometric Functions exercise 3.4 mathematics

EXERCISE 3.4 Page No 78: Question 1: Find the principal and general solutions of the equation  Answer: Therefore, the principal solutions are x =and. Therefore, the general solution is Question 2: Find the principal and general solutions of the equation  Answer: Therefore, the principal solutions are x =and. Therefore, the general solution is, where n ∈ Z Question 3: Find the principal …

## NCERT solution class 11 chapter 3 Trigonometric Functions exercise 3.3 mathematics

EXERCISE 3.3 Page No 73: Question 1: Answer: L.H.S. =  Question 2: Prove that  Answer: L.H.S. =  Question 3: Prove that  Answer: L.H.S. = Question 4: Prove that  Answer: L.H.S = Question 5: Find the value of: (i) sin 75° (ii) tan 15° Answer: (i) sin 75° = sin (45° + 30°) = sin 45° …

## NCERT solution class 11 chapter 3 Trigonometric Functions exercise 3.2 mathematics

EXERCISE 3.2 Page No 63: Question 1: Find the values of other five trigonometric functions if , x lies in third quadrant. Answer: Since x lies in the 3rd quadrant, the value of sin x will be negative. Question 2: Find the values of other five trigonometric functions if , x lies in second quadrant. Answer: Since x lies in the 2nd quadrant, the value of cos x will be negative …

## NCERT solution class 11 chapter 3 Trigonometric Functions exercise 3.1 mathematics

EXERCISE 3.1 Page No 54: Question 1: Find the radian measures corresponding to the following degree measures: (i) 25° (ii) – 47° 30′ (iii) 240° (iv) 520° Answer: (i) 25° We know that 180° = π radian (ii) –47° 30′ –47° 30′ = degree [1° = 60′]  degree Since 180° = π radian (iii) 240° We …

## NCERT solution class 11 chapter 2 Sets exercise 2.4 mathematics

EXERCISE 2.4 Page No 46: Question 1: The relation f is defined by  The relation g is defined by  Show that f is a function and g is not a function. Answer: The relation f is defined as It is observed that for 0 ≤ x < 3, f(x) = x2 3 < x ≤ 10, f(x) = 3x Also, at x = 3, f(x) = 32 = 9 or f(x) = 3 × 3 = 9 …

## NCERT solution class 11 chapter 2 Sets exercise 2.3 mathematics

EXERCISE 2.3 Page No 44: Question 1: Which of the following relations are functions? Give reasons. If it is a function, determine its domain and range. (i) {(2, 1), (5, 1), (8, 1), (11, 1), (14, 1), (17, 1)} (ii) {(2, 1), (4, 2), (6, 3), (8, 4), (10, 5), (12, 6), (14, 7)} (iii) …

## NCERT solution class 11 chapter 2 Sets exercise 2.2 mathematics

EXERCISE 2.2 Page No 35: Question 1: Let A = {1, 2, 3, … , 14}. Define a relation R from A to A by R = {(x, y): 3x – y = 0, where x, y ∈ A}. Write down its domain, codomain and range. Answer: The relation R from A to A is given as R = {(x, y): 3x – y = 0, …

## NCERT solution class 11 chapter 2 Sets exercise 2.1 mathematics

EXERCISE 2.1 Page No 33: Question 1: If, find the values of x and y. Answer: It is given that. Since the ordered pairs are equal, the corresponding elements will also be equal. Therefore,  and.   ∴ x = 2 and y = 1 Question 2: If the set A has 3 elements and the set B = {3, 4, 5}, then …

## NCERT solution class 11 chapter 1 Sets exercise 1.7 mathematics

EXERCISE 1.7 Page No 26: Question 1: Decide, among the following sets, which sets are subsets of one and another: A = {x: x ∈ R and x satisfy x2 – 8x + 12 = 0}, B = {2, 4, 6}, C = {2, 4, 6, 8…}, D = {6}. Answer: A = {x: x ∈ R and x satisfies x2 – 8x + 12 = 0} 2 and …

## NCERT solution class 11 chapter 1 Sets exercise 1.6 mathematics

EXERCISE 1.6 Page No 24: Question 1: If X and Y are two sets such that n(X) = 17, n(Y) = 23 and n(X ∪ Y) = 38, find n(X ∩Y). Answer: It is given that: n(X) = 17, n(Y) = 23, n(X ∪ Y) = 38 n(X ∩ Y) = ? We know that: Question 2: If X and Y …

## NCERT solution class 11 chapter 1 Sets exercise 1.5 mathematics

EXERCISE 1.5 Page No 20: Question 1: Let U ={1, 2, 3; 4, 5, 6, 7, 8, 9}, A = {1, 2, 3, 4}, B = {2, 4, 6, 8} and C = {3, 4, 5, 6}. Find (i)  (ii)  (iii)  (iv)  (v)  (vi)  Answer: U ={1, 2, 3, 4, 5, 6, 7, 8, 9} …

## NCERT solution class 11 chapter 1 Sets exercise 1.4 mathematics

EXERCISE 1.4 Page No 17: Question 1: Find the union of each of the following pairs of sets: (i) X = {1, 3, 5} Y = {1, 2, 3} (ii) A = {a, e, i, o, u} B = {a, b, c} (iii) A = {x: x is a natural number and multiple of 3} B = {x: x is a natural number less than 6} (iv) A = …

## NCERT solution class 11 chapter 1 Sets exercise 1.3 mathematics

EXERCISE 1.3 Page No 12: Question 1: Make correct statements by filling in the symbols ⊂ or ⊄ in the blank spaces: (i) {2, 3, 4} … {1, 2, 3, 4, 5} (ii) {a, b, c} … {b, c, d} (iii) {x: x is a student of Class XI of your school} … {x: x student of your school} (iv) {x: x is a circle in the plane} … …

## NCERT solution class 11 chapter 1 Sets exercise 1.2 mathematics

EXERCISE 1.2 Page No 8: Question 1: Which of the following are examples of the null set (i) Set of odd natural numbers divisible by 2 (ii) Set of even prime numbers (iii) {x:x is a natural numbers, x < 5 and x > 7 } (iv) {y:y is a point common to any two parallel lines} Answer: (i) A set of …

## NCERT solution class 11 chapter 1 Sets exercise 1.1 mathematics

EXERCISE 1.1 Page No 4: Question 1: Which of the following are sets? Justify our answer. (i) The collection of all months of a year beginning with the letter J. (ii) The collection of ten most talented writers of India. (iii) A team of eleven best-cricket batsmen of the world. (iv) The collection of all boys in your class. …

## NCERT solution class 10 chapter 15 Statistics exercise 15.4 mathematics

EXERCISE 15.4 Page No 293: Question 1: The following distribution gives the daily income of 50 workers of a factory. Daily income (in Rs) 100 − 120 120 − 140 140 − 160 160 − 180 180 − 200 Number of workers 12 14 8 6 10 Convert the distribution above to a less than …

## NCERT solution class 10 chapter 15 Statistics exercise 15.3 mathematics

EXERCISE 15.3 Page No 287: Question 1: The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them. Monthly consumption (in units) Number of consumers 65 − 85 4 85 − 105 5 105 − 125 13 125 …

## NCERT solution class 10 chapter 15 Statistics exercise 15.2 mathematics

EXERCISE 15.2 Page No 275: Question 1: The following table shows the ages of the patients admitted in a hospital during a year: age (in years) 5 − 15 15 − 25 25 − 35 35 − 45 45 − 55 55 − 65 Number of patients 6 11 21 23 14 5 Find the …

## NCERT solution class 10 chapter 15 Statistics exercise 15.1 mathematics

EXERCISE 15.1 Page No 270: Question 1: A survey was conducted by a group of students as a part of their environment awareness programme, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house. Number of plants 0­ − …

## NCERT solution class 10 chapter 14 Probability exercise 14.2 mathematics

EXERCISE 14.2 Question 1: Two customers Shyam and Ekta are visiting a particular shop in the same week (Tuesday to Saturday). Each is equally likely to visit the shop on any day as on another day. What is the probability that both will visit the shop on (i) the same day? (ii) consecutive days? (iii) …

## NCERT solution class 10 chapter 14 Probability exercise 14.1 mathematics

EXERCISE 14.1 Page No 308: Question 1: Complete the following statements: (i) Probability of an event E + Probability of the event ‘not E’ = _______. (ii) The probability of an event that cannot happen is _________. Such as event is called _________. (iii) The probability of an event that is certain to happen is …

## NCERT solution class 10 chapter 13 Surface Areas and Volumes exercise 13.5 mathematics

EXERCISE 13.5 Page No 258: Question 1: A copper wire, 3 mm in diameter, is wound about a cylinder whose length is 12 cm, and diameter 10 cm, so as to cover the curved surface of the cylinder. Find the length and mass of the wire, assuming the density of copper to be 8.88 g …

## NCERT solution class 10 chapter 13 Surface Areas and Volumes exercise 13.4 mathematics

EXERCISE 13.4 Page No 257: Question 1: A drinking glass is in the shape of a frustum of a cone of height 14 cm. The diameters of its two circular ends are 4 cm and 2 cm. Find the capacity of the glass.  Answer: Radius (r1) of  Radius (r2) of  Capacity of glass = Volume …

## NCERT solution class 10 chapter 13 Surface Areas and Volumes exercise 13.3 mathematics

EXERCISE 13.3 Page No 251: Question 1: A metallic sphere of radius 4.2 cm is melted and recast into the shape of a cylinder of radius 6 cm. Find the height of the cylinder. Answer: Radius (r1) of hemisphere = 4.2 cm Radius (r2) of cylinder = 6 cm Let the height of the cylinder …

## NCERT solution class 10 chapter 13 Surface Areas and Volumes exercise 13.2 mathematics

EXERCISE 13.2 Page No 247: Question 1: A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid in terms of π. Answer: Given that, Height (h) …

## NCERT solution class 10 chapter 13 Surface Areas and Volumes exercise 13.1 mathematics

EXERCISE 13.1 Page No 244: Question 1: 2 cubes each of volume 64 cm3 are joined end to end. Find the surface area of the resulting cuboids. Answer: Given that, Volume of cubes = 64 cm3 (Edge) 3 = 64 Edge = 4 cm If cubes are joined end to end, the dimensions of the resulting cuboid will …

## NCERT solution class 10 chapter 12 Areas Related to Circles exercise 12.3 mathematics

EXERCISE 12.3 Page No 234: Question 1: Find the area of the shaded region in the given figure, if PQ = 24 cm, PR = 7 cm and O is the centre of the circle.  Answer: It can be observed that RQ is the diameter of the circle. Therefore, ∠RPQ will be 90º. By applying …

## NCERT solution class 10 chapter 12 Areas Related to Circles exercise 12.2 mathematics

EXERCISE 12.2 Page No 230: Question 1: Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60°.  Answer: Let OACB be a sector of the circle making 60° angle at centre O of the circle. Area of sector of angle θ = Area of sector …

## NCERT solution class 10 chapter 12 Areas Related to Circles exercise 12.1 mathematics

EXERCISE 12.1 Page No 225: Question 1: The radii of two circles are 19 cm and 9 cm respectively. Find the radius of the circle which has circumference equal to the sum of the circumferences of the two circles. Answer: Radius (r1) of 1st circle = 19 cm Radius (r2) or 2nd circle = 9 cm Let …

## NCERT solution class 10 chapter 11 Constructions exercise 11.2 mathematics

EXERCISE 11.2 Page No 221: Question 1: Draw a circle of radius 6 cm. From a point 10 cm away from its centre, construct the pair of tangents to the circle and measure their lengths. Give the justification of the construction. Answer: A pair of tangents to the given circle can be constructed as follows. …

## NCERT solution class 10 chapter 11 Constructions exercise 11.1 mathematics

EXERCISE 11.1 Page No 219: Question 1: Draw a line segment of length 7.6 cm and divide it in the ratio 5:8. Measure the two parts. Give the justification of the construction. Answer: A line segment of length 7.6 cm can be divided in the ratio of 5:8 as follows. Step 1 Draw line segment AB …

## NCERT solution class 10 chapter 10 Circles exercise 10.2 mathematics

EXERCISE 10.2 Page No 213: Question 1: From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is (A) 7 cm (B) 12 cm (C) 15 cm (D) 24.5 cm Answer: Let O be …

## NCERT solution class 10 chapter 10 Circles exercise 10.1 mathematics

EXERCISE 10.1 Page No 209: Question 1: How many tangents can a circle have? Answer: A circle can have infinite tangents. Question 2: Fill in the blanks: (i) A tangent to a circle intersects it in _______ point (s). (ii) A line intersecting a circle in two points is called a __________. (iii) A circle …

## NCERT solution class 10 chapter 9 Some Applications of Trigonometry exercise 9.1 mathematics

EXERCISE 9.1 Page No 203: Question 1: A circus artist is climbing a 20 m long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole, if the angle made by the rope with the ground level is 30 °. Answer: It …

## NCERT solution class 10 chapter 8 Introduction to Trigonometry exercise 8.4 mathematics

EXERCISE 8.4 Page No 193: Question 1: Express the trigonometric ratios sin A, sec A and tan A in terms of cot A. Answer: We know that,  will always be positive as we are adding two positive quantities. Therefore,  We know that,  However,  Therefore,  Also, Question 2: Write all the other trigonometric ratios of ∠A …

## NCERT solution class 10 chapter 8 Introduction to Trigonometry exercise 8.3 mathematics

EXERCISE 8.3 Page No 189: Question 1: Evaluate (I)  (II)  (III) cos 48° − sin 42° (IV)cosec 31° − sec 59° Answer: (I) (II) (III)cos 48° − sin 42° = cos (90°− 42°) − sin 42° = sin 42° − sin 42° = 0 (IV) cosec 31° − sec 59° = cosec (90° − 59°) …

## NCERT solution class 10 chapter 8 Introduction to Trigonometry exercise 8.2 mathematics

EXERCISE 8.2 Page No 187: Question 1: Evaluate the following (i) sin60° cos30° + sin30° cos 60° (ii) 2tan245° + cos230° − sin260° (iii)  (iv)  (v) Answer: (i) sin60° cos30° + sin30° cos 60° (ii) 2tan245° + cos230° − sin260° (iii)  (iv)  (v)  Question 2: Choose the correct option and justify your choice. (i)  (A). …

## NCERT solution class 10 chapter 8 Introduction to Trigonometry exercise 8.1 mathematics

EXERCISE 8.1 Page No 181: Question 1: In ΔABC right angled at B, AB = 24 cm, BC = 7 cm. Determine (i) sin A, cos A (ii) sin C, cos C Answer: Applying Pythagoras theorem for ΔABC, we obtain AC2 = AB2 + BC2 = (24 cm)2 + (7 cm)2 = (576 + 49) cm2 = 625 …

## NCERT solution class 10 chapter 7 Coordinate Geometry exercise 7.4 mathematics

EXERCISE 7.4 Page No 171: Question 1: Determine the ratio in which the line 2x + y − 4 = 0 divides the line segment joining the points A(2, − 2) and B(3, 7) Answer: Let the given line divide the line segment joining the points A(2, −2) and B(3, 7) in a ratio k : 1. Coordinates of the …

## NCERT solution class 10 chapter 7 Coordinate Geometry exercise 7.3 mathematics

EXERCISE 7.3 Page No 170: Question 1: Find the area of the triangle whose vertices are: (i) (2, 3), (− 1, 0), (2, − 4) (ii) (− 5, − 1), (3, − 5), (5, 2) Answer: (i) Area of a triangle is given by (ii)  Question 2: In each of the following find the value …

## NCERT solution class 10 chapter 7 Coordinate Geometry exercise 7.2 mathematics

EXERCISE 7.2 Page No 167: Question 1: Find the coordinates of the point which divides the join of (− 1, 7) and (4, − 3) in the ratio 2:3. Answer: Let P(x, y) be the required point. Using the section formula, we obtain Therefore, the point is (1, 3). Question 2: Find the coordinates of the …

## NCERT solution class 10 chapter 7 Coordinate Geometry exercise 7.1 mathematics

EXERCISE 7.1 Page No 161: Question 1: Find the distance between the following pairs of points: (i) (2, 3), (4, 1) (ii) (−5, 7), (−1, 3) (iii) (a, b), (− a, − b) Answer: (i) Distance between the two points is given by (ii) Distance between is given by (iii) Distance between is given by Question 2: Find the distance …

## NCERT solution class 10 chapter 6 Triangles exercise 6.6 mathematics

EXERCISE 6.6 Page No 152: Question 1: In the given figure, PS is the bisector of ∠QPR of ΔPQR. Prove that . Answer: Let us draw a line segment RT parallel to SP which intersects extended line segment QP at point T. Given that, PS is the angle bisector of ∠QPR. ∠QPS = ∠SPR … (1) …

## NCERT solution class 10 chapter 6 Triangles exercise 6.5 mathematics

EXERCISE 6.5 Page No 150: Question 1: Sides of triangles are given below. Determine which of them are right triangles? In case of a right triangle, write the length of its hypotenuse. (i) 7 cm, 24 cm, 25 cm (ii) 3 cm, 8 cm, 6 cm (iii) 50 cm, 80 cm, 100 cm (iv) 13 …

## NCERT solution class 10 chapter 6 Triangles exercise 6.4 mathematics

EXERCISE 6.4 Page No 143: Question 1: Let  and their areas be, respectively, 64 cm2 and 121 cm2. If EF = 15.4 cm, find BC. Answer: Video Solution Question 2: Diagonals of a trapezium ABCD with AB || DC intersect each other at the point O. If AB = 2CD, find the ratio of the areas of …

## NCERT solution class 10 chapter 6 Triangles exercise 6.3 mathematics

EXERCISE 6.3 Page No 138: Question 1: State which pairs of triangles in the following figure are similar? Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form: (i) (ii) (iii) (iv) (v) (vi) Answer: (i) ∠A = ∠P = 60° ∠B …

## NCERT solution class 10 chapter 6 Triangles exercise 6.2 mathematics

EXERCISE 6.2 Page No 128: Question 1: In figure.6.17. (i) and (ii), DE || BC. Find EC in (i) and AD in (ii). (i) (ii) Answer: (i) Let EC = x cm It is given that DE || BC. By using basic proportionality theorem, we obtain (ii) Let AD = x cm It is given that DE || BC. …

## NCERT solution class 10 chapter 6 Triangles exercise 6.1 mathematics

EXERCISE 6.1 Page No 122: Question 1: Fill in the blanks using correct word given in the brackets:− (i) All circles are __________. (congruent, similar) (ii) All squares are __________. (similar, congruent) (iii) All __________ triangles are similar. (isosceles, equilateral) (iv) Two polygons of the same number of sides are similar, if (a) their corresponding …

## NCERT solution class 10 chapter 5 Arithmetic Progressions exercise 5.4 mathematics

EXERCISE 5.4 Page No 115: Question 1: Which term of the A.P. 121, 117, 113 … is its first negative term? [Hint: Find n for an < 0] Answer: Given A.P. is 121, 117, 113 … a = 121 d = 117 − 121 = −4 an = a + (n − 1) d = 121 + (n − 1) (−4) = 121 − 4n + 4 = …

## NCERT solution class 10 chapter 5 Arithmetic Progressions exercise 5.3 mathematics

EXERCISE 5.3 Page No 112: Question 1: Find the sum of the following APs. (i) 2, 7, 12 ,…., to 10 terms. (ii) − 37, − 33, − 29 ,…, to 12 terms (iii) 0.6, 1.7, 2.8 ,…….., to 100 terms (iv)  ,………, to 11 terms Answer: (i)2, 7, 12 ,…, to 10 terms For this …

## NCERT solution class 10 chapter 5 Arithmetic Progressions exercise 5.2 mathematics

EXERCISE 5.2 Page No 105: Question 1: Fill in the blanks in the following table, given that a is the first term, d the common difference and an the nth term of the A.P. a d n an I 7 3 8 …… II − 18 ….. 10 0 III ….. − 3 18 − 5 IV − 18.9 2.5 ….. 3.6 V …

## NCERT solution class 10 chapter 5 Arithmetic Progressions exercise 5.1 mathematics

EXERCISE 5.1 Page No 99: Question 1: In which of the following situations, does the list of numbers involved make as arithmetic progression and why? (i) The taxi fare after each km when the fare is Rs 15 for the first km and Rs 8 for each additional km. (ii) The amount of air present …

## NCERT solution class 10 chapter 4 Quadratic Equations exercise 4.4 mathematics

EXERCISE 4.4 Question 1: Find the nature of the roots of the following quadratic equations. If the real roots exist, find them; (I) 2×2 −3x + 5 = 0 (II)  (III) 2×2 − 6x + 3 = 0 Answer: We know that for a quadratic equation ax2 + bx + c = 0, discriminant is b2 − 4ac. (A) If b2 − 4ac > 0 → two distinct real roots …

## NCERT solution class 10 chapter 4 Quadratic Equations exercise 4.3 mathematics

EXERCISE 4.3 Page No 87: Question 1: Find the roots of the following quadratic equations, if they exist, by the method of completing the square: Answer: Question 2: Find the roots of the quadratic equations given in Q.1 above by applying the quadratic formula. Answer: Page No 88: Question 3: Find the roots of the …

## NCERT solution class 10 chapter 4 Quadratic Equations exercise 4.2 mathematics

EXERCISE 4.2 Page No 76: Question 1: Find the roots of the following quadratic equations by factorisation: Answer: Roots of this equation are the values for which= 0 ∴ = 0 or = 0 i.e., x = 5 or x = −2 Roots of this equation are the values for which= 0 ∴ = 0 or = 0 i.e., x = −2 or x = Roots of …

## NCERT solution class 10 chapter 4 Quadratic Equations exercise 4.1 mathematics

EXERCISE 4.1 Page No 73: Question 1: Check whether the following are quadratic equations: Answer: It is of the form . Hence, the given equation is a quadratic equation. It is of the form . Hence, the given equation is a quadratic equation. It is not of the form . Hence, the given equation is not a quadratic …

## NCERT solution class 10 chapter 3 Pair of Linear Equations in Two Variables exercise 3.7 mathematics

EXERCISE 3.7 Page No 68: Question 1: The ages of two friends Ani and Biju differ by 3 years. Ani’s father Dharam is twice as old as Ani and Biju is twice as old as his sister Cathy. The ages of Cathy and Dharam differs by 30 years. Find the ages of Ani and Biju. …

## NCERT solution class 10 chapter 3 Pair of Linear Equations in Two Variables exercise 3.6 mathematics

EXERCISE 3.6 Page No 67: Question 1: Solve the following pairs of equations by reducing them to a pair of linear equations:    Answer: Let  and, then the equations change as follows. Using cross-multiplication method, we obtain Putting and in the given equations, we obtain Multiplying equation (1) by 3, we obtain 6p + 9q = 6 (3) Adding equation …

## NCERT solution class 10 chapter 3 Pair of Linear Equations in Two Variables exercise 3.5 mathematics

EXERCISE 3.5 Page No 62: Question 1: Which of the following pairs of linear equations has unique solution, no solution or infinitely many solutions? In case there is a unique solution, find it by using cross multiplication method. Answer: Therefore, the given sets of lines are parallel to each other. Therefore, they will not intersect …

## NCERT solution class 10 chapter 3 Pair of Linear Equations in Two Variables exercise 3.4 mathematics

EXERCISE 3.4 Page No 56: Question 1: Solve the following pair of linear equations by the elimination method and the substitution method:   Answer: (i) By elimination method Multiplying equation (1) by 2, we obtain Subtracting equation (2) from equation (3), we obtain Substituting the value in equation (1), we obtain By substitution method From equation …

## NCERT solution class 10 chapter 3 Pair of Linear Equations in Two Variables exercise 3.3 mathematics

EXERCISE 3.3 Page No 53: Question 1: Solve the following pair of linear equations by the substitution method. Answer: (i) x + y = 14 (1) x − y = 4 (2) From (1), we obtain x = 14 − y (3) Substituting this value in equation (2), we obtain Substituting this in equation (3), we obtain (ii)  From (1), we obtain Substituting this value …

## NCERT solution class 10 chapter 3 Pair of Linear Equations in Two Variables exercise 3.2 mathematics

EXERCISE 3.2 Page No 49: Question 1: Form the pair of linear equations in the following problems, and find their solutions graphically. (i) 10 students of Class X took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who …

## NCERT solution class 10 chapter 3 Pair of Linear Equations in Two Variables exercise 3.1 mathematics

EXERCISE 3.1 Page No 44: Question 1: Aftab tells his daughter, “Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be.” (Isn’t this interesting?) Represent this situation algebraically and graphically. Answer: Let the present age of …

## NCERT solution class 10 chapter 2 polynomials exercise 2.4 mathematics

EXERCISE 2.4 Question 1: Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also verify the relationship between the zeroes and the coefficients in each case: Answer: (i)  Therefore, , 1, and −2 are the zeroes of the given polynomial. Comparing the given polynomial with , we obtain a = 2, b = 1, c = −5, d = 2 …

## NCERT solution class 10 chapter 2 polynomials exercise 2.3 mathematics

EXERCISE 2.3 Page No 36: Question 1: Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following: (i)  (ii)  (iii)  Answer: Quotient = x − 3 Remainder = 7x − 9 Quotient = x2 + x − 3 Remainder = 8 Quotient = −x2 − 2 Remainder = −5x +10 Question 2: Check whether the first polynomial …

## NCERT solution class 8 sanskrit chapter 15 प्रहेलिका

Page No 104: Question 1 श्लोकांशेषु रिक्तस्थानानि पूरयत − (क) सीमन्तिनीषु का ———————राजा ——————— गुणोत्तम: । (ख) कं सञ्जघान ——————— का ——————— गङ्गा? (ग) के ——————— कं ——————— न बाधते शीतम् ।। (घ) वृक्षाग्रवासी न च ———————, ——————— न च शूलपाणि । Answer: (क) सीमन्तिनीषु का शान्ता राजा कोऽभूत् गुणोत्तम: । (ख) कं सञ्जघान कृष्ण: का शीतलवाहिनी गङ्गा? (ग) के दारपोषणरता: कं बलवन्तं न बाधते शीतम् ।। (घ) वृक्षाग्रवासी …

## NCERT solution class 8 sanskrit chapter 14 आर्यभटः

Page No 97: Question 1 एकपदेन उत्तरत − (क) सूर्य कस्यां दिशायाम् उदेति? (ख) आर्यभटस्य वेधशाला कुत्र आसीत्? (ग) महान गणितज्ञ: ज्योतिविच्च क: अस्ति? (घ) आर्यभटेन क: ग्रन्थ: रचित: ? (ङ) अस्माकं प्रथमोपग्रहस्य नाम किम् अस्ति? Answer: (क) पूर्वस्याम् । (ख) पाटलीपुत्रे। (ग) आर्यभट: । (घ) आर्यभटीयम्। (ङ) आर्यभट: । Question 2 सन्धिविच्छेद कुरूत − …

## NCERT solution class 8 sanskrit chapter 13 हिमालयः (सन्धि:)

Page No 90: Question 1 अधोलिखितानां प्रश्नानाम् एकपदेन उत्तराणि लिखत − (क) पृथिव्या: मानदण्ड: क:? (ख) हिमालय: भारतस्य कस्यां दिशि वर्त्तते? (ग) इन्दो: किरणेषु क: निमज्जति? (घ) कुमारसंभव महाकाव्यस्य रचयिता क:? (ङ) हिमालय: गुहासु क रक्षति? Answer: (क) हिमालय:। (ख) उत्तरस्याम् । (ग) अङ्क:। (घ) कालिदास:। (ङ) अन्धकारम् । Page No 91: Question 2 अधोरेखाङ्कितेभ्य: …

## NCERT solution class 8 sanskrit chapter 12 कः रक्षति कः रक्षितः

Page No 85: Question 2 अधोलिखितेषु पदेषु प्रयुक्तां विभक्तिं वचनं च लिखत − पदानि विभक्ति: वचनम् यथा-भ्रमणाय चतुर्थी: एकवचनम् (क) वस्तूनि ————————— ————————— (ख) प्लाटिकेन ————————— ————————— (ग) विकीर्णानि ————————— ————————— (घ) मसियष्ट्या ————————— ————————— (ङ) स्यूतकेषु ————————— ————————— (च) काष्ठपीठे ————————— ————————— (छ) पृथिव्याम् ————————— ————————— Answer: पदानि विभक्ति: वचनम् यथा-भ्रमणाय चतुर्थी: एकवचनम् (क) …

## NCERT solution class 8 sanskrit chapter 11 सावित्री बाई फुले

Page No 77: Question 1 अधोलिखितानां प्रश्नानाम् उत्तराणि लिखत − (क) महाराष्ट्रस्य प्रथमा महिला शिक्षिका का आसीत्? (ख) कस्य समुदायस्य बालिकानां कृते सावित्री अपरं विद्यालयं प्रारब्धवती? (ग) कीदृशीनां कुरीतानां सावित्री विरोधम् अकरोत्। (घ) किमर्थं शीर्णवस्त्रावृता: नार्य: कूपात् जलं ग्रहीतुं वारिता:? (ङ) सावित्र्या: मृत्यु: कथम् अभवत्? (च) तस्या: द्वयो: काव्यसङ्कलनयो नामनी के? Answer: (क) सावित्री बाई …

## NCERT solution class 8 sanskrit chapter 10 अशोक वनिका (सर्वनाम, तत्, एतत्-प्रयोग:)

Page No 70: Question 3 श्लोकेषु रिक्तस्थानानि पूरयत- (क) अशोक ——————– तु तस्यां वानर ——————–। स ददर्शाविदूरस्थं ——————– प्रासादमूर्जितम्‌|| (ख) कर्णिकारै: किंशुकैश्च ——————–। स देश: ——————– तेषां प्रदीप्त इव ——————–। Answer: (क) अशोक वनिकायां तु तस्यां वानर पुंगव:। स ददर्शाविदूरस्थं चैत्य प्रासादमूर्जितम्‌|| (ख) कर्णिकारै: किंशुकैश्च कुसुमितै: सुपुष्पितै:। स देश: प्रभया तेषां प्रदीप्त इव सर्वत:। Page No 71: Question 4 श्लोकांशान्‌ योजयत- क …

## NCERT solution class 8 sanskrit chapter 9 सप्तभगिन्यः (प्रत्यय:)

Page No 64: Question 2 प्रश्नानाम् उत्तराणि एकपदेन लिखत − (क) अस्माकं देशे कति राज्यानि सन्ति? (ख) प्राचीनेतिहासे का: स्वाधीना आसन्? (ग) केषां समवाय: ‘सप्तभगिन्य:’ इति कथ्यते? (घ) अस्माकं देशे कति केन्द्रशासित प्रदेशा: सन्ति? (ङ) सप्तभगिनी प्रदेशे क: उद्योग: सर्वप्रमुख: ? Answer: (क) अष्टविंशति: (ख) सप्तभगिन्य: (ग) सप्तराज्यानाम्। (घ) सप्त। (ङ) वंशोद्योग:। Question 3 अधोलिखितपदेषु …

## NCERT solution class 8 sanskrit chapter 8 संसारसागरस्य नायकाः

Page No 54: Question 1 एकपदेन उत्तरत- (क) कस्य राज्यस्य भागेषु गजधर: शब्द: प्रयुज्यते? (ख) गजपरिमाणं क: धारयति? (ग) कार्यसमाप्तौ वेतनानि अतिरिच्य गजधरेभ्य: किं प्रदीयते स्म? (घ) के शिल्पिरूपेण न समादृता: भवन्ति? Answer: (क) राजस्थानस्य। (ख) गजधर:। (ग) सम्मानमपि। (घ) गजधरा:। Question 2 अधोलिखितानां प्रश्नानामुत्तराणि लिखत- (क) तडागा: कुत्र निर्मीयन्ते स्म? (ख) गजधरा: कस्मिन्‌ रूपे …

## NCERT solution class 8 sanskrit chapter 7 जलवाहिनी (विशेष्य-विशेषणप्रयोग:)

Page No 45: Question 2 अधोलिखितानां प्रश्नानामुत्तराणि लिखत- (क) कूप: कीदृश: अस्ति? (ख) महास्फोट: कुत्र आस्ते? (ग) बाल कुत्र विलपिष्यति? (घ) जलवाहिनी यदि द्रुतं गच्छेत्‌ तर्हि किं स्यात्‌? Answer: (क) गंभीर (महालम्बस्तले)। (ख) करे। (ग) गृहे। (घ)  वसनं आर्द्र भवेत्। Question 3 अधोलिखितेषु पदेषु यथापेक्षितं सन्धिं/विच्छेदं कुरुत- (क) तस्य + अस्ति = ——————–। (ख) ——————– …

## NCERT solution class 8 sanskrit chapter 6 प्रेमलस्य प्रेमल्याश्च कथा (ऋकारान्त-स्त्रीलिङ्ग:)

Page No 40: Question 1 एकपदेन उत्तरत- (क) प्रेमल: कथं श्रान्त: आसीत्‌? (ख) प्रेमल: कदा गृहम्‌ आगच्छत्‌? (ग) प्रेमल: कानि प्राज्वालयत्‌? (घ) चुल्लीं प्रज्वालयितुं प्रेमल: किं कृतवान्‌? (ङ) प्रेमली कस्य पत्नी आसीत्‌? Answer: (क) काष्ठच्छेदेन्। (ख) सायं काले। (ग) काष्ठानि (घ) फूत्कारम् (ङ) प्रेमलस्य। Question 2 अधोलिखितेषु पदेषु उपसर्गं चित्वा पृथक् कुरुत- यथा-अपगमिष्याति = अप …

## NCERT solution class 8 sanskrit chapter 5 धर्मे धमनं पापे पुण्यम (ऋकारान्त-स्त्रीलिङ्ग:)

Page No 33: Question 1 एकपदेन उत्तरं लिखत- (क) व्याधस्य नाम किम्‌ आसीत्‌? (ख) चञ्चल: व्याघ्रं कुत्र दृष्टवान्‌? (ग) विस्तृते जाले क: बद्ध: आसीत्‌। (घ) बदरी-गुल्मानां पृष्ठे का निलीना आसीत्‌? (ङ) अनारतं कूर्दनेन क: श्रान्त: अभवत्‌? Answer: (क) चञ्चल:। (ख) वने। (ग) व्याघ्र:। (घ) लोमशिका। (ङ) व्याघ्र:। Question 2 संस्कृतेन उत्तरत- (क) चञ्चलेन वने किं …

## NCERT solution class 8 sanskrit chapter 4 सदैव पुरतो निधेहि चरणम् (लोट्-विधिलिङ्ग-प्रयोग:)

Page No 25: Question 2 अधोलिखितानां प्रश्नानाम्‌ उत्तराणि एकपदेन लिखत- (क) स्वकीयं साधनं किं भवति? (ख) पथि के विषमा: प्रखरा:? (ग) सततं किं करणीयम्‌? (घ) एतस्य गीतस्य रचयिता क:? (ङ) स: कीदृश: कवि: मन्यते? Answer: (क) बलम्। (ख)  पाषाणा:। (ग) ध्येय स्मरणम्। (घ) श्रीधर भास्कर वर्णेकर:। (ङ) राष्ट्रवादी। Question 3 मञ्जूषात: क्रियापदानि चित्वा रिक्तस्थानानि पूरयत- …

## NCERT solution class 8 sanskrit chapter 3 भगवदज्जुकम (नकारान्तपुँल्लिङ्ग:)

Page No 20: Question 1 अधोलिखितानां प्रश्नानाम्‌ उत्तराणि एकपदेन लिखत- (क) गणिकाया: नाम किम्‌? (ख) परिव्राजकस्य शिष्य: क: आसीत्‌? (ग) यमदूत: गणिकाया: जीवं कस्य शरीरे निदधति? (घ) परहितनिरता के भवन्तु? Answer: (क) वसन्तसेना। (ख) शाण्डिल्य:। (ग) परिव्राजकस्य। (घ) भूतगणा:। Question 2 सन्धिविच्छेदं कुरुत- यथा – तत्रैव तत्र + एव भगवन्नयम्‌ ——————– + ——————– श्वासश्च ——————– …

## NCERT solution class 8 sanskrit chapter 2 बिलस्य वाणी न कदापि मे श्रुता (अव्ययप्रयोग:)

Page No 9: Question 2 एकपदेन उत्तरं लिखत- (क) सिंहस्य नाम किम्‌? (ख) गुहाया: स्वामी क: आसीत्‌? (ग) सिंह: कस्मिन्‌ समये गुहाया: समीपे आगत:? (घ) हस्तपादादिका: क्रिया: केषां न प्रवर्तन्ते? (ङ) गुहा केन प्रतिध्वनिता? Answer: (क) खरनखर:। (ख) दधिपुच्छ: शृगाल:। (ग) सूर्यास्तसमये। (घ) भयसन्त्रस्तमनसाम् (ङ) सिंहस्य गर्जनेन Question 3 मञ्जूषात: अव्ययपदं चित्वा वाक्यानि पूरयत- सदा …

## NCERT solution class 8 sanskrit chapter 1 सूभाषितानि

Page No 3: Question 2 श्लोकांशेषु रिक्तस्थानानि पूरयत- (क) समुद्रमासाद्य ———————। (ख) ——————— वच: मधुरसूक्तरसं सृजन्ति। (ग) तद्भभागधेयं ——————— पशूनाम्‌। (घ) विद्याफलं ——————— कृपणस्य सौख्यम्‌। (घ) स्त्रियां ——————— सर्वं तद् ———————कुलम्‌। Answer: (क) समुद्रमासाद्य भवन्त्यपेया:। (ख) श्रुत्वा वच: मधुरसूक्तरसं सृजन्ति। (ग) तद्भागधेयं परमं पशूनाम्‌। (घ) विद्याफलं व्यसनिनं कृपणस्य सौख्यम्‌। (घ) स्त्रियां रोचमानायां सर्वं तद् रोचते कुलम्‌। Question 3 प्रश्नानाम्‌ उत्तराणि एकपदेन लिखत- (क) व्यसनिन: किं नश्यति? …

## NCERT solution class 7 chapter 4 simple equations exercise 4.3 mathematics

Exercise 4.3 QUESTION 1 Solve the following equations. (a) Sol :   {Transposing to R.H.S } 2y = 16 [dividing both sides by 2] y = 8   (b) 5t + 28 = 10  Sol : 5t = 10 – 28 5t = -18 [transposing 28 to R.H.S]   (c) Sol : [transposing 3 to …

## NCERT solution class 7 chapter 4 simple equations exercise 4.2 mathematics

EXERCISE 4.2 QUESTION 1 Give f‌irst the step you will use to separate the variable and then solve the equation: (a) x+ 1 = 0 Sol : Adding 1 to both sides of the given equation, we obtain x – 1 + 1 = 0 + 1 x = 1   (b) x + 1 …

## NCERT solution class 10 chapter 2 polynomials

Exercise 2.2 Question 1 Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients. (i) Sol : or or Therefore , the zeroes of are 4 and -2 . Sum of zeroes Product of zeroes   (ii) Sol : Therefore , the zeroes of are and Sum …

## NCERT Solutions for class 8 Maths chapter 3 Understanding Quadrilaterals

Exercise 3.1  Exercise 3.2  Exercise 3.3  Exercise 3.4 Exercise 3.3 Question 1 Given a parallelogram ABCD. Complete each statement along with the definition or property used. (i) AD = … (ii) ∠DCB = … (iii) OC = … (iv) m∠DAB + m∠CDA = …   Sol : (i) In a parallelogram, opposite sides are equal …

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