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Class 11

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 …

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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 …

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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 6.3 mathematics part 2 Read More »

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 …

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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 …

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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). …

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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 …

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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 …

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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. …

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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 …

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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, …

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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 …

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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. …

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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) …

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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 …

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NCERT solution class 12 chapter 3 Differential Equations exercise 3.6 mathematics part 2

EXERCISE 3.6 Page No 413: Question 1: Answer: The given differential equation is  This is in the form of  The solution of the given differential equation is given by the relation, Therefore, equation (1) becomes: This is the required general solution of the given differential equation. Question 2: Answer: The given differential equation is  The …

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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, …

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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 …

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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 3.3 mathematics part 2 Read More »

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: …

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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 …

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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 …

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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 …

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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 = …

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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: …

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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, …

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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 …

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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 …

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NCERT solution class 12 chapter 1 Integrals exercise 1.7 mathematics part 2

EXERCISE 1.7 Page No 330: Question 1: Answer:     Question 2: Answer: Question 3: Answer: Question 4: Answer: Question 5: Answer: Question 6: Answer: Question 7: Answer: Question 8: Answer: Question 9: Answer: Question 10: is equal to A.  B.  C.  D.  Answer: Hence, the correct answer is A. Question 11: is equal to …

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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, …

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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 = …

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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 …

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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 …

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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 …

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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, …

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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 …

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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 − …

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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 = …

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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. …

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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 …

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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 …

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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: …

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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) …

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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: …

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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. …

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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 …

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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 …

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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 …

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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 …

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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 …

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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, Δ 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: …

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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 …

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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 …

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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 = …

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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 4.3 mathematics part 1 Read More »

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 …

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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 + …

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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, …

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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 …

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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: …

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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: …

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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 …

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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 …

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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: …

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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 …

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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 …

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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 * …

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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} → …

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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 …

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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 …

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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 …

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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 …

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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 …

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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 = …

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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 …

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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 …

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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 …

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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 …

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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, …

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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 = …

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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 …

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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 12.4 mathematics Read More »

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 12.3 mathematics Read More »

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 …

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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. …

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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 …

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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 11.4 mathematics Read More »

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 …

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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 11.2 mathematics Read More »

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 …

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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 10.4 mathematics Read More »

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 10.3 mathematics Read More »

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 …

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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 …

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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 9.5 mathematics Read More »

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 9.3 mathematics Read More »

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 9.2 mathematics Read More »

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, …

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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 8.3 mathematics Read More »

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 8.2 mathematics Read More »

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 …

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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. …

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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 …

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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 …

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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 …

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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 6.3 mathematics Read More »

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 6.2 mathematics Read More »

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, …

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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 5.4 mathematics Read More »

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 …

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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 …

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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) …

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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., …

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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:  …

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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 …

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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° …

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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 …

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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 …

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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 …

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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} … …

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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 …

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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. …

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Circle

EXERCISE 24.3 QUESTION 1 Find the equation of the circle, the end points of whose diameter are $(2,-3)$ and $(-2,4)$ . Find its center and radius. Sol : $(2,-3)$ and $(-2,4)$ are the ends points of the diameter of a circle. The equation of this circle is $(x-2)(x+2)+(y+3)(y-4)=0$ $\Rightarrow x^{2}-4+y^{2}-4 y+3 y-12=0$ $\Rightarrow x^{2}+y^{2}-y-16=0 \quad …

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Circle

EXERCISE 24.2 QUESTION 1 Find the co-ordinates of the center and radius of each of the following circles: (i) Sol : The given equation can be rewritten as Center And, radius = 7   (ii) Sol : The given equation can be rewritten as Center Radius   (iii) Sol : The given equation can be …

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Circle

EXERCISE 24.1 QUESTION 1 Find the equation of the circle with : (i) Center $(-2,3)$ and radius 4 Sol : Let $(h, k)$ be the centre of a circle with radius a. Thus, its equation will be $(x-h)^{2}+(y-k)^{2}=a^{2}$ Here, $h=-2, k=3$ and $a=4$ $\therefore$ Required equation of the circle: $\Rightarrow (x+2)^{2}+(y-3)^{2}=4^{2}$ $\Rightarrow(x+2)^{2}+(y-3)^{2}=16$   (ii) Centre …

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Hydrocarbons

HYDROCARBONS Organic compound are composed of only carbons and hydrogen are called hydrocarbons. ON THE BASIS OF STRUCTURE HYDROCARBONS ARE OF TWO TYPES  1. Acyclic or open chain hydrocarbons 2. Cyclic or closed chain hydrocarbons   Acyclic or open chain hydrocarbons These hydrocarbons contains open chains of carbon atoms in their molecules. These hydrocarbons are …

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REDOX REACTIONS

REDOX REACTIONS Chemical reaction in which electron transfer from one chemical substance to another are termed as oxidation-reduction reactions or redox reactions. You can see many examples from your daily life like production of electricity in batteries , production of heat by burning chemical substances , electroplating , manufacturing of useful products caustic soda, potassium …

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Thermodynamics

Thermodynamics The word “thermodynamics” is derived from Greek words therme (heat) and dinamics (flow or motion) . Thermodynamics mainly deals with the transformation of heat into mechanical energy and vice versa .   LIMITATIONS OF THERMODYNAMICS (i) It does not give any direct information about the nature or structure of matter. (ii) It does not give …

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straight line

Straight line EXERCISE 23.1 1. Find the slopes of the lines which make the following angles with the positive direction of x-axis : (i) ​\( -\dfrac{\pi}{4} \)​ sol: m=tan​\( \theta \)​ =tan(​\( -\dfrac{\pi}{4} \)​)  = -1 (ii) ​\( \dfrac{2\pi}{3} \)​ sol: m=tan​\( \theta \)​ =tan(​\( \dfrac{2\pi}{3} \)​)  = tan(​\( \pi-\dfrac{\pi}{3} \)​)  = ​\( -\sqrt{3} \)​ (iii) ​\( \dfrac{3\pi}{4} \)​ …

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Chapter classification of elements and periodicity in properties

CLASSIFICATION OF ELEMENTS AND PERIODICITY IN  PROPERTIES Brief Historical Development Of Periodic Table 1.Johann Dobereiner’s (Law of triads) In between 1815 and 1829 he gave law of triads.According to which group of three elements which posses similar chemical properties, then the central element mass is equal to the arithmetic mean of atomic masses of the …

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