In locus of a moving point, we will learn; Show
Locus and Equation to a Locus: If a point moves on a plane satisfying some given geometrical condition then the path trace out by the point in the plane is called its locus. By definition, a locus is determined if some geometrical condition are given. Evidently, the co-ordinate of all points on the locus will satisfy the given geometrical condition. The algebraic form of the given geometrical condition which is satisfy by the co-ordinate of all points on the locus is called the equation to the locus of the moving point. Thus, the co-ordinates of all points on the locus satisfy its equation of locus: but the co-ordinates of a point which does not lie on the locus, do not satisfy the equation of locus. Conversely, the points whose co-ordinates satisfy the equation of locus lie on the locus of the moving point. 1. A point moving in such a manner that three times of distance from the x-axis is grater by 7 than 4 times of its distance form the y-axis; find the equation of its locus. Solution: Let P (x, y) be any position of the moving point on its locus. Then the distance of P from the x-axis is y and its distance from the y-axis is x. By problem, 3y – 4x = 7, Which is the required equation to the locus of the moving point. 2. Find the equation to the locus of a moving point which is always equidistant from the points (2, -1) and (3, 2). What curve does the locus represent? Solution:Let A (2, -1) and B (3, 2) be the given points and (x, y) be the co-ordinates of a point P on the required locus. Then, PA2 = (x - 2)2 + (y + 1)2 and PB2 = (x - 3)2 + (y - 2)2By problem, PA = PB or, PA2 = PB2 or, (x - 2)2 + (y + 1)2 = (x - 3)2 + (y - 2)2 or, x2 - 4x + 4 + y2 + 2y + 1 = x2 – 6x + 9 + y2 – 4y + 4 or, 2x + 6y = 8 or, x + 3y = 4 ……… (1) Which is the required equation to the locus of the moving point. Clearly, equation (1) is a first degree equation in x and y; hence, the locus of P is a straight line whose equation is x + 3y = 4. 3. A and B are two given point whose co-ordinates are (-5, 3) and (2, 4) respectively. A point P moves in such a manner that PA : PB = 3 : 2. Find the equation to the locus traced out by P. what curve does it represent? Solution: Let (h, k) be the co-ordinates of any position of the moving point on its locus. By question, PA/PB = 3/2 or, 3 ∙ PB = 2 ∙ PAor, 9 ∙ PB2 = 4 ∙ PA2 Or, 9[(h - 2)2 + (k - 4)2] = 4[(h + 5)2 + (k - 3)2] or, 9 [h2 - 4h + 4 + k2 - 8k + 16] = 4[h2 + 10h + 25 + k2 - 6k + 9] Or, 5h2 + 5k2 – 76h – 48k + 44 = 0 Therefore , the required equation to the locus traces out by P is5x2 + 5y2 – 76x – 48y + 44 = 0 ……….. (1) We see that the equation (1) is a second degree equation in x, y and its coefficients of x2 and y2 are equal and coefficients of xy is zero. Therefore, equation (1) represents a circle. Therefore, the locus of P represents the equation of a circle.4. Find the locus of a moving point which forms a triangle of area 21 square units with the point (2, -7) and (-4, 3). Solution: Let the given point be A (2, -7) and B (-4, 3) and the moving point P (say), which forms a triangle of area 21 square units with A and B, have co-ordinates (x, y). Thus, by question area of the triangle PAB is 21 square units. Hence, we have, Therefore, the required equation to the locus of the moving point is 5x + 3y = 10 or, 5x + 3y + 21 = 0. ½ | (6 – 4y - 7x) – ( 28 + 3x + 2y) | = 21 or, |6 – 28 - 4y – 2y - 7x – 3x | = 42 or, 10x + 6y + 22 = ±42 Therefore, either, 10x + 6y + 22 = 42 i.e., 5x + 3y = 10 or, 10x + 6y + 22 = - 42 i.e., 5x + 3y + 32 = 05. The sum of the distance of a moving point from the points (c,0) and (-c, 0) is always 2a units. Find the equation to the locus of the moving point. Solution: Let P be the moving point and the given points be A (c,0) and B (-c, 0). If (h, k) be the co-ordinates of any position of P on its locus then by question, PA + PB = 2aor, PA = 2a - PB or, PA2 = 4a2 + PB2 – 4a ∙ PB or, PA2 – PB2 = 4a2 – 4a ∙ PB or, [(h - c)2 +(k - 0)2] - [(h + c)2 +(k - 0)2] = 4a2 – 4a. PB or, -4hc = 4a2 – 4a∙PB or, a ∙ PB = a2 + hc or, a2 ∙ PB2 = (a2 + hc)2 (squaring both sides) or, a2 [(h + c)2 + (k - 0)2] = (a2 + hc)2 or, a2 [h2 + c2 + 2hc + k2] = a4 + 2a2hc + h2c2 or, a2h2 – h2c2 + a2k2 = a4 – a2c2 or, (a2 – c2)h2 + a2k2 = a2 (a2 – c2) or, h2/a2 + k2/a2 – c2 = 1 Therefore, the required equation to the locus of P is x2/a2 + y2/(a2 – c2) = 1 ● Locus 11 and 12 Grade Math From Locus of a Moving Point to Home Page
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Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 5 Locus and Straight Line Ex 5.1 Questions and Answers. Question 1.
If A(1, 3) and B(2, 1) are points, find the equation of the locus of point P such that PA = PB.
Solution:
Let P(x, y) be any point on the required locus.
Given, A(1, 3) and B(2, 1).
PA = PB
∴ PA2 = PB2
∴ The required equation of locus is 2x – 4y + 5 = 0.  Question 2. A(-5, 2) and B(4, 1). Find the equation of the locus of point P, which is equidistant from A and B. Solution: Let P(x, y) be any point on the required locus. P is equidistant from A(-5, 2) and B(4, 1). ∴ PA = PB ∴ PA2 = PB2 ∴ (x + 5)2 + (y – 2)2 = (x – 4)2 + (y – 1)2 ∴ x2 + 10x + 25 + y2 – 4y + 4 = x2 – 8x + 16 + y2 – 2y + 1 ∴ 10x – 4y + 29 = -8x – 2y + 17 ∴ 18x – 2y + 12 = 0 ∴ 9x – y + 6 = 0 ∴ The required equation of locus is 9x – y – 6 = 0 Question 3. If A(2, 0) and B(0, 3) are two points, find the equation of the locus of point P such that AP = 2BP. Solution: Let P(x, y) be any point on the required locus. Given, A(2, 0), B(0, 3) and AP = 2BP ∴ AP2 = 4BP2 ∴ (x – 2)2 + (y – 0)2 = 4[(x – 0)2 + (y – 3)2] ∴ x2 – 4x + 4 + y2 = 4(x2 + y2 – 6y + 9) ∴ x2 – 4x + 4 + y2 = 4x2 + 4y2 – 24y + 36 ∴ 3x2 + 3y2 + 4x – 24y + 32 = 0 ∴ The required equation of locus is 3x2 + 3y2 + 4x – 24y + 32 = 0 Question 4. Given, A(4, 1), B(5, 4) and PA2 = 3PB2 ∴ (x – 4)2 + (y – 1)2 = 3[(x – 5)2 + (y – 4)2] ∴ x2 – 8x + 16 + y2 – 2y + 1 = 3(x2 – 10x + 25 + y2 – 8y + 16) ∴ x2 – 8x + y2 – 2y + 17 = 3x2 – 30x + 75 + 3y2 – 24y + 48 ∴ 2x2 + 2y2 – 22x – 22y + 106 = 0 ∴ x2 + y2 – 11x – 11y + 53 = 0 ∴ The required equation of locus is x2 + y2 – 11x – 11y + 53 = 0. Question 5. Given, A(2, 4), B(5, 8) and PA2 – PB2 = 13 ∴ [(x – 2)2 + (y – 4)2] – [(x – 5)2 + (y – 8)2] = 13 ∴ (x2 – 4x + 4 + y2 – 8y + 16) – (x2 – 10x + 25 + y2 – 16y + 64) = 13 ∴ 6x + 8y – 69 = 13 ∴ 6x + 8y – 82 = 0 ∴ 3x + 4y – 41 = 0 ∴ The required equation of locus is 3x + 4y – 41 = 0 Question 6. A(1, 6) and B(3, 5), find the equation of the locus of point P such that segment AB subtends a right angle at P. (∠APB = 90°) Solution: Let P(x. y) be any point on the required locus. Given, A(1, 6) and B(3, 5), ∠APB = 90° ∴ ΔAPB is a right-angled triangle.
AP2 + PB2 = AB2 ∴ [(x – 1)2 + (y – 6)2] + [(x – 3)2 + (y – 5)2] = (1 – 3)2 + (6 – 5)2 ∴ x2 – 2x + 1 + y2 – 12y + 36 + x2 – 6x + 9 + y2 – 10y + 25 = 4 + 1 ∴ 2x2 + 2y2 – 8x – 22y + 66 = 0 ∴ x2 + y2 – 4x – 11y + 33 = 0 ∴ The required equation of locus is x2 + y2 – 4x – 11y + 33 = 0 Question 7. If the origin is shifted to the point O'(2, 3), the axes remaining parallel to the original axes, find the new co-ordinates of the points (a) A(1, 3) (b) B(2, 5) Solution: Origin is shifted to (2, 3) = (h, k) Let the new co-ordinates be (X, Y). ∴ x = X + h and y = Y + k ∴ x = X + 2 and y = Y + 3 …..(i) (a) Given, A(x, y) = A(1, 3) x = X + 2 and y = Y + 3 …..[From (i)] ∴ 1 = X + 2 and 3 = Y + 3 ∴ X = -1 and Y = 0 ∴ the new co-ordinates of point A are (-1, 0). (b) Given, B(x, y) = B(2, 5) x = X + 2 andy = Y + 3 ……[From (i)] ∴ 2 = X + 2 and 5 = Y + 3 ∴ X = 0 and Y = 2 ∴ the new co-ordinates of point B are (0, 2). Question 8. If the origin is shifted to the point O'(1, 3), the axes remaining parallel to the original axes, find the old co-ordinates of the points (a) C(5, 4) (b) D(3, 3) Solution: Origin is shifted to (1, 3) = (h, k) Let the new co-ordinates be (X, Y) x = X + h and y = Y + k ∴ x = X + 1 and 7 = Y + 3 …..(i) (a) Given, C(X, Y) = C(5, 4) ∴ x = X + 1 andy = Y + 3 …..[From(i)] ∴ x = 5 + 1 = 6 and y = 4 + 3 = 7 ∴ the old co-ordinates of point C are (6, 7). (b) Given, D(X, Y) = D(3, 3) ∴ x = X + 1 and y = Y + 3 …..[From (i)] ∴ x = 3 + 1 = 4 and y = 3 + 3 = 6 ∴ the old co-ordinates of point D are (4, 6). Question 9. If the co-ordinates (5, 14) change to (8, 3) by the shift of origin, find the co-ordinates of the point, where the origin is shifted. Solution: Let the origin be shifted to (h, k). Given, (x,y) = (5, 14), (X, Y) = (8, 3) Since, x = X + h and y = Y + k ∴ 5 = 8 + h and 14 = 3 + k ∴ h = -3 and k = 11 ∴ the co-ordinates of the point, where the origin is shifted are (-3, 11). Question 10. Obtain the new equations of the following loci if the origin is shifted to the point O'(2, 2), the direction of axes remaining the same: (a) 3x – y + 2 = 0 (b) x2 + y2 – 3x = 7 (c) xy – 2x – 2y + 4 = 0 Solution: Given, (h, k) = (2, 2) Let (X, Y) be the new co-ordinates of the point (x, y). ∴ x = X + h and y = Y + k ∴ x = X + 2 and y = Y + 2 (a) Substituting the values of x and y in the equation 3x – y + 2 = 0, we get 3(X + 2) – (Y + 2) + 2 = 0 ∴ 3X + 6 – Y – 2 + 2 = 0∴ 3X – Y + 6 = 0, which is the new equation of locus. (b) Substituting the values of x and y in the equation x2 + y2 – 3x = 7, we get (c) Substituting the values of x and y in the equation xy – 2x – 2y + 4 = 0, we get (X + 2) (Y + 2) – 2(X + 2) – 2(Y + 2) + 4 = 0 ∴ XY + 2X + 2Y + 4 – 2X – 4 – 2Y – 4 + 4 = 0 ∴ XY = 0, which is the new equation of locus.
Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 4 Sequences and Series Miscellaneous Exercise 4 Questions and Answers. Maharashtra State Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Miscellaneous Exercise 4Question 1. In a G.P., the fourth term is 48 and the eighth term is 768. Find the tenth term. Solution:
Question 2.
Question 3. The sequence (tn) is a G.P., if \(\frac{5^{n-2}}{7^{n-3}}\) = constant, for all n ∈ N. ∴ first term = t1 = \(\frac{5^{1-2}}{7^{1-3}}=\frac{5^{-1}}{7^{-2}}=\frac{7^{2}}{5}=\frac{49}{5}\) Question 4. Find three numbers in G.P., such that their sum is 35 and their product is 1000. Solution: Let the three numbers in G.P. be \(\frac{a}{r}\), a, ar. According to the first condition,
∴ the three numbers in G.P. are 20, 10, 5 or 5, 10, 20. Question 5. Find 4 numbers in G. P. such that the sum of the middle 2 numbers is \(\frac{10}{3}\) and their product is 1. Solution: Let the four numbers in G.P. be \(\frac{a}{r^{3}}, \frac{a}{r}, a r, a r^{3}\). According to the second condition, \(\frac{\mathrm{a}}{\mathrm{r}^{3}}\left(\frac{\mathrm{a}}{\mathrm{r}}\right)(\mathrm{ar})\left(\mathrm{ar}^{3}\right)=1\) ∴ a4 = 1 ∴ a = 1 According to the first condition,
Question 6. Find five numbers in G.P. such that their product is 243 and the sum of the second and fourth numbers is 10. Solution: Let the five numbers in G.P. be \(\frac{a}{r^{2}}, \frac{a}{r}, a, a r, a r^{2}\) According to the first condition,
Question 7.
Question 8. Find 2 + 22 + 222 + 2222 + …… upto n terms. Solution: Sn = 2 + 22 + 222 +….. upto n terms = 2(1 + 11 + 111 +…… upto n terms) = \(\frac{2}{9}\) (9 + 99 + 999 + … upto n terms) = \(\frac{2}{9}\) [(10 – 1) + (100 – 1) + (1000 – 1) +…… upto n terms] = \(\frac{2}{9}\) [(10 + 100 + 1000 + … upto n terms) – (1 + 1 + 1 + ….. n times)] Since, 10, 100, 1000, …… n terms are in G.P. with a = 10, r = \(\frac{100}{10}\) = 10
Question 9. Find the nth term of the sequence 0.6, 0.66, 0.666, 0.6666,….. Solution: 0.6, 0.66, 0.666, 0.6666, …… ∴ t1 = 0.6 t2 = 0.66 = 0.6 + 0.06 t3 = 0.666 = 0.6 + 0.06 + 0.006 Hence, in general tn = 0.6 + 0.06 + 0.006 + …… upto n terms. The terms are in G.P.with a = 0.6, r = \(\frac{0.06}{0.6}\) = 0.1∴ tn = the sum of first n terms of the G.P. Question 10. Find \(\sum_{r=1}^{n}\left(5 r^{2}+4 r-3\right)\). Solution:
Question 11. Find \(\sum_{\mathbf{r}=1}^{\mathbf{n}} \mathbf{r}(\mathbf{r}-\mathbf{3})(\mathbf{r}-\mathbf{2})\). Solution:
Question 12. Find \(\sum_{r=1}^{n} \frac{1^{2}+2^{2}+3^{2}+\ldots+r^{2}}{2 r+1}\) Solution: We know that,
Question 13. Find \(\sum_{r=1}^{n} \frac{1^{3}+2^{3}+3^{3}+\ldots+r^{3}}{(r+1)^{2}}\) Solution:
Question 14. Find 2 × 6 + 4 × 9 + 6 × 12 + …… upto n terms. Solution: 2, 4, 6, … are in A.P. ∴ rth term = 2 + (r – 1)2 = 2r 6, 9, 12, … are in A.P. ∴ rth term = 6 + (r – 1) (3) = (3r + 3) ∴ 2 × 6 + 4 × 9 + 6 × 12 +…… upto n terms
= 2n(n + 1)(n + 2) Question 15. 122 + 132 + 142 + 152 + …… + 202 = (12 + 22 + 32 + 42 + ……. + 202) – (12 + 22 + 32 + 42 + …… + 112) = 2364 Question 16. (502 – 492) + (482 – 472) + (462 – 452) + …… + (22 – 12) = (502 + 482 + 462 + …… + 22) – (492 + 472 + 452 + …… + 12) = \(\sum_{r=1}^{25}(2 r)^{2}-\sum_{r=1}^{25}(2 r-1)^{2}\)
= 1275 Question 17.
Question 18. Find k so that k – 1, k, k + 2 are consecutive terms of a G.P. Solution: Since k – 1, k, k + 2 are consecutive terms of a G.P. ∴ \(\frac{k}{k-1}=\frac{k+2}{k}\) ∴ k2 = k2 + k – 2 ∴ k – 2 = 0∴ k = 2 Question 19. If pth, qth and rth terms of a G.P. are x, y, z respectively, find the value of \(x^{q-r} \cdot y^{r-p} \cdot z^{p-q}\). Solution: Let a be the first term and R be the common ratio of the G.P. ∴ tn = \(\text { a. } R^{n-1}\) ∴ x = \(\text { a. } R^{p-1}\), y = \(\text { a. } R^{q-1}\), z = \(\text { a. } R^{r-1}\)
Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 4 Sequences and Series Ex 4.5 Questions and Answers. Maharashtra State Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Ex 4.5Question 1. Find the sum \(\sum_{r=1}^{n}(r+1)(2 r-1)\). Solution:
Question 2. Find \(\sum_{r=1}^{n}\left(3 r^{2}-2 r+1\right)\). Solution:
Question 3. Find \(\sum_{r=1}^{n} \frac{1+2+3+\ldots+r}{r}\). Solution:
Question 4. Find \(\sum_{r=1}^{n} \frac{1^{3}+2^{3}+\ldots+r^{3}}{r(r+1)}\). Solution: We know that,
Question 5. Find the sum 5 × 7 + 7 × 9 + 9 × 11 + 11 × 13 + …… upto n terms. Solution: 5 × 7 + 7 × 9 + 9 × 11 + 11 × 13 + ….. upto n terms Now, 5, 7, 9, 11, … are in A.P. rth term = 5 + (r – 1) (2) = 2r + 3 7, 9, 11,. … are in A.P. rth term = 7 + (r – 1) (2) = 2r + 5 ∴ 5 × 7 + 7 × 9 + 9 × 11 + 11 × 13 + …… upto n terms
Question 6. 22 + 42 + 62 + 82 + …… upto n terms = (2 × 1)2 + (2 × 2)2 + (2 × 3)2 + (2 × 4)2 + …… Question 7. Let S = (702 – 692) + (682 – 672) + …… +(22 – 12) ∴ S = (22 – 12) + (42 – 32) + …… + (702 – 692) Here, 2, 4, 6,…, 70 is an A.P. with rth term = 2r and 1, 3, 5,….., 69 in A.P. with rth term = 2r – 1
Question 8. Find the sum 1 × 3 × 5 + 3 × 5 × 7 + 5 × 7 × 9 + …… + (2n – 1) (2n + 1) (2n + 3) Solution: 1 × 3 × 5 + 3 × 5 × 7 + 5 × 7 × 9 + …… + (2n – 1) (2n + 1) (2n + 3) Now, 1, 3, 5, 7, … are in A.P. with a = 1 and d = 2. ∴ rth term = 1 + (r – 1)2 = 2r – 1 3, 5, 7, 9, … are in A.P. with a = 3 and d = 2 ∴ rth term = 3 + (r – 1)2 = 2r + 1 and 5, 7, 9, 11, … are in A.P. with a = 5 and d = 2 ∴ rth term = 5 + (r – 1)2 = 2r + 3 ∴ 1 × 3 × 5 + 3 × 5 × 7 + 5 × 7 × 9 + …… upto n terms
= n(n + l)(2n2 + 6n + 1) – 3n = n(2n3 + 8n2 + 7n + 1 – 3) = n(2n3 + 8n2 + 7n – 2) Question 9. Find n, if \(\frac{1 \times 2+2 \times 3+3 \times 4+4 \times 5+\ldots+\text { upto } n \text { terms }}{1+2+3+4+\ldots+\text { upto } n \text { terms }}\) = \(\frac{100}{3}\) Solution:
Question 10.
Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 4 Sequences and Series Ex 4.4 Questions and Answers. Maharashtra State Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Ex 4.4Question 1. Verify whether the following sequences are H.P. (i) \(\frac{1}{3}, \frac{1}{5}, \frac{1}{7}, \frac{1}{9}, \ldots\) (ii) \(\frac{1}{3}, \frac{1}{6}, \frac{1}{9}, \frac{1}{12}, \ldots \ldots \ldots \ldots\) (iii) \(\frac{1}{7}, \frac{1}{9}, \frac{1}{11}, \frac{1}{13}, \frac{1}{15}, \ldots\) Solution: (i) \(\frac{1}{3}, \frac{1}{5}, \frac{1}{7}, \frac{1}{9}, \ldots\) Here, the reciprocal sequence is 3, 5, 7, 9, … ∴ t1 = 3, t2 = 5, t3 = 7, ….. ∵ t2 – t1 = t3 – t2 = t4 – t3 = 2, constant ∴ The reciprocal sequence is an A.P. ∴ the given sequence is H.P. (ii) \(\frac{1}{3}, \frac{1}{6}, \frac{1}{9}, \frac{1}{12}, \ldots \ldots \ldots \ldots\) Here, the reciprocal sequence is 3, 6, 9, 12 … ∴ t1 = 3, t2 = 6, t3 = 9, t4 = 12, … ∵ t2 – t1 = t3 – t2 = t4 – t3 = 3, constant ∴ The reciprocal sequence is an A.P. ∴ The given sequence is H.P. (iii) \(\frac{1}{7}, \frac{1}{9}, \frac{1}{11}, \frac{1}{13}, \frac{1}{15}, \ldots\) Here, the reciprocal sequence is 7, 9, 11, 13, 15, …… ∴ t1 = 7, t2 = 9, t3 = 11, t4 = 13, ….. ∵ t2 – t1 = t3 – t2 = t4 – t3 = 2, constant ∴ The reciprocal sequence is an A.P. ∴ The given sequence is H.P. Question 2. Find the nth term and hence find the 8th term of the following H.P.s: (i) \(\frac{1}{2}, \frac{1}{5}, \frac{1}{8}, \frac{1}{11}, \ldots \ldots \ldots\) (ii) \(\frac{1}{4}, \frac{1}{6}, \frac{1}{8}, \frac{1}{10}, \ldots \ldots \ldots \ldots\) (iii) \(\frac{1}{5}, \frac{1}{10}, \frac{1}{15}, \frac{1}{20}, \cdots \cdots \cdots\) Solution:
Question 3. Find A.M. of two positive numbers whose G.M. and H.M. are 4 and \(\frac{16}{5}\). Solution: G.M. = 4, H.M. = \(\frac{16}{5}\) ∵ (G.M.)2 = (A.M.) (H.M.) ∴ 16 = A.M. × \(\frac{16}{5}\)∴ A.M. = 5 Question 4. Find H.M. of two positive numbers whose A.M. and G.M. are \(\frac{15}{2}\) and 6. Solution: A.M. = \(\frac{15}{2}\), G.M. = 6 Now, (G.M.)2 = (A.M.) (H.M.) ∴ 62 = \(\frac{15}{2}\) × H.M. ∴ H.M. = 36 × \(\frac{2}{15}\) ∴ H.M. = \(\frac{24}{5}\) Question 5. Find G.M. of two positive numbers whose A.M. and H.M. are 75 and 48. Solution: A.M. = 75, H.M. = 48 (G.M.)2 = (A.M.) (H.M.) ∵ (G.M.)2 = 75 × 48 ∵ (G.M.)2 = 25 × 3 × 16 × 3 ∵ (G.M.)2 = 52 × 42 × 32 ∴ G.M. = 5 × 4 × 3 ∴ G.M. = 60 Question 6. Insert two numbers between \(\frac{1}{7}\) and \(\frac{1}{13}\) so that the resulting sequence is a H.P. Solution: Let the required numbers be \(\frac{1}{\mathrm{H}_{1}}\) and \(\frac{1}{\mathrm{H}_{2}}\). ∴ \(\frac{1}{7}, \frac{1}{\mathrm{H}_{1}}, \frac{1}{\mathrm{H}_{2}}, \frac{1}{13}\) are in H.P. ∴ 7, H1, H2 and 13 are in A.P. ∴ t1 = a = 7 and t4 = a + 3d = 13 ∴ 7 + 3d = 13 ∴ 3d = 6 ∴ d = 2 ∴ H1 = t2 = a + d = 7 + 2 = 9 and H2 = t3 = a + 2d = 7 + 2(2) = 11 ∴ \(\frac{1}{9}\) and \(\frac{1}{11}\) are the required numbers to be inserted between \(\frac{1}{7}\) and \(\frac{1}{13}\) so that the resulting sequence is a H.P. Question 7. Insert two numbers between 1 and -27 so that the resulting sequence is a G.P. Solution: Let the required numbers be G1 and G2. ∴ 1, G1, G2, -27 are in G.P. ∴ t1 = 1, t2 = G1, t3 = G2, t4 = -27 ∴ t1 = a = 1 tn = arn-1 ∴ t4 = (1) r4-1 ∴ -27 = r3 ∴ r3 = (-3)3 ∴ r = -3 ∴ G1 = t2 = ar = 1(-3) = -3 ∴ G2 = t3 = ar = 1(-3)2 = 9 ∴ -3 and 9 are the required numbers to be inserted between 1 and -27 so that the resulting sequence is a G.P. Question 8. Find two numbers whose A.M. exceeds their G.M. by \(\frac{1}{2}\) and their H.M. by \(\frac{25}{26}\). Solution: Let a, b be the two numbers.
∴ a2 – 13a + 36 = 0 ∴ (a – 4)(a – 9) = 0 ∴ a = 4 or a = 9 When a = 4, b = 13 – 4 = 9 When a = 9, b = 13 – 9 = 4∴ the two numbers are 4 and 9. Question 9. Find two numbers whose A.M. exceeds G.M. bv 7 and their H.M. by \(\frac{63}{5}\). Solution: Let a, b be the two numbers.
∴ ab = 282 = 784 ∴ a(70 – a) = 784 ……[From (ii)]∴ 70a – a2 = 784 ∴ a2 – 70a + 784 = 0 ∴ a2 – 56a – 14a + 784 = 0 ∴ (a – 56) (a – 14) = 0 ∴ a = 14 or a = 56 When a = 14, b = 70 – 14 = 56 When a = 56, b = 70 – 56 = 14 ∴ the two numbers are 14 and 56.
Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 4 Sequences and Series Ex 4.3 Questions and Answers. Maharashtra State Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Ex 4.3Question 1. Determine whether the sum to infinity of the following G.P’.s exist. If exists, find it. (i) \(\frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \frac{1}{16}, \ldots\) (ii) \(2, \frac{4}{3}, \frac{8}{9}, \frac{16}{27}, \ldots\) (iii) \(-3,1, \frac{-1}{3}, \frac{1}{9}, \ldots\) (iv) \(\frac{1}{5}, \frac{-2}{5}, \frac{4}{5}, \frac{-8}{5}, \frac{16}{5}, \ldots\) Solution:
Question 2. Express the following recurring decimals as a rational number. (i) \(0 . \overline{32}\) (ii) 3.5 (iii) \(4 . \overline{18}\) (iv) \(0.3 \overline{45}\) (v) \(3.4 \overline{56}\) Solution: (i) \(0 . \overline{32}\) = 0.323232….. = 0.32 + 0.0032 + 0.000032 + ….. Here, 0.32, 0.0032, 0.000032, … are in G.P. with a = 0.32 and r = 0.01 Since, |r| = |0.01| < 1 ∴ Sum to infinity exists. ∴ Sum to infinity = \(\frac{a}{1-r}\)
(ii) 3.5 = 3.555… = 3 + 0.5 + 0.05 + 0.005 + … Here, 0.5, 0.05, 0.005, … are in G.P. with a = 0.5 and r = 0.1 Since, |r| = |0.1| < 1 ∴ Sum to infinity exists.
(iii) \(4 . \overline{18}\) = 4.181818….. = 4 + 0.18 + 0.0018 + 0.000018 + ….. Here, 0.18, 0.0018, 0.000018, … are in G.P. with a = 0.18 and r = 0.01 Since, |r| = |0.01| < 1 ∴ Sum to infinity exists.
(iv) 0.345 = 0.3454545….. = 0.3 + 0.045 + 0.00045 + 0.0000045 + ….. Here, 0.045, 0.00045, 0.0000045, … are in G.P. with a = 0.045, r = 0.01 Since, |r| = |0.01| < 1 ∴ Sum to infinity exists.
(v) \(3.4 \overline{56}\) = 3.4565656 ….. = 3.4 + 0.056 + 0.00056 + 0.0000056 + …. Here, 0.056, 0.00056, 0.0000056, … are in G.P. with a = 0.056 and r = 0.01 Since, |r| = |0.01| < 1 ∴ Sum to infinity exists.
Question 3. If the common ratio of a G.P. is \(\frac{2}{3}\) and sum of its terms to infinity is 12. Find the first term. Solution: r = \(\frac{2}{3}\), sum to infinity = 12 … [Given] Sum to infinity = \(\frac{a}{1-r}\) ∴ 12 = \(\frac{a}{1-\frac{2}{3}}\) ∴ a = 12 × \(\frac{1}{3}\) ∴ a = 4 Question 4. If the first term of a G.P. is 16 and sum of its terms to infinity is \(\frac{176}{5}\), find the common ratio. Solution: a = 16, sum to infinity = \(\frac{176}{5}\) … [Given] Sum to infinity = \(\frac{a}{1-r}\) ∴ \(\frac{176}{5}=\frac{16}{1-r}\) ∴ \(\frac{11}{5}=\frac{1}{1-r}\) ∴ 11 – 11r = 5 ∴ 11r = 6 ∴ r = \(\frac{6}{11}\) Question 5. The sum of the terms of an infinite G.P. is 5 and the sum of the squares of those terms is 15. Find the G.P. Solution: Let the required G.P. be a, ar, ar2, ar3, ….. Sum to infinity of this G.P. = 5 ∴ 5 = \(\frac{a}{1-r}\) ∴ a = 5(1 – r) ……(i) Also, the sum of the squares of the terms is 15.
Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 4 Sequences and Series Ex 4.2 Questions and Answers. Maharashtra State Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Ex 4.2Question 1.
Question 2. For a G.P., if (i) a = 2, r = \(-\frac{2}{3}\), find S6. (ii) S5 = 1023, r = 4, find a. Solution:
Question 3. For a G. P., if (i) a = 2, r = 3, Sn = 242, find n. (ii) sum of the first 3 terms is 125 and the sum of the next 3 terms is 27, find the value of r. Solution:(i) a = 2, r = 3, Sn = 242 Sn = \(a\left(\frac{r^{n}-1}{r-1}\right)\), for r > 1
Question 4. For a G. P., (i) if t3 = 20, t6 = 160, find S7. (ii) if t4 = 16, t9 = 512, find S10. Solution: (i) t3 = 20, t6 = 160 tn = arn-1 ∴ t3 = ar3-1 = ar2 ∴ ar2 = 20 ∴ a = \(\frac{20}{\mathrm{r}^{2}}\) ……(i)
Question 5. Find the sum to n terms: (i) 3 + 33 + 333 + 3333 + …… (ii) 8 + 88 + 888 + 8888 + …….. Solution: (i) Sn = 3 + 33 + 333 +….. upto n terms = 3(1 + 11 + 111 +….. upto n terms) = \(\frac{3}{9}\)(9 + 99 + 999 + … upto n terms) = \(\frac{3}{9}\)[(10 – 1) + (100 – 1) + (1000 – 1) +… upto n terms] = \(\frac{3}{9}\)[(10 + 100 + 1000 + … upto n terms) – (1 + 1 + 1 + … n times)] But 10, 100, 1000, … n terms are in G.P. with a = 10, r = \(\frac{100}{10}\) = 10
(ii) Sn = 8 + 88 + 888 + … upto n terms = 8(1 + 11 + 111 + … upto n terms) = \(\frac{8}{9}\) (9 + 99 + 999 + … upto n terms) = \(\frac{8}{9}\) [(10 – 1) + (100 – 1) + (1000 – 1) +… upto n terms] = \(\frac{8}{9}\) [(10 + 100 + 1000 + … upto n terms) – (1 + 1 + 1 + … n times)] But 10, 100, 1000, … n terms are in G.P. with a = 10, r = \(\frac{100}{10}\) = 10
Question 6. Find the sum to n terms: (i) 0.4 + 0.44 + 0.444 + …… (ii) 0.7 + 0.77 + 0.777 + ….. Solution: (i) Sn = 0.4 + 0.44 + 0.444 + ….. upto n terms = 4(0.1 + 0.11 + 0.111 + …. upto n terms) = \(\frac{4}{9}\) (0.9 + 0.99 + 0.999 + … upto n terms) = \(\frac{4}{9}\) [(i – 0.1) + (1 – 0.01) + (1 – 0.001) … upto n terms] = \(\frac{4}{9}\) [(1 + 1 + 1 + …n times) – (0.1 + 0.01 + 0.001 +… upto n terms)] But 0.1, 0.01, 0.001, … n terms are in G.P. with a = 0.1, r = \(\frac{0.01}{0.1}\) = 0.1∴ Sn = \(\frac{4}{9}\left\{\mathrm{n}-0.1\left[\frac{1-(0.1)^{\mathrm{n}}}{1-0.1}\right]\right\}\) (ii) Sn = 0.7 + 0.77 + 0.777 + … upto n terms = 7(0.1 + 0.11 + 0.111 + … upto n terms) = \(\frac{7}{9}\) (0.9 + 0.99 + 0.999 + … upto n terms) = \(\frac{7}{9}\) [(1 – 0.1) + (1 – 0.01) + (1 – 0.001) +… upto n terms] = \(\frac{7}{9}\) [(1 + 1 + 1 +… n times) – (0.1 + 0.01 + 0.001 +… upto n terms)] But 0.1, 0.01, 0.001, … n terms are in G.P. with a = 0.1, r = \(\frac{0.01}{0.1}\) = 0.1
Question 7. Find the nth terms of the sequences: (i) 0.5, 0.55, 0.555,….. (ii) 0.2, 0.22, 0.222,….. Solution: (i) Let t1 = 0.5, t2 = 0.55, t3 = 0.555 and so on. t1 = 0.5 t2 = 0.55 = 0.5 + 0.05 t3 = 0.555 = 0.5 + 0.05 + 0.005 ∴ tn = 0.5 + 0.05 + 0.005 + … upto n terms But 0.5, 0.05, 0.005, … upto n terms are in G.P. with a = 0.5 and r = 0.1 ∴ tn = the sum of first n terms of the G.P. (ii) Let t1 = 0.2, t2 = 0.22, t3 = 0.222 and so on ∴ tn = the sum of first n terms of the G.P. Question 8.
Question 9. ∴ the G.P. is a, ar, ar2, ar3, …, arn-1 Question 10.
Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 4 Sequences and Series Ex 4.1 Questions and Answers. Maharashtra State Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Ex 4.1Question 1. Verify whether the following sequences are G.P. If so, write tn. (i) 2, 6, 18, 54, …… (ii) 1, -5, 25, -125, ……. (iii) \(\sqrt{5}, \frac{1}{\sqrt{5}}, \frac{1}{5 \sqrt{5}}, \frac{1}{25 \sqrt{5}}, \ldots\) (iv) 3, 4, 5, 6, …… (v) 7, 14, 21, 28, ….. Solution: (i) 2, 6, 18, 54, ……. t1 = 2, t2 = 6, t3 = 18, t4 = 54, ….. Here, \(\frac{t_{2}}{t_{1}}=\frac{t_{3}}{t_{2}}=\frac{t_{4}}{t_{3}}=3\) Since, the ratio of any two consecutive terms is a constant, the given sequence is a geometric progression. Here, a = 2, r = 3tn= arn-1 ∴ tn = 2(3n-1) (ii) 1, -5, 25, -125, …… tn = arn-1 ∴ tn = (-5)n-1 (iii) \(\sqrt{5}, \frac{1}{\sqrt{5}}, \frac{1}{5 \sqrt{5}}, \frac{1}{25 \sqrt{5}}, \ldots\)
(iv) 3, 4, 5, 6,…… ∴ the given sequence is not a geometric progression. (v) 7, 14, 21, 28, ….. ∴ the given sequence is not a geometric progression. Question 2. For the G.P., (i) if r = \(\frac{1}{3}\), a = 9, find t7. (ii) if a = \(\frac{7}{243}\), r = \(\frac{1}{3}\), find t3. (iii) if a = 7, r = -3, find t6. (iv) if a = \(\frac{2}{3}\), t6 = 162, find r. Solution:
Question 3. Which term of the G. P. 5, 25, 125, 625, ….. is 510? Solution:
Question 4. For what values of x, \(\frac{4}{3}\), x, \(\frac{4}{27}\) are in G. P.? Solution:
Question 5. If for a sequence, \(t_{n}=\frac{5^{n-3}}{2^{n-3}}\), show that the sequence is a G. P. Find its first term and the common ratio. Solution: The sequence (tn) is a G.P., if \(\frac{t_{n}}{t_{n-1}}\) = constant, for all n ∈ N
First term, t1 = \(\frac{5^{\mathrm{l}-3}}{2^{1-3}}=\frac{2^{2}}{5^{2}}=\frac{4}{25}\) Question 6. Find three numbers in G. P. such that their sum is 21 and sum of their squares is 189. Solution: Let the three numbers in G. P. be \(\frac{a}{\mathrm{r}}\), a, ar. According to the first condition,
122 + 62 + 32 = 144 + 36 + 9 = 189 Thus, both the conditions are satisfied. Question 7. Find four numbers in G. P. such that sum of the middle two numbers is \(\frac{10}{3}\) and their product is 1. Solution: Let the four numbers in G.P. be \(\frac{a}{r^{3}}, \frac{a}{r}, a r, a r^{3}\). According to the second condition, \(\frac{\mathrm{a}}{\mathrm{r}^{3}}\left(\frac{\mathrm{a}}{\mathrm{r}}\right)(\mathrm{ar})\left(\mathrm{ar}^{3}\right)=1\) ∴ a4 = 1 ∴ a = 1 According to the first condition,
Question 8. Find five numbers in G. P. such that their product is 1024 and the fifth term is square of the third term. Solution: Let the five numbers in G. P. be \(\frac{a}{r^{2}}, \frac{a}{r}, a, a r, a r^{2}\) According to the given conditions,
\(\frac{a}{r^{2}}\) = 1, \(\frac{a}{r}\) = -2, a = 4, ar = -8, ar2 = 16 ∴ the five numbers in G.P. are 1, 2, 4, 8, 16 or 1, -2, 4, -8, 16. Question 9.
Question 10. If p, q, r, s are in G. P., show that p + q, q + r, r + s are also in G.P. Solution: p, q, r, s are in G.P.
∴ p + q, q + r, r + s are in G.P.
Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 3 Complex Numbers Ex 3.3 Questions and Answers. Maharashtra State Board 11th Commerce Maths Solutions Chapter 3 Complex Numbers Ex 3.3Question 1. If ω is a complex cube root of unity, show that (i) (2 – ω)(2 – ω2) = 7 (ii) (2 + ω + ω2)3 – (1 – 3ω + ω2)3 = 65 (iii) \(\frac{\left(\mathbf{a}+\mathbf{b} \omega+\mathbf{c} \omega^{2}\right)}{\mathbf{c}+\mathbf{a} \omega+\mathbf{b} \omega^{2}}\) = ω2 Solution: ω is the complex cube root of unity. ∴ ω3 = 1 and 1 + ω + ω2 = 0 Also, 1 + ω2 = -ω, 1 + ω = -ω2 and ω + ω2 = -1 (i) L.H.S. = (2 – ω)(2 – ω2) = 4 – 2ω2 – 2ω + ω3 = 4 – 2(ω2 + ω) + 1 = 4 – 2(-1) + 1 = 4 + 2 + 1 = 7 = R.H.S. (ii) L.H.S. = (2 + ω + ω2)3 – (1 – 3ω + ω2)3 = R.H.S. (iii) L.H.S. =\(\frac{\left(\mathbf{a}+\mathbf{b} \omega+\mathbf{c} \omega^{2}\right)}{\mathbf{c}+\mathbf{a} \omega+\mathbf{b} \omega^{2}}\) = ω2 = R.H.S. Question 2. If ω is a complex cube root of unity, find the value of (i) ω + \(\frac{1}{\omega}\) (ii) ω2 + ω3 + ω4 (iii) (1 + ω2)3 (iv) (1 – ω – ω2)3 + (1 – ω + ω2)3 (v) (1 + ω)(1 + ω2)(1 + ω4)(1 + ω8) Solution: ω is the complex cube root of unity. ∴ ω3 = 1 and 1 + ω + ω2 = 0 Also, 1 + ω2 = -ω, 1 + ω = -ω2 and ω + ω2 = -1 (i) ω + \(\frac{1}{\omega}\) = \(\frac{\omega^{2}+1}{\omega}\) = \(\frac{-\omega}{\omega}\) = -1 (ii) ω2 + ω3 + ω4 (iii) (1 + ω2)3 (iv) (1 – ω – ω2)3 + (1 – ω + ω2)3 = 0 (v) (1 + ω)(1 + ω2)(1 + ω4)(1 + ω8) Question 3.
L.H.S. = α2 + β2 + αβ = α2 + 2αβ + β2 + αβ – 2αβ ……[Adding and subtracting 2αβ] = (α2 + 2αβ + β2) – αβ = (α + β)2 – αβ = (-1)2 – 1 = 1 – 1 = 0 = R.H.S. Question 4.
Question 5. If ω is a complex cube root of unity, then prove the following: (i) (ω2 + ω – 1)3 = -8 (ii) (a + b) + (aω + bω2) + (aω2 + bω) = 0 Solution: ω is the complex cube root of unity. ∴ ω3 = 1 and 1 + ω + ω2 = 0 Also, 1 + ω2 = -ω, 1 + ω = -ω2 and ω + ω2 = -1 (i) L.H.S. = (ω2 + ω – 1)3 = (-1 – 1)3 = (-2)3 = -8 = R.H.S. (ii) L.H.S. = (a + b) + (aω + bω2) + (aω2 + bω) = R.H.S.
Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 3 Complex Numbers Ex 3.2 Questions and Answers. Maharashtra State Board 11th Commerce Maths Solutions Chapter 3 Complex Numbers Ex 3.2Question 1. Find the square root of the following complex numbers: (i) -8 – 6i Solution: Let \(\sqrt{-8-6 i}\) = a + bi, where a, b ∈ R Squaring on both sides, we get -8 – 6i = (a + bi)2 -8 – 6i = a2 + b2i2 + 2abi -8 – 6i = (a2 – b2) + 2abi …..[∵ i2 = -1] Equating real and imaginary parts, we get
(ii) 7 + 24i Solution: Let \(\sqrt{7+24 i}\) = a + bi, where a, b ∈ R Squaring on both sides, we get 7 + 24i = (a + bi)2 7 + 24i = a2 + b2i2 + 2abi 7 + 24i = (a2 – b2) + 2abi …..[∵ i2 = -1] Equating real and imaginary parts, we get
(iii) 1 + 4√3i Solution: Let \(\sqrt{1+4 \sqrt{3} i}\) = a + bi, where a, b ∈ R Squaring on both sides, we get 1 + 4√3i = (a + bi)2 1 + 4√3i = a2 + b2i2 + 2abi 1 +4√3i = (a2 – b2) + 2abi ……[∵ i2 = -1] Equating real and imaginary parts, we get
(iv) 3 + 2√10i Solution: Let \(\sqrt{3+2 \sqrt{10}} i\) = a + bi, where a, b ∈ R Squaring on both sides, we get 3 + 2√10i = (a + bi)2 3 + 2√10i = a2 + b2i2 + 2abi 3 + 2√10i = (a2 – b2) + 2abi …..[∵ i2 = -1] Equating real and imaginary parts, we get a2 – b2 = 3 and 2ab = 2√10 a2 – b2 = 3 and b = \(\frac{\sqrt{10}}{a}\) (v) 2(1 – √3i) Solution: Let \(\sqrt{2(1-\sqrt{3} i)}\) = a + bi, where a, b ∈ R Squaring on both sides, we get 2(1 – √3i) = (a + bi)2 2(1 – √3i) = a2 + b2i2 + 2abi 2 – 2√3i = (a2 – b2) + 2abi …..[∵ i2 = -1] Equating real and imaginary parts, we get
Question 2. Solve the following quadratic equations. (i) 8x2 + 2x + 1 = 0 Given equation is 8x2 + 2x + 1 = 0 Comparing with ax2 + bx + c = 0, we get a = 8, b = 2, c = 1 Discriminant = b2 – 4ac = (2)2 – 4 × 8 × 1 = 4 – 32 = -28 < 0 So, the given equation has complex roots. These roots are given by
∴ the roots of the given equation are \(\frac{-1+\sqrt{7} \mathrm{i}}{8}\) and \(\frac{-1-\sqrt{7} \mathrm{i}}{8}\) (ii) 2x2 – √3x + 1 = 0 Solution: Given equation is 2x2 – √3x + 1 = 0 Comparing with ax2 + bx + c = 0, we get a = 2, b = -√3, c = 1 Discriminant = b2 – 4ac = (-√3)2 – 4 × 2 × 1 = 3 – 8 = -5 < 0 So, the given equation has complex roots. These roots are given by
∴ the roots of the given equation are \(\frac{\sqrt{3}+\sqrt{5} i}{4}\) and \(\frac{\sqrt{3}-\sqrt{5} i}{4}\) (iii) 3x2 – 7x + 5 = 0 Solution: Given equation is 3x2 – 7x + 5 = 0 Comparing with ax2 + bx + c = 0, we get a = 3, b = -7, c = 5 Discriminant = b2 – 4ac = (-7)2 – 4 × 3 × 5 = 49 – 60 = -11 < 0 So, the given equation has complex roots. These roots are given by
∴ the roots of the given equation are \(\frac{7+\sqrt{11} i}{6}\) and \(\frac{7-\sqrt{11} i}{6}\) (iv) x2 – 4x + 13 = 0 Solution: Given equation is x2 – 4x + 13 = 0 Comparing with ax2 + bx + c = 0, we get a = 1, b = -4, c = 13 Discriminant = b2 – 4ac = (-4)2 – 4 × 1 × 13 = 16 – 52 = -36 < 0 So, the given equation has complex roots. These roots are given by
∴ the roots of the given equation are 2 + 3i and 2 – 3i. Question 3. Solve the following quadratic equations. (i) x2 + 3ix + 10 = 0 Solution:Given equation is x2 + 3ix + 10 = 0 Comparing with ax2 + bx + c = 0, we get a = 1, b = 3i, c = 10 Discriminant = b2 – 4ac = (3i)2 – 4 × 1 × 10 = 9i2 – 40 = -9 – 40 …..[∵ i2 = -1] = -49 So, the given equation has complex roots. These roots are given by
L.H.S. = x2 + 3ix + 10 = (2i)2 + 3i(2i) + 10i = 4i2 + 6i2 + 10 = 10i2 + 10 = -10 + 10 ……[∵ i2 = -1] = 0 = R.H.S. L.H.S. = x2 + 3ix + 10 = (-5i)2 + 3i(-5i) + 10 = 25i2 – 15i2 + 10 = 10i2 + 10 = -10 + 10 …..[∵ i2 = -1] = 0 = R.H.S. Thus, our answer is correct. (ii) 2x2 + 3ix + 2 = 0 Solution: Given equation is 2x2 + 3ix + 2 = 0 Comparing with ax2 + bx + c = 0, we get a = 2, b = 3i, c = 2 Discriminant = b2 – 4ac = (3i)2 – 4 × 2 × 2 = 9i2 – 16 = -9 – 16 = -25 < 0 So, the given equation has complex roots. These roots are given by
∴ the roots of the given equation are \(\frac{1}{2}\)i and -2i. (iii) x2 + 4ix – 4 = 0 Solution: Given equation is x2 + 4ix – 4 = 0 Comparing with ax2 + bx + c = 0, we get a = 1, b = 4i, c = -4 Discriminant = b2 – 4ac = (4i)2 – 4 × 1 × -4 = 16i2 + 16 = -16 + 16 …..[∵ i2 = -1] = 0 So, the given equation has equal roots. These roots are given by
∴ the roots of the given equation are -2i and -2i. (iv) ix2 – 4x – 4i = 0 Solution: ix2 – 4x – 4i = 0 Multiplying throughout by i, we geti2x2 – 4ix – 4i2 = 0 ∴ -x2 – 4ix + 4 = 0 ……[∵ i2 = -1] ∴ x2 + 4ix – 4 = 0 Comparing with ax2 + bx + c = 0, we get a = 1, b = 4i, c = -4 Discriminant = b2 – 4ac = (4i)2 – 4 × 1 × -4 = 16i2 + 16 = -16 + 16 …..[∵ i2 = -1] = 0 So, the given equation has equal roots. These roots are given by
∴ the roots of the given equation are -2i and -2i. Question 4. Solve the following quadratic equations. (i) x2 – (2 + i) x – (1 – 7i) = 0 Solution:Given equation is x2 – (2 + i)x – (1 – 7i) = 0 Comparing with ax2 + bx + c = 0, we get a = 1, b = -(2 + i), c = -(1 – 7i) Discriminant = b2 – 4ac = [-(2 + i)]2 – 4 × 1 × -(1 – 7i) = 4 + 4i + i2 + 4 – 28i = 4 + 4i – 1 + 4 – 28i …….[∵ i2 = -1] = 7 – 24i So, the given equation has complex roots. These roots are given by
(ii) x2 – (3√2 + 2i) x + 6√2i = 0 Solution: Given equation is x2 – (3√2 + 2i) x + 6√2i = 0 Comparing with ax2 + bx + c = 0, we get a = 1, b = -(3√2 + 2i), c = 6√2i Discriminant = b2 – 4ac = [-(3√2 + 2i)]2 – 4 × 1 × 6√2i = 18 + 12√2i + 4i2 – 24√2i = 18 – 12√2i – 4 …..[∵ i2 = -1] = 14 – 12√2i So, the given equation has complex roots. These roots are given by
(iii) x2 – (5 – i) x + (18 + i) = 0 Solution: Given equation is x2 – (5 – i)x + (18 + i) = 0 Comparing with ax2 + bx + c = 0, we get a = 1, b = -(5 – i), c = 18 + i Discriminant = b2 – 4ac = [-(5 – i)]2 – 4 × 1 × (18 + i) = 25 – 10i + i2 – 72 – 4i = 25 – 10i – 1 – 72 – 4i …..[∵ i2 = -1] = -48 – 14i So, the given equation has complex roots. These roots are given by
(iv) (2 + i) x2 – (5 – i) x + 2(1 – i) = 0 Solution: Given equation is (2 + i) x2 – (5 – i) x + 2(1 – i) = 0 Comparing with ax2 + bx + c = 0, we get a = 2 + i, b = -(5 – i), c = 2(1 – i) Discriminant = b2 – 4ac = [-(5 – i)]2 – 4 × (2 + i) × 2(1 – i) = 25 – 10i + i2 – 8(2 + i)(1 – i) = 25 – 10i + i2 – 8(2 – 2i + i – i2) = 25 – 10i – 1 – 8(2 – i + 1) …..[∵ i2 = -1] = 25 – 10i – 1 – 16 + 8i – 8 = -2i So, the given equation has complex roots. These roots are given by
Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 3 Complex Numbers Ex 3.1 Questions and Answers. Maharashtra State Board 11th Commerce Maths Solutions Chapter 3 Complex Numbers Ex 3.1Question 1. Write the conjugates of the following complex numbers: (i) 3 + i (ii) 3 – i (iii) -√5 – √7i (iv) -√-5 (v) 5i (vi) √5 – i (vii) √2 + √3i Solution: (i) Conjugate of (3 + i) is (3 – i) (ii) Conjugate of (3 – i) is (3 + i) (iii) Conjugate of (-√5 – √7i) is (-√5 + √7i) (iv) -√-5 = -√5 × √-1 = -√5i Conjugate of -√-5 is √5i (v) Conjugate of 5i is -5i (vi) Conjugate of √5 – i is √5 + i (vii) Conjugate of √2 + √3i is √2 – √3i Question 2. Express the following in the form of a + ib, a, b ∈ R, i = √-1. State the values of a and b: (i) (1 + 2i)(-2 + i) (ii) \(\frac{\mathrm{i}(4+3 \mathrm{i})}{(1-\mathrm{i})}\) (iii) \(\frac{(2+i)}{(3-i)(1+2 i)}\) (iv) \(\frac{3+2 i}{2-5 i}+\frac{3-2 i}{2+5 i}\) (v) \(\frac{2+\sqrt{-3}}{4+\sqrt{-3}}\) (vi) (2 + 3i)(2 – 3i) (vii) \(\frac{4 i^{8}-3 i^{9}+3}{3 i^{11}-4 i^{10}-2}\) Solution:
Question 3. (-1 + √3i)3 = (-1)3 + 3(-1)2 (√3i) + 3(-1)(√3i)2 +(√3i)3 [∵ (a + b)3 = a3 + 3a2b + 3ab2 + b3] = -1 + 3√3i – 3(3i2) + 3√3 i3 = -1 + 3√3i – 3(-3) – 3√3i [∵ i2 = -1, i3 = -1] = -1 + 9 = 8, which is a real number. Question 4. Evaluate the following: (i) i35 (ii) i888 (iii) i93 (iv) i116 (v) i403 (vi) \(\frac{1}{i^{58}}\) (vii) i30 + i40 + i50 + i60 Solution:We know that, i2 = -1, i3 = -i, i4 = 1 (i) i35 = (i4)8 (i2) i = (1)8 (-1) i = -i (ii) i888 = (i4)222 = (1)222 = 1 (iii) i93 = (i4)23 . i = (1)23 . i = i (iv) i116 = (i4)29 = (1)29 = 1 (v) i403 = (i4)100 (i2) i = (1)100 (-1) i = -i (vi) \(\frac{1}{i^{88}}=\frac{1}{\left(i^{4}\right)^{14} \cdot i^{2}}=\frac{1}{(1)^{14}(-1)}=-1\) (vii) i30 + i40 + i50 + i60 = (i4)7 i2 + (i4)10 + (i4)12 i2 + (i4)15 = (1)7 (-1) + (1)10 + (1)12 (-1) + (1)15 = -1 + 1 – 1 + 1 = 0 Question 5. 1 + i10 + i20 + i30 = 1 + (i4)2 . i2 + (i4)5 + (i4)7 . i2 = 1 + (1)2 (-1) + (1)5 + (1)7 (-1) [∵ i4 = 1, i2 = -1] = 1 – 1 + 1 – 1 = 0, which is a real number. Question 6. Find the value of (i) i49 + i68 + i89 + i110 (ii) i + i2 + i3 + i4 Solution: (i) i49 + i68 + i89 + i110 = (i4)12 . i + (i4)17 + (i4)22 . i + (i4)27 . i2 = (1)12 . i + (1)17 + (1)22 . i + (1)27(-1) ……[∵ i4 = 1, i2 = -1] = i + 1 + i – 1 = 2i (ii) i + i2 + i3 + i4 Question 7. 1 + i2 + i4 + i6 + i8 + ….. + i20 = 1 + (i2 + i4) + (i6 + i8) + (i10 + i12) + (i14 + i16) + (i18 + i20) = 1 + [i2 + (i2)2] + [(i2)3 + (i2)4] + [(i2)5 + (i2)6] + [(i2)7 + (i2)8] + [(i2)9 + (i2)10] = 1 + [-1 + (- 1)2] + [(-1)3 + (-1)4] + [(-1)5 + (-1)6] + [(-1)7 + (-1)8] + [(-1)9 + (-1)10] [∵ i2 = -1] = 1 + (-1 + 1) + (-1 + 1) + (-1 + 1) + (-1 + 1) + (-1 + 1) = 1 + 0 + 0 + 0 + 0 + 0 = 1 Question 8. Find the values of x and y which satisfy the following equations (x, y ∈ R): (i) (x + 2y) + (2x – 3y)i + 4i = 5 (ii) \(\frac{x+1}{1+\mathrm{i}}+\frac{y-1}{1-\mathrm{i}}=\mathrm{i}\) Solution: (i) (x + 2y) + (2x – 3y)i + 4i = 5 ∴ (x + 2y) + (2x – 3y)i = 5 – 4i Equating real and imaginary parts, we get x + 2y = 5 ……..(i) and 2x – 3y = -4 ………(ii) Equation (i) × 2 – equation (ii) gives 7y = 14 ∴ y = 2 Putting y- 2 in (i), we get x + 2(2) = 5 ∴ x + 4 = 5 ∴ x = 1 ∴ x = 1 and y = 2 Check: If x = 1 and y = 2 satisfy the given condition, then our answer is correct. L.H.S. = (x + 2y) + (2x – 3y)i + 4i = (1 + 4) + (2 – 6)i + 4i = 5 – 4i + 4i = 5 = R.H.S. Thus, our answer is correct. (ii) \(\frac{x+1}{1+\mathrm{i}}+\frac{y-1}{1-\mathrm{i}}=\mathrm{i}\) ∴ x = -2 and y = 2 Question 9. Find the value of: (i) x3 – x2 + x + 46, if x = 2 + 3i (ii) 2x3 – 11x2 + 44x + 27, if x = \(\frac{25}{3-4 i}\) Solution: (i) x = 2 + 3i ∴ x – 2 = 3i ∴ (x – 2)2 = 9i2 ∴ x2 – 4x + 4 = 9(-1) …..[∵ i2 = -1] ∴ x2 – 4x + 13 = 0 ……(i) ∴ x3 – x2 + x + 46 = (x2 – 4x + 13)(x + 3) + 7 = 0(x + 3) + 7 ……[From (i)] = 7 (ii) x = \(\frac{25}{3-4 i}\) ∴ (x – 3)2 = 16i2 ∴ x2 – 6x + 9 = 16(-1) …….[∵ i2 = -1] ∴ x2 – 6x + 25 = 0 …….(i) ∴ 2x3 – 11x2 + 44x + 27 = (x2 – 6x + 25) (2x + 1) + 2 = 0 . (2x + 1) + 2 ……[From (i)] = 0 + 2 = 2 |
A(2, 0) and B(0, 3) are two points, find the equation of the locus of point P such that AP = 2BP
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