Balbharati Maharashtra State BoardStd 12 Commerce Statistics Part 1 Digest PdfChapter 5 Integration Ex 5.6 Questions and Answers.
Maharashtra State Board 12th Commerce Maths Solutions Chapter 5 Integration Ex 5.6
Evaluate:
Question 1
Maharashtra Board Solution
Solution & Step-by-Step Answer:
Let I = Let ∴ 2x + 1 = A(x – 2) + B(x + 1) Put x + 1 = 0, i.e. x = -1, we get 2(-1) + 1 = A(-3) + B(0) ∴ A = Put x – 2 = 0, i.e. x = 2, we get 2(2) + 1 = A(0) + B(3) ∴ B =
Question 2
Maharashtra Board Solution
Solution & Step-by-Step Answer:
Let I = Let ∴ 2x + 1 = A(x – 1)(x – 4) + Bx(x – 4) + Cx(x – 1) Put x = 0, we get 2(0) + 1 = A(-1)(-4) + B(0)(-4) + C(0)(-1) ∴ 1 = 4A ∴ A = Put x – 1 = 0, i.e. x = 1, we get 2(1) + 1 = A(0)(-3) + B(1)(-3) + C(1)(0) ∴ 3 = -3B ∴ B = -1 Put x – 4 = 0, i.e x = 4, we get 2(4) + 1 = A(3)(0) + B(4)(0) + C(4)(3) ∴ 9 = 12C ∴ C =
Question 3
Maharashtra Board Solution
Solution & Step-by-Step Answer:
∴ 1 = A(x – 2) + B(x + 3) Put x + 3 = 0, i.e. x = -3, we get 1 = A(-5) + B (0) ∴ A = Put x – 2 = 0, i.e. x = 2, we get 1 = A(0) + B(5) ∴ B =
Question 4
Maharashtra Board Solution
Solution & Step-by-Step Answer:
Let I = Let ∴ x = A(x – 1)(x + 2) + B(x + 2) + C(x – 1)2 Put x – 1 = 0, i.e. x = 1, we get 1 = A(0)(3) + B(3) + C(0) ∴ B = Put x + 2 = 0, i.e. x = -2, we get -2 = A (-3)(0) + B(0) + C(9) ∴ C = Put x = -1, we get, -1 = A(-2)(1) + B(1) + C(4) But B = and C =
Question 5
Maharashtra Board Solution
Solution & Step-by-Step Answer:
Let I = Let ∴ 3x – 2 = A(x + 1)(x + 3) + B(x + 3) + C(x + 1)2 Put x + 1 = 0, i.e. x = -1, we get 3(-1) – 2 = A(0)(2) + B(2) + C(0) ∴ -5 = 2B ∴ B = Put x + 3 = 0, i.e. x = -3, we get 3(-3) – 2 = A(-2)(0) + B(0) + C(4) ∴ -11 = 4C ∴ C = Put x = 0, we get 3(0) – 2 = A(1)(3) + B(3) + C(1) ∴ -2 = 3A + 3B + C But B = and C =
Question 6
Maharashtra Board Solution
Solution & Step-by-Step Answer:
Let I = =
Question 7
Maharashtra Board Solution
Solution & Step-by-Step Answer:
Question 8
Maharashtra Board Solution
Solution & Step-by-Step Answer:
Let ∴ 5x2 + 20x + 6 = A(x + 1)2 + Bx(x + 1) + Cx Put x = 0, we get 0 + 0 + 6 = A(1) + B(0)(1) + C(0) ∴ A = 6 Put x + 1 = 0, i.e. x = -1, we get 5(1) + 20(-1) + 6 = A(0) + B(-1)(0) + C(-1) ∴ -9 = -C ∴ C = 9 Put x = 1, we get 5(1) + 20(1) + 6 = A(4) + B(1)(2) + C(1) But A = 6 and C = 9 ∴ 31 = 24 + 2B + 9 ∴ B = -1
