Exercise 2.2 Class 9 maths

#rajansir #mathsclass9 Exercise 2.2 Class 9 maths Exercise 2.2 Class 9 maths by @rajansir07 for Online Class Call- 9811757843 Class 9 Chapter 2 – Polynomials Exercise 2.2 1. Find the value of the polynomial (x)=5x−4x2+3 (i) x = 0 (ii) x = – 1 (iii) x = 2 Solution: Let f(x) = 5x−4x2+3 (i) When x = 0 f(0) = 5(0)-4(0)2+3 = 3 (ii) When x = -1 f(x) = 5x−4x2+3 f(−1) = 5(−1)−4(−1)2+3 = −5–4+3 = −6 (iii) When x = 2 f(x) = 5x−4x2+3 f(2) = 5(2)−4(2)2+3 = 10–16+3 = −3 2. Find p(0), p(1) and p(2) for each of the following polynomials. (i) p(y)=y2−y+1 Solution: p(y) = y2–y+1 ∴p(0) = (0)2−(0)+1=1 p(1) = (1)2–(1)+1=1 p(2) = (2)2–(2)+1=3 (ii) p(t)=2+t+2t2−t3 Solution: p(t) = 2+t+2t2−t3 ∴p(0) = 2+0+2(0)2–(0)3=2 p(1) = 2+1+2(1)2–(1)3=2+1+2–1=4 p(2) = 2+2+2(2)2–(2)3=2+2+8–8=4 (iii) p(x)=x3 Solution: p(x) = x3 ∴p(0) = (0)3 = 0 p(1) = (1)3 = 1 p(2) = (2)3 = 8 (iv) P(x) = (x−1)(x+1) Solution: p(x) = (x–1)(x+1) ∴p(0) = (0–1)(0+1) = (−1)(1) = –1 p(1) = (1–1)(1+1) = 0(2) = 0 p(2) = (2–1)(2+1) = 1(3) = 3 For, x = -1/√3 , 2/√3 ; p(x) = 3x2−1 ∴p(-1/√3) = 3(-1/√3)2-1 = 3(1/3)-1 = 1-1 = 0 ∴p(2/√3 ) = 3(2/√3)2-1 = 3(4/3)-1 = 4−1=3 ≠ 0 ∴-1/√3 is a zero of p(x) but 2/√3 is not a zero of p(x). (viii) p(x) =2x+1, x = 1/2 Solution: For, x = 1/2 p(x) = 2x+1 ∴ p(1/2)=2(1/2)+1 = 1+1 = 2≠0 ∴1/2 is not a zero of p(x). ∴x = 0 is a zero polynomial of the polynomial p(x). (vii) p(x) = cx+d, c ≠ 0, c, d are real numbers. Solution: p(x) = cx + d ⇒ cx+d =0 ⇒ x = -d/c ∴ x = -d/c is a zero polynomial of the polynomial p(x).