Class IX Math NCERT Solutions For Polynomials Exercise– 2.2

Exercise– 2.2
1.  Find the value of the polynomial 5x – 4x2 + 3 at
         (i)   x = 0
         (ii)   x = –1
         (iii)   x = 2
Ans. (i)   ∵ p(x) = 5x – 4x2 + 3 = 5(x) – 4(x)2 + 3
                  ∴ p(0) = 5(0) – 4(0) + 3 = 0 – 0 + 3 = 3
                  Thus, the value of 5x – 4x2 + 3 at x = 0 is 3.
         (ii)   ∵ p(x) = 5x – 4x2 + 3 = 5(x) – 4(x)2 + 3
                 ∴ p(–1) = 5(–1) – 4(–1)2 + 3 = – 5 – 4(1) + 3
                 = –5 – 4 + 3 = –9 + 3 = –6
                 ∴ The value of 5x – 4x2 + 3 at x = –1 is –6.
         (iii)  ∵ p(x) = 5x – 4x2 + 3 = 5(x) – 4(x)2 + 3
                 ∴ p(2) = 5(2) – 4(2)2 + 3 = –10 – 4(4) + 3
                 = 10 – 16 + 3 = –3
                 Thus the value of 5x – 4x2 + 3 at x = 2 is –3
2.  Find p(0), p(1) and p(2) for each of the following polynomials:
         (i)     p(y) = y2 – y + 1
         (ii)    p(t) = 2 + t + 2t2 – t3
         (iii)   p(x) = x3
         (iv)   p(x) = (x – 1) (x + 1)
Ans. (i)   p(y) = y2 – y + 1
                  ∵ p(y) = y2 – y + 1 = (y)2 – y + 1
                  ∴ p(0) = (0)2 – (0) + 1 = 0 – 0 + 1 = 1
                  p(1) = (1)2 – (1) + 1 = 1 – 1 + 1 = 1
                  p(2) = (2)2 – 2 + 1 = 4 – 2 + 1 = 3
         (ii)   p(t) = 2 + t + 2t2 – t3
                ∵ p(t) = 2 + t + 2t2 – t3 = 2 + t + 2(t)2 – (t)3
                ∴ p(0) = 2 + (0) + 2(0)2 – (0)3
                = 2 + 0 + 0 – 0 = 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
                ∵ p(x) = x3 = (x)3
                ∴ p(0) = (0)3 = 0
                p(1) = (1)3 = 1
                p(2) = (2)3 = 8                [∵ 2 × 2 × 2 = 8]
         (iv)   p(x)= (x – 1)(x + 1)
                ∵ 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)(2) = 3
3.  Verify whether the following are zeros of the polynomial, indicated against them.
       
Ans. (i)   ∵ p(x) = 3x + 1
       
         (iii)   Since, p(x) = x2 – 1
                ∴ p(1) = (1)2 – 1 = 1 – 1 = 0
                Since, p(1) = 0,
                ∴ x = 1 is a zero of x2 – 1.
                Also p(–1) = (–1)2 – 1 = 1 – 1 = 0
                i.e. p(–1) = 0,
                ∴ x = –1 is also a zero of x2 – 1.
         (iv)   We have p(x) = (x + 1)(x – 2)
                ∴ p(–1) = (–1 + 1)(–1 – 2) = (0)(–3) = 0
                Since p(–1) = 0,
                ∴ x = –1 is a zero of (x + 1)(x – 1).
                ∴ Also, p(2) = (2 + 1)(2 – 2) = (3)(0) = 0
                ∴ Since p(2) = 0,
                ∴ x = 2 is also a zero of (x + 1)(x – 1).
         (v)   We have p(x) = x2
                ∴ p(0) = (0)2 = 0
                Since p(0) = 0,
                ∴ 0 is a zero of x2.
         (vi)   We have p(x) = lx + m
                   
         (vii)   We have p(x) = 3x2 – 1
                   
         (viii)   We have p(x) = 2x + 1
                   
                    is not a zero of 2x + 1.
4.  Find the zero of the polynomial in each of the following cases:
         (i)   p(x) = x + 5
         (ii)   p(x) = x – 5
         (iii)   p(x) = 2x + 5
         (iv)   p(x) = 3x – 2
         (v)   p(x) = 3x
         (vi)   p(x) = ax, a ≠ 0
         (vii)   p(x) = cx + d, c ≠ 0, c, d are real numbers.
Ans. (i)   We have p(x) = x + 5
                ∴ p(x) = 0
                ⇒ x + 5 = 0
                ∴ or x = –5
                ∴ Thus, a zero of x + 5 is (–5).
         (ii)   We have p(x) = x – 5
                ∴ p(x) = 0
                ⇒ x – 5 = 0
                ∴ or x = 5
                ∴ Thus, a zero of x – 5 is 5.
         (iii)   We have p(x) = 2x + 5
                ∴ p(x) = 0
                ⇒ 2x + 5 = 0
                or 2x = –5
                
                

Thus, a zero of 3x – 2 is 
         (v)   Since p(x) = 3x
                ∴ p(x) = 0
                ⇒ 3x = 0
                
                Thus, a zero of 3x is 0.
         (vi)   Since, p(x) = ax, a ≠ 0
                ∴ p(x) = 0
                ⇒ ax = 0
                or 
                Thus, a zero of ax is 0.
         (viii)   Since, p(x) = cx + d
                ∴ p(x) = 0
                ⇒ cx + d = 0
                 or cx = –d
                or 
                Thus, a zero of cx + d is 
5.  If p(x) = x2 – 4x + 3, evaluate: 
Ans. We have p(x) = x2 4x + 3
                ∴ p(–1) = (–1)2 – 4(–1) + 3
                = 1 + 4 + 3 = 8
                and p(2) = (2)2 – 4(2) + 3
                = 4 – 8 + 3 = –1
                 

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