Jharkhand Board JAC Class 9 Maths Solutions Chapter 2 Polynomials Ex 2.2 Textbook Exercise Questions and Answers.
JAC Board Class 9th Maths Solutions Chapter 2 Polynomials Ex 2.2
Page – 34
Question 1.
Find the value of the polynomial 5x + 4x2 + 3 at
(i) x = 0
(ii) x = – 1
(iii) x = 2 ,
Answer:
(i) p(x) = 5x + 4x2 + 3
p(0) = 5(0) + 4(0)2 + 3 = 3
(ii) p(x) = 5x + 4x2 + 3
P(-1) = 5(-1) + 4(-1)2 + 3
= -5 + 4(1) + 3 = 2
(iii) p(x) = 5x + 4x2 + 3
p(2) = 5(2) + 4(2)2 + 3
= 10 + 16 + 3 = 29
Question 2.
Find p(0), p(1) and p(2) for each of the following polynomials:
(i) p(y) = y2 – y + 1
(ii) p(x) = x3
(iii) p(t) = 2 + t + 2t2 – t3
(iv) p(x) = (x – 1) (x + 1)
Answer:
(i) 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(x) = x3
P(0) = (0)3 = 0
p(1) = (1)3 = 1
p(2) = (2)3 = 8
(iii) 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
(iv) 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
Page – 35
Question 3.
Verify whether the following are zeroes of the polynomial, indicated against them.
(i) p(x) = 3x + 1, x = \(– \frac{1}{3}\)
(ii) p(x) = 5x – π, x = \(\frac{4}{5}\)
(iii) p(x) = x2 – 1, x = 1, – 1
(iv) p(x) = (x + 1) (x – 2), x = – 1, 2
(v) p(x) = x2, x = 0
(vi) p(x) = lx + m, x = \(– \frac{m}{l}\)
(vii) p(x) = 3x2 – 1, x = \(-\frac{1}{\sqrt{3}}, \frac{2}{\sqrt{3}}\)
(viii) p(x) = 2x + 1, x = \(\frac{1}{2}\)
Answer:
(i) If x = \(– \frac{1}{3}\) is a zero of polynomial p(x) then p(\(– \frac{1}{3}\)) = 0
p(\(– \frac{1}{3}\)) = 3(\(– \frac{1}{3}\)) + 1 = -1 + 1 = 0
Therefore, x = \(– \frac{1}{3}\) is a zero of polynomial p(x) = 3x + 1.
(ii) If x = \(\frac{4}{5}\) is a zero of polynomial
p(x) = 5x – π then p(\(\frac{4}{5}\)) should be 0.
p(\(\frac{4}{5}\)) = 5(\(\frac{4}{5}\)) – π = 4 – π ≠ 0
Therefore, x = \(\frac{4}{5}\) is not a zero of given polynomial p(x) = 5x – π.
(iii) If x = 1 and x = – 1 are zeroes of polynomial p(x) = x2 – 1, then p(1) and p(-1) each should be 0.
p(1) = (1)2 – 1 = 0 and
P(-1) = (-1)2 – 1 = 0
Hence, x = 1 and – 1 are zeroes of the polynomial p(x) = x2 – 1.
(iv) If x = -1 and x = 2 are zeroes of polynomial p(x) = (x + 1) (x – 2), then p( -1) and p(2) each should be 0.
p(-1) = (-1 + 1) (-1 – 2) = 0 (-3) = 0, and
p(2) = (2 + 1) (2 – 2) = 3 (0) = 0
Therefore, x = – 1 and x = 2 are zeroes of the polynomial p(x) = (x+1) (x – 2).
(v) If x = 0 is a zero of polynomial p(x) = x2, then p(0) should be zero.
Here, p(0) = (0)2 = 0
Hence, x = 0 is a zero of the polynomial p(x) = x2.
(vi) If x = \(– \frac{m}{l}\) is a zero of polynomial p(x) = lx + m, then p(\(– \frac{m}{l}\))= 0 should be 0.
p(\(– \frac{m}{l}\)) = l(\(– \frac{m}{l}\)) + m = -m + m = 0
Therefore x = (\(– \frac{m}{l}\)) is a zero of polynomial p(x) = lx + m
(vii) If x = \(-\frac{1}{\sqrt{3}}\) and x = \( \frac{2}{\sqrt{3}}\) are zeros of polynomial p(x) = 3x2 – 1, then p(\(-\frac{1}{\sqrt{3}}\)) and p(\( \frac{2}{\sqrt{3}}\))should be 0.
p(\( \frac{-1}{\sqrt{3}}\)) = \(3\left(\frac{-1}{\sqrt{3}}\right)^2\) – 1 = 3(\(\frac{1}{3}\))-1 = 1 – 1 = 0
p(\( \frac{2}{\sqrt{3}}\)) = \(3\left(\frac{2}{\sqrt{3}}\right)^2\) – 1 = 3(\(\frac{4}{3}\)) – 1 = 4 – 1 = 3 ≠ 0
Therefore, x = \( \frac{-1}{\sqrt{3}}\) is a zero of polynomial p(x) = 3x2 – 1 but, x = \( \frac{2}{\sqrt{3}}\) is not a zero of this polynomial.
(viii) If x = \(\frac{1}{2}\) is a zero of polynomial p(x) = 2x + 1then p(\(\frac{1}{2}\)) should be 0.
p(\(\frac{1}{2}\)) = 2(\(\frac{1}{2}\)) + 1 = 1 + 1 = 2 ≠ 0
Therefore, x = \( \frac{1}{2}\) is not a zero of given polynomial p(x) = 2x + 1
Question 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, d, are real numbers.
Answer:
(i) p(x) = x + 5
p(x) = 0
x + 5 = 0
x = – 5
Therefore, x = – 5 is a zero of polynomial p(x) = x + 5 .
(ii) p(x) = x – 5
p(x) =0
x – 5 = 0
x = 5
Therefore, x = 5 is a zero of polynomial p(x) = x – 5.
(iii) p(x) = 2x + 5
p(x) = 0
2x + 5 = 0
2x = – 5
x = \(– \frac{5}{2}\)
Therefore, x = \(– \frac{5}{2}\) is a zero of polynomial p(x) = 2x + 5.
(iv) p(x) = 3x – 2
p(x) = 0
3x – 2 = 0
x = \( \frac{2}{3}\)
Therefore, x = \( \frac{2}{3}\) is a zero of polynomial p(x) = 3x – 2.
(v) p(x) = 3x
p(x) = 0
3x = 0
x = 0
Therefore, x = 0 is a zero of polynomial p(x) = 3x.
(vi) p(x) = ax, a ≠ 0
p(x) = 0
ax = 0
x = 0
Therefore, x = 0 is a zero of polynomial p(x) = ax.
(vii) p(x) = cx + d, c ≠ 0
p(x) = 0
cx + d = 0
x = \(– \frac{d}{c}\)
Therefore, x = \(– \frac{d}{c}\) is a zero of polynomial p(x) = cx + d.
