(i) f(x) = 3x + 1, x = \(\frac{-1}{3}\)
f(x) = 3x + 1
Substitute x = \(\frac{-1}{3}\) in f(x)
\(f (\frac{-1}{3})\)= \(3(\frac{-1}{3}) + 1\)
= -1 + 1
= 0
Since, the result is 0, so x = \(\frac{-1}{3}\) is the root of 3x + 1
(ii) f(x) = x2 – 1, x = 1,−1
f(x) = x2 – 1
Given that x = (1 , -1)
Substitute x = 1 in f(x)
f(1) = 12 – 1
= 1 – 1
= 0
Now, substitute x = (-1) in f(x)
f(-1) = (−1)2 – 1
= 1 – 1
= 0
Since , the results when x = 1 and x = -1 are 0, so (1 , -1) are the roots of the polynomial f(x) = x2 – 1
(iii) g(x) = 3x2 – 2 , \(x = \frac{2}{\sqrt3}\) ,\(\frac{-2}{\sqrt3}\)
g(x) = 3x2 – 2
Substitute \(x = \frac{2}{\sqrt3}\) in g(x)
g\((\frac{2}{\sqrt3})\) = \(3(\frac{2}{\sqrt3})^2 - 2\)
= \(3 (\frac{4}{3}) - 2\)
= 4 – 2
= 2 ≠ 0
Now, Substitute x = \(\frac{-2}{\sqrt3}\) in g(x)
\(g(\frac{2}{\sqrt3}) = 3 (\frac{-2}{\sqrt3})^2 - 2\)
= \(3 (\frac{4}{3}) - 2\)
= 4 – 2
= 2 ≠ 0
Since, the results when x = \(\frac{2}{\sqrt3}\) and x = \(\frac{-2}{\sqrt3}\) are not 0. Therefore (\(\frac{2}{\sqrt3}\) ,\(\frac{-2}{\sqrt3}\) ) are not zeros of 3x2 – 2.
(iv) p(x) = x3 – 6x2 + 11x – 6 , x = 1, 2, 3
p(1) = 13 – 6(1)2 + 11 x 1 – 6
= 1 – 6 + 11 – 6
= 0
p(2) = 23 – 6(2)2 + 11 x 2 – 6
= 8 – 24 – 22 – 6
= 0
p(3) = 33 – 6(3)2 + 11 x 3 – 6
= 27 – 54 + 33 – 6
= 0
Therefore, x = 1, 2, 3 are zeros of p(x).
(v) f(x) = 5x – π, x = \(\frac{4}{5}\)
f( \(\frac{4}{5}\) ) = 5 x \(\frac{4}{5}\) – π
= 4 – π ≠ 0
Therefore, x = \(\frac{4}{5}\) is not a zeros of f(x).
(vi) f(x) = x2 , x = 0
f(0) = 02 = 0
Therefore, x = 0 is a zero of f(x).
(vii) f(x) = lx + m, x = \(\frac{-m}{l}\)
f( \(\frac{-m}{l}\) ) = l x \(\frac{-m}{l}\) + m
= -m + m
= 0
Therefore, x = \(\frac{-m}{l}\) is a zero of f(x).
(viii) f(x) = 2x + 1, x = \(\frac{1}{2}\)
f(\(\frac{1}{2}\)) = 2 x \(\frac{1}{2}\) + 1
= 1 + 1
= 2 ≠ 0
Therefore, x = \(\frac{1}{2}\) is not a zero of f(x).
