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Verify whether the indicated numbers are zeros of the polynomials corresponding to them in the following cases:(i) f(x) = 3x + 1; x = \(-\frac{1}{3}\)(ii) f(x) = x2 - 1; x = 1, -1(iii) g(x) = 3x2 - 2; x = \(\frac{2}{\sqrt{3}},-\frac{2}{\sqrt{.3}}\)(iv) p(x) = x3 - 6x2 + 11x - 6, x = 1,2,3(v) f(x) = 5x-π, \(x=\frac{4}{5}\) |
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Answer» (i) f(x) = 3x + 1 Put x = -1/3 f (-1/3) = 3 x (-1/3) + 1 = -1 + 1 = 0 Therefore, x = -1/3 is a root of f (x) = 3x + 1 (ii) We have, f (x) = x2 – 1 Put x = 1 and x = -1 f (1) = (1)2 – 1 and f (-1) = (-1)2 – 1 = 1 – 1 = 1- 1 = 0 = 0 Therefore, x = -1 and x = 1 are the roots of f(x) = x2 – 1 (iii) g (x) = 3x2 – 2 Put x = \(\frac{2}{\sqrt{3}}\) and x = \(\frac{-2}{\sqrt{3}}\) g \((\frac{2}{\sqrt{3}}) \) = \(3(\frac{2}{\sqrt{3}})^{2}-2\) and g \((\frac{-2}{\sqrt{3}})\) = 3 \((\frac{-2}{\sqrt{3}})^{2}-2\) = 3 x \(\frac{4}{3}\) - 2 = 3 x \(\frac{4}{3}\) - 2 = 2\(\neq\) 0 = 2\(\neq\) 0 Therefore, x = \(\frac{2}{\sqrt{3}}\) and x \(\frac{-2}{\sqrt{3}}\) are not the roots of g (x) = 3x2 – 2 (iv) p (x) = x3 – 6x2 + 11x – 6 Put x = 1 p (1) = (1)3 – 6 (1)2 + 11 (1) – 6 = 1 – 6 + 11 – 6 = 0 Put x = 2 p (2) = (2)3 – 6 (2)2 + 11 (2) – 6 = 8 – 24 + 22 – 6 = 0 Put x = 3 p (3) = (3)3 – 6 (3)2 + 11 (3) – 6 = 27 – 54 + 33 – 6 = 0 Therefore, x = 1, 2, 3 are roots of p (x) = x3 – 6x2 + 11x – 6 (v) f (x) = 5x - π Put x = \(\frac{4}{5}\) f \((\frac{4}{5})\) = = 5 x \(\frac{4}{5}-\) π = 4 - π \(\neq0\) Therefore, x = \(\frac{4}{5}\) is not a root of f (x) = 5x - π |
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