Answer :

(i) f(x) = 3x + 1

_{Put x = -1/3}

_{f (-1/3) = 3 * (-1/3) + 1}

_{= -1 + 1}

_{= 0}

_{Therefore, x = -1/3 is a root of f (x) = 3x + 1}

_{(ii) We have,}

_{f (x) = x}^{2} – 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) = x^{2} – 1

(iii) g (x) = 3x^{2} – 2

Put x = and x =

g () = 3 ()^{2} – 2 and g () = 3 ()^{2} – 2

= 3 * – 2 = 3 * – 2

= 2 0 = 2 0

Therefore, x = and x = are not the roots of g (x) = 3x^{2} – 2

(iv) p (x) = x^{3} – 6x^{2} + 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) = x^{3} – 6x^{2} + 11x – 6

(v) f (x) = 5x –

Put x =

f () = 5 * –

= 4 – 0

Therefore, x = is not a root of f (x) = 5x –

(vi) f (x) = x^{2}

Put x = 0

f (0) = (0)^{2}

= 0

Therefore, x = 0 is not a root of f (x) = x^{2}

(vii) f (x) = lx + m

Put x =

f () = l * () + m

= -m + m

= 0

Therefore, x = is a root of f (x) = lx + m

(viii) f (x) = 2x + 1

Put x =

f () = 2 * + 1

= 1 + 1

= 2 0

Therefore, x = is not a root of f (x) = 2x + 1

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