Q. 24.7( 7 Votes )

# Verify whether th

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) = 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 = 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) = 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 – Put x = f ( ) = 5 *  = 4 – 0

Therefore, x = is not a root of f (x) = 5x – (vi) f (x) = x2

Put x = 0

f (0) = (0)2

= 0

Therefore, x = 0 is not a root of f (x) = x2

(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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