A cubic polynomial f(x) is such that f(1) = 1, f(2) = 2, f(3) = 3 and

1.	A cubic polynomial f(x) is such that f(1) = 1, f(2) = 2, f(3) = 3 and f(4) = 5, then f(6) equals :(a) 7 (b) 6 (c) 10 (d) 13
Answer» (b) 6 Let the cubic polynomial be : f(x) = ax3 + bx2 + cx + d. Given, f(1) = 1 ⇒ a + b + c + d = 1 ....(i) f(2) = 2 ⇒ 8a + 4b + 2c + d = 2 ...(ii) f(3) = 4 ⇒ 27a + 9b + 3c + d = 3 ...(iii) f(4) = 5 ⇒ 125a + 25b + 5c + d = 5 ...(iv) (ii) – (i) ⇒ 7a + 3b + c = 1 ...(v) (iii) – (ii) ⇒ 19a + 5b + c = 1 ...(vi) (iv) – (iii) ⇒ 98a + 16b + 2c = 2 ...(vii) (vi) – (v) ⇒ 12a + 2b = 0 ⇒ 6a + b = 0 ...(viii) (vii) – 2 (vi) ⇒ 60a + 6b = 0 ⇒ 10a + b = 0 ...(ix) Solving (viii) and (ix), we get a = 0 ⇒ b = 0 Putting a = 0, b = 0 in (v), we, get c = 1 Also from (i), a = 0, b = 0, c = 1 ⇒ d = 0. Putting values of a, b, c, d in f(x) = ax³ + bx² + cx + d, we get the polynomial f(x) = x ⇒ f(6) = 6.

A cubic polynomial f(x) is such that f(1) = 1, f(2) = 2, f(3) = 3 and f(4) = 5, then f(6) equals :(a) 7 (b) 6 (c) 10 (d) 13

Answer»

(b) 6

Let the cubic polynomial be :

f(x) = ax3 + bx2 + cx + d.

Given, f(1) = 1 ⇒ a + b + c + d = 1 ....(i)

f(2) = 2 ⇒ 8a + 4b + 2c + d = 2 ...(ii)

f(3) = 4 ⇒ 27a + 9b + 3c + d = 3 ...(iii)

f(4) = 5 ⇒ 125a + 25b + 5c + d = 5 ...(iv)

(ii) – (i) ⇒ 7a + 3b + c = 1 ...(v)

(iii) – (ii) ⇒ 19a + 5b + c = 1 ...(vi)

(iv) – (iii) ⇒ 98a + 16b + 2c = 2 ...(vii)

(vi) – (v) ⇒ 12a + 2b = 0 ⇒ 6a + b = 0 ...(viii)

(vii) – 2 (vi) ⇒ 60a + 6b = 0 ⇒ 10a + b = 0 ...(ix)

Solving (viii) and (ix), we get a = 0 ⇒ b = 0

Putting a = 0, b = 0 in (v), we, get c = 1

Also from (i), a = 0, b = 0, c = 1 ⇒ d = 0.

Putting values of a, b, c, d in f(x) = ax³ + bx² + cx + d, we get the polynomial f(x) = x ⇒ f(6) = 6.