Question

Suppose f(x) is polynomial of degree 5 and with leading coefficient 2009. If further that f(1) = 1. f(2)= 3, f(3) = 5, f(4) = 7, f(5) 9. The value of (6) is..?

Answer is not 11

Options:-
A) 242091
B) 241091
C) 241031

Answer 1

Answer:

Option(B)

Step-by-step explanation:

Given:

(i) f(1) = 1

⇒ f(1) - 1 = 0

(ii) f(2) = 3

⇒ f(2) - 3 = 0

(iii) f(3) = 5

⇒ f(3) - 5 = 0

(iv) f(4) = 7

⇒ f(4) - 7 = 0

(v) f(5) = 9

⇒ f(5) - 9 = 0

Now,

Let the polynomial q(x) = f(x) - (2x - 1).

q(x) has x = 1, x = 2, x = 3, x = 4, x = 5 as its factors.

∴ q(x) = (x - 1)(x - 2)(x - 3)(x - 4)(x - 5)

         = f(x) - (2x - 1)

∴ Given, f(x) is leading coefficient is 2009,

q(x) leading coefficient will be 2009.

Now,

q(x) = 2009(x - 1)(x - 2)(x - 3)(x - 4)(x - 5)

      = f(x) - (2x - 1)

We have to find the value of f(6).

f(6) = 2009 * (6 - 1)! + [2(6) - 1]

      = 2009 * 5! + 11

      = 241091

∴ The value of 6 = 241091

Hope it helps!

Answer 2

Since p(a)=p(b)=p(c)=0, and p is 3rd degree, p(x) = k(x-a)(x-b)(x-c) for some constant k.  

Since p(x) has leading coefficient 1, k = 1 and so p(x) = (x-a)(x-b)(x-c).  

Thus, (2-a)(2-b)(2-c) = p(2) = 2^3 - 3(2^2) + 2(2) + 5 = 5.  

2. Let f(x) = g(x) + 2x - 1. Since f(1)=1, f(2)=3, f(3)=5, f(4)=7, f(5)=9, it then follows that g(1) = g(2) = g(3) = g(4) = g(5) = 0.  

Since f(x) is 5th degree with leading coefficient 2009, the same is true for g(x).  

Therefore, g(x) = 2009(x - 1)(x - 2)(x - 3)(x - 4)(x - 5).  

Finally, we have  

f(x) = 2009(x - 1)(x - 2)(x - 3)(x - 4)(x - 5) + 2x - 1  

f(6) = 2009(5)(4)(3)(2)(1) + 12 - 1 = 241091.