If f(x) is a polynomial of degree 4, f(n) = n + 1 & f(0) = 25, then find f(5)...

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Question Id : 11531 | Context :NIMCET 2023

If f(x) is a polynomial of degree 4, f(n) = n + 1 & f(0) = 25, then find f(5) = ?

🎥 Video solution / Text Solution of this question is given below:

Step 1: Define a helper polynomial:

$g(x) = f(x) - (x + 1)$

Given: $f(1) = 2, f(2) = 3, f(3) = 4, f(4) = 5 \Rightarrow g(1) = g(2) = g(3) = g(4) = 0$

So, $g(x) = A(x - 1)(x - 2)(x - 3)(x - 4) \quad \Rightarrow \quad f(x) = A(x - 1)(x - 2)(x - 3)(x - 4) + (x + 1)$

Step 2: Use $f(0) = 25$ to find A:

$f(0) = A(-1)(-2)(-3)(-4) + (0 + 1) = 24A + 1 = 25 \Rightarrow A = 1$

Step 3: Compute $f(5)$ :

$f(5) = (5 - 1)(5 - 2)(5 - 3)(5 - 4) + (5 + 1) = 4 \cdot 3 \cdot 2 \cdot 1 + 6 = 24 + 6 = \boxed{30}$

✅ Final Answer: $\boxed{f(5) = 30}$

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