Question

If \(\underset{x\rightarrow 1}{lim}\frac{x^3+ax^2+bx+c}{x^2-2x+1} = 2026\) then the value of \(a-c\) is...

• A. \(2\)
• B. \(1\)
• C. \(-1\)
• D. \(-2\)

Hint:

For a finite limit the numerator needs a double root at x = 1, so write it as (x-1)^2 (x + k).

Answer 1

The Correct Option is D

Step 1: Understand the concept
The denominator \(x^2 - 2x + 1 = (x-1)^2\) is zero at \(x = 1\). For the limit to be finite, the numerator must also vanish at \(x = 1\) to the second order, so \((x-1)^2\) is a factor of it.

Step 2: Factorise the numerator
The cubic is monic, so \(x^3 + ax^2 + bx + c = (x-1)^2(x + k) = x^3 + (k-2)x^2 + (1-2k)x + k\). Therefore \(a = k - 2\), \(b = 1 - 2k\) and \(c = k\).

Step 3: Use the limit value
Cancelling \((x-1)^2\) gives
\[ \lim_{x\to1}(x + k) = 1 + k = 2026 \Rightarrow k = 2025 \]

Step 4: Find a - c
\(a = 2023\) and \(c = 2025\), so \(a - c = (k-2) - k = -2\). Option (D). This does not depend on the value 2026.

Final Answer:
a - c equals -2. This is option (D). \[ \boxed{\text{(D) }-2} \]