Let f(x) be a polynomial function such that f(x) + f'(x) + f''(x) = x⁵ + 64. Then, the value of lim _(x → 1) f(x)/(x - 1) is equal to:

Solution

Maths sir · 4 steps

Find f by matching coefficients, check that f(1) = 0, and then the limit is just f'(1).

1. Degree of f. f' and f'' have lower degree than f, so the x⁵ on the right must come from f itself. Take f(x) = x⁵ + ax⁴ + bx³ + cx² + dx + e.
2. Compare coefficients. Add f, f' and f'', then match each power with x⁵ + 64. From x⁴: a + 5 = 0, so a = -5. From x³: b + 4a + 20 = 0, so b = 0. From x²: c + 3b + 12a = 0, so c = 60. From x: d + 2c + 6b = 0, so d = -120. From the constant: e + d + 2c = 64, so e = 64.

   f(x) = x⁵ - 5x⁴ + 60x² - 120x + 64

3. Check the form. f(1) = 1 - 5 + 60 - 120 + 64 = 0, so the limit is of the form 0/0. Because f(1) = 0, f(x)/(x-1) = (f(x) - f(1))/(x - 1), and its limit is f'(1) by the definition of the derivative.
4. So. f'(x) = 5x⁴ - 20x³ + 120x - 120, so f'(1) = 5 - 20 + 120 - 120 = -15. The answer is (a).

The idea to keep. When f(a) = 0, the limit of f(x)/(x-a) as x → a is just f'(a). For f + f' + f'' = a polynomial, start with the same leading term and match coefficients from the top power down.

Answer from the NTA final answer key. Solution by PrepWiser.