Let f(x)=x³+x² f′(1)+x f^(′ ′)(2)+f^(′ ′ ′)(3), x ∈ R. Then f′(10) is equal to ____________.

Solution

Maths sir · 4 steps

Those derivative values are just constants: call them a, b, c and let the cubic fix them.

1. Name the constants. Let a = f'(1), b = f''(2) and c = f'''(3). These are fixed numbers, so f(x) = x³ + ax² + bx + c is an ordinary cubic.
2. Differentiate. f'(x) = 3x² + 2ax + b, f''(x) = 6x + 2a and f'''(x) = 6.
3. Match the definitions. c = f'''(3) = 6. Next, b = f''(2) = 12 + 2a. Then a = f'(1) = 3 + 2a + b = 3 + 2a + 12 + 2a, so a = 15 + 4a, which gives a = -5 and b = 12 - 10 = 2.

   f'(x) = 3x² - 10x + 2

4. So. f'(10) = 300 - 100 + 2 = 202.

The common slip: 422. Solving a = 15 + 4a as a = 5 instead of -5 makes b = 22, and f'(10) comes out as 422.

The idea to keep. When f'(1), f''(2) and so on appear inside f(x), treat them as unknown constants. Differentiate, put each one back at its own point, and solve the small system.

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