Let f(x)=a x³+b x²+c x+41 be such that f(1)=40, f′(1)=2 and f^(′ ′)(1)=4. Then a²+b²+c² is equal to:

Solution

Maths sir · 4 steps

Three conditions, three unknowns: turn each one into a linear equation and solve.

1. Differentiate twice. f'(x) = 3ax² + 2bx + c and f''(x) = 6ax + 2b. The constant 41 disappears after the first derivative.
2. Use the conditions. f(1) = 40 gives a + b + c + 41 = 40, so a + b + c = -1. f'(1) = 2 gives 3a + 2b + c = 2. f''(1) = 4 gives 6a + 2b = 4, so 3a + b = 2.
3. Solve the system. Subtract the first equation from the second: 2a + b = 3. Subtract this from 3a + b = 2: a = -1. Then b = 3 - 2a = 5 and c = -1 - a - b = -5.

   a² + b² + c² = (-1)² + 5² + (-5)² = 1 + 25 + 25 = 51

4. So. With a = -1, b = 5, c = -5, the sum of squares is 51, option (b).

The idea to keep. Each value of f, f' or f'' at a point is one linear equation in the coefficients. Start with the highest derivative, since it has the fewest unknowns.

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