If f(x)=x²+g′(1) x+g^(′ ′)(2) and g(x)=f(1) x²+x f′(x)+f^(′ ′)(x), then the value of f(4)-g(4) is equal to ____________.

Solution

Maths sir · 5 steps

Call the two unknown constants a and b, write g in terms of them, then make g agree with its own definition.

1. Name the constants. g'(1) and g''(2) are just numbers. Call them a and b, so f(x) = x² + ax + b. Then f(1) = 1 + a + b, f'(x) = 2x + a and f''(x) = 2.
2. Build g. Substitute these into the rule for g: g(x) = (1 + a + b)x² + x(2x + a) + 2. Collecting terms gives the line below.

   g(x) = (3 + a + b)x² + ax + 2

3. Match the definitions. g''(x) = 2(3 + a + b) is constant, and it must equal b. Also g'(x) = 2(3 + a + b)x + a, so g'(1) = 6 + 3a + 2b, and this must equal a. That gives a + b = -3, so g''(2) = 0 and therefore b = 0, a = -3.
4. Evaluate. Now f(x) = x² - 3x and g(x) = -3x + 2, because its x² coefficient is 3 + a + b = 0. So f(4) = 16 - 12 = 4 and g(4) = -12 + 2 = -10.
5. So. f(4) - g(4) = 4 - (-10) = 14.

The common slip: -6. Subtracting carelessly as 4 - 10 ignores the sign of g(4) = -10 and gives -6, not 14.

The idea to keep. When a function is defined using values like g'(1) or f(1), treat each one as an unknown constant, build the function, then demand that it reproduces those same constants.

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