Let f: ℝ → ℝ be defined as f(x)=a e²ˣ+b eˣ+c x. If f(0)=-1, f′(log ₑ 2)=21 and ∫₀^(log ₑ 4)(f(x)-c x) d x=39/2, then the value of |a+b+c| equals

Solution

Maths sir · 5 steps

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

1. Use f(0). f(0) = a + b + 0 = -1, so a + b = -1.
2. Use the derivative. f'(x) = 2ae²ˣ + beˣ + c. At x = logₑ 2, eˣ = 2 and e²ˣ = 4, so 8a + 2b + c = 21.

   ∫₀^(logₑ 4)(ae²ˣ + beˣ)dx = a/2(16 - 1) + b(4 - 1) = 15a/2 + 3b

3. Use the integral. Removing cx leaves only the exponentials. Setting 15a/2 + 3b = 39/2 gives 15a + 6b = 39, that is, 5a + 2b = 13.
4. Solve. Put b = -1 - a into 5a + 2b = 13: 3a - 2 = 13, so a = 5 and b = -6. Then c = 21 - 8(5) - 2(-6) = -7.
5. So. a + b + c = 5 - 6 - 7 = -8, so |a + b + c| = 8, option (c).

The idea to keep. Every condition on f, f' or a definite integral is just one linear equation in the constants. Write e^(logₑ k) = k early and the algebra stays clean.

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