If c is a point at which Rolle's theorem holds for the function,
f(x) = log ₑ( (x² + α)/7x ) in the interval [3, 4], where a ∈ R, then ƒ''(c) is equal to

Solution

Maths sir · 4 steps

Use f(3) = f(4) to fix α first. Then f' = 0 gives c, and f'' at c is a quick substitution.

1. Find alpha. Rolle's theorem needs f(3) = f(4). Since ln is one-one, the arguments must be equal: (9+α)/21 = (16+α)/28. Cross-multiplying gives 252 + 28α = 336 + 21α, so α = 12.
2. Split the log. Write f(x) = ln(x²+12) - ln 7 - ln x. Then f'(x) = 2x/(x²+12) - 1/x = (x²-12)/x(x²+12).
3. Locate c. f'(c) = 0 gives c² = 12, so c = 2√3 ≈ 3.46. This lies inside (3, 4).

   f''(x) = (24 - 2x²)/(x²+12)² + 1/x²

4. So. At c² = 12 the first term is (24-24)/24² = 0, so f''(c) = 1/12. The answer is (a).

The idea to keep. With an unknown constant in a Rolle's question, use f(a) = f(b) to find it first. Splitting a log of a quotient into separate logs makes the derivatives much cleaner.

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