Let k and m be positive real numbers such that the function f(x)={3 x²+k √(x+1),, 0 < x < 1; m x²+k²,, x ≥ 1 is differentiable for all x > 0. Then (8 f′(8))/f′(1/8) is equal to ____________.

Solution

Maths sir · 5 steps

Differentiable at x = 1 gives two equations: the pieces must meet, and their slopes must match.

1. Continuity at 1. Both pieces must give the same value at x = 1: 3 + k√2 = m + k².
2. Slopes match. The left piece has derivative 6x + k/2√(x+1) and the right piece has 2mx. At x = 1 they must be equal: 6 + k/2√2 = 2m, so m = 3 + k/4√2.

   k√2 - k/4√2 = k² ⇒ k = (8 - 1)/4√2 = 7/4√2

3. Find k and m. Putting m into the continuity equation cancels the 3 on both sides and leaves the line above. We may divide by k because k > 0. Then m = 3 + 7/32 = 103/32.
4. Both derivatives. f'(8) = 16m = 103/2, so 8f'(8) = 412. At x = 1/8, √(x+1) = 3/2√2, so f'(1/8) = 3/4 + k√2/3 = 3/4 + 7/12 = 4/3.
5. So. 8f'(8)/f'(1/8) = 412 × 3/4 = 309.

The common slip: 38.625. Dropping the factor 8 in the numerator gives (103/2)/(4/3) = 309/8 = 38.625, not 309.

The idea to keep. For a piecewise function to be differentiable at the joint, write two conditions: equal values and equal derivatives there. Two equations give you the two unknowns.

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