If f(x) = { 1/|x|, ; |x| ≥ 1; ax² + b, ; |x| < 1 is differentiable at every point of the domain, then the values of a and b are respectively :

Solution

Maths sir · 4 steps

Check the join at x = 1 twice: the values must meet, and then the slopes must meet.

1. Where to check. Both pieces are smooth on their own, so the only risky points are the joins x = ± 1. The function is even, so if it is differentiable at x = 1, it is differentiable at x = -1 too.
2. Continuity at 1. For x ≥ 1, f(x) = 1/x, so f(1) = 1. From the left, ax² + b tends to a + b. So a + b = 1.
3. Slopes at 1. On the right, d/dx(1/x) = -1/x², which is -1 at x = 1. On the left, d/dx(ax² + b) = 2ax, which is 2a at x = 1.

   2a = -1 ⇒ a = -1/2, b = 1 - a = 3/2

4. So. a = -1/2 and b = 3/2, which is option (d).

The idea to keep. At a join between two pieces, use two equations: equal values (continuity) and equal derivatives (differentiability). For an even function, checking one side is enough.

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