If f(x)={x³ sin (1/x),, x ≠ 0; 0, , x=0 , then

Solution

Maths sir · 5 steps

At x = 0 use the limit definition; away from 0, differentiate twice with the usual rules.

1. First derivative. For x ≠ 0, the product and chain rules give f'(x) = 3x² sin 1/x - x cos 1/x. At x = 0 use the definition: f'(0) = lim(h → 0) (h³ sin(1/h))/h = lim(h → 0) h² sin 1/h = 0.
2. Check f''(0). f''(0) = lim(h → 0) (f'(h) - 0)/h = lim(h → 0) (3h sin 1/h - cos 1/h). The first term goes to 0, but cos 1/h keeps swinging between -1 and 1. So f''(0) does not exist, and (a) and (b) are both false.
3. Differentiate again. For x ≠ 0, differentiate f'(x) term by term and collect the cos 1/x terms.

   f''(x) = 6x sin 1/x - 4 cos 1/x - 1/x sin 1/x

4. Put in the point. At x = 2/π we have 1/x = π/2, so sin 1/x = 1 and cos 1/x = 0. This gives f''(2/π) = 12/π - π/2 = (24 - π²)/2π.
5. So. The correct statement is (c): f''(2/π) = (24 - π²)/2π.

The idea to keep. For piecewise functions with sin 1/x, find the derivative at the joining point from the limit definition. A leftover cos 1/x with no shrinking factor means the limit does not exist.

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