The left hand derivative of at , is an integer, is
- A.
- B.
- C.
- D.
Show answer and solution
Correct option: A
Solution
The left hand derivative is lim(h→0⁻) [f(k+h) - f(k)]/h.
For h→0⁻, x = k+h is just below k, so [k+h] = k-1. Also f(k) = [k]sin(kπ) = k·0 = 0.
So the limit = lim(h→0⁻) [(k-1)sin(π(k+h)) - 0]/h = (k-1)·lim(h→0⁻) sin(kπ + πh)/h.
Now sin(kπ + πh) = sin(kπ)cos(πh) + cos(kπ)sin(πh) = 0 + (-1)k sin(πh).
So the limit = (k-1)·(-1)k · lim(h→0⁻) sin(πh)/h = (k-1)·(-1)k · π.
This gives (-1)k (k-1)π.
For h→0⁻, x = k+h is just below k, so [k+h] = k-1. Also f(k) = [k]sin(kπ) = k·0 = 0.
So the limit = lim(h→0⁻) [(k-1)sin(π(k+h)) - 0]/h = (k-1)·lim(h→0⁻) sin(kπ + πh)/h.
Now sin(kπ + πh) = sin(kπ)cos(πh) + cos(kπ)sin(πh) = 0 + (-1)k sin(πh).
So the limit = (k-1)·(-1)k · lim(h→0⁻) sin(πh)/h = (k-1)·(-1)k · π.
This gives (-1)k (k-1)π.
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