Option : (A)
f(x) = x2 + \(\frac{250}{x}\)
Differentiating f(x) with respect to x, we get
f'(x) = 2x - \(\frac{250}{x^2}\)
Differentiating f’(x) with respect to x, we get
f''(x) = 2 + \(\frac{500}{x^3}\)
For minima at x = c,
f’(c) = 0 and f’’(c)>0 f’(x) = 0
⇒ x3 = 125 or x = 5
f’’(5) = 7>0
Hence,
x = 5 is a point of minima for f(x) and f(5) = 75 is the minimum value of (x).