Option : (B)
xy = 1, x > 0, y > 0
⇒ y = \(\frac{1}{x}\)
x+y = x + \(\frac{1}{x}\)
Let f(x) = x + \(\frac{1}{x}\),x > 0
Differentiating f(x) with respect to x, we get
f'(x) = 1 - \(\frac{1}{x^2}\)
Also,
Differentiating f’(x) with respect to x, we get
f''(x) = \(\frac{2}{x^3}\)
For minima at x = c,
f’(c) = 0 and f’’(c) < 0
⇒ 1 - \(\frac{1}{x^2}\) = 0
or x = 1 (Since x>0)
f’’(1) = 2 > 0
Hence,
x = 1 is a point of minima for f(x) and f(1) = 2 is the minimum value of f(x) for x > 0.