Option : (A)
f(x) = \(\frac{1}{4x^2+2x+1}\) achieves it’s maximum value when g(x) = 4x2+2x+1 achieves it’s minimum value.
Differentiating g(x) with respect to x, we get
g’(x) = 8x+2
Differentiating g’(x) with respect to x, we get
g’’(x) = 8
For minima at x = c,
g’(c) = 0 and g’’(c) > 0 g’(x) = 0
⇒ x = \(-\frac{1}{4}\)
g''(\(-\frac{1}{4}\)) = 8>0
Hence,
x = \(-\frac{1}{4}\) is a point of minima for g(x) and g(\(-\frac{1}{4}\)) = \(\frac{3}{4}\)is the minimum value of g(x).
Hence the maximum value of f(x) = \(\frac{1}{g(x)}\) = \(\frac{4}{3}\)