If f(x) = \(\frac{1}{4x^2+2x+1}\), then its maximum value is :A. \(\

1.	If f(x) = \(\frac{1}{4x^2+2x+1}\), then its maximum value is :A. \(\frac{4}{3}\)B. \(\frac{2}{3}\)C. 1 D. \(\frac{3}{4}\)
Answer» 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) = \(\frac{1}{g(x)}\) = \(\frac{4}{3}\)

If f(x) = \(\frac{1}{4x^2+2x+1}\), then its maximum value is :A. \(\frac{4}{3}\)B. \(\frac{2}{3}\)C. 1 D. \(\frac{3}{4}\)

Answer»

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) = \(\frac{1}{g(x)}\) = \(\frac{4}{3}\)