Let $f: R→R$ be defined such that

$f(x) = 16x^2 -16x+12 $

(A) Maximum value of f(x) is 8

(B) Minimum value of f(x) is 8

(C) Minimum value of f(x) is 16

(D) No maximum value of f(x)

Choose the correct answer from the options given below :

(B), (D) Only

The correct answer is Option (2) → (B), (D) Only

$f(x)=16x^2-16x+12$

$⇒f'(x)=32x-16$

for critical points,

$(32x-16)=0$

$x=\frac{16}{32}=\frac{1}{2}$

Now,

$f''(x)=32>0$

∴ f only has minima

$f(\frac{1}{2})=16(\frac{1}{2})-16(\frac{1}{2})+12$

$=8$