Let f: R→ R be defined as f(x)=4x^2-16x+5, then A. Maximum value of f…

The correct answer is Option (1) → B and D only

$f: R→ R,f(x)=4x^2-16x+5$

It''s critical point is, $f'(c)=0$

$⇒8c-16=0$

$⇒c=2$

Now, for $x=2$ to be minima, $f''(x)>0$

$⇒f''(2)=8>0$

∴ at $x=2$, $f(x)$ has minima

$f(2)=4(2)^2-16(2)+5$

$=-11$

and,

thus it has no maxima value.