Let f: R→ R be a function defined as f(x)=2x^3-21x^2+36x-20, then : (…

The correct answer is Option (1) → (A), (B) Only

$f(x)=2x^3-21x^2+36x-20$

$f'(x)=6x^2-42x+36$

∴ for critical points $f'(c)=0$

$⇒6x^2-42x+36=0$

$⇒2x^2-14x+12=0$

$⇒x^2-7x+6=0$

$⇒x^2-6x-x+6=0$

$⇒(x-6)(x-1)=0$

$⇒x=6\,or\,1$

and,

for maximum value $f''(c)<0$,

$⇒f''(1)<0$

$⇒12(1)-42=-30<0$

∴ Maximum value at $x = 1$

Minimum value at $x = 6$

$f(1)=2(1)^3-21(1)^2+36(1)-20$

$=2-21+36-20=-3$ → Maximum Value

$f(6)=2(6)^3-21(6)^2+36(6)-20$

$=-128$ → Minimum Value