Match List I with List II

Choose the correct answer from the options given below :

A-III, B-IV, C-II, D-I

The correct answer is Option (3) → A-III, B-IV, C-II, D-I

(A) $f(x)=x^2-8x+17$

for min. value $f'(c)=0$

$⇒2c-8=0$

$⇒c=\frac{8}{2}=4$

Now, $x=4$ to have min. value $f''(c)>0$

$⇒2>0$

$∴f(4)=16-8×4+17=33-32=1$ (III)

(B) $f(x)=x+\frac{1}{x}$

for max. and min. value $f'(c)=0$

$⇒1-\frac{1}{c^2}=0$

$⇒c^2-1=0$

$⇒(c-1)(c+1)=0$

$⇒c=1\,or\,-1$

for min. value $f''(c)<0$,

$f''(c)=\frac{2}{c^3}$

$f''(1)=2>0$

∴ At $f(1)$ we have min. value

$∴f(1)=1+\frac{1}{1}=2$ (IV)

(C) for max. value,

$f(-1)=-1-1=-2$ (II)

(D) Maximum value of $e^x$ does not exist as it will continue till $+∞$. (I)