Match List I with List II
| List I | List II | ||
| A | If x is real, minimum value of $f(x)=x^2-8x+17$ | I. | Does not exist |
| B | Minimum value of $f(x)=x+\frac{1}{x};x>0$ | II. | -2 |
| C | Maximum value of $f(x)=x+\frac{1}{x}; x< 0$ | III. | 1 |
| D | Maximum value of $f(x)=e^x$ is : | IV. | 2 |
Choose the correct answer from the options given below :
Answer & explanation
Correct answer: option 3
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)