Match List I with List II LIST I LIST II A. The maximum value if the…

The correct answer is Option (4) → A-III, B-I, C-II, D-IV

(A) $f(x)=25x-\frac{5x^2}{2}+7$

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

$⇒25-5x=0$

$⇒x=5$

$f(5)=125-\frac{125}{2}+7=\frac{139}{2}$

(B) $f(x)=2x^3-15x^2+36x+1$

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

$⇒f'(c)=6x^2-30x+36=0$

$⇒x^2-5x+6=0$

$⇒(x-3)(x-2)=0$

$x=3\,or\,2$

$⇒f''(c)>0⇒2x-5>0$

for $f''(3)=2×3-5>0$

$∴f(3)=2×27-15×9+36×3+1=24$

(C) $f(x)=\frac{x}{2}-x^2$

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

$⇒\frac{1}{2}-2x=0$

$⇒x=\frac{1}{4}$

$∴f(\frac{1}{4})=\frac{1}{8}-\frac{1}{16}=\frac{1}{16}$

(D) $f(x)=\frac{9}{(x+3)}+x$

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

$⇒\frac{-9}{(x+3)^2}+1=0$

$⇒(x+3)^2=9$

$⇒x+3=3$ or $x+3=-3$

$⇒x=0$, $x=-6$

$f(-6)=\frac{9}{-6+3}+(-6)$

$=\frac{9}{-3}+(-6)$

$=-9=\frac{-36}{4}$