A) 12
B) 1
C) 9
D) 8
Correct Answer: C
Solution :
Given \[f(x)=\frac{[(5+x)(2+x)]}{[1+x]}\] \[f(x)=1+\frac{4}{1+x}+(5+x)=(6+x)+\frac{4}{(1+x)}\] Þ \[f'(x)=1-\frac{4}{{{(1+x)}^{2}}}=0\]; \[{{x}^{2}}+2x-3=0\]Þ\[x=-3,\ 1\] Now \[{f}''\,(x)=\frac{8}{{{(1+x)}^{3}}}\], \[{f}''\,(-3)=-ve\], \[{f}''\,(1)=+ve\] Hence minimum value at \[x=1\] \[f(1)=\frac{(5+1)(2+1)}{(1+1)}=\frac{6\times 3}{2}=9\].You need to login to perform this action.
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