question_answer

The area between \[y={{x}^{2}}-3\] and the line \[y=kx\] + 2 is least when k =

A)  0                                            

B)  1

C)  2                                            

D)  None of these

Correct Answer: A

Solution :

\[{{x}^{2}}-3=kx+2\] \[\Rightarrow \,\,{{x}^{2}}-kx-5=0\] \[\Rightarrow \,\,x=\frac{k\pm \,\sqrt{{{k}^{2}}+20}}{2}\] \[f(k)=\int\limits_{\frac{k-\sqrt{{{k}^{2}}+20}}{2}}^{\frac{k+\sqrt{{{k}^{2}}+20}}{2}}{(kx+5-{{x}^{2}})dx}\] \[f(k)=\frac{k}{2}\,\left( \frac{4k\sqrt{{{k}^{2}}+20}}{4} \right)+5\left( \frac{2\sqrt{{{k}^{2}}+20}}{2} \right)\] \[-\frac{1}{3}\,\left( \frac{2\sqrt{{{k}^{2}}+20}(2{{k}^{2}}+2{{k}^{2}}+40-20)}{8} \right)\] \[=\sqrt{{{k}^{2}}+20}\left( \frac{1}{2}{{k}^{2}}+5-\frac{{{k}^{2}}+5}{3} \right)\] \[=\left( \frac{{{k}^{2}}+20}{6} \right)\sqrt{{{k}^{2}}+20}=\frac{{{({{k}^{2}}+20)}^{3/2}}}{6}\]