A) -4 and 2
B) 2 and 4
C) -2 and -4
D) 8 and -8
Correct Answer: A
Solution :
[a] \[y=\int_{0}^{4/a}{(a\cdot x-\sqrt{4a\cdot x})dx}\] \[\frac{1}{3}=\int_{0}^{4/a}{ax\,\,dx-\int_{0}^{4/a}{\sqrt{4ax}\,\,dx}}\] \[\frac{1}{3}=\left[ \frac{a{{x}^{2}}}{2} \right]_{0}^{4/a}-2\left[ \frac{{{(4ax)}^{3/2}}}{3} \right]_{0}^{4/a}\] \[\frac{1}{3}=\frac{\frac{16a}{{{a}^{2}}}}{2}-\frac{2}{3}\left[ 4a{{\left( \frac{4}{a} \right)}^{3/2}} \right],a=8\]. Putting the value of a in \[{{x}^{2}}+2x-a=0\], we get its roots i.e., -4 and 2.
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