A) \[{{(8-4\sqrt{2})}^{1/3}}\]
B) \[{{(8+\sqrt{2})}^{1/3}}\]
C) no value of\[a\]
D) None of these
Correct Answer: A
Solution :
Curve\[y={{x}^{2}}(8-{{x}^{3}}),\]intersect\[x-\]axis at\[x=0\] and\[x=2\]. Here, \[y>0,0<x<2\] and \[y<0,\text{ }x>2\] Area bounded by the curve and\[x-\]axis \[=\int_{0}^{2}{(8{{x}^{2}}-{{x}^{5}})}dx=\left[ \frac{8{{x}^{3}}}{3}-\frac{{{x}^{6}}}{6} \right]_{0}^{2}\] \[=\frac{8}{3}.8-\frac{64}{6}=\frac{64}{3}-\frac{32}{3}\] \[=\frac{32}{3}>\frac{16}{3}\] \[\therefore \] \[a<2\] and area\[=\int_{0}^{a}{(8{{x}^{2}}-{{x}^{5}})}dx=\frac{16}{3}\] \[\Rightarrow \] \[\frac{8{{a}^{3}}}{3}-\frac{{{a}^{6}}}{6}=\frac{16}{3}\] \[\Rightarrow \] \[16{{a}^{3}}-{{a}^{6}}=32\] \[\Rightarrow \] \[{{a}^{6}}-16{{a}^{3}}+32=0\] \[\Rightarrow \] \[{{a}^{3}}=\frac{+16\pm \sqrt{16\times 16-4\times 32}}{2}\] \[\Rightarrow \] \[{{a}^{3}}=\frac{16\pm 8\sqrt{2}}{2}\] \[\Rightarrow \] \[{{a}^{3}}=8\pm 4\sqrt{2}\] \[a<2\] \[\Rightarrow \] \[{{a}^{3}}<8\] \[\therefore \] \[{{a}^{3}}=8-4\sqrt{2}\] \[\Rightarrow \] \[a={{(8-4\sqrt{2})}^{1/3}}\]
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