A) \[{{\left( {{a}^{\frac{3}{2}}}+{{b}^{\frac{3}{2}}} \right)}^{\frac{2}{3}}}\]
B) \[{{\left( {{a}^{\frac{2}{3}}}+{{b}^{\frac{2}{3}}} \right)}^{\frac{3}{2}}}\]
C) \[{{\left( {{a}^{\frac{2}{3}}}+{{b}^{\frac{2}{3}}} \right)}^{3}}\]
D) \[{{\left( {{a}^{\frac{3}{2}}}+{{b}^{\frac{3}{2}}} \right)}^{3}}\]
Correct Answer: B
Solution :
[b] |
From the figure, |
\[PC=b\,\,\cos ec\theta \] and \[AP=a\,\,sec\theta \] |
\[AC=PC+AP\] |
or \[AC=b\,\,\cos ec\theta +a\,\,\sec \theta ....(1)\] |
\[\therefore \frac{d(AC)}{d\theta }=-b\,\,\cos ec\theta \cot \theta +a\,\,\sec \theta \tan \theta \] |
For minimum length, \[\frac{d(AC)}{d\theta }=0\] |
or |
\[a\sec \theta \tan \theta =b\cos ec\theta \cot \theta \] or \[\tan \theta ={{\left( \frac{b}{a} \right)}^{1/3}}\] |
\[\therefore \sin \theta =\frac{{{(b)}^{1/3}}}{\sqrt{{{a}^{2/3}}+{{b}^{2/3}}}}\] and |
\[\cos \theta =\frac{{{(a)}^{1/3}}}{\sqrt{{{a}^{2/3}}+{{b}^{2/3}}}}\] |
Also, \[\theta \in (0,\pi /2)\] |
\[\underset{\theta \to 0}{\mathop{\lim }}\,(a\,\sec \theta +b\,\cos ec\theta )\to \infty \] |
and \[\underset{\theta \to \pi /2}{\mathop{\lim }}\,(a\,\sec \theta +b\,\cos ec\theta )\to \infty \] |
Therefore, \[\theta ={{\tan }^{-1}}{{\left( \frac{b}{a} \right)}^{1/3}}\] is a point of minima. |
For this value of \[\theta \], |
\[AC=\frac{b\sqrt{{{a}^{2/3}}+{{b}^{2/3}}}}{{{b}^{1/3}}}+\frac{a\sqrt{{{a}^{2/3}}+{{b}^{2/3}}}}{{{a}^{1/3}}}\] |
[Using (1) and (2)] |
\[=\sqrt{{{a}^{2/3}}+{{b}^{2/3}}}({{b}^{2/3}}+{{a}^{2/3}})={{({{a}^{2/3}}+{{b}^{2/3}})}^{3/2}}\] |
Hence, the minimum length of the hypotenuse is |
\[{{({{a}^{2/3}}+{{b}^{2/3}})}^{3/2}}.\] |
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