A) \[=ab(a+b)-6abh+8{{h}^{3}}=0\]
B) \[{{a}^{2}}b+ab+6abh+8h=0\]
C) \[{{a}^{2}}b+a{{b}^{2}}-3abh+8{{h}^{3}}=0\]
D) \[{{a}^{2}}b+a{{b}^{2}}-6abh-8{{h}^{3}}=0\]
Correct Answer: A
Solution :
Here, \[{{m}_{1}}=m_{2}^{2}\,\Rightarrow m_{2}^{2}+{{m}_{2}}\,=\frac{-2h}{b}\] ?(i) and \[m_{2}^{2}{{m}_{2}}=\frac{a}{b}\] \[\Rightarrow \] \[{{m}_{2}}={{\left( \frac{a}{b} \right)}^{1/3}}\] ?(ii) Putting this value of \[{{m}_{2}}\]in Eq. (i) we get \[{{\left\{ {{\left( \frac{a}{b} \right)}^{1/3}} \right\}}^{2}}+{{\left( \frac{a}{b} \right)}^{1/3}}=\frac{-2h}{b}\] On cubing both sides, we get \[{{\left( \frac{a}{b} \right)}^{2}}+\frac{a}{b}+3{{\left( \frac{a}{b} \right)}^{2/3}}.{{\left( \frac{a}{b} \right)}^{1/3}}\] \[\left\{ {{\left( \frac{a}{b} \right)}^{2/3}}+{{\left( \frac{a}{b} \right)}^{1/3}} \right\}=\frac{-8{{h}^{3}}}{{{b}^{3}}}\] \[\Rightarrow \] \[{{\left( \frac{a}{b} \right)}^{2}}+\frac{a}{b}-\frac{6ah}{{{b}^{2}}}=\frac{-8{{h}^{3}}}{{{b}^{3}}}\] \[\left\{ \therefore \,{{\left( \frac{a}{b} \right)}^{2/3}}+{{\left( \frac{a}{b} \right)}^{1/3}}=\frac{-2h}{b} \right\}\] \[=ab(a+b)-6abh+8{{h}^{3}}=0\]
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