A) \[b-a\]
B) \[ab(b-a)\]
C) \[a(b-a)\]
D) \[ab(a-b)\]
Correct Answer: B
Solution :
By the given condition \[(1-ab){{m}^{2}}-({{a}^{2}}+{{b}^{2}})m-(1+ab)=0\] \[\Rightarrow \]\[m({{a}^{2}}+{{b}^{2}})+({{m}^{2}}+1)ab={{m}^{2}}-1\] ......(i) Now \[{{H}_{1}}=\]First H.M. between \[a\] and \[b\] \[=\frac{(m+1)ab}{a+mb}\] and \[{{H}_{m}}=\frac{(m+1)ab}{b+ma}\] \[\therefore \]\[{{H}_{m}}-{{H}_{1}}=(m+1)ab\left[ \frac{1}{b+ma}-\frac{1}{a+mb} \right]\] \[=(m+1)ab\frac{[(m-1)(b-a)]}{(b+ma)(a+mb)}\]\[=\frac{({{m}^{2}}-1)ab(b-a)}{m({{a}^{2}}+{{b}^{2}})+({{m}^{2}}+1)ab}\] \[=\frac{({{m}^{2}}-1)ab(b-a)}{{{m}^{2}}-1}\] [by (i)] \[=ab(b-a)\].
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