Question

If $ {{y}^{1/m}}+{{y}^{-1/m}}=2x, $ then the value of $ ({{x}^{2}}-1){{y}_{2}}+x{{y}_{1}}, $ is

• A. $ {{m}^{2}}y $
• B. $ -{{m}^{2}}y $
• C. $ \pm {{m}^{2}}y $
• D. None of these

Answer 1

Answer: (A)

Given, $ {{y}^{1/m}}+{{y}^{-1/m}}=2x $
$ \Rightarrow $ $ {{y}^{1/m}}+\frac{1}{{{y}^{1/m}}}=2x $
$ \Rightarrow $ $ {{({{y}^{1/m}})}^{2}}+1=2x{{y}^{1/m}} $
$ \Rightarrow $ $ {{({{y}^{1/m}})}^{2}}-2x{{y}^{1/m}}+1=0, $ which is a quadratic equation.
So, $ {{y}^{1/m}}=\frac{2x\pm \sqrt{4{{x}^{2}}-4}}{2} $
$ \Rightarrow $ $ {{y}^{1/m}}=x\pm \sqrt{{{x}^{2}}-1} $
$ \Rightarrow $ $ y={{[x\pm \sqrt{{{x}^{2}}-1}]}^{m}} $ ...(i)
$ \therefore $ $ \frac{dy}{dx}={{y}_{1}} $
$ =m{{[x\pm \sqrt{{{x}^{2}}-1}]}^{m-1}} $
$ \left[ 1\pm \frac{1}{2\sqrt{{{x}^{2}}-1}}.2x \right] $
$ \Rightarrow $ $ {{y}_{1}}=m{{[x\pm \sqrt{{{x}^{2}}-1}]}^{m-1}}\left[ \frac{\sqrt{{{x}^{2}}-1}\pm x}{\sqrt{{{x}^{2}}-1}} \right] $
$ \Rightarrow $ $ {{y}_{1}}=\frac{m{{[x\pm \sqrt{{{x}^{2}}-1}]}^{m}}}{\sqrt{{{x}^{2}}-1}} $
$ \Rightarrow $ $ {{y}_{1}}=\frac{my}{\sqrt{{{x}^{2}}-1}} $ [from Eq.(i)] $ \Rightarrow $
$ \sqrt{{{x}^{2}}-1}.{{y}_{1}}=my $
On squaring both sides, we get
$ ({{x}^{2}}-1)y_{1}^{2}={{m}^{2}}{{y}^{2}} $
On differentiating both sides, we get
$ ({{x}^{2}}-1)2{{y}_{1}}{{y}_{2}}+2xy_{1}^{2}=2{{m}^{2}}y{{y}_{1}} $
$ \Rightarrow $ $ {{y}_{1}}{{y}_{2}}({{x}^{2}}-1)+xy_{1}^{2}={{m}^{2}}y{{y}_{1}} $
$ \Rightarrow $ $ ({{x}^{2}}-1){{y}_{2}}+x{{y}_{1}}={{m}^{2}}y $