A) \[{{m}^{2}}y\]
B) \[-{{m}^{2}}y\]
C) \[\pm {{m}^{2}}y\]
D) None of these
Correct Answer: A
Solution :
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 \[({{x}^{2}}-1)2{{y}_{1}}{{y}_{2}}+2xy_{1}^{2}=2{{m}^{2}}y{{y}_{1}}\] \[\Rightarrow \] \[{{y}_{1}}{{y}_{2}}({{x}^{2}}-1)+xy_{1}^{2}={{m}^{2}}y{{y}_{1}}\] \[\Rightarrow \] \[({{x}^{2}}-1){{y}_{2}}+x{{y}_{1}}={{m}^{2}}y\]
You need to login to perform this action.
You will be redirected in
3 sec
