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JAMIA MILLIA ISLAMIA Jamia Millia Islamia Solved Paper-2012

  • question_answer
        Let\[P(a\sec \theta ,b\tan \theta )\]and\[Q(a\sec \phi ,b\tan \phi ),\]where\[\theta +\phi =\frac{\pi }{2}\]be two points on the hyperbola\[\frac{{{x}^{2}}}{{{a}^{2}}}-\frac{{{y}^{2}}}{{{b}^{2}}}=1\]. If\[(h,k)\]is the point on intersection of the normals at P and Q, then k is equal to

    A)  \[\frac{{{a}^{2}}+{{b}^{2}}}{a}\]                              

    B)  \[-\left( \frac{{{a}^{2}}+{{b}^{2}}}{a} \right)\]

    C)  \[\frac{{{a}^{2}}+{{b}^{2}}}{b}\]                              

    D)  \[-\left( \frac{{{a}^{2}}+{{b}^{2}}}{b} \right)\]

    Correct Answer: D

    Solution :

                    Let\[y=f(x)\] \[\Rightarrow \]               \[y=\frac{{{10}^{x}}-{{10}^{x}}}{{{10}^{x}}+{{10}^{-x}}}\] \[\Rightarrow \]               \[y({{10}^{x}}+{{10}^{-x}})={{10}^{x}}-{{10}^{-x}}\] \[\Rightarrow \]               \[y{{.10}^{2x}}-{{10}^{+2x}}=-1-y\] \[\Rightarrow \]               \[{{10}^{2x}}(1-y)=1+y\] \[\Rightarrow \]               \[{{10}^{2x}}=\frac{1+y}{1-y}\] \[\Rightarrow \]               \[2x={{\log }_{10}}\frac{1+y}{1-y}\] \[\Rightarrow \]               \[x=\frac{1}{2}{{\log }_{10}}\left( \frac{1+y}{1-y} \right)\] \[\Rightarrow \]               \[{{f}^{-1}}(y)=\frac{1}{2}{{\log }_{10}}\left( \frac{1+y}{1-y} \right)\]                                     \[[\because y=f(x)\Rightarrow {{f}^{-1}}(y)=x]\] \[\Rightarrow \]               \[{{f}^{-1}}(x)=\frac{1}{2}{{\log }_{10}}\left( \frac{1+x}{1-x} \right)\]


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