Question-25
Let \(\displaystyle f\) be a twice differentiable, real-valued function. Let \(\displaystyle g( x) =f^{\prime }( x)\) and \(\displaystyle g^{\prime }( x) =-f( x)\). If \(\displaystyle f( 5) =f^{\prime }( 5) =2\), find \(\displaystyle f^{2}( 10) +g^{2}( 10)\).
Let \(h(x) = f^2(x) + g^2(x)\). Now differentiate.
\(8\)
Let \(\displaystyle h( x) =f^{2}( x) +g^{2}( x)\), then:
\[\begin{equation*} \begin{aligned} h^{\prime }( x) & =2f( x) f^{\prime }( x) +2g( x) g^{\prime }( x)\\ & =2f( x) f^{\prime }( x) -2f^{\prime }( x) f( x)\\ & =0 \end{aligned} \end{equation*}\]
Hence, \(\displaystyle h( x) =c\), where \(\displaystyle c\) is some real number.
\[\begin{equation*} \begin{aligned} f^{2}( 10) +g^{2}( 10) & =h( 10)\\ & =h( 5)\\ & =f^{2}( 5) +g^{2}( 5)\\ & =f^{2}( 5) +f{^{\prime }}^{2}( 5)\\ & =4+4\\ & =8 \end{aligned} \end{equation*}\]