Let f:R R be a twice differentiable function such that the quadratic equation f(x)m^2-2f^'…

Question

Let $f:R	o R$ be a twice differentiable function such that the quadratic equation $f(x){m}^{2}-2{f}^{'}(x)m+{f}^{''}(x)=0$ in m , has two equal roots for every $x\in R$. If $f(0)=1,{f}^{'}(0)=2$, and ( $\alpha ,\beta $ ) is the largest interval in which the function $f({\log }_{e}x-x)$ is increasing, then $\alpha +\beta $ is equal to

$____$ .

Answer: 1

Accepted Answer

For equal roots, discriminant is 0 .

$4{({f}^{'}(x))}^{2}-4(f(x))({f}^{''}(x))=0$

Let $y=f(x)$

$\begin{gathered} {({y}^{'})}^{2}=y{y}^{''} \ \Rightarrow \frac{{y}^{'}}{y}=\frac{{y}^{''}}{{y}^{'}} \end{gathered}$

$\begin{gathered} \Rightarrow \ln |y|=\ln |{y}^{'}|+\ln K \ \ln (1)=\ln |2|+\ln (K)\Rightarrow \ \ln K=-\ln 2 \end{gathered}$

$\begin{gathered} \Rightarrow \, \ln |y|=\ln |{y}^{'}|-\ln 2 \ \Rightarrow \, \ln |\frac{y}{{y}^{'}}|=-\ln 2=\ln (\frac{1}{2}) \ \Rightarrow \, 2y={y}^{'} \ \Rightarrow \, 2=|\frac{{y}^{'}}{y}|\Rightarrow 2x=\ln |y|+\ln \lambda \ \, 0=\ln |1|+\ln \lambda \ \Rightarrow \, \ln \lambda =0 \ \Rightarrow \, \ln |y|=2x \ y={e}^{2x} \end{gathered}$

$\begin{gathered} {f}^{(\ln x-x)}={e}^{2(\ln x-x)}={e}^{2(\ln {x}^{2}-2x)} \ =\frac{{x}^{2}}{{e}^{2x}} \ \frac{{x}^{2}}{{e}^{2x}}(0,1) \ \Rightarrow \alpha +\beta =1 \end{gathered}$

Solutions · 1