Let f: R -> R be a twice differentiable function such that the quadratic equation f( x) m^2-2 f^prime( x) m+ f^prime( x)=0 in m , has two equal roots for every x in R. If f(0)=1, f^prime(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

Given quadratic equation has equal roots, thus \ D=0 => ( f^prime( x))^2= f^prime prime( x) * f( x) fracf^prime(x)f(x)=fracf^prime prime(x)f^prime(x) Integrate \ l n( f( x))=l n( f^prime( x))+l nC => f( x)= c . f^prime( x) Put = x=0 1= c .2 => c=(1 / 2) Now 2 f( x)= f^prime( x) => frac f^prime( x) f( x)=2 Integrate \ l n( f( x))=2 x+ d => d=0 => ln ( f( x))=2 x => f( x)= e^2 x Now let g( x)= f(ln x- x)= e^2(nx x) Therefore g^prime( x)=2 e^2(l n x- x)(frac1 x-1) >= 3 => (1-x / x) >= 0 => x in (0,1] => alpha=0, beta= alpha+beta=1.

Let f R arrow R be a twice differentiable function such... | null (Mathematics)

Given quadratic equation has equal roots, thus

\\ \mathrm{D}=0 \Rightarrow\left(\mathrm{f}^{\prime}(\mathrm{x})\right)^{2}=\mathrm{f}^{\prime \prime}(\mathrm{x}) \cdot \mathrm{f}(\mathrm{x})\\

\frac{f^{\prime}(x)}{f(x)}=\frac{f^{\prime \prime}(x)}{f^{\prime}(x)}\\

Integrate

\\ \ell \mathrm{n}(\mathrm{f}(\mathrm{x}))=\ell \mathrm{n}\left(\mathrm{f}^{\prime}(\mathrm{x})\right)+\ell \mathrm{nC} \Rightarrow \mathrm{f}(\mathrm{x})=\mathrm{c} . \mathrm{f}^{\prime}(\mathrm{x})\\

Put

=\mathrm{x}=0\\

1=\mathrm{c} .2 \Rightarrow \mathrm{c}=\frac{1}{2}\\

Now

2 \mathrm{f}(\mathrm{x})=\mathrm{f}^{\prime}(\mathrm{x})\\

\Rightarrow \frac{\mathrm{f}^{\prime}(\mathrm{x})}{\mathrm{f}(\mathrm{x})}=2\\

Integrate

\\ \ell \mathrm{n}(\mathrm{f}(\mathrm{x}))=2 \mathrm{x}+\mathrm{d}\\

\Rightarrow \mathrm{d}=0\\

\Rightarrow \ln (\mathrm{f}(\mathrm{x}))=2 \mathrm{x} \Rightarrow \mathrm{f}(\mathrm{x})=\mathrm{e}^{2 \mathrm{x}}\\

Now let

\mathrm{g}(\mathrm{x})=\mathrm{f}(\ln \mathrm{x}-\mathrm{x})=\mathrm{e}^{2(\operatorname{nx} \mathrm{x})}\\

\therefore \mathrm{g}^{\prime}(\mathrm{x})=2 \mathrm{e}^{2(\ell \mathrm{n} \mathrm{x}-\mathrm{x})}\left(\frac{1}{\mathrm{x}}-1\right) \geq 3\\

\Rightarrow \frac{1-x}{x} \geq 0\\

\Rightarrow \mathrm{x} \in(0,1]\\

\Rightarrow \alpha=0, \beta=\\

\alpha+\beta=1

Explanation

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