Let f: RR be a twice differentiable function such that the quadratic equation prime prime (x)=0 in m, has two equal roots for every . If f(0) = 1 f'(0) = 2 and is the largest interval in which the function is increasing, then is equal to

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Solution

Q.
Let $f : R \to R$ 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 $(α, β)$ is the largest interval in which the function $f (\log_{e} x - x)$ is increasing, then $α + β$ is equal to [2026]

Ans.
(1)

Given quadratic equation has equal roots, thus

$D = 0 \Rightarrow (f' {(x))}^{2} = f'' (x) \cdot f (x)$

$\frac{f' (x)}{f (x)} = \frac{f'' (x)}{f' (x)}$

Integrate,

$l n (f (x)) = l n (f' (x)) + \ln C \Rightarrow f (x) = c f' (x)$

Put $x = 0$ ,

$1 = c \cdot 2 \Rightarrow c = \frac{1}{2}$

Now, $2 f (x) = f' (x)$

$\Rightarrow \frac{f' (x)}{f (x)} = 2$

Integrate,

$l n (f (x)) = 2 x + d$

$\Rightarrow d = 0$

$\Rightarrow l n (f (x)) = 2 x \Rightarrow f (x) = e^{2 x}$

Now let $g (x) = f (l n x - x) = e^{2 (l n x - x)}$

$∴$ $g' (x) = 2 e^{2 (l n x - x)} (\frac{1}{x} - 1) \geq 3$

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

$\Rightarrow x \in (0, 1]$

$\Rightarrow α = 0, β = 1$

$α + β = 1 .$

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