Let f: R to R be a twice differentiable funct | vedclass

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(A) The given quadratic equation $f(x)m^{2}-2f^{\prime}(x)m+f^{\prime\prime}(x)=0$ has equal roots, so its discriminant $D=0$. $D = (-2f^{\prime}(x))^

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(A) The given quadratic equation $f(x)m^{2}-2f^{\prime}(x)m+f^{\prime\prime}(x)=0$ has equal roots, so its discriminant $D=0$. $D = (-2f^{\prime}(x))^{2} - 4f(x)f^{\prime\prime}(x) = 0 \Rightarrow 4(f^{\prime}(x))^{2} = 4f(x)f^{\prime\prime}(x) \Rightarrow (f^{\prime}(x))^{2} = f(x)f^{\prime\prime}(x)$. This implies $\frac{f^{\prime\prime}(x)}{f^{\prime}(x)} = \frac{f^{\prime}(x)}{f(x)}$. Integrating both sides with respect to $x$, we get $\ln(f^{\prime}(x)) = \ln(f(x)) + C_1$, which simplifies to $f^{\prime}(x) = c f(x)$. Given $f(0)=1$ and $f^{\prime}(0)=2$, we have $2 = c(1) \Rightarrow c=2$. Thus, $f^{\prime}(x) = 2f(x) \Rightarrow \frac{f^{\prime}(x)}{f(x)} = 2$. Integrating again, $\ln(f(x)) = 2x + C_2$. Since $f(0)=1$, $\ln(1) = 0 + C_2 \Rightarrow C_2 = 0$. So, $f(x) = e^{2x}$. Now, $g(x) = f(\ln x - x) = e^{2(\ln x - x)}$. For $g(x)$ to be increasing, $g^{\prime}(x) \geq 0$. $g^{\prime}(x) = e^{2(\ln x - x)} \cdot 2(\frac{1}{x} - 1) \geq 0$. Since $e^{2(\ln x - x)} > 0$, we must have $\frac{1-x}{x} \geq 0$. This inequality holds for $x \in (0, 1]$. Thus, the interval $(\alpha, \beta)$ is $(0, 1)$, so $\alpha=0$ and $\beta=1$. Therefore, $\alpha+\beta = 0+1 = 1$.

$(a)$ $x\|x\|$	$(i)$ continuous in $(-1, 1)$
$(b)$ $\sqrt{\|x\|}$	$(ii)$ differentiable in $(-1, 1)$
$(c)$ $x+[x]$	$(iii)$ strictly increasing in $(-1, 1)$
$(d)$ $\|x-1\|+\|x+1\|$	$(iv)$ not differentiable at,at least one point in $(-1, 1)$

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