Question

If $f(x)$ be a twice differentiable function from R → R such that $t^2 f(x) - 2tf'(x) + f''(x) = 0$ has two equal roots of $t$, $\forall x$ and $f(0) = 1$, $f'(0) = 3$, then the value of $\lim_{x \to 0} \left( \frac{f(x)-1}{x} - \frac{t}{3} \right)$ is

Answer 1

2

Answer 2

The condition that the quadratic equation $t^2 f(x) - 2tf'(x) + f''(x) = 0$ has two equal roots in $t$ implies that its discriminant is zero. The discriminant is $(-2f'(x))^2 - 4(f(x))(f''(x))$. Setting this to zero gives $4(f'(x))^2 - 4f(x)f''(x) = 0$, which simplifies to $(f'(x))^2 = f(x)f''(x)$.

This differential equation can be rewritten as $\frac{f''(x)}{f'(x)} = \frac{f'(x)}{f(x)}$. Integrating both sides with respect to $x$ yields $\ln|f'(x)| = \ln|f(x)| + C_1$, which leads to $f'(x) = kf(x)$ for some constant $k$.

The solution to $f'(x) = kf(x)$ is $f(x) = Ae^{kx}$. Using the initial condition $f(0) = 1$, we get $A e^{k \cdot 0} = 1$, so $A=1$. Thus, $f(x) = e^{kx}$. The derivative is $f'(x) = ke^{kx}$. Using the initial condition $f'(0) = 3$, we get $ke^{k \cdot 0} = 3$, so $k=3$. Therefore, the function is $f(x) = e^{3x}$.

The value of the repeated root $t$ for the quadratic equation $t^2 f(x) - 2tf'(x) + f''(x) = 0$ is given by $t = \frac{2f'(x)}{2f(x)} = \frac{f'(x)}{f(x)}$. Substituting $f(x)=e^{3x}$ and $f'(x)=3e^{3x}$, we find $t = \frac{3e^{3x}}{e^{3x}} = 3$. So, $t$ is a constant value of $3$.

The limit to evaluate is $\lim_{x \to 0} \left( \frac{f(x)-1}{x} - \frac{t}{3} \right)$. Substituting $f(x) = e^{3x}$ and $t=3$, we get: $\lim_{x \to 0} \left( \frac{e^{3x}-1}{x} - \frac{3}{3} \right) = \lim_{x \to 0} \left( \frac{e^{3x}-1}{x} - 1 \right)$ We know the standard limit $\lim_{y \to 0} \frac{e^y-1}{y} = 1$. By setting $y=3x$, we have $\lim_{x \to 0} \frac{e^{3x}-1}{x} = \lim_{x \to 0} \frac{e^{3x}-1}{3x} \cdot 3 = 1 \cdot 3 = 3$. Alternatively, $\lim_{x \to 0} \frac{f(x)-1}{x}$ is the definition of $f'(0)$. Given $f'(0)=3$, the limit is $f'(0) - \frac{t}{3} = 3 - \frac{3}{3} = 3 - 1 = 2$.