Number of Roots of Derivative Combination Using Rolles Theorem

To determine the number of solutions to the equation $f'(x) - 3g'(x) = 0$ in the given intervals, let us define an auxiliary function $h: [-1, 2] \to \mathbb{R}$ by: $h(x) = f(x) - 3g(x)$

1. Continuity and Differentiability

Since $f$ and $g$ are continuous on $[-1, 2]$ and twice differentiable on $(-1, 2)$, the function $h$ is also:

• Continuous on $[-1, 2]$
• Twice differentiable on $(-1, 2)$

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2. Value of $h(x)$ at the Given Points

Using the given table of values for $f(x)$ and $g(x)$:

• At $x = -1$: $h(-1) = f(-1) - 3g(-1) = 3 - 3(0) = 3$

• At $x = 0$: $h(0) = f(0) - 3g(0) = 6 - 3(1) = 3$

• At $x = 2$: $h(2) = f(2) - 3g(2) = 0 - 3(-1) = 3$

Thus, we have: $h(-1) = h(0) = h(2) = 3$

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3. Existence of Solutions Using Rolle's Theorem

• In the interval $[-1, 0]$: The function $h(x)$ is continuous on $[-1, 0]$, differentiable on $(-1, 0)$, and $h(-1) = h(0) = 3$. By Rolle's Theorem, there exists at least one $c_1 \in (-1, 0)$ such that: $h'(c_1) = 0 \implies f'(c_1) - 3g'(c_1) = 0$

• In the interval $[0, 2]$: The function $h(x)$ is continuous on $[0, 2]$, differentiable on $(0, 2)$, and $h(0) = h(2) = 3$. By Rolle's Theorem, there exists at least one $c_2 \in (0, 2)$ such that: $h'(c_2) = 0 \implies f'(c_2) - 3g'(c_2) = 0$

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4. Uniqueness of Solutions

We are given that $h''(x) = (f - 3g)''(x) \neq 0$ for all $x \in (-1, 0)$ and $x \in (0, 2)$.

• Uniqueness in $(-1, 0)$: Suppose $h'(x) = 0$ has two or more solutions in $(-1, 0)$. Let $a, b \in (-1, 0)$ with $a < b$ be two distinct roots such that $h'(a) = 0$ and $h'(b) = 0$. By Rolle's Theorem applied to $h'(x)$ on $[a, b]$, there must exist some $k \in (a, b) \subset (-1, 0)$ such that: $h''(k) = 0$ This contradicts the given condition that $h''(x)$ never vanishes in $(-1, 0)$. Thus, $h'(x) = 0$ can have at most one solution in $(-1, 0)$. Since it has at least one solution, it must have exactly one solution in $(-1, 0)$.

• Uniqueness in $(0, 2)$: By the exact same reasoning, if $h'(x) = 0$ had two or more solutions in $(0, 2)$, then $h''(x)$ would vanish at least once in $(0, 2)$, which contradicts the given premise. Therefore, $h'(x) = 0$ has exactly one solution in $(0, 2)$.

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Conclusion

• In $(-1, 0)$, $f'(x) - 3g'(x) = 0$ has exactly one solution. (Option B is correct)
• In $(0, 2)$, $f'(x) - 3g'(x) = 0$ has exactly one solution. (Option C is correct)
• The total number of solutions in $(-1, 0) \cup (0, 2)$ is $1 + 1 = 2$ (hence Options A and D are incorrect).

Correct Options: (B) and (C)