Let f, g : [-1, 2] → R be continuous functions which are twice differentiable on (-1, 2). Let the va

Q: Why do we define $h(x) = f(x) - 3g(x)$? Can't we just work with $f$ and $g$ separately?

Great question! While you could work with them separately, defining $h(x)$ is a powerful strategy to simplify the problem. By bundling the expression $f(x) - 3g(x)$ into a single function, we transform a complex relationship into a simple search for the roots of $h'(x) = 0$. It turns two functions into one, making the application of theorems like Rolle's much more intuitive.

Q: Why does $h''(x) \neq 0$ imply exactly one root?

Think of $h'(x)$ as a function itself. If its derivative, $h''(x)$, is never zero, then $h'(x)$ is strictly monotonic—it is either always increasing or always decreasing. A function that is strictly moving in one direction can cross the x-axis (i.e., equal zero) at most once. Since Rolle's Theorem guarantees at least one root, and monotonicity limits it to at most one, we are left with exactly one!

Q: Wait, why do we reject the possibility of multiple roots in the interval?

We don't reject it; we prove it's impossible! If there were two roots in an interval, say $c_1$ and $c_2$, then by Rolle's Theorem applied to $h'(x)$, there would have to be a point between them where the derivative of $h'(x)$—which is $h''(x)$—is zero. But the problem explicitly tells us $h''(x)$ is never zero. Therefore, having two roots is mathematically impossible.

Question

Let $f, g : [-1, 2] 	o \mathbb{R}$ be continuous functions which are twice differentiable on $(-1, 2)$. Let the values of $f$ and $g$ at the points $-1, 0$ and $2$ be as given in the following table: 
\begin{center}
\begin{array}{|c|c|c|c|}
\hline
 & $x = -1$ & $x = 0$ & $x = 2$ \ \hline
$f(x)$ & 3 & 6 & 0 \ \hline
$g(x)$ & 0 & 1 & -1 \ \hline
\end{array}
\end{center}
In each of the intervals $(-1, 0)$ and $(0, 2)$ the function $(f - 3g)''$ never vanishes. Then the correct statement(s) is(are)

Options:

(A) $f'(x) - 3g'(x) = 0$ has exactly three solutions in $(-1, 0) \cup (0, 2)$

(B) $f'(x) - 3g'(x) = 0$ has exactly one solution in $(-1, 0)$

(C) $f'(x) - 3g'(x) = 0$ has exactly one solution in $(0, 2)$

(D) $f'(x) - 3g'(x) = 0$ has exactly two solutions in $(-1, 0)$ and exactly two solutions in $(0, 2)$

Sigma Percentile · Accepted Answer

### The Detective's Approach to Calculus

Welcome, future engineer! Today, we are going to dive into a problem that might look like a dry table of values, but is actually a beautiful puzzle waiting to be solved.

When you see a problem asking about the roots of a derivative, your first instinct should be to look for the **Mean Value Theorem** or **Rolle's Theorem**. Let's peel back the layers of this problem together.

### The Auxiliary Function: Our Secret Weapon

We are given information about $f(x)$ and $g(x)$ at $x = -1, 0, 2$. The question asks us to analyze the roots of $f'(x) - 3g'(x) = 0$.

Let's define a new auxiliary function:
$$h(x) = f(x) - 3g(x)$$

Now, our target equation is simply $h'(x) = 0$. Because $f$ and $g$ are continuous and twice differentiable, our new function $h(x)$ inherits these wonderful properties. We have essentially transformed the problem into finding the horizontal tangents of $h(x)$.

### The Table of Truth

Now, let's evaluate $h(x)$ at the points provided in the table. This is where the magic happens.

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

Moving to $x = 0$:
$$h(0) = f(0) - 3g(0) = 6 - 3(1) = 3$$

Finally, at $x = 2$:
$$h(2) = f(2) - 3g(2) = 0 - 3(-1) = 3$$

Look at that! $h(-1) = h(0) = h(2) = 3$. All three points lie on the horizontal line $y = 3$. This is not a coincidence; it is the core of the problem.

### The Rolle's Theorem Insight

Imagine you are drawing a smooth, continuous curve that passes through these three points at the same height. To get from $x = -1$ to $x = 0$ and stay smooth, the curve must turn.

**Rolle's Theorem** states that if a function is continuous and differentiable on an interval $[a, b]$ and $h(a) = h(b)$, there must be at least one point $c$ in $(a, b)$ where $h'(c) = 0$.

Applying this to our intervals, we know there is at least one root in $(-1, 0)$ and at least one root in $(0, 2)$.

### The Second Derivative Constraint

But wait, the question asks for the **exact** number of solutions. We know there is at least one, but could there be more?

This is where the final, crucial piece of information comes in: $(f - 3g)''(x) 
eq 0$, which means $h''(x) 
eq 0$. Physically, the second derivative represents the rate of change of the slope.

If $h''(x)$ is never zero, the slope of $h(x)$ is never constant and, more importantly, it never stops changing in the same direction. This means $h'(x)$ is strictly monotonic.

A strictly monotonic function can cross the zero line at most once. If it crossed twice, it would have to turn around, which would require the derivative to be zero at some point—but we just established that $h''(x)$ is never zero!

### The Final Victory

We have combined the power of Rolle's Theorem, which guarantees at least one root, with the constraint of the second derivative, which guarantees at most one root.

The conclusion is inescapable: there is **exactly one root in $(-1, 0)$ and exactly one root in $(0, 2)$**.

You have just navigated a complex calculus problem using nothing but fundamental theorems and logical deduction. Keep this detective mindset, and no JEE problem will ever be able to stand in your way!

Select Answer:

Visualized Solution

Analyzing the Target Equation

Defining the Auxiliary Function

Evaluating $h (x)$ at $x = - 1$

Evaluating $h (x)$ at $x = 0$

Evaluating $h (x)$ at $x = 2$

Visualizing the Function $h (x)$

Applying Rolle's Theorem

Roots in the Interval $(- 1, 0)$

Roots in the Interval $(0, 2)$

Analyzing the Second Derivative

Monotonicity of $h^{'} (x)$

Final Conclusion

The Sigma Insight: Mean Value Theorems

The Detective's Approach to Calculus

The Auxiliary Function

The Table of Truth

The Rolle's Theorem Insight

The Second Derivative Constraint

The Final Victory

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Select Answer:

Visualized Solution

Analyzing the Target Equation

Defining the Auxiliary Function

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } h(x) at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=−1

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } h(x) at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=0

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Visualizing the Function .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } h(x)