Let h(x) = f(x) - (f(x))² + (f(x))³ for every real number x. Then (A) h is increasing whenever f is

Q: Wait, why do we need to calculate the discriminant of the quadratic part? Can't we just look at the expression?

That is a great question! While you might be tempted to guess, in mathematics, especially in JEE Advanced, we need rigorous proof. By calculating the discriminant $D = (-2)^2 - 4(3)(1) = -8$, we prove that the quadratic $3y^2 - 2y + 1$ has no real roots and, because the leading coefficient $3$ is positive, it is strictly positive for all real $y$. This guarantees that the sign of $h'(x)$ is always the same as the sign of $f'(x)$.

Q: Does this mean $h(x)$ is always increasing if $f(x)$ is increasing?

Exactly! Since we proved that the quadratic factor $3(f(x))^2 - 2f(x) + 1$ is always positive, the derivative $h'(x)$ will have the same sign as $f'(x)$. If $f(x)$ is increasing, $f'(x) > 0$, which forces $h'(x) > 0$, making $h(x)$ increasing as well. They are locked in sync.

Q: What if the quadratic had a negative leading coefficient? Would the answer change?

That is a brilliant hypothetical! If the leading coefficient were negative, the parabola would open downwards. If the discriminant were still negative, the entire expression would be strictly negative. In that case, $h'(x)$ would have the opposite sign of $f'(x)$, meaning $h(x)$ would be decreasing whenever $f(x)$ is increasing. Always check the leading coefficient and the discriminant!

Question

Let $h(x) = f(x) - (f(x))^2 + (f(x))^3$ for every real number $x$. Then

Options:

(A) $h$ is increasing whenever $f$ is increasing

(B) $h$ is increasing whenever $f$ is decreasing

(C) $h$ is decreasing whenever $f$ is increasing

(D) nothing can be said in general.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing at the edge of a vast mathematical landscape. You have been given a function $h(x) = f(x) - (f(x))^2 + (f(x))^3$ and asked to determine its behavior based on the behavior of $f(x)$.

At first glance, this might look like a daunting algebraic mess, but let us peel back the layers together. This problem is not about brute force; it is about understanding the elegant relationship between a function and its composition.

### The Derivative: Our Compass

To understand if a function is increasing or decreasing, we must look at its rate of change. We need the derivative, $h'(x)$.

Applying the derivative operator to our definition, we get:

$$h'(x) = \frac{d}{dx}[f(x) - (f(x))^2 + (f(x))^3]$$

This is where the Chain Rule becomes our best friend. When we differentiate each term, we must remember to multiply by the derivative of the inner function, $f'(x)$.

The result is:

$$h'(x) = f'(x) - 2f(x)f'(x) + 3(f(x))^2 f'(x)$$

### The Power of Factoring

Now, look closely at that expression. Do you see the hidden pattern? The term $f'(x)$ is common to every single part of the equation.

By factoring it out, we transform our expression into:

$$h'(x) = f'(x) [3(f(x))^2 - 2f(x) + 1]$$

This is a pivotal moment. We have separated the derivative into two distinct parts: the rate of change of the original function, $f'(x)$, and a quadratic expression in terms of $f(x)$. Let us define this quadratic as $Q(y) = 3y^2 - 2y + 1$, where $y = f(x)$.

### The Quadratic Mystery

The behavior of $h'(x)$ now hinges entirely on the sign of $Q(y)$. Can this quadratic ever be negative?

To find out, we calculate the discriminant $D = b^2 - 4ac$. With $a = 3$, $b = -2$, and $c = 1$, we find:

$$D = (-2)^2 - 4(3)(1) = 4 - 12 = -8$$

Because the discriminant is negative and the leading coefficient $a = 3$ is positive, the parabola $Q(y)$ never touches the horizontal axis and stays strictly above it. This means $Q(y) > 0$ for all real values of $y$.

### The Final Synthesis

We have reached the climax of our journey. We know that $h'(x) = f'(x) \cdot Q(f(x))$.

Since $Q(f(x))$ is always positive, it acts like a positive multiplier that does not change the sign of $f'(x)$. Therefore, the sign of $h'(x)$ is identical to the sign of $f'(x)$.

If $f(x)$ is increasing, $f'(x) > 0$, which forces $h'(x) > 0$, meaning $h(x)$ is also increasing. They move in perfect harmony.

We have successfully navigated the complexity and found that **$h$ is increasing whenever $f$ is increasing**. Mathematics is not just about solving for $x$; it is about finding the beautiful, underlying connections that govern the behavior of functions.

Animated Solution for Mathematics - Differentiation: Let $h (x) = f (x) - (f (x))^{2} + (f (x))^{3}$ for every real number $x$ . Then

Select Answer:

Visualized Solution

Defining the Function $h (x)$

The Monotonicity Condition

Differentiating $h (x)$

Applying the Chain Rule

Factoring the Derivative

Rearranging the Quadratic Factor

Introducing the Quadratic Function

Analyzing the Roots of $Q (y)$

Calculating the Discriminant

Geometric Interpretation of $D < 0$

The Quadratic is Always Positive

Connecting Back to $h^{'} (x)$

Final Conclusion

The Sigma Insight: Monotonicity

Analyzing the Setup

The Derivative

The Power of Factoring

The Quadratic Mystery

The Final Synthesis

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Animated Solution for Mathematics - Differentiation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let h(x)=f(x)−(f(x))2+(f(x))3 for every real number x. Then