Question

Let domain and range of \(f(x)\) and \(g(x)\) is \([0,\infty)\). If \(f(x)\) is an increasing function, \(g(x)\) is a decreasing function, \(h(x)=f\{g(x)\}\), \(h(0)=0\) and \(p(x)=h(x^{3}-2x^{2}+2x)-h(4)\) then for all \(x\in(0,2)\):

• A. $p(x)=-3$
• B. $p(x)=0$
• C. $0
• D. $0\le p(x)\le-h(4)$

Hint:

Remember this handy composite rule: composing two functions with the same monotonicity (both increasing or both decreasing) creates an *increasing* function, while composing two functions with different monotonicity creates a *decreasing* function!

Answer 1

The Correct Option is C

Concept: The composite function formed by nesting a decreasing function inside an increasing function results in an overall decreasing function. We solve this by finding the range of the internal polynomial argument across the interval domain and tracking it through the function layers. Step 1: Determine the monotonicity of the composite function \(h(x)\).
We are given that $g(x)$ is a decreasing function and $f(x)$ is an increasing function.
• Since $g(x)$ is decreasing: $x_1 < x_2 \implies g(x_1) > g(x_2)$.
• Since $f(x)$ is increasing, it preserves this flipped direction: $f(g(x_1)) > f(g(x_2))$. Therefore, the composite function $h(x) = f(g(x))$ behaves as a strictly decreasing function.

Step 2: Analyze the behavior of the internal polynomial argument.
Let our internal expression be defined as $\phi(x) = x^3 - 2x^2 + 2x$. Let us find its first derivative to check its behavior: $$\phi'(x) = 3x^2 - 4x + 2$$ Calculate the discriminant of this quadratic derivative function: $D = (-4)^2 - 4(3)(2) = 16 - 24 = -8$. Since the discriminant is negative ($D < 0$) and the leading coefficient is positive ($3 > 0$), the derivative $\phi'(x)$ is strictly greater than zero for all real inputs. This confirms that $\phi(x)$ is a strictly increasing function.

Step 3: Find the boundaries of the polynomial across the interval domain.
Evaluate the values of our increasing polynomial function $\phi(x)$ at the limits of the open interval $x \in (0, 2)$:
• Lower limit: $\phi(0) = 0^3 - 2(0)^2 + 2(0) = 0$
• Upper limit: $\phi(2) = 2^3 - 2(2)^2 + 2(2) = 8 - 8 + 4 = 4$ Since the function increases continuously between these endpoints, its values are bounded by: $$0 < x^3 - 2x^2 + 2x < 4$$

Step 4: Track the inequalities through the decreasing function \(h(x)\).
Apply the decreasing function $h(x)$ across our inequality chain. Because $h(x)$ is decreasing, it flips the direction of the inequality signs: $$h(0) > h(x^3 - 2x^2 + 2x) > h(4)$$ Substitute the initial boundary condition $h(0) = 0$ given in the problem statement: $$0 > h(x^3 - 2x^2 + 2x) > h(4) \quad \Rightarrow \quad h(4) < h(x^3 - 2x^2 + 2x) < 0$$ Subtract the constant term $h(4)$ from all parts of the inequality to isolate our target function $p(x)$: $$0 < h(x^3 - 2x^2 + 2x) - h(4) < -h(4) \quad \Rightarrow \quad 0 < p(x) < -h(4)$$ This matches option (C) perfectly.