The rate R at which a chemical reaction in a certain industrial process proceeds is a function of time t...

Understanding the Question

We need to determine if there exists a positive value of t such that \(\mathrm{R} = 0\), where \(\mathrm{R} = \mathrm{a}\mathrm{t}^3 + \mathrm{b}\mathrm{t}^2 + \mathrm{c}\).

This is a yes/no question about whether the cubic function R ever crosses the horizontal axis for positive values of t.

What "sufficient" means here: We need enough information to give a definitive YES or NO answer. Either we can prove a positive root exists (YES) or we can prove no positive root exists (NO).

Given information:

• \(\mathrm{R} = \mathrm{a}\mathrm{t}^3 + \mathrm{b}\mathrm{t}^2 + \mathrm{c}\) (cubic function)
• a, b, and c are constants
• \(\mathrm{t} > 0\) (we only consider positive values)

Key insight: For a continuous function to equal zero, it must cross the horizontal axis. This happens when the function has different signs at different points. Since we're looking at \(\mathrm{t} > 0\):

• As t approaches \(0^+\), R approaches c
• As t approaches infinity, R's sign is dominated by \(\mathrm{a}\mathrm{t}^3\)

Analyzing Statement 1

Statement 1 tells us: \(\mathrm{a} > \mathrm{b}\)

This gives us the relative relationship between a and b, but crucially, we don't know their individual signs. Let's test different scenarios:

Scenario 1: \(\mathrm{a} = 2\), \(\mathrm{b} = 1\), \(\mathrm{c} = 1\) (both a and b positive)

• Near \(\mathrm{t} = 0\): \(\mathrm{R} \approx 1\) (positive)
• For large t: R is dominated by \(2\mathrm{t}^3\) (positive)
• Since R stays positive throughout, no zero exists ✗

Scenario 2: \(\mathrm{a} = 1\), \(\mathrm{b} = -2\), \(\mathrm{c} = 1\) (a positive, b negative)

• Near \(\mathrm{t} = 0\): \(\mathrm{R} \approx 1\) (positive)
• For large t: R is dominated by \(\mathrm{t}^3\) (positive)
• But the negative \(\mathrm{b}\mathrm{t}^2\) term creates a dip. Let's check: At \(\mathrm{t} = 1\), \(\mathrm{R} = 1 - 2 + 1 = 0\)
• A zero exists at t = 1 ✓

Without knowing the sign of c or the specific signs of a and b, we get different possible answers (YES in Scenario 2, NO in Scenario 1).

Statement 1 is NOT sufficient.

[STOP - Not Sufficient!] This eliminates choices A and D.

Analyzing Statement 2

Now let's forget Statement 1 completely and analyze Statement 2 independently.

Statement 2 tells us: \(\mathrm{c} > 0\)

This means when t is very close to \(0^+\), \(\mathrm{R} \approx \mathrm{c}\) which is positive. But what happens as t increases depends entirely on the unknown values of a and b.

Scenario 1: \(\mathrm{a} = 1\), \(\mathrm{b} = 0\), \(\mathrm{c} = 1\)

• \(\mathrm{R} = \mathrm{t}^3 + 1\)
• Since \(\mathrm{t}^3 > 0\) for all \(\mathrm{t} > 0\), we have \(\mathrm{R} > 1\) for all positive t
• No zero exists ✗

Scenario 2: \(\mathrm{a} = -1\), \(\mathrm{b} = 0\), \(\mathrm{c} = 1\)

• \(\mathrm{R} = -\mathrm{t}^3 + 1\)
• Near \(\mathrm{t} = 0\): \(\mathrm{R} \approx 1\) (positive)
• At \(\mathrm{t} = 1\): \(\mathrm{R} = -1 + 1 = 0\)
• For large t: R becomes negative
• Since R is continuous and changes from positive to negative, it must cross zero ✓

Different values of a lead to different answers about whether \(\mathrm{R} = 0\) has a positive solution.

Statement 2 is NOT sufficient.

[STOP - Not Sufficient!] This eliminates choices B and D (already eliminated).

Combining Statements

Using both statements together, we know:

• \(\mathrm{a} > \mathrm{b}\)
• \(\mathrm{c} > 0\)

The key question remains: What is the sign of a?

Case 1: If a > 0 (which allows \(\mathrm{a} > \mathrm{b}\) regardless of b's sign)

• Near \(\mathrm{t} = 0\): \(\mathrm{R} \approx \mathrm{c} > 0\) (positive)
• For large t: R is dominated by \(\mathrm{a}\mathrm{t}^3 > 0\) (positive)
• Even with a negative b term, if \(\mathrm{a} > 0\), the positive \(\mathrm{a}\mathrm{t}^3\) term eventually dominates
• Example: \(\mathrm{a} = 3\), \(\mathrm{b} = -1\), \(\mathrm{c} = 1\)
• \(\mathrm{R} = 3\mathrm{t}^3 - \mathrm{t}^2 + 1\)
• At \(\mathrm{t} = 0.5\): \(\mathrm{R} = 3(0.125) - 0.25 + 1 = 0.375 - 0.25 + 1 = 1.125 > 0\)
• The function might dip but stays positive for all \(\mathrm{t} > 0\)

Case 2: If a < 0 but still a > b (meaning b is even more negative)

• Near \(\mathrm{t} = 0\): \(\mathrm{R} \approx \mathrm{c} > 0\) (positive)
• For large t: R is dominated by \(\mathrm{a}\mathrm{t}^3 < 0\) (negative)
• Since R starts positive and ends negative, it must cross zero ✓

Let's verify with concrete examples:

Example with no positive root: \(\mathrm{a} = 2\), \(\mathrm{b} = 1\), \(\mathrm{c} = 1\)

• Satisfies \(\mathrm{a} > \mathrm{b}\) (2 > 1) ✓ and \(\mathrm{c} > 0\) ✓
• \(\mathrm{R} = 2\mathrm{t}^3 + \mathrm{t}^2 + 1\)
• All positive terms mean \(\mathrm{R} > 1\) for all \(\mathrm{t} > 0\)
• Answer: NO

Example with a positive root: \(\mathrm{a} = -1\), \(\mathrm{b} = -2\), \(\mathrm{c} = 1\)

• Satisfies \(\mathrm{a} > \mathrm{b}\) (-1 > -2) ✓ and \(\mathrm{c} > 0\) ✓
• \(\mathrm{R} = -\mathrm{t}^3 - 2\mathrm{t}^2 + 1\)
• At \(\mathrm{t} = 0.5\): \(\mathrm{R} = -0.125 - 0.5 + 1 = 0.375 > 0\)
• At \(\mathrm{t} = 1\): \(\mathrm{R} = -1 - 2 + 1 = -2 < 0\)
• R crosses zero between \(\mathrm{t} = 0.5\) and \(\mathrm{t} = 1\)
• Answer: YES

Since we can construct valid scenarios that give different answers, the statements together are NOT sufficient.

[STOP - Not Sufficient!] This eliminates choices C.

The Answer: E

Even with both pieces of information, we cannot definitively answer whether \(\mathrm{R} = 0\) has a positive solution because the sign of a remains undetermined, and this is crucial for determining the function's behavior as t increases.

Answer Choice E: "The statements together are not sufficient."