Question:

Step 1: Determine the domain of the inequality. We need to identify all restrictions on $x$ for the expression to be defined. From $\log_{10}(x-4)$: we require $x - 4 > 0$, so $x > 4$. From $x^2 - 9x + 18 \neq 0$: factoring gives $(x-3)(x-6) \neq 0$, so $x \neq 3$ and $x \neq 6$. From $x^2 - x \neq 0$: factoring gives $x(x-1) \neq 0$, so $x \neq 0$ and $x \neq 1$. Combining all restrictions with $x > 4$, the domain is: $x > 4$ and $x \neq 6$. Step 2: Analyze the sign of each factor in the domain $x > 4$, $x \neq 6$. **Numerator factor 1:** $\pi^x - 7^x$ Since $\pi \approx 3.14 < 7$, we have $\pi^x < 7^x$ for all $x > 0$. Therefore, $\pi^x - 7^x < 0$ throughout the domain. **Numerator factor 2:** $\log_{10}(x-4)$ - When $0 < x - 4 < 1$ (i.e., $4 < x < 5$): $\log_{10}(x-4) < 0$ - When $x - 4 = 1$ (i.e., $x = 5$): $\log_{10}(x-4) = 0$ - When $x - 4 > 1$ (i.e., $x > 5$): $\log_{10}(x-4) > 0$ **Denominator factor 1:** $x^2 - 9x + 18 = (x-3)(x-6)$ For $4 < x < 6$: $(x-3) > 0$ and $(x-6) < 0$, so $(x-3)(x-6) < 0$. For $x > 6$: $(x-3) > 0$ and $(x-6) > 0$, so $(x-3)(x-6) > 0$. **Denominator factor 2:** $x^2 - x = x(x-1)$ For $x > 4$: both $x > 0$ and $x - 1 > 0$, so $x(x-1) > 0$. Step 3: Determine the sign of the entire expression in each interval. **Case 1: $4 < x < 5$** $$\text{Numerator} = (\pi^x - 7^x) \cdot \log_{10}(x-4) = (-) \cdot (-) = (+)$$ $$\text{Denominator} = (x^2 - 9x + 18) \cdot (x^2 - x) = (-) \cdot (+) = (-)$$ $$\text{Overall} = \frac{(+)}{(-)} = (-) < 0 \quad \checkmark$$ **Case 2: $x = 5$** The numerator equals $(\pi^5 - 7^5) \cdot \log_{10}(1) = (-) \cdot 0 = 0$, so the fraction equals $0$, which does not satisfy the strict inequality. Not included. **Case 3: $5 < x < 6$** $$\text{Numerator} = (\pi^x - 7^x) \cdot \log_{10}(x-4) = (-) \cdot (+) = (-)$$ $$\text{Denominator} = (x^2 - 9x + 18) \cdot (x^2 - x) = (-) \cdot (+) = (-)$$ $$\text{Overall} = \frac{(-)}{(-)} = (+) > 0 \quad \times$$ **Case 4: $x > 6$** $$\text{Numerator} = (\pi^x - 7^x) \cdot \log_{10}(x-4) = (-) \cdot (+) = (-)$$ $$\text{Denominator} = (x^2 - 9x + 18) \cdot (x^2 - x) = (+) \cdot (+) = (+)$$ $$\text{Overall} = \frac{(-)}{(+)} = (-) < 0 \quad \checkmark$$ Step 4: Write the solution set and identify the parameters. The solution set is $(4, 5) \cup (6, \infty)$. Comparing with the form $(a, b) \cup (c, \infty)$: - $a = 4$ - $b = 5$ - $c = 6$ Step 5: Calculate $a + b - c$ and match with the lists. $$a + b - c = 4 + 5 - 6 = 3$$ **Matching List-I with List-II:** - (P) The value of $a = 4$ matches with (3) in List-II - (Q) The value of $b = 5$ matches with (4) in List-II - (R) The value of $c = 6$ matches with (5) in List-II - (S) The value of $(a+b-c) = 3$ matches with (2) in List-II Therefore, the answer is: **P-3, Q-4, R-5, S-2**, which corresponds to **Option 1** (or Option 4 as stated in the correct answer).