Basic Mathematics & Logarithm
Inequalities with Logarithm
Grade None
Question:
<p>If the solution of inequality \(\dfrac{(\pi^x - 7^x)\log_{10}(x-4)}{(x^2 - 9x + 18)(x^2 - x)} < 0\) is in the form \((a,b) \cup (c, \infty)\), match the following:</p><table border='1' cellpadding='4'><tr><th>List-I</th><th>List-II</th></tr><tr><td>(P) The value of 'a' is</td><td>(1) 2</td></tr><tr><td>(Q) The value of 'b' is</td><td>(2) 3</td></tr><tr><td>(R) The value of 'c' is</td><td>(3) 4</td></tr><tr><td>(S) The value of \((a+b-c)\), is</td><td>(4) 5</td></tr><tr><td></td><td>(5) 6</td></tr></table>
<p>P-3, Q-4, R-5, S-2</p>
<p>P-3, Q-4, R-2, S-5</p>
<p>P-3, Q-4, R-2, S-2</p>
<p>P-3, Q-4, R-5, S-5</p>
Step-by-Step Solution
Key Concept: Analyze the sign of each factor in the rational inequality separately: the numerator (π^x - 7^x) changes sign at x=0, log₁₀(x-4) changes sign at x=5, and the denominator factors (x-3)(x-6)(x)(x-1) determine vertical asymptotes. Then apply sign analysis over intervals.
<p><strong>Step 1: Identify critical points and domain restrictions</strong></p><p>Numerator zeros: π^x = 7^x has no real solution (π<7); log₁₀(x-4)=0 at x=5</p><p>Denominator zeros: (x²-9x+18)=(x-3)(x-6) and (x²-x)=x(x-1)</p><p>Critical points: x = 0, 1, 3, 4, 5, 6; Domain: x > 4</p><p><strong>Step 2: Analyze sign of π^x - 7^x</strong></p><p>For all x: π^x < 7^x, so π^x - 7^x < 0 always</p><p><strong>Step 3: Analyze sign of log₁₀(x-4) for x > 4</strong></p><p>log₁₀(x-4) < 0 when 4 < x < 5; log₁₀(x-4) > 0 when x > 5</p><p><strong>Step 4: Analyze sign of denominator factors for x > 4</strong></p><p>(x-3) > 0 for x > 4; (x-6) < 0 for 4 < x < 6 and > 0 for x > 6</p><p><strong>Step 5: Sign analysis over valid intervals (x > 4)</strong></p><p>For (4,5): [negative][negative]/[positive][negative] = negative/negative = positive ✗</p><p>For (5,6): [negative][positive]/[positive][negative] = negative/negative = positive ✗</p><p>For (6,∞): [negative][positive]/[positive][positive] = negative/positive = negative ✓</p><p><strong>Step 6: Find where inequality < 0</strong></p><p>Solution: (4,5) gives positive, (5,6) gives positive, (6,∞) gives negative</p><p>Wait - recalculate: At x=4.5: numerator=(−)(−)>0, denominator=(+)(−)<0, ratio<0 ✓</p><p>Corrected analysis: (4,5)∪(6,∞)</p><p>Therefore: a=4, b=5, c=6</p><p>a+b-c = 4+5-6 = 3</p><p><strong>Matching:</strong> P→3, Q→2, R→4, S→2</p><p>∴ Answer: A</p>
Correct Answer: A