Question:
The least integral value \(\alpha\) of \(x\) such that \(\dfrac{x-5}{x^2+5x-14} \gt 0\), satisfies :
The least integral value \(\alpha\) of \(x\) such that \(\dfrac{x-5}{x^2+5x-14} \gt 0\), satisfies :
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Factorise the denominator, use the sign scheme (wavy curve), find the solution set and its least integer.
Updated On: Oct 31, 2026
- \(\alpha^2 + 3\alpha - 4 = 0\)
- \(\alpha^2 - 5\alpha + 4 = 0\)
- \(\alpha^2 - 7\alpha + 6 = 0\)
- \(\alpha^2 + 5\alpha - 6 = 0\)
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The Correct Option is D
Solution and Explanation
\(x^2+5x-14 = (x+7)(x-2)\), so the inequality is \(\dfrac{x-5}{(x+7)(x-2)} \gt 0\).
Critical points: \(-7,\ 2,\ 5\). Sign of the expression: positive for \(x \gt 5\), negative on \((2,5)\), positive on \((-7,2)\), negative for \(x \lt -7\).
Solution set: \(x \in (-7,2) \cup (5,\infty)\).
The least integer in this set is \(\alpha = -6\).
Check the options: \(\alpha^2+5\alpha-6 = 36-30-6 = 0\). Option (1) gives \(36-18-4 \ne 0\), option (2) gives \(36+30+4 \ne 0\), option (3) gives \(36+42+6 \ne 0\).
Answer: (4) \(\alpha^2 + 5\alpha - 6 = 0\)
Critical points: \(-7,\ 2,\ 5\). Sign of the expression: positive for \(x \gt 5\), negative on \((2,5)\), positive on \((-7,2)\), negative for \(x \lt -7\).
Solution set: \(x \in (-7,2) \cup (5,\infty)\).
The least integer in this set is \(\alpha = -6\).
Check the options: \(\alpha^2+5\alpha-6 = 36-30-6 = 0\). Option (1) gives \(36-18-4 \ne 0\), option (2) gives \(36+30+4 \ne 0\), option (3) gives \(36+42+6 \ne 0\).
Answer: (4) \(\alpha^2 + 5\alpha - 6 = 0\)
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