<p>Number of integral values of \(x\) satisfying the inequality \(\dfrac{x^2+6x-7}{|x+2||x+3|} < 0\) is</p>
Step-by-Step Solution
Key Concept: The inequality involves absolute values in the denominator, which creates critical points where the expression changes sign. Factor the numerator to identify roots, then test sign variations across intervals determined by critical points x = -3, -2, and roots of the numerator.
<p><strong>Step 1:</strong> Factor the numerator: x² + 6x - 7 = (x+7)(x-1)</p><p><strong>Step 2:</strong> The inequality becomes: <span style='font-family:monospace'>(x+7)(x-1)/[|x+2||x+3|] < 0</span></p><p>Since |x+2| and |x+3| are always positive (where defined), we need: (x+7)(x-1) < 0</p><p><strong>Step 3:</strong> This gives us: -7 < x < 1, BUT x ≠ -3 and x ≠ -2 (denominator undefined)</p><p><strong>Step 4:</strong> Integral values in (-7, 1) excluding -3 and -2 are: {-6, -5, -4, -1, 0}</p><p><strong>Step 5:</strong> Count = 5 integral values</p><p>∴ Answer: A</p>
Correct Answer: A