<p><strong>782.</strong> Find the number of integral values of \(x\) satisfying the inequality</p><p>\[\frac{\left(2^{\tan^{-1}x} - 4\right)(x-4)(x-10)}{x! - (x-1)!} < 0\]</p>
Step-by-Step Solution
Key Concept: The inequality requires simultaneous analysis of three factors in the numerator and the denominator's sign, where x! - (x-1)! = (x-1)!(x-1). The critical insight is recognizing that 2^(tan⁻¹x) ranges over (1,2) for all real x, making the first factor's sign dependent on comparing with 4.
<p><strong>Step 1: Determine the sign of 2^(tan⁻¹x) - 4</strong></p><p>Since tan⁻¹x ∈ (-π/2, π/2) for all real x, we have 2^(tan⁻¹x) ∈ (2^(-π/2), 2^(π/2)) ≈ (0.33, 4.77).</p><p>Therefore, 2^(tan⁻¹x) < 4 always, so (2^(tan⁻¹x) - 4) < 0 for all x.</p><p><strong>Step 2: Analyze the denominator</strong></p><p>x! - (x-1)! = (x-1)!(x - 1) = (x-1)!(x-1)</p><p>For integer x ≥ 0: denominator is positive when x ≥ 2, undefined at x = 1, and for x = 0 gives negative value.</p><p><strong>Step 3: Apply sign condition</strong></p><p>Given (negative)(x-4)(x-10) ≥ 0, with positive denominator (x ≥ 2, x ≠ 1):</p><p>-(x-4)(x-10) ≥ 0 ⟹ (x-4)(x-10) ≤ 0 ⟹ 4 ≤ x ≤ 10</p><p><strong>Step 4: Apply domain restrictions</strong></p><p>Need x ∈ ℤ, x ≥ 2 (for denominator positive), x ≠ 1, and 4 ≤ x ≤ 10</p><p>Valid integers: x ∈ {4, 5, 6, 7, 8, 9, 10}</p><p><strong>∴ Answer: 7 integral values</strong></p>
Correct Answer: 7