Question:

<p><strong>Step 1: Define auxiliary polynomial.</strong></p><p>Let g(x) = xf(x) - 1.</p><p>Since f(r) = 1/r for r = 1,2,3,...,9, we have:</p><p>g(r) = r·f(r) - 1 = r·(1/r) - 1 = 1 - 1 = 0</p><p>Therefore, g(x) has roots at x = 1, 2, 3, ..., 9 (nine roots).</p><p><strong>Step 2: Determine the form of g(x).</strong></p><p>Since f(x) is a polynomial of degree 8, g(x) = xf(x) - 1 is a polynomial of degree 9.</p><p>Since g(x) has degree 9 and we know all 9 roots, we can write:</p><p>g(x) = c(x - 1)(x - 2)(x - 3)···(x - 9)</p><p>where c is a constant (the leading coefficient).</p><p><strong>Step 3: Find the leading coefficient c.</strong></p><p>Since f(x) has degree 8 with leading coefficient a₈, then xf(x) has degree 9 with leading coefficient a₈.</p><p>Thus g(x) = xf(x) - 1 also has leading coefficient a₈.</p><p>Comparing with g(x) = c(x - 1)(x - 2)···(x - 9), we have c = a₈.</p><p>However, we need another condition. Consider: g(0) = 0·f(0) - 1 = -1.</p><p>From g(x) = c(x - 1)(x - 2)···(x - 9):</p><p>g(0) = c(-1)(-2)(-3)···(-9) = c·(-1)⁹·9! = -c·9!</p><p>Therefore: -c·9! = -1, which gives c = 1/9!</p><p><strong>Step 4: Express g(x) explicitly.</strong></p><p>g(x) = (1/9!)·(x - 1)(x - 2)(x - 3)···(x - 9)</p><p><strong>Step 5: Calculate g(10).</strong></p><p>g(10) = (1/9!)·(10 - 1)(10 - 2)(10 - 3)···(10 - 9)</p><p>g(10) = (1/9!)·(9)(8)(7)(6)(5)(4)(3)(2)(1)</p><p>g(10) = (1/9!)·9!</p><p>g(10) = 1</p><p><strong>Step 6: Find f(10).</strong></p><p>Since g(x) = xf(x) - 1, we have:</p><p>g(10) = 10·f(10) - 1</p><p>1 = 10·f(10) - 1</p><p>10·f(10) = 2</p><p>f(10) = 1/5</p><p><strong>Step 7: Calculate 1/f(10).</strong></p><p>1/f(10) = 1/(1/5) = 5</p><p><strong>∴ Answer: 1/f(10) = 5</strong></p>