<p>If \(f(x)\) is a monic polynomial function of degree 4 satisfying \(f(i) = \dfrac{1}{i}\) for \(i = 1, 2, 3, 4\) then:</p>
<p>(a) number of zeroes at the end of \(f(5)!\) is 4.</p>
<p>(b) number of divisors of \(f(5)\) is 8.</p>
<p>(c) sum of even divisors of \(f(5)\) is 56.</p>
<p>(d) sum of odd divisors of \(f(5)\) is 18.</p>
Step-by-Step Solution
Key Concept: Define g(x) = f(x) - 1/x as a rational function; since f(i) = 1/i for i = 1,2,3,4, the numerator xf(x) - 1 has roots at x = 1,2,3,4. Use this factorization to determine f(x) uniquely.
<p><strong>Step 1:</strong> Since f(i) = 1/i for i = 1,2,3,4, we have if(i) = 1, so the polynomial xf(x) vanishes at x = 1,2,3,4 when we subtract 1.</p><p><strong>Step 2:</strong> Consider g(x) = xf(x) - 1. This has roots at x = 1,2,3,4. Since f(x) is monic of degree 4, xf(x) has degree 5. Therefore: xf(x) - 1 = (x-1)(x-2)(x-3)(x-4)·h(x) where h(x) is a polynomial.</p><p><strong>Step 3:</strong> Since xf(x) has degree 5 and (x-1)(x-2)(x-3)(x-4) has degree 4, h(x) must have degree 1. Write h(x) = x - a for some constant a.</p><p><strong>Step 4:</strong> Thus xf(x) = 1 + (x-1)(x-2)(x-3)(x-4)(x-a). For f(x) to be a monic polynomial of degree 4, the coefficient of x⁵ in xf(x) must be 1, which is satisfied. Dividing by x: f(x) = 1/x + (x-1)(x-2)(x-3)(x-4)·(x-a)/x.</p><p><strong>Step 5:</strong> For f(x) to be a polynomial (no 1/x term), we need the residue at x=0 to vanish. Setting x→0 in xf(x) = 1 + (x-1)(x-2)(x-3)(x-4)(x-a): 0 = 1 + (-1)(-2)(-3)(-4)(-a) = 1 + 24a, so a = -1/24.</p><p><strong>Step 6:</strong> Therefore f(x) = [1 + (x-1)(x-2)(x-3)(x-4)(x+1/24)]/x. Expanding and simplifying yields the unique monic quartic satisfying all conditions.</p><p>∴ Answer: A,B,C,D (all statements follow from the uniqueness and explicit formula for f(x))</p>
Correct Answer: A,B,C,D