<p>The number of integral values of \(a\) for which the quadratic equation \((x - a)(x - 1991) + 1 = 0\) has integral roots are</p>
Step-by-Step Solution
Key Concept: Expand the equation to $(x-a)(x-1991) = -1$, then recognize that if $r$ and $s$ are integral roots, the product $(r-a)(r-1991) = -1$ forces $(r-a)$ and $(r-1991)$ to be divisors of $-1$ (i.e., $±1$), severely constraining possibilities.
<p><strong>Step 1:</strong> Rewrite the equation as $(x-a)(x-1991) = -1$.</p><p><strong>Step 2:</strong> Let $r$ be an integral root. Then $(r-a)(r-1991) = -1$. Since $r$, $a$, and $1991$ are integers, both $(r-a)$ and $(r-1991)$ are integers whose product is $-1$.</p><p><strong>Step 3:</strong> The only integer factor pairs of $-1$ are $(1, -1)$ and $(-1, 1)$.</p><p><strong>Step 4:</strong> <strong>Case 1:</strong> $(r-a) = 1$ and $(r-1991) = -1$. This gives $r = 1990$ and $a = 1989$.</p><p><strong>Step 5:</strong> <strong>Case 2:</strong> $(r-a) = -1$ and $(r-1991) = 1$. This gives $r = 1992$ and $a = 1993$.</p><p><strong>Step 6:</strong> Verify both values work. For $a = 1989$: $(x-1989)(x-1991) + 1 = (x-1990)^2 = 0$ gives $x = 1990$ (integral). For $a = 1993$: $(x-1993)(x-1991) + 1 = (x-1992)^2 = 0$ gives $x = 1992$ (integral).</p><p><strong>Step 7:</strong> Both roots in each case are identical (double roots), and both yield integral roots. Thus there are exactly <strong>2</strong> integral values of $a$.</p><p>∴ Answer: D</p>
Correct Answer: D