Question:
<p>Let S be the set of integral values of ‘a’ for which one root of quadratic equation (a<sup>2</sup> - 2a + 3)x<sup>2</sup> - 6ax + 4 = 0 is less than 1 while the other is greater than it, then the number of point(s) P (a<sup>2</sup>, a) where a <span class="math-tex">\(\in\)</span> S lies in the region between the lines x + y = 6 and x + y - 12 is (are) :</p>
<p style="display:inline">4</p>
<p style="display:inline">0</p>
<p style="display:inline">5</p>
<p style="display:inline">3</p>
Step-by-Step Solution
Key Concept: For a quadratic equation f(x) = 0 with positive leading coefficient to have roots on opposite sides of x = 1, we need f(1) < 0. Here, the coefficient (a² - 2a + 3) is always positive since its discriminant is negative, so the condition becomes (a² - 2a + 3)(1)² - 6a(1) + 4 < 0, simplifying to a² - 8a + 7 < 0, which gives 1 < a < 7.
<p>0</p>
Correct Answer: B