Let S be the set of positive integral values of a for which (a x²+2(a+1) x+9 a+4)/(x²-8 x+32) < 0, ∀ x ∈ ℝ. Then, the number of elements in S is :

Solution

Maths sir · 4 steps

Denominator is always positive, so the numerator must be negative for all x: no positive a can do that.

1. Denominator sign. For x² - 8x + 32 the discriminant is 64 - 128 = -64 < 0 and the leading coefficient is positive. So the denominator is positive for every real x, since x² - 8x + 32 = (x-4)² + 16.
2. Reduce the condition. The fraction is negative for all x exactly when the numerator is negative for all x. A quadratic stays below zero everywhere only if it opens downward and has no real roots.

   a < 0 and 4(a+1)² - 4a(9a+4) < 0

3. Check positive a. Every positive integer a breaks the first condition. With a > 0 the numerator opens upward, so it becomes positive for large |x|. The discriminant condition reduces to (4a-1)(2a+1) > 0, and together with a < 0 it only allows a < -1/2.
4. So. No positive integer works, so S is empty and it has 0 elements, option (a).

The idea to keep. A quadratic is negative for every real x only if its leading coefficient and its discriminant are both below zero. Check the leading coefficient first, because it often settles the question.

Answer from the NTA final answer key. Solution by PrepWiser.