Consider the function.
f(x)={a(7 x-12-x²)/(b|x²-7 x+12|), , x<3; 2^((sin (x-3))/(x-[x])), , x>3; b, , x=3,
where [x] denotes the greatest integer less than or equal to x. If S denotes the set of all ordered pairs (a, b) such that f(x) is continuous at x=3, then the number of elements in S is :

Solution

Maths sir · 4 steps

Factor the quadratic and fix its sign just left of 3; then the left side becomes a constant.

1. Left-hand limit. x²-7x+12=(x-3)(x-4). Just left of 3, both factors are negative, so the product is positive and |x²-7x+12| = x²-7x+12. The numerator is -a(x²-7x+12), so f(x) = -a/b there, and the left-hand limit is -a/b.
2. Right-hand limit. Just right of 3, [x]=3, so x-[x]=x-3. The exponent is sin(x-3)/(x-3) → 1, so the right-hand limit is 2¹ = 2.

   -a/b = 2 = b

3. Match all three. For continuity, the left-hand limit, the right-hand limit and f(3)=b must all be equal. From f(3)=2 we get b=2. Then -a/2=2 gives a=-4.
4. So. Only one pair works, (a,b)=(-4,2). The set S has exactly 1 element.

The idea to keep. With a modulus near a point, factor it and decide its sign on that side before taking the limit. Continuity needs LHL, RHL and f(a) all equal, and every one of these conditions can fix an unknown.

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