The domain of the function
f(x) = (cos ⁻¹( (x² - 5x + 6)/(x² - 9) ))/(log ₑ(x² - 3x + 2)) is :

Solution

Maths sir · 5 steps

Stack three conditions: cos⁻¹ input in [-1,1], log input positive, and the log itself not zero.

1. Simplify the fraction. Factor: (x²-5x+6)/(x²-9) = (x-2)(x-3)/(x-3)(x+3) = (x-2)/(x+3). This cancelling is valid only for x ≠ 3. At x = 3 the original fraction is 0/0, so x = 3 must be removed. x = -3 is also excluded.
2. Inverse cosine condition. We need -1 ≤ (x-2)/(x+3) ≤ 1. The right part gives -5/(x+3) ≤ 0, so x > -3. The left part gives (2x+1)/(x+3) ≥ 0, so x < -3 or x ≥ -1/2. Together these give x ≥ -1/2.
3. Log input positive. We need x²-3x+2 = (x-1)(x-2) > 0, so x < 1 or x > 2. Combining this with x ≥ -1/2 gives [-1/2, 1) ∪ (2, ∞).
4. Denominator not zero. logₑ(x²-3x+2) = 0 when x²-3x+2 = 1, that is x²-3x+1 = 0, so x = (3 ± √5)/2. These are about 0.38 and 2.62, and both lie inside our set, so both must be removed.
5. So. The domain is [-1/2, 1) ∪ (2, ∞) - {3, (3+√5)/2, (3-√5)/2}, which is option (d).

The idea to keep. For a quotient of functions, list every condition separately: each function's own domain, then denominator not zero. When you cancel a factor, remove the value that made it zero.

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