▶️ Answer/Explanation
Solution
Correct Answer: A
1. For continuity at x=3, left and right limits must be equal:
\[k^3 + 3 = \frac{16}{k^2 – 3}\]
2. Solving for k > 0 gives k ≈ 2.081
3. With this k, \(\lim_{x\to3}f(x)\) exists and equals f(3)
4. Therefore, f is continuous at x=3 when k=2.081
▶️ Answer/Explanation
Solution
Correct Answer: A
1. Factor numerator and denominator:
Numerator: \(2(x^2 + 7x – 8) = 2(x+8)(x-1)\)
Denominator: \((x-8)(x-1)\)
2. Discontinuities occur at x=1 and x=8
3. At x=1: Common factor (x-1) cancels → removable discontinuity
4. At x=8: No cancellation → non-removable discontinuity
5. Therefore, only x=1 has a removable discontinuity
Question
Let f be the function defined above, where α is a constant. For what values of a, if any, is f continuous at x=3?
A 0 only
B 3 only
C 0 and 3
D There is no such a
▶️Answer/Explanation
Ans:B
f is continuous at x=3 if f(3) and both exist and are equal to each other.
=\(a^{2}+3^{2}=a^{2}+9\)
a(3+3)=6a
f(3)=a(3+3)=6a Setting these equal to each other and solving for a yields the following. \(a^{2}+9=6a\Rightarrow a^{2}-6a+9=0\Rightarrow (a-3)^{2}=0\Rightarrow a=3\)
▶️ Answer/Explanation
Solution
Correct Answer: A
1. Factor numerator: \(2x^2 + 14x – 16 = 2(x^2 + 7x – 8) = 2(x+8)(x-1)\)
2. Factor denominator: \(x^2 – 9x + 8 = (x-8)(x-1)\)
3. Simplified form: \(f(x) = \frac{2(x+8)}{x-8}\) for \(x \neq 1\)
4. Removable discontinuity at x=1 (common factor cancels)
5. Non-removable discontinuity at x=8 (vertical asymptote)