AP Calculus BC 1.11 Defining Continuity at a Point - MCQs -Exam Style Questions
No-CalcQuestion
Let \( f(x)=\begin{cases} \dfrac{\tan(5x)}{x}, & x\ne 0\\[4pt] b, & x=0 \end{cases}\). For what value of \(b\) is \(f\) continuous at \(x=0\)?
(A) \(0\)
(B) \(\tfrac{1}{5}\)
(C) \(1\)
(D) \(5\)
(B) \(\tfrac{1}{5}\)
(C) \(1\)
(D) \(5\)
▶️ Answer/Explanation
Continuity at \(x=0\) requires \(b=\displaystyle\lim_{x\to 0}\frac{\tan(5x)}{x}\).
Use \(\displaystyle\lim_{u\to 0}\frac{\tan u}{u}=1\):
\[ \lim_{x\to 0}\frac{\tan(5x)}{x} = \lim_{x\to 0}\Big(\frac{\tan(5x)}{5x}\Big)\cdot 5 = 1\times 5 = 5. \] ✅ Correct: (D) \(5\)
Use \(\displaystyle\lim_{u\to 0}\frac{\tan u}{u}=1\):
\[ \lim_{x\to 0}\frac{\tan(5x)}{x} = \lim_{x\to 0}\Big(\frac{\tan(5x)}{5x}\Big)\cdot 5 = 1\times 5 = 5. \] ✅ Correct: (D) \(5\)
No-Calc Question
\( f(x)=\begin{cases} x^{2}-4x, & x<2\\[4pt] x^{3}-12, & x\ge 2 \end{cases} \). Which of the following statements about \(f\) is false?
(A) \( \displaystyle \lim_{x\to2} f(x)\) exists
(B) \( \displaystyle \lim_{x\to2} f'(x)\) exists
(C) \( f\) is continuous at \(x=2\)
(D) \( f\) is differentiable at \(x=2\)
(B) \( \displaystyle \lim_{x\to2} f'(x)\) exists
(C) \( f\) is continuous at \(x=2\)
(D) \( f\) is differentiable at \(x=2\)
▶️ Answer/Explanation
Detailed solution
Continuity test
\(f(2^-)=2^{2}-4\times 2=-4\)
\(f(2^+)=2^{3}-12=-4\)
\(f(2)=-4\)
Therefore \(\lim_{x\to 2} f(x)\) exists and equals \(f(2)\); \(f\) is continuous at \(x=2\).
Derivative one-sided limits
For \(x<2\): \(f'(x)=2x-4 \Rightarrow f’_{-}(2)=0\).
For \(x>2\): \(f'(x)=3x^{2} \Rightarrow f’_{+}(2)=12\).
Since \(f’_{-}(2)\ne f’_{+}(2)\), \(f\) is not differentiable at \(x=2\), and \(\lim_{x\to 2} f'(x)\) does not exist.
Among the choices, the statement “\(f\) is differentiable at \(x=2\)” is false.
✅ Answer: (D)