Question

Let \( f(x) = \begin{cases 1 + \frac{2x}{a}, & 0 \le x \le 1
ax, & 1<x \le 2 \end{cases} \). If \( \lim_{x \to 1} f(x) \) exists, then the sum of the cubes of the possible values of \( a \) is: }

• A. 1
• B. 5
• C. 7
• D. 9

Hint:

When dealing with piecewise functions, ensure that the limits from both sides at the boundary point are equal to ensure continuity.

Answer 1

The Correct Option is C

For \( \lim_{x \to 1} f(x) \) to exist, the left
-hand limit and the right
-hand limit must be equal. Left
-hand limit: \[ \lim_{x \to 1^
-} f(x) = \lim_{x \to 1^
-} \left( 1 + \frac{2x}{a} \right) = 1 + \frac{2(1)}{a} = 1 + \frac{2}{a} \] Right
-hand limit: \[ \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (ax) = a(1) = a \] For the limit to exist, the left
-hand limit must equal the right
-hand limit: \[ 1 + \frac{2}{a} = a \] \[ a + 2 = a^2 \] \[ a^2
- a
- 2 = 0 \] We can factor the quadratic equation: \[ (a
- 2)(a + 1) = 0 \] Thus, the possible values of \( a \) are \( a = 2 \) and \( a =
-1 \). The sum of the cubes of the possible values of \( a \) is: \[ 2^3 + (
-1)^3 = 8 + (
-1) = 7 \] Therefore, the sum of the cubes of the possible values of \( a \) is 7. Final Answer: (3) 7