Question:
Define \(f(x)=\left\{\begin{array}{cl}b-a x & , \text { if } x<2 \\ 3 & , \text { if } x=2 \\ a+2 b x & , \text { if } x>2\end{array}\right.\) and if \(\lim _{x \rightarrow 2} f(x)\) exists, then \(\frac{a}{b}=\)
Define \(f(x)=\left\{\begin{array}{cl}b-a x & , \text { if } x<2 \\ 3 & , \text { if } x=2 \\ a+2 b x & , \text { if } x>2\end{array}\right.\) and if \(\lim _{x \rightarrow 2} f(x)\) exists, then \(\frac{a}{b}=\)
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The limit at x = 2 exists only if the left-hand and right-hand limits are equal; the value 3 at x = 2 does not matter.
Updated On: Nov 13, 2026
1
-1
\(\frac{2}{3}\)
\(\frac{3}{2}\)
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept
For \(\lim_{x\to2}f(x)\) to exist, \(\lim_{x\to2^{-}}f(x)=\lim_{x\to2^{+}}f(x)\). The value \(f(2)=3\) plays no role.
Step 2: Left and right limits
Left: \(b-a(2)=b-2a\). Right: \(a+2b(2)=a+4b\).
Step 3: Calculation
\(b-2a=a+4b\), so \(-3a=3b\), giving \(\dfrac{a}{b}=-1\).
Final Answer:
The ratio a/b equals -1. \[ \boxed{\text{(B) }-1} \]
For \(\lim_{x\to2}f(x)\) to exist, \(\lim_{x\to2^{-}}f(x)=\lim_{x\to2^{+}}f(x)\). The value \(f(2)=3\) plays no role.
Step 2: Left and right limits
Left: \(b-a(2)=b-2a\). Right: \(a+2b(2)=a+4b\).
Step 3: Calculation
\(b-2a=a+4b\), so \(-3a=3b\), giving \(\dfrac{a}{b}=-1\).
Final Answer:
The ratio a/b equals -1. \[ \boxed{\text{(B) }-1} \]
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