Question

If \( f(x) \) is given as: \( f(x) = \begin{cases} 3ax - 2b, & x<1 ax + b + 1, & x<1 \end{cases} \) and \( \lim_{x \to 1} f(x) \) exists, then the relation between \( a \) and \( b \) is:

• A. \( 3a - 2b = 1 \)
• B. \( 2a - 3b = 1 \)
• C. \( 2a + 3b = 1 \)
• D. \( 2a + 3b = -1 \)

Hint:

For a function \( f(x) \) to be continuous at \( x = c \), it must satisfy: \( \lim\limits_{x \to c^-} f(x) = \lim\limits_{x \to c^+} f(x) = f(c). \)

Answer 1

The Correct Option is B

Step 1: Condition for the existence of \( \lim\limits_{x \to 1} f(x) \) For the limit of \( f(x) \) to exist at \( x = 1 \), the left-hand limit (LHL) and right-hand limit (RHL) must be equal, i.e., \( \lim\limits_{x \to 1^-} f(x) = \lim\limits_{x \to 1^+} f(x). \) 

Step 2: Compute \( \lim\limits_{x \to 1^-} f(x) \) For \( x<1 \), we use the function: \( f(x) = ax + b + 1. \) Substituting \( x = 1 \), \( \lim\limits_{x \to 1^-} f(x) = a(1) + b + 1 = a + b + 1. \) 

Step 3: Compute \( \lim\limits_{x \to 1^+} f(x) \) For \( x<1 \), we use the function: \( f(x) = 3ax - 2b. \) Substituting \( x = 1 \), \( \lim\limits_{x \to 1^+} f(x) = 3a(1) - 2b = 3a - 2b. \) 

Step 4: Equating LHL and RHL Since the limit must exist, we equate both limits: \( a + b + 1 = 3a - 2b. \) 

Step 5: Solve for \( a \) and \( b \) Rearranging the equation: \( a + b + 1 - 3a + 2b = 0. \) \( -2a + 3b + 1 = 0. \) \( 2a - 3b = 1. \) Thus, the required relation between \( a \) and \( b \) is: \( \boxed{2a - 3b = 1.} \)