Question

Assertion (A):
\( f(x) = \begin{cases} 3x - 8, & x \leq 5 \\ 2k, & x > 5 \end{cases} \)
is continuous at \( x = 5 \) for \( k = \frac{5}{2} \).

Reason (R):
For a function \( f \) to be continuous at \( x = a \),
\[ \lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = f(a) \]

• A. Both Assertion (A) and Reason (R) are true and the Reason (R) is the correct explanation of the Assertion (A).
• B. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
• C. Assertion (A) is true, but Reason (R) is false.
• D. Assertion (A) is false, but Reason (R) is true.

Answer 1

The Correct Option is A

For the function to be continuous at \( x = 5 \), we need to check whether the left-hand limit (\( \lim\limits_{x \to 5^-} f(x) \)), the right-hand limit (\( \lim\limits_{x \to 5^+} f(x) \)), and the function value at \( x = 5 \) are all equal.

Step-by-Step Evaluation:

1. Left-hand limit:
   \[ \lim\limits_{x \to 5^-} f(x) = 3(5) - 8 = 15 - 8 = 7 \]
2. Right-hand limit:
   \[ \lim\limits_{x \to 5^+} f(x) = 2k \]
3. Function value at \( x = 5 \):
   \[ f(5) = 3(5) - 8 = 15 - 8 = 7 \]

Condition for Continuity:

For \( f(x) \) to be continuous at \( x = 5 \), the following must hold:

\[ \lim\limits_{x \to 5^-} f(x) = \lim\limits_{x \to 5^+} f(x) = f(5) \]

So we equate:

\[ 2k = 7 \Rightarrow k = \frac{7}{2} \]

Conclusion:

The given assertion claims continuity at \( x = 5 \) for \( k = \frac{5}{2} \), which is incorrect. Hence:

• Assertion (A) is false
• Reason (R) is true (it correctly states the definition of continuity)

Final Answer: Assertion is false, but Reason is true.