Question

If a real-valued function $$ f(x) = \begin{cases} \frac{2x^2 + k(2x) + 9}{3x^2 - 7x - 6}, & \text{for } x \neq 3, \\ l, & \text{for } x = 3 \end{cases} $$ is continuous at \( x = 3 \) and \( l \) is a finite value, then \( l - k = \): 

• A. \( \frac{31}{11} \)
• B. \( \frac{124}{11} \)
• C. \( 24 \)
• D. \( 32 \)

Hint:

For continuity at a point \( x = a \), ensure that the function is well-defined at \( x = a \) and that both the left-hand and right-hand limits are equal to \( f(a) \).

Answer 1

The Correct Option is B

Step 1: Condition for continuity 
For a function to be continuous at \( x = 3 \), we must have: \[ \lim\limits_{x \to 3} f(x) = f(3) = l. \] That is: \[ \lim\limits_{x \to 3} \frac{2x^2 + (k+2)x + 9}{3x^2 - 7x - 6} = l. \] 

Step 2: Factorizing denominator 
We factorize \( 3x^2 - 7x - 6 \) as: \[ 3x^2 - 7x - 6 = (3x + 2)(x - 3). \] Since the denominator becomes zero at \( x = 3 \), the numerator must also be zero at \( x = 3 \) to avoid discontinuity. 

Step 3: Finding \( k \) 
Substituting \( x = 3 \) in the numerator: \[ 2(3)^2 + (k+2)(3) + 9 = 0. \] \[ 18 + 3k + 6 + 9 = 0. \] \[ 3k + 33 = 0. \] \[ k = -11. \] 

Step 4: Computing limit 
Now, we find: \[ \lim\limits_{x \to 3} \frac{2x^2 + (-11+2)x + 9}{3x^2 - 7x - 6}. \] Factorizing the numerator: \[ 2x^2 - 9x + 9 = (2x - 3)(x - 3). \] Canceling \( (x - 3) \) from numerator and denominator: \[ \lim\limits_{x \to 3} \frac{(2x - 3)(x - 3)}{(3x + 2)(x - 3)} = \lim\limits_{x \to 3} \frac{2x - 3}{3x + 2}. \] Substituting \( x = 3 \): \[ \frac{2(3) - 3}{3(3) + 2} = \frac{6 - 3}{9 + 2} = \frac{3}{11}. \] Thus, \( l = \frac{124}{11} \). 

Step 5: Computing \( l - k \) 
\[ l - k = \frac{124}{11} - (-11) = \frac{124}{11} + \frac{121}{11} = \frac{124}{11}. \] 

Step 6: Conclusion 
Thus, the final answer is: \[ \boxed{\frac{124}{11}}. \]