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AP EAMCET · Maths · Continuity and Differentiability

If a real valued function
\(f(x)=\left\{\begin{array}{cl}\frac{2 x^2+(k+2) x+9}{3 x^2-7 x-6} & , \text { for } x \neq 3 \\ l & , \text { for } x=3\end{array}\right.\)
\(x=3\) and \(l\) is a finite value, then \(l-k=\)

  1. A \(\frac{31}{11}\)
  2. B \(\frac{124}{11}\)
  3. C \(24\)
  4. D \(32\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\frac{124}{11}\)

Step-by-step Solution

Detailed explanation

Since, \(f(x)\) is continuous at \(x=3 \Rightarrow \lim _{x \rightarrow 3} f(x)=f(3)\) Now, \(\lim _{x \rightarrow 3} f(x)=\lim _{x \rightarrow 3} \frac{2 x^2+(k+2) x+9}{3 x^2-7 x-6}\) For this limit to exist, we must have…