TS EAMCET · Maths · Continuity and Differentiability
If ' \(a\) ' is the point of discontinuity of the function \[ f(x)=\left\{\begin{array}{cl} \cos 2 x, & \text { for }-\infty < x < 0 \ e^{3 x}, & \text { for } 0 \leq x < 3 \ x^2-4 x+3, & \text { for } 3 \leq x \leq 6 \ \frac{\log (15 x-89)}{x-6}, & \text { for } x>6 \end{array}\right. \] Then, \(\lim _{x \rightarrow a} \frac{x^2-9}{x^3-5 x^2+9 x-9}=\)
- A 1
- B 0
- C 6
- D 3
Answer & Solution
Correct Answer
(A) 1
Step-by-step Solution
Detailed explanation
We have, \[ \begin{aligned} & f(x)=\left\{\begin{array}{rc} \cos 2 x, & -\infty 6 \end{array}\right. \\ & \because \lim _{x \rightarrow 3^{-}} e^{3 x}=e^9 \text { and } \lim _{x \rightarrow 3^{+}} x^2-4 x+3=9-12+3=0 \end{aligned} \] Clearly,…
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