AP EAMCET · Maths · Continuity and Differentiability
\(f(x)=\left\{\begin{array}{cc}\frac{\left(2 x^2-a x+1\right)-\left(a x^2+3 b x+2\right)}{x+1} & ; \text { if } x \neq-1 \\ k & , \text { if } x=-1\end{array}\right.\)
is a real valued function. If \(a, b, k \in \mathrm{R}\) and \(f\) is continuous on \(\mathbf{R}\) then \(k=\)
- A \(-\frac{1}{3}\)
- B \(6\)
- C \(a-2\)
- D \(a-3\)
Answer & Solution
Correct Answer
(D) \(a-3\)
Step-by-step Solution
Detailed explanation
Given the function \(f(x)=\left\{\begin{array}{cc}\frac{\left(2 x^2-a x+1\right)-\left(a x^2+3 b x+2\right)}{x+1} & \text { if } x \neq-1 \\ k & \text { if } x=1\end{array}\right.\) Now, \(\lim _{x \rightarrow-1} \frac{(2-a) x^2-(a+3 b) x-1}{x+1}\) For existence of limit…
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