TS EAMCET · Maths · Limits
If \(\quad f:[-2,2] \rightarrow R \quad\) is defined by \(f(x)=\left\{\begin{array}{cc}\frac{\sqrt{1+c x}-\sqrt{1-c x}}{x} & \text { for }-2 \leq x < 0 \ \frac{x+3}{x+1} & \text { for } 0 \leq x \leq 2\end{array}\right.\) continuous on \([-2,2]\), then \(c\) is equal to
- A \(\frac{2}{\sqrt{3}}\)
- B \(3\)
- C \(\frac{3}{2}\)
- D \(\frac{3}{\sqrt{2}}\)
Answer & Solution
Correct Answer
(B) \(3\)
Step-by-step Solution
Detailed explanation
Given, \(f:[-2,2] \rightarrow R\) \[ f(x)=\left\{\begin{array}{cc} \frac{\sqrt{1+c x}-\sqrt{1-c x}}{x}, & -2 \leq x < 0 \\ \frac{x+3}{x+1}, & 0 \leq x \leq 2 \end{array}\right. \] Now, \(\quad\) LHL \(=\lim _{x \rightarrow 0^{-}} f(x)\)…
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