Question:
for which value of \(\lambda\) is the function ,\(f(x) = \begin{cases} \lambda(x^2-2x) & \text{if } x \leq 0 \\ 4x+1& \text{if } x > 0 \end{cases}\) continuous at \(x=0 ?\)
for which value of \(\lambda\) is the function ,\(f(x) = \begin{cases} \lambda(x^2-2x) & \text{if } x \leq 0 \\ 4x+1& \text{if } x > 0 \end{cases}\) continuous at \(x=0 ?\)
Updated On: May 21, 2024
- \(\lambda=0\)
- \(\lambda=1\)
- \(\lambda=-1\)
- for no value of \(\lambda\)
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The Correct Option is D
Solution and Explanation
The correct option is (D):for no value of \(\lambda\)
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