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प्रश्न
For what value of λ is the function
\[f\left( x \right) = \begin{cases}\lambda( x^2 - 2x), & \text{ if } x \leq 0 \\ 4x + 1 , & \text{ if } x > 0\end{cases}\]continuous at x = 0? What about continuity at x = ± 1?
उत्तर
The given function f is \[f\left( x \right) = \begin{cases}\lambda( x^2 - 2x), & \text{ if } x \leq 0 \\ 4x + 1 , & \text{ if } x > 0\end{cases}\]
If f is continuous at x = 0, then
Therefore, there is no value of λ for which f(x) is continuous at x = 0.
At x = 1,
f (1) = 4x + 1 = 4 × 1 + 1 = 5
Therefore, for any values of λ, f is continuous at x = 1
At x = -1, we have
f (-1) =
\[\lim_{x \to - 1} \lambda\left( 1 + 2 \right) = 3\lambda\]
\[ \therefore \lim_{x \to - 1} f\left( x \right) = f\left( - 1 \right)\]
Therefore, for any values of λ, f is continuous at x = -1
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