If function f (x) given by f(x)=(sin⁡x)1/(π−2x),x≠π/2λ,x=π/2is continuous at x=π2, then, λ=

Solution:

For $f (x)$ to be continuous at $x = \frac{π}{2}$ we must have

$\begin{array}{l} lim_{x \to π / 2} f (x) = f (\frac{π}{2}) \\ \Rightarrow lim_{x \to π / 2} (\sin x)^{1 / (π - 2 x)} = λ \\ \Rightarrow lim_{x \to π / 2} {1 + (\sin x - 1)}^{1 \cdot (π - 2 x)} = λ \\ \Rightarrow lim_{x \to π / 2} \frac{\sin x - 1}{π - 2 x} = λ \\ \Rightarrow e^{- \frac{1}{2} x \to π / 2} \frac{1 - \cos (π / 2 - x)}{(π / 2 - x)} = λ \Rightarrow e^{0} = λ \Rightarrow λ = 1 \end{array}$

If function $f (x)$ given by
$f (x) = \{\begin{array}{cc} (\sin x)^{1 / (π - 2 x)}, & x \neq π / 2 \\ λ & , x = π / 2 \end{array}$
is continuous at $x = \frac{π}{2},$ then, $λ =$

Solution:

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