If $f (α) = \lim_{x \to 2} {(\sin^{x} α + \cos^{x} α)}^{\frac{1}{(x - 2)}} for α \in [0, \frac{π}{2}]$ , then

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f(0) = 1

$f (\frac{π}{2}) = 1$

$f (α) = {(cosα)}^{\cos^{2} α} . {(sinα)}^{\sin^{2} a} if α \in (0, \frac{π}{2})$

$f (α) = \frac{{(sinα)}^{\sin^{2} α}}{{(cosα)}^{\cos^{2} α}} if α \in (0, \frac{π}{2})$

answer is A, B, C.

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Detailed Solution

$\begin{array}{l} f (α) = \lim_{x \to 2} {(\sin^{x} α + \cos^{x} α)}^{\frac{1}{(x - 2)}} (1^{\infty} form) \\ = \{\begin{cases} e^{\lim_{x \to 2} \frac{\sin^{x} α + \cos^{x} α - 1}{x - 2}}, α \in (0, \frac{π}{2}) \\ 1, α = 0, \frac{π}{2} \end{cases} \\ Now, e^{\lim_{x \to 2} \frac{\sin^{x} α + \cos^{c} a - 1}{x - 2}} = e^{\lim_{x \to 2} . \frac{\sin^{x} α + \cos^{x} α - \sin^{2} α - \cos^{2} α}{x - 2}} \\ = e^{\lim_{x \to 2} \frac{\sin^{x} α (\sin^{x - 2} α - 1) + \cos^{2} α (\cos^{x - 2} α - 1)}{x - 2}} \\ = e^{\sin^{2} {αlog}_{e} sinα + \cos^{2} {αlog}_{e} cosα} \\ = e^{\log_{e} {(sinα)}^{\sin^{2} α} + \log_{e} {(cosα)}^{\cos^{2} α}} \\ = e^{\log_{e} {(sinα)}^{\sin^{2} α} {(cosα)}^{\cos^{2} α}} \\ = {(sinα)}^{\sin^{2} α} {(cosα)}^{\cos^{2} α} \\ ∴ f (x) = \{\begin{cases} {(\cos α)}^{\cos^{2} α .} {(sinα)}^{\sin^{2} α}, a \in (0, \frac{π}{2}) \\ 1, α = 0, \frac{π}{2} \end{cases} \end{array}$