If `f(a)=lim_(xto2)(sin^(x)a+cos^(x)a)^((1)/((x-2)))" for "ain[0,(pi)/(2)],` then

Options

A. `-np`
B. `np`
C. `n^(2)p`
D. `np^(2)`

Answer

A, B, C

Explanation

`f(a)=underset(xto2)lim(sin^(x)alpha+cos^(x)alpha)^((1)/((x-2)))" "(1^(oo)" form")`
`={{:(e^(underset(xto2)lim(sin^(x)alpha+cos^(x)alpha-1)/(x-2))","alphain(0","(pi)/(2))),(1", "alpha=0", "(pi)/(2)):}`
Now `e^(underset(xto2)lim(sin^(x)alpha+cos^(x)alpha-1)/(x-2))=e^(underset(xto2)lim(sin^(x)alpha+cos^(x)alpha-sin^(2)alpha-cos^(2)alpha)/(x-2))`
`=e^(underset(xto2)lim(sin^(2)alpha(sin^(x-2)a-1)+cos^(2)alpha(cos^(x-2)alpha-1))/(x-2))`
`=e^(sin^(2)alphalog_(e)sinalpha+cos^(2)alphalog_(e)cosalpha)`
`=e^(log_(e)(sinalpha)^(sin^(2)alpha)+log_(e)(cosalpha)^(cos^(2)alpha))`
`=e^(log_(e)(sinalpha)^(sin^(2)alpha)(cosalpha)^(cos^(2)alpha))`
`=(sinalpha)^(sin^(2)alpha)(cosalpha)^(cos^(2)alpha)`
`:.f(x)={{:((cosalpha)^(cos^(2)alpha).(sinalpha)^(sin^(2)alpha)", "alphain(0", "(pi)/(2))),(1", "alpha=0", "(pi)/(2)):}`