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If f(a)=lim(xto2)(sin^(x)a+cos^(x)a)^((1...

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

A

`f(0)=1`

B

`f(pi/2)=1`

C

`f(alpha)=(cos alpha)^(cos^2 alpha) *(sin alpha)^(sin^2 alpha)` if `alpha in (0,pi/2)`

D

`f(alpha)=(sin alpha)^(sin^2 alpha)/(cos alpha)^(cos^2 alpha)` if `alpha in (0,pi/2)`

Text Solution

Verified by Experts

The correct Answer is:
A, B, C

`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)):}`
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