InterviewSolution
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InterviewSolution
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If `f(a)=lim_(xto2)(sin^(x)a+cos^(x)a)^((1)/((x-2)))" for "ain[0,(pi)/(2)],` thenA. `-np`B. `np`C. `n^(2)p`D. `np^(2)` |
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Answer» Correct Answer - 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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