InterviewSolution
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InterviewSolution
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If `{(sin{cosx})/(x-pi/2),x!=pi/2 and 1,x=pi/2,` where {.} represents the fractional part function, then `f(x)` isA. `-1`B. 1C. non-existantD. none of these |
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Answer» Correct Answer - C We have ` lim_(xtopi//2^-)f(x)=lim_(xto pi//2^-)(sin{cosx})/(x-pi//2)` ` rArr lim_(xtopi//2^-)f(x)=lim_(hto0)(sin{cos((pi)/(2)-h)})/(pi//2-h-pi//2)=lim_(hto0) (sin{sin h})/(-h)` ` rArr lim_(xtopi//2^-)f(x)=lim_(hto0) (sin(sin h))/(-h)=-lim_(hto0) (sin(sin h))/(sin h)xx(sinh)/(h)=-1` and, ` lim_(xtopi//2^+)f(x)=lim_(hto0) (sin{-sin h})/(x-pi//2)=lim_(hto0) (sin{cos (pi//2+h)})/((pi)/(2)+h-(pi)/(2))` ` rArr lim_(xtopi//2^+)f(x)=lim_(hto0) (sin{-sinh})/(h)=lim_(hto0) (sin(1-sinh))/(h)to oo` Hence, `lim_(xto pi//2)f(x)` does not exist. |
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