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If {(sin{cosx})/(x-pi/2),x!=pi/2 and 1,x...

If `{(sin{cosx})/(x-pi/2),x!=pi/2 and 1,x=pi/2,` where {.} represents the fractional part function, then `f(x)` is

A

`-1`

B

1

C

non-existant

D

none of these

Text Solution

Verified by Experts

The correct Answer is:
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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