If `{(sin{cosx})/(x-pi/2),x!=pi/2 and 1,x=pi/2,` where {.} represents the fractional part function, then `f(x)` isA. continuous at `x=pi//2`B. `underset(x to pi//2)lim f(x)` but f(x) is not continuous at `x=pi//2`C. `underset(x to pi//2)lim f(x)` does not existD. `underset(x to pi//2^(-))lim f(x) =-1`

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

1.	If `{(sin{cosx})/(x-pi/2),x!=pi/2 and 1,x=pi/2,` where {.} represents the fractional part function, then `f(x)` isA. continuous at `x=pi//2`B. `underset(x to pi//2)lim f(x)` but f(x) is not continuous at `x=pi//2`C. `underset(x to pi//2)lim f(x)` does not existD. `underset(x to pi//2^(-))lim f(x) =-1`
Answer» Correct Answer - B We have `underset(xto pi//2^(-))lim f(x)=underset(h to 0)lim f(pi//2-h)` `Rightarrow underset(xto pi//2^(-))lim f(x)=underset(h to 0)lim (sin {cos (pi//2-h)})/(-h)` `Rightarrow underset(xto pi//2^(-))lim f(x)=underset(h to 0)lim (sin{sin h})/(-h)2` `Rightarrow underset(xto pi//2^(-))lim f(x)=underset(h to 0)lim (sin(sin h))/(-h)` `Rightarrow underset(xto pi//2^(-))lim f(x)=underset(h to 0)lim (sin(sin h))/(sin h)xx(sin h)/(h)=-1` and `underset(xto pi//2^(-))lim f(x)=underset(h to 0)lim f(pi//2+h)` `Rightarrow underset(xto pi//2^(+))lim f(x)=underset(h to 0)lim (sin {cos (pi//2+h)})/((pi)/(2)+h-(pi)/(2))` `Rightarrow underset(xto pi//2^(+))lim f(x)=underset(h to 0)lim (sin{-sin h})/(h)` `Rightarrow underset(xto pi//2^(+))lim f(x)=underset(h to 0)lim (sin{sin h})/(h)xx(sin h)/(h)=-1` `"Clearly", underset(x to pi//2^(-))lim f(x)=underset(x to pi//2^(+))lim f(x) ne f((pi)/(2))` `"So", underset(x to pi//2)lim` f(x) exists but f(x) is not continuous at `x=pi//2`.