Question 1209178
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<pre>

Consider function  


    sin(x)*(sin(x)-2)+1) = sin^2(x) - 2sin(x) + sin(x) = sin^2(x) - sin(x) = sin(x)*(sin(x) -1).


In vicinity of  x= {{{pi/2}}}  this function is negative (except of its value at x= {{{pi/2}}}, where it is zero).


Therefore, in vicinity of  x = {{{pi/2}}}  the given expression, which is the square root of sin(x)*(sin(x)-2)+1),
is not defined, at all, as a real function, so the question  "is it continuous or not"

even can not be posed: this function does not exist in real domain.
</pre>

Solved.



It is a mathematical joke problem (a kind of TRAP) to check your alertness.