18. If f(x)=[xsin πx] {where [x] denotes greatest integer function}, then f(x) is:
since x denotes the greatest integers which could the negative or the positive values, also x has a domain of all real numbers, and has no discontinuous point, then x is continuous in (-1,0).
Answer: B]
20. Given that g(x)=1/(x^2+x-1) and f(x)=1/(x-3), then to evaluate the discontinuous point in g(f(x)) we consider the denominator of g(x) and f(x). g(x) has no discontinuous point while f(x) is continuous at all points but x=3. Hence we shall say that g(f(x)) will also be discontinuous at x=3. Hence the answer is:
C] 3
21. Given that f(x)=[tan² x] where [.] is greatest integer function, from this we can see that tan x is continuous at all points apart from the point 180x+90, where x=0,1,2,3....
This implies that since some points are not continuous, then the limit does not exist.
Answer is:
A]
3.8 repeating or 3 8/9 i think
Answer:
Step-by-step explanation:
First convert them into improper fractions:
and
. Now we add:
or
.
.........Phitagors theory
Hi there! Yes, x = 2 is a solution to this equation.
Let's solve this equation step by step!

Subtract 2 from both sides.

Now divide both sides of the equation by -4 (and remember that when dividing two negatives we end up with a positive).

And therefore x = 2 is a solution to this equation.
~ Hope this helps you!