The Answer is (D. No solution)
(√10)/(√3)*(√3)/(<span>√3); multiplied by the conjugate which is = (</span><span>√30/3)</span>
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]
Answer:
C. 8
Step-by-step explanation:
![\because \: {s}^{3} = 64 \\ s = \sqrt[3]{64} \\ s = 4 \\ side \: of \: cube = 4 \: units \\ when \: side \: is \: halved \\ new \: side \: length = \frac{4}{2} = 2 \\ \: new \: volume = {2}^{3} = 8 \: cubic \: units](https://tex.z-dn.net/?f=%20%5Cbecause%20%5C%3A%20%20%7Bs%7D%5E%7B3%7D%20%20%3D%2064%20%5C%5C%20s%20%3D%20%20%5Csqrt%5B3%5D%7B64%7D%20%20%5C%5C%20s%20%3D%204%20%5C%5C%20side%20%5C%3A%20of%20%5C%3A%20cube%20%3D%204%20%5C%3A%20units%20%5C%5C%20when%20%5C%3A%20side%20%5C%3A%20is%20%5C%3A%20halved%20%5C%5C%20new%20%5C%3A%20side%20%5C%3A%20length%20%3D%20%20%5Cfrac%7B4%7D%7B2%7D%20%20%3D%202%20%5C%5C%20%20%5C%3A%20new%20%5C%3A%20volume%20%3D%20%20%7B2%7D%5E%7B3%7D%20%20%3D%208%20%5C%3A%20cubic%20%5C%3A%20units)
