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]
A * 4 = E........A = E / 4
B / 4 = E........B = 4E
C + 4 = E......C = E - 4
D - 4 = E.......D = E + 4
A + B + C + D = 100
(E/4) + (4E) + (E - 4) + (E + 4) = 100
E/4 + 6E = 100 ....multiply by 4
E + 24E = 400
25E = 400
E = 400/25
E = 16 <===
Answer:
Circumference of circle =2πr=2π×10.5=21πm
Step-by-step explanation:
Answer:
x=0
x=-1
x=-5
Step-by-step explanation:
To find the roots, we set the equation equal to zero and factor.
x(x²+6x+5)=0 [factor]
x(x+1)(x+5)=0 [set each factor equal to 0]
x=0
x+1=0 [subtract both sides by 1]
x=-1
x+5=0 [subtract both sides by 5]
x=-5
Now, we have found three roots, x=0, x=-1, x=-5.
