Q1)Let g(x) be be function defined on{-1, 1].If the area of the equilateral triangle of two of vertices at(0,0) & (x,g(x)) is √3/4, then the function may be
a)g(x)=±√(1-x2)
b)+√(1-x2)
c)-√(1-x2)
Q2)find domain:f(x)=√(sin x) -√(16-x^2)
Q3)pls explain this:
(1) for domain h(x)={f(x)} g(x), conventionally f(x)>0 and g(x) must be real
(2)for domain h(x)=f(x) C g(x) or f(x) P g(x) conventional conditions of domain are f(x)≥g(x)
and f(x) is a natural no , g(x) is a whole no.
Q4)Is x6 = y3 a function.If not why?
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2 Answers
62The area of a triangle is dependent on the side...
So the side of equiliateral triangle is constant...
Now can you think of the rest for the 1st question...
FOr the 2nd, I would like to see your attempt.. because as I see it, it is quite simple...
62x6=y3
It is a function if you go from R->R.. because for every value of x, you have a unique value of y... (I am presuming that y is the dependent variable) (Ie y is a function of x and not the other way round)
but not a function if you go from Complex to Complex numbers....