Math Homework Help
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Math Homework Help

Identity Function

published on March 22nd, 2008 . by Vanaja

The function that associates each real number to itself is called the identity function and is usually denoted by I.

So, the function f:R->R defined by f(x)=x for all x in R is called the identity function.

From the knowledge of coordinate geometry, y=x represents a straight line passing through the origin and inclined at angle 45° with the x axis.

identiti-function.JPG

Clearly the domain and range of the identity function are both equal to R.

We can observe that it is a bijection.

Graph of Functions (Constant Function)

published on March 15th, 2008 . by Vanaja

A function of the type y=f(x)=k where k is a fixed real number.

Graph of constant function.
The graph of the constant function is a straight line parallel to x axis, which is above or below according to k is positive or negative. That is if k>0 the graph will be above x axis and at a distance k units above it. If k<0, then it will be k units below it. If k=0, then the graph will coincide with the x axis.

The domain of the constant function f(x)=k is the set R of all real numbers and range of the function is the singleton set {k}

So, we can see a constant function is a many-one into function.

Two Similar Cylinders

published on March 5th, 2008 . by Vanaja


The two cylindrical pans are similar. The diameter of the smaller pan is equal to the radius of the larger pan. How many of these smaller cans could fill the larger can?

similar-cylinders.JPG

Hint: Since the two cylinders are similar, their dimensions are in the same ratio. It is given that the diameter of the smaller pan is same as the radius of the larger pan. That is the radius of the two pans are in the ratio 2:1. In other words we can say if the radius of the larger pan is ‘r’ the radius of the smaller pan is r/2 and since they are similar their heights are also in the same ratio 2:1. So, if h is the height of the larger triangle, h/2 is the height of the smaller triangle.

Now,

Volume of the larger cylinder Vl= pi r2h

Volume of the smaller cylinder Vs=pi(r/2)2(h/2) =(pir2h )/8 = Vl/8

i.e Vs =Vl /8

Therefore 8 smaller pans can fill one larger pan.