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If f(x)=f(-x), then f(x) is even function
and if f(x) =-f(-x) then f(x) is odd function
For example Sin(x) is odd function while cos(x) is even function. This means that sin(x)=-sin(-x) and cos(x)=cos(-x)

With this basic understanding, let us see this question.

Problem : If function f satisfies the relation f(x+1)+f(1-x)=2f(x) for all real values of x and f(0)< or >0, then prove that f(x) is even.
Replacing x with -x
Comparing both we get, 2 f(x) = 2 f(-x)
Hence f(x) = f(-x)
Hence proved that, f(x) is an even function.


Let f : A à B, and  g : B à C, be two functions, then gof : A à C
This is known as the product function or composite of f and g, given by gof(x)=g{f(x)} for all real values of x.
The symbols  f : A à B is read as Function F maps A to B.

It is  to be noted that fog = f{g (x)} and (f ± g) x = f(x) ± g(x) and (f/g)x=f(x)/g(x)

If F : RàR and G : RàR, be two mappings such that f(x) = sin x and g(x)=x2
Then show that fog gof
Solution : Let x R,
So (fog) x = f {g(x)} = sin(x2)         -----(1)
While (gof) x = g {f(x)} = g {sinx} = (sinx)2 = sin2x    -----(2)
From (1) & (2), (fog) x (gof) x

A function f(x) is said to be periodic function of x, if there exists a positive real number T such that f(x+T) = f(x). Then the smallest value of T is called the Period of the function.
For example Sin x is periodic because

Sin(2π+x)= Sin(x) and also sin(4π+x)=Sin(x) = Sin(6π+x) and so on…. But 2π being the smallest value is the period of the function.
Example: Show tha cos(x) is non-periodic.
Solution: Let Cos(x) be periodic with T as the period.
So, Cos(x) = Cos{ √ (T+x) }
à √ (T+x)  = 2nπ ± x
Putting x = 0, we get,  √T= 2nπ    -----(1)
Putting x=T, we get  √2T = 2nπ ± √T     ----(2)
From (1) and (2) we get, √2T = √T ± √T
Or √T x √2 = √T (1 ± 1)
Or √2 = 1 ± 1, which is not possible, Hence Cos√x is not periodic function.

IIT JEE EXAM 2011- EXAM DATE 10 April 2011, Last date to Apply = 15 December 2010

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