Question

F(x) is given by. (1-cos2x/2x^2) when x is not equal to 0. And (K) when x=0 . What will be the value of k?

Answer 1

K = 1 when x = 0

Explanation:

A trigonometric identity that is relevant here is:

1-`cos(2x)` = 2`sin^2(x)`

It follows that

`(1-cos(2x))/(2x^2) = sin^2(x)/x^2 = (sin(x)/x)^2`

It is known that

`lim sin(x)/x = 1` as `x -> 0`

The given function, therefore, will approach 1 as x approaches 0, and therefore K = 1.

Useful links

https://www.khanacademy.org/math/ap-calculus-ab/ab-derivative-rules/ab-derivtive-rules-opt-vids/v/sinx-over-x-as-x-approaches-0

http://www.alcyone.com/max/reference/maths/trigonometry.html

Answer 2

Unless you forgot to tell us something, `k` may be any real number (or none at all)

Explanation:

If we do not select a `k`, then the domain of `F` is `(-oo,0)uu(0,oo)`. (All reals except `0`.)

If we want the domain of `F` to be all reals, then `k` may be any real number.

If we want `F` to be continuous at `0` , then we need to have `lim_(xrarr0)F(x) = F(0) = k`.

So we need to find `lim_(xrarr0)F(x) = lim_(xrarr0)(1-cos2x)/(2x^2)`.

If we attempt substitution, we get the indeterminate form `0/0`.

We'll change the way the function is written so we can use `lim_(trarr0)sint/t=1`.

Recall that `cos2x = cos^2x-sin^2x` and, since `cos^2x = 1-sin^2x`, we also have

`cos2x = 1-2sin^2x`

Therefore

`lim_(xrarr0)F(x) = lim_(xrarr0)(1-cos2x)/(2x^2)`.

`= lim_(xrarr0)(1-(1-2sin^2x))/(2x^2)`

`= lim_(xrarr0)(2sin^2x)/(2x^2)`

`= lim_(xrarr0)sin^2x/x^2`

`= (lim_(xrarr0)sin x/x)^2 = 1^2 =1`

So, make `k = 1` to make `F` continuous at `0`.