00:01
To find the time at which the concentration is maximum, we need to find the critical points of the concentration function k of x.
00:09
The maximum concentration will corresponds to the highest value of this concentration function k of x at that critical point.
00:17
First part of a question is to find the time at which the concentration is maximum.
00:22
So time is question mark.
00:23
To find the critical points, we need to find where the derivative of k of x is equal to and k of x is equal to twice x divided by x square plus 4, this value is given in the question.
00:38
Now we will take the derivative with respect to x.
00:41
So k dash of x it is equal to d by dx of twice x divided by x square plus 4.
00:50
This is nothing but equal to the denominator that is x square plus 4 as it is derivative of numerator is 2 minus twice x that is numerator as it is derivative of denominator is nothing but twice x divided by denominator square x square plus 4 its square.
01:12
This is k dash of x.
01:14
Solving this we get here 8 minus twice x square and divided by x square plus 4 whole square.
01:24
Now we will set this k dash of x it is equal to 0 for the critical points...