00:01
This question, there are three questions.
00:02
Question number 11 talks about the horizontal phase shift of this particular graph is what? first of all, let's represent it in the standard form.
00:09
And what is the standard form? that is why is equal to twice of cost of this three should be taken as a common factor.
00:16
So it's three is taken as a common factor.
00:18
We are left for theta plus pi over six, pie over six.
00:22
That's what it comes here.
00:24
And now since this is in standard form, we can easily find its horizontal shift.
00:31
So the horizontal shift is actually pi over six.
00:34
But whether it's in the right or to the left, so since this is plus, that should be towards the left.
00:39
Remember, it works opposite.
00:41
So since this is plus, this is a horizontal shift towards the left.
00:45
So we are going to say that it is pi over six units towards the left.
00:50
For question number 12, they are saying that this is a particular function and it has a range of minimum value of four and maximum value of 10.
00:58
So which of the following is the value of a and d? now we know that cosine maximum, cosine maximum value is 1 and cosine minimum value is minus 1.
01:11
If cosine maximum value is 1, so if we place this as 1, then a times 1 plus d will be the maximum value of function, which is a plus d.
01:21
And this is already given to us as 10.
01:25
And the minimum value is cosine should be minus 1 like we said.
01:28
So the minimum value is minus a plus d, which is already given to us as 4...