Find the maximun and minimum value of :
4 sin2 x + 3 cos2 x + sin x2 + cos x2
3414 \sin^2 x + 3 \cos^2 x + \sin \frac{x}{2} + \cos \frac{x}{2} = 3 + \sin^2 x + \sin \frac{x}{2} + \cos \frac{x}{2}
Now if we let t = \sin \frac{x}{2} + \cos \frac{x}{2} we have \sin^2x = (t^2-1)^2
So, the expression in terms of t is 3+ (t^2-1)^2+t
Note that -\sqrt 2 \le t \le \sqrt 2. Its obvious that the maximum occurs when t>0 and also that both (t2-1)2 and t both attain their maximum when t = √2. So that the max is 4 + √2
I am not quite sure of the minimum though, is it 2?
1The minimum value, as written in the answer, is (3-√2) ; but I'm not satisfied with the solution given in the book. They have just separately calculated the minimum value of (3+sin2x) and (sinx2+cosx2) and added them up to get the solution.
Will taking derivative of the given expression be worth?
62no the book is wrong mayukh...
have no doubt about it :)
What the book has probably given is a upper bound to the lowest value...
341From Nishant sir's post I gather that the original question probably asked to prove that the expression is greater than 3-√2. This is easily done as we have to prove that
3+(t^2-1)^2 + t > 3- \sqrt 2 or (t^2-1)^2 > -(t+ \sqrt 2) which is true as -\sqrt 2 \le t \Rightarrow -(t+\sqrt 2) \le 0
62@Mayukh.. (I hope you are not confused by what prophet sir and I have been trying to write)
see this
(x-1)2+(x-2)2>0
We know that individually, both are greater than zero...
but the expression above will never attain a value zero..
The lowest value of the expression will be at x=3/2 and the value will be given by 1/2
so even though you can easily prove (x-1)2+(x-2)2>0, it is not the best lower limit. :)