if f(x)=sin4x + cos4x, then what is the range of f(x)

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if f(x)=sin⁴x + cos⁴x, then what is the range of f(x)

Chirag Jain , 9 Years ago

Grade 12

4 Answers

Vijay Mukati

Last Activity: 9 Years ago

Dear Student,

Differnetiate the functions and equate the result with zero to get the point of minima and maximua. You will get x = 0, 45 and 90 degree.

From here we can easily find the range and it will come out to be [1/2, 1]

Thanks.

Chirag Jain

Last Activity: 9 Years ago

Sir, the answer is [3/4,1] as stated in the book and the book is Objective Maths by Purshottam Kumar Sharma

Chirag Jain

Last Activity: 9 Years ago

The service of askiitians is very slow, posted a question 3 days ago, but till now haven’t got any satisfactory rreply. How can someone invest in such a slow service? I am sorry if you think I have been rude in my comment.

arun

Last Activity: 9 Years ago

in this question we have to find the range for $\sin ^4x + \cos ^4x$

now, $\sin ^4x + \cos ^4x = (\sin ^2x)^2+(\cos ^2x)^2$

$\Rightarrow \sin ^4x + \cos ^4x = (\sin ^2x + \cos ^2x)^2-2\sin ^2x\cos ^2x$

$\Rightarrow \sin ^4x + \cos ^4x = 1-\frac{1}{2}(4\sin^2x\cos^2x)$

$\Rightarrow \sin ^4x + \cos ^4x = 1-\frac{1}{2}\sin^22x$

as $\sin^2\theta$ have a maximum value of 1 and a minimum value of 0

so, minimum value of $\sin ^4x + \cos ^4x = 1-\frac{1}{2}(1) = \frac{1}{2}$

and maximum value of $\sin ^4x + \cos ^4x = 1-\frac{1}{2}(0)=1$

so, range of f(x)=[1/2,1]

I think there is any mistake in the book. As you can see the detailed solution of the question and if there is any other doubt related to this question you can ask