Q4

Geometry Level 2

Find the minimum value of f ( x ) = sin ⁡ 4 ( x ) + cos ⁡ 4 ( x ) f(x) = \sin^4 (x) + \cos^4(x) where 0 ≤ x ≤ π 2 0 \le x \le \frac\pi2 .

2 2 \frac{\sqrt2}2 1 4 \frac14 − 1 2 -\frac12 1 2 \frac12

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Sahil Bansal by Sahil Bansal , 21, India

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2 solutions

Akhil Bansal
Oct 12, 2015

⇒ f ( x ) = sin ⁡ 4 x + cos ⁡ 4 x = ( sin ⁡ 2 x ) 2 + ( cos ⁡ 2 x ) 2 \Rightarrow f(x) = \sin^4x + \cos^4x = (\sin^2x)^2 + (\cos^2x)^2 ⇒ f ( x ) = ( sin ⁡ 2 x + cos ⁡ 2 x ) 2 − 2 sin ⁡ 2 x cos ⁡ 2 x \Rightarrow f(x) = (\sin^2x + \cos^2x)^2 - 2\sin^2x\cos^2x ⇒ f ( x ) = 1 − ( sin ⁡ 2 x ) 2 2 \Rightarrow f(x) = 1 - \dfrac{(\sin 2x)^2}{2} Maximum of ( sin ⁡ 2 x ) 2 = 1 at x = π 4 \text{Maximum of} (\sin 2x)^2 = 1 \ \text{at} \ x = \dfrac{\pi}{4} ∴ Minimum of f ( x ) = 1 − 1 2 = 1 2 \therefore \text{Minimum of} \ f(x) = 1 - \dfrac{1}{2} = \boxed{\color{#3D99F6}{\dfrac{1}{2}}}

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We can solve this algebraically using AM-GM inequality.

Kushagra Sahni - 5 years, 8 months ago

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Yes, that are many methods to solve this,
It can also be solved by finding critical points.

Akhil Bansal - 5 years, 8 months ago
Denver Dsouza
Oct 22, 2015

An intuitive solution is to put x= 45°. Graphs of sin and cos interect at it. It would be nice if someone could explain why this works .

1 Helpful 0 Interesting 0 Brilliant 0 Confused

We can differentiate the function and equate it to 0.(for finding critical points) The result will be x=π/4.(among the solutions, π/4 will yield the minimum value.

Ankur Mukherjee - 5 years, 7 months ago

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