Question -29

Geometry Level 3

For real values x x , consider the expression

2 − cos ⁡ x + sin ⁡ 2 x 2 - \cos x + \sin^2 x

Find the ratio of the greatest value of the expression to its least value.

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17 4 \frac{17}{4} 13 2 \frac{13}{2} 13 4 \frac{13}{4} 9 4 \frac{9}{4}

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Sandeep Bhardwaj by Sandeep Bhardwaj , 28, India

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1 solution

Let f ( x ) = 2 − cos ⁡ x + sin ⁡ 2 x = 2 − cos ⁡ x + 1 − cos ⁡ 2 x = 3 − cos ⁡ x − cos ⁡ 2 x = 3 − ( cos ⁡ 2 x + cos ⁡ x + 1 4 − 1 4 ) = 13 4 − ( cos ⁡ x + 1 2 ) 2 \begin{aligned} \text{Let } f(x) & = 2-\cos{x}+\sin^2{x} \\ & = 2-\cos{x}+1 - \cos^2{x} \\ & = 3-\cos{x} - \cos^2{x} \\ & = 3-\left( \cos^2{x} + \cos{x} + \frac{1}{4} - \frac{1}{4} \right) \\ & = \frac {13}{4}-\left( \cos{x} + \frac{1}{2} \right)^2 \end{aligned}

We note that: g ( x ) = ( cos ⁡ x + 1 2 ) 2 > 0 g(x) = \left( \cos{x} + \frac{1}{2} \right)^2 > 0 and:

⇒ f ( x ) \Rightarrow f(x) is minimum when g ( x ) g(x) is maximum:

⇒ f m i n ( x ) = 13 4 − ( 1 + 1 2 ) 2 = 13 4 − 9 4 = 1 \Rightarrow f_{min} (x) = \frac {13}{4} - \left( 1 + \frac{1}{2} \right)^2 = \frac {13}{4} - \frac {9}{4} = 1

⇒ f ( x ) \Rightarrow f(x) is maximum when g ( x ) g(x) is minimum:

⇒ f m a x ( x ) = 13 4 − ( − 1 2 + 1 2 ) 2 = 13 4 − 0 = 13 4 \Rightarrow f_{max} (x) = \frac {13}{4} - \left( -\frac{1}{2} + \frac{1}{2} \right)^2 = \frac {13}{4} - 0 = \frac{13}{4}

⇒ f m a x ( x ) f m i n ( x ) = 13 4 \Rightarrow \dfrac {f_{max}(x)}{f_{min}(x)} = \boxed{\dfrac{13}{4}}

8 Helpful 0 Interesting 0 Brilliant 0 Confused

Here's my approach, f ( x ) = 2 − cos ⁡ x + sin ⁡ 2 x f\left( x \right) =2-\cos { x } +\sin ^{ 2 }{ x } Then, f ′ ( x ) = sin ⁡ x + 2 sin ⁡ x cos ⁡ x f^{ ' }\left( x \right) =\sin { x } +2\sin { x } \cos { x } Equating f ′ ( x ) f^{ ' }\left( x \right) to zero , sin ⁡ x ( 1 + 2 cos ⁡ x ) = 0 \sin { x } (1+2\cos { x } )=0 ⇒ sin ⁡ x = 0 cos ⁡ x = − 1 2 \Rightarrow \sin { x } =0\quad \cos { x } =\frac { -1 }{ 2 } ⇒ x = 0 , π x = 2 π 3 , 4 π 3 \Rightarrow x=0,\pi \quad \quad \quad x=\frac { 2\pi }{ 3 } ,\frac { 4\pi }{ 3 }

Now,

at x = 0 x=0 and x = π x=\pi , f ′ ′ ( x ) > 0 f^{ '' }\left( x \right) >0

at x = 2 π 3 , 4 π 3 x=\frac { 2\pi }{ 3 } ,\frac { 4\pi }{ 3 } , f ′ ′ ( x ) < 0 f^{ '' }\left( x \right) <0

By checking,

minimum value is at x = 0 x=0 and the minimum value is 1 1

Maximum value is at x = 2 π 3 , 4 π 3 x=\frac { 2\pi }{ 3 } ,\frac { 4\pi }{ 3 } and the maximum value is 13 4 \frac { 13 }{ 4 }

Therefore the ratio is 13 4 \boxed{\frac { 13 }{ 4 } }

Anandhu Raj - 5 years, 10 months ago

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