Value of a composite function

Geometry Level 3

If g ( x ) = ( 4 cos ⁡ 4 x − 2 cos ⁡ ( 2 x ) − 1 2 cos ⁡ ( 4 x ) − x 7 ) 1 7 g(x)=\left( 4\cos^4x-2\cos(2x)-\frac{1}{2}\cos(4x)-x^7 \right)^{\frac{1}{7}} , then find the value of g ( g ( 100 ) ) g(g(100)) .

If you're looking to skyrocket your preparation for JEE-2015, then go for solving this set of questions .


The answer is 100.

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Sandeep Bhardwaj by Sandeep Bhardwaj , 28, India

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3 solutions

Consider A = 4. c o s 4 x − 2. c o s 2 x − 1 2 . c o s 4 x A=4.cos^{ 4 }x-2.cos2x-\frac { 1 }{ 2 } .cos4x

d d x ( 4. c o s 4 x − 2. c o s 2 x − 1 2 . c o s 4 x ) = − 16 cos ⁡ 3 x sin ⁡ x + 4 sin ⁡ 2 x + 2 sin ⁡ 4 x = − 16 cos ⁡ 3 x sin ⁡ x + 4 sin ⁡ 2 x + 4 sin ⁡ 2 x cos ⁡ 2 x = − 16 cos ⁡ 3 x sin ⁡ x + 8 cos ⁡ 2 x sin ⁡ 2 x = 0 \frac { d }{ dx } (4.cos^{ 4 }x-2.cos2x-\frac { 1 }{ 2 } .cos4x)\quad =\quad -16\cos ^{ 3 }{ x } \sin { x } +4\sin { 2x } +2\sin { 4x } \\ =-16\cos ^{ 3 }{ x } \sin { x } +4\sin { 2x } +4\sin { 2x } \cos { 2x } \\ =-16\cos ^{ 3 }{ x } \sin { x } +8\cos ^{ 2 }{ x } \sin { 2x } \\ =\quad 0

This means that A A is a constant. It's easy to see that A = 3 / 2 A=3/2 (just put x = 0 x=0 ).

⇒ g ( x ) = ( 1.5 − x 7 ) 1 / 7 \Rightarrow g(x)= (1.5-x^{7})^{1/7}

⇒ g ( g ( x ) ) = x \Rightarrow g(g(x))= x

⇒ g ( g ( 100 ) ) = 100 \Rightarrow g(g(100))= \boxed{100}

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One another way by using complex numbers : cos ⁡ ( n x ) = e i n x + e − i n x 2 ( L e t e i x = y ) 2 cos ⁡ ( n x ) = ( y n + 1 y n ) . . . ( 1 ) 4 cos ⁡ 4 x = 1 4 ( y + 1 y ) 4 = 1 4 ( ( y 4 + 1 y 4 ) + 4 ( y 2 + 1 y 2 ) + 6 ) 4 cos ⁡ 4 x = 1 4 ( 2 cos ⁡ ( 4 x ) + 8 cos ⁡ ( 2 x ) + 6 ) 4 cos ⁡ 4 x − 2 cos ⁡ ( 2 x ) − 1 2 cos ⁡ ( 4 x ) = 3 2 . . . . ( 2 ) g ( x ) = ( 1.5 − x ) 1 7 \displaystyle{\cos { (nx) } =\cfrac { { e }^{ inx }+{ e }^{ -inx } }{ 2 } \quad (Let\quad { e }^{ ix }=y)\\ 2\cos { (nx) } =\left( { y }^{ n }+\cfrac { 1 }{ { y }^{ n } } \right) \quad .\quad .\quad .(1)\\ 4\cos ^{ 4 }{ x } =\cfrac { 1 }{ 4 } { \left( y+\cfrac { 1 }{ y } \right) }^{ 4 }=\cfrac { 1 }{ 4 } \left( ({ y }^{ 4 }+\cfrac { 1 }{ { y }^{ 4 } } )+4({ y }^{ 2 }+\cfrac { 1 }{ { y }^{ 2 } } )+6 \right) \\ 4\cos ^{ 4 }{ x } =\cfrac { 1 }{ 4 } \left( 2\cos { (4x) } +8\cos { (2x) } +6 \right) \\ 4\cos ^{ 4 }{ x } -2\cos { (2x) } -\cfrac { 1 }{ 2 } \cos { (4x) } =\cfrac { 3 }{ 2 } \quad .\quad .\quad .\quad .(2)\\ \boxed { g\left( x \right) ={ \left( 1.5-x \right) }^{ \cfrac { 1 }{ 7 } } } }

Deepanshu Gupta - 6 years, 3 months ago

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Yeah!! I learn't this technique from a comment in your note Complex number techniques ... :)

Nishu sharma - 6 years, 1 month ago
Ravi Dwivedi
Jul 4, 2015

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Moderator note:

Simple standard approach.

L e t f ( x ) = 4. c o s 4 x − 2. c o s 2 x − 1 2 . c o s 4 x . f ( x ) < 7 S o g ( 100 ) i t i s a s g o o d a s = ( − 10 0 7 ) 1 7 = − 100. ∴ g ( g ( 100 ) ) = g ( { ( − ( − 100 ) ) 7 } 1 7 ) 100 Let~ f(x)=4.cos^4x-2.cos2x-\frac{1}{2}.cos4x. ~~f(x)<7 \\So ~ g(100)~ it ~is~ as~ good~ as=(-100^7)^{\frac{1}{7}}=-100.\\\therefore g(g(100))=g \left(\color{#D61F06}{ \{ ( -(-100))^7 \}}^{\frac{1}{7} } \right)\\\boxed{\color{#3D99F6}{100}}

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