Fun With Calculus-10

Calculus Level 3

If u ( x ) u(x) is the larger of x + x 3 6 x+\frac{x^3}{6} and tan ⁡ − 1 x {\tan^{-1}x} for 0 < x < 1 0<x<1 ,

then what is the value of 144 u ( 1 2 ) ? 144 u(\frac{1}{2})?

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The answer is 75.

Akshay Sharma by Akshay Sharma , 21, India

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1 solution

Akshay Sharma
May 10, 2016

Let f ( x ) = x + x 3 6 − tan ⁡ − 1 x f(x)=x+\frac{x^3}{6}-\tan^{-1}x

⟹ f ′ ( x ) = 1 + x 2 2 − 1 1 + x 2 \large \implies f'(x)=1+\frac{x^2}{2}-\frac{1}{1+x^2} = x 2 ( 3 + x 2 ) 2 ( 1 + x 2 ) > 0 =\frac{x^2(3+x^2)}{2(1+x^2)}>0

⟹ \implies f(x) is increasing

since x > 0 x>0 therefore f ( x ) > f ( 0 ) f(x)>f(0) x + x 3 6 − tan ⁡ − 1 x > 0 \large x+\frac{x^3}{6}-\tan^{-1}x>0

⟹ x + x 3 6 > tan ⁡ − 1 x \large \implies x+\frac{x^3}{6}>\tan^{-1}x therefore u ( x ) = x + x 3 6 u(x)=x+\frac{x^3}{6}

Then, 144 u ( 1 2 ) = 144 ( 1 2 + 1 48 ) 144u(\frac{1}{2})=144\left (\frac{1}{2}+\frac{1}{48}\right ) = 144 ( 25 48 ) =144(\frac{25}{48}) = 75 \boxed {=75}

5 Helpful 0 Interesting 1 Brilliant 0 Confused

This is never a Level 5 one!

Andreas Wendler - 5 years, 1 month ago

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@Akshay Sharma Please set a resonable level for your problems. If you think that this is of difficulty level 5, that would cause me to question if you are really level 5, or if I have to reset your levels.

Calvin Lin Staff - 5 years ago

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I will take care of that @Calvin Lin

Akshay Sharma - 5 years ago

You should have gone for options because many of the people would have just entered the two possibilities and would have got the answer.

Rudraksh Shukla - 5 years, 1 month ago

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