Common Tangent Line 2

Calculus Level 3

Let n n be a positive integer and

{ f ( x ) = − x 2 n g ( x ) = x 2 n \large \begin{cases} f(x) = -x^{2n} \\ g(x) = \sqrt[2n]x\ \end{cases}

Find the value of n n for which the common tangent line to both curves above passes through the point below, ( ( − n + 3 + 51 2 n ) ( 1 2 n ) 1 2 n − 1 , ( − n 2 − 20 − 50 2 n ) ( 1 2 n ) 1 2 n − 1 ) \left(\left(-n + 3 + \dfrac{51}{2n}\right)\left(\frac{1}{2n}\right)^{\frac{1}{2n - 1}}, \left(-\frac{n}{2} -20 - \dfrac{50}{2n}\right)\left(\frac{1}{2n}\right)^{\frac{1}{2n - 1}}\right)


The answer is 50.

Rocco Dalto by Rocco Dalto , 67, USA

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1 solution

Rocco Dalto
Dec 6, 2018

f ′ ( a ) = − 2 n a 2 n − 1 = g ′ ( b ) = 1 2 n b 2 n − 1 2 n f'(a) = -2na^{2n - 1} = g'(b) = \dfrac{1}{2n b^{\frac{2n - 1}{2n}}} ⟹ a = − 1 α n b 1 2 n \implies a = -\dfrac{1}{\alpha_{n} b^{\frac{1}{2n}}} , where α n = ( 2 n ) 2 2 n − 1 ⟹ \alpha_{n} = (2n)^{\frac{2}{2n - 1}} \implies

A : ( a , a 2 n ) = ( − 1 α n b 1 2 n , − 1 α n 2 n b ) A:(a,a^{2n}) = (-\dfrac{1}{\alpha_{n} b^{\frac{1}{2n}}}, -\dfrac{1}{\alpha_{n}^{2n} b}) and B : ( b , b 1 2 n ) B:(b,b^{\frac{1}{2n}}) ⟹ m A B = ( α n 2 n b 2 n + 1 2 n + 1 α n b 2 n + 1 2 n + 1 ) ( 1 α 2 n − 1 b 2 n − 1 2 n ) = \implies m_{AB} = (\dfrac{\alpha_{n}^{2n} b^{\frac{2n + 1}{2n}} + 1}{\alpha_{n} b^{\frac{2n + 1}{2n}} + 1})(\dfrac{1}{\alpha^{2n - 1} b^{\frac{2n - 1}{2n}}}) = 1 α n 2 n − 1 2 b 2 n − 1 2 n \dfrac{1}{\alpha_{n}^{\frac{2n - 1}{2}} b^{\frac{2n - 1}{2n}}}

⟹ α n n b 2 n + 1 2 n = 1 α n ⟹ \implies {\alpha_{n}}^{n} b^{\frac{2n + 1}{2n}} = \dfrac{1}{\sqrt{\alpha_{n}}} \implies

b = ( 1 α n n α n ) 2 n 2 n + 1 ⟹ b = (\dfrac{1}{\alpha_{n}^{n}\sqrt{\alpha_{n}}})^{\frac{2n}{2n + 1}} \implies a = − ( α n n α n ) 1 2 n + 1 α n a = -\dfrac{(\alpha_{n}^{n}\sqrt{\alpha_{n}})^{\frac{1}{2n + 1}}}{\alpha_{n}} and α n = ( 2 n ) 2 2 n − 1 ⟹ m A B = − 2 n a 2 n − 1 = ( α n ) 2 n − 1 ( α n α n n + 1 ) 2 n − 1 2 n + 1 = ( 2 n ) 2 ( n + 1 ) 2 n + 1 ( 2 n ) 2 ( n + 1 ) 2 n + 1 = 1 \alpha_{n} = (2n)^{\frac{2}{2n - 1}} \implies m_{AB} = -2n a^{2n - 1} = (\sqrt{\alpha_{n}})^{2n - 1} (\dfrac{\sqrt{\alpha_{n}}}{\alpha_{n}^{n +1}})^{\frac{2n - 1}{2n + 1}} = \dfrac{(2n)^{\frac{2(n +1)}{2n + 1}}}{(2n)^{\frac{2(n +1)}{2n + 1}}} = 1 .

Using B : ( ( 1 2 n ) 2 n 2 n − 1 , ( 1 2 n ) 1 2 n − 1 ) ⟹ B:((\dfrac{1}{2n})^{\frac{2n}{2n - 1}}, (\dfrac{1}{2n})^{\frac{1}{2n - 1}}) \implies y − x = ( 2 n − 1 2 n ) ( 1 2 n ) 1 2 n − 1 y - x = (\dfrac{2n - 1}{2n})(\dfrac{1}{2n})^{\frac{1}{2n - 1}}

The above common tangent line to both curves passes thru the point ( ( − n + 3 + 51 2 n ) ( 1 2 n ) 1 2 n − 1 , ( − n 2 − 20 − 50 2 n ) ( 1 2 n ) 1 2 n − 1 ) \left(\left(-n + 3 + \dfrac{51}{2n}\right)\left(\frac{1}{2n}\right)^{\frac{1}{2n - 1}}, \left(-\frac{n}{2} -20 - \dfrac{50}{2n}\right)\left(\frac{1}{2n}\right)^{\frac{1}{2n - 1}}\right)

⟹ ( 1 2 n ) 1 2 n − 1 ( n 2 − 23 − 101 2 n ) = ( 2 n − 1 2 n ) ( 1 2 n ) 1 2 n − 1 ⟹ n 2 − 46 n − 101 = 2 n − 1 ⟹ \implies (\dfrac{1}{2n})^{\frac{1}{2n - 1}}(\dfrac{n}{2} - 23 - \dfrac{101}{2n}) = (\dfrac{2n - 1}{2n})(\dfrac{1}{2n})^{\frac{1}{2n - 1}} \implies n^2 - 46n - 101= 2n - 1 \implies

n 2 − 48 n − 100 = 0 ⟹ ( n − 50 ) ( n + 2 ) = 0 n^2 - 48n - 100 = 0 \implies (n - 50)(n + 2) = 0 and n ≠ − 2 ⟹ n = 50 . n \neq -2 \implies n = \boxed{50}.

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