Extremas of a Function!

Calculus Level 5

f ( cot ⁡ ( x ) ) = sin ⁡ ( 2 x ) + cos ⁡ ( 2 x ) \large{f(\cot(x)) = \sin(2x) + \cos(2x)}

Let f f be a function defined on the set of real numbers R \mathbb R , taking the values in R \mathbb R , and satisfying the above condition for ∀ x ∈ ( 0 , π ) \forall \ x \in (0,\pi) .

If the sum of the least and greatest values of the function g ( x ) = f ( x ) ⋅ f ( 1 − x ) g(x) = f(x) \cdot f(1-x) on the closed interval [ − 1 , 1 ] [-1,1] can be expressed as:

A B − C \large{\dfrac{A}{B} - \sqrt{C}}

such that A , B , C ∈ Z + ; gcd ⁡ ( A , B ) = 1 A,B,C \in \mathbb Z^+\ ; \ \gcd(A,B)=1 and C C has no perfect n t h n^{th} power factor, n ∈ Z + , n ≥ 2 n \in \mathbb Z^+, n \geq 2 . Submit the value of A + B + C A+B+C as your answer.


The answer is 160.

Satyajit Mohanty by Satyajit Mohanty , 24, India

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1 solution

Mark Hennings
Jan 3, 2016

Since sin ⁡ 2 x = 2 tan ⁡ x 1 + tan ⁡ 2 x = 2 cot ⁡ x cot ⁡ 2 x + 1 cos ⁡ 2 x = ; 1 − tan ⁡ 2 x 1 + tan ⁡ 2 x = cot ⁡ 2 x − 1 cot ⁡ 2 x + 1 \sin2x \; =\; \frac{2\tan x}{1 + \tan^2x} \; = \; \frac{2\cot x}{\cot^2x + 1} \qquad \qquad \cos2x \; =\ ; \frac{1 - \tan^2x }{1 + \tan^2x} \; = \; \frac{\cot^2x - 1}{\cot^2x + 1} we see that f ( u ) = u 2 + 2 u − 1 u 2 + 1 f(u) \; = \; \frac{u^2 + 2u - 1}{u^2+1} so that g ( u ) = ( u 2 + 2 u − 1 ) ( u 2 − 4 u + 2 ) ( u 2 + 1 ) ( u 2 − 2 u + 2 ) g(u) \; = \; \frac{(u^2+2u-1)(u^2 - 4u + 2)}{(u^2+1)(u^2-2u+2)} and hence g ′ ( u ) = 2 ( 2 u − 1 ) ( 5 u 4 − 10 u 3 + 9 u 2 − 4 u − 6 ) ( u 2 + 1 ) 2 ( u 2 − 2 u + 1 ) 2 g'(u) \; =\; \frac{2(2u-1)(5u^4 - 10u^3 + 9u^2 - 4u - 6)}{(u^2+1)^2(u^2-2u+1)^2} The numerator of g ′ ( u ) g'(u) factorizes as 2 ( 2 u − 1 ) ( 5 u 4 − 10 u 3 + 9 u 2 − 4 u − 6 ) = 2 5 ( 2 u − 1 ) ( 5 u 2 − 5 u + 2 + 34 ) ( 5 u 2 − 5 u + 2 − 34 ) 2(2u-1)(5u^4 - 10u^3 + 9u^2 - 4u - 6) \; = \; \tfrac25(2u-1)(5u^2 - 5u + 2 + \sqrt{34})(5u^2 - 5u + 2 - \sqrt{34}) and so the turning points of g ( u ) g(u) in the interval [ − 1 , 1 ] [-1,1] occur at u = 1 2 u = 1 2 ( 1 − 1 5 ( 4 34 − 3 ) ) . u \; =\; \tfrac12 \qquad \qquad u \; = \; \tfrac12 \left(1 - \sqrt{\tfrac15 (4 \sqrt{34} - 3)}\right) \;. The maximum value of g ( u ) g(u) in the interval [ − 1 , 1 ] [-1,1] is g ( 1 2 ) = 1 25 , g(\tfrac12) \; = \; \tfrac{1}{25} \;, while the maximum value of g ( u ) g(u) in this interval is g ( 1 2 ( 1 − 1 5 ( 4 34 − 3 ) ) ) = 4 − 34 . g\left(\tfrac12 \left(1 - \sqrt{\tfrac15 (4 \sqrt{34} - 3)}\right)\right) \; = \; 4 - \sqrt{34} \;. and so the sum of the maximum and minimum values of g ( u ) g(u) over the interval [ − 1 , 1 ] [-1,1] is 1 25 + ( 4 − 34 ) = 101 25 − 34 \tfrac{1}{25} + \big(4 - \sqrt{34}\big) \; = \; \tfrac{101}{25} - \sqrt{34} leading to the answer 101 + 25 + 34 = 160 101 + 25 + 34 \,=\, 160 .

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Actually, you need to verify that the values of function at boundary are not minima/maxima. However, that's not the case.

Perathorn Pooksombat - 5 years, 5 months ago

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The real roots of g ′ g' are simple, so tracking the sign changes of g ′ g' is elementary.

Mark Hennings - 5 years, 5 months ago

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