Domainomania

Geometry Level 4

f ( x ) = log x cos ( 2 π x ) \large f(x)=\sqrt {\log_x \cos(2\pi x)}

If the domain of f ( x ) f(x) can be represented as ( a , y z ) ( b c , d ) { e , f , g , } \left(a,\dfrac {y}{z} \right)\cup \left (\dfrac{b}{c},d \right )\cup\{e,f,g, \ldots\} , where y y and z z , and b b and c c are coprime integers, find the value of a + y + z + b + c + d a+y+z+b+c+d .


The answer is 13.

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2 solutions

Arjun Mishra
Sep 22, 2015

Before you start, keep following properties in your mind: 1. log cannot have negative values. 2. The base of the log cannot be negative. 3. Cosine is only positive in first and fourth quadrant.

Now, for the first quadrant, cos is +ve for 0 to pie/2. Therefore, value of x varies from 0 to 1/4. Therefore, first range is (0,1/4). Similarly, there are two ways you can reach the last quadrant. The first range for x is (-1/4,0) and the second range is (3/4,1). Now, since x cannot be negative therefore, the (3/4,1) is the accepted range. Now, on comparing, the values, a=0 y=1 z=4 b=3 c=4 d=1, so on adding these we get 13.

did the same way, I also get that{e,f,g...}is {2,3,4...}, is it correct?

Kelvin Hong - 5 years, 8 months ago

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Yeah i too got the same @kelvin hong 方

Prakhar Bindal - 5 years ago
Aditya Chauhan
May 30, 2016

For the function to be defined

log x cos 2 π x 0 \log_x \cos2 \pi x \geq 0

Case 1: When x > 1 x>1

cos 2 π x x 0 \cos2 \pi x \geq x^{0}

cos 2 π x 1 \cos2 \pi x \geq 1

cos 2 π x = 1 \cos2 \pi x = 1

x = 1 , 2 , 3 , 4 , . . . x = {1,2,3,4,...}

Case 2: When 0 < x < 1 0 < x < 1

0 < cos 2 π x < 1 0 < \cos2 \pi x < 1

0 < 2 π x < π 2 o r 3 π 2 < 2 π x < 2 π 0 < 2 \pi x < \dfrac{\pi}{2} or \dfrac{3 \pi}{2} < 2 \pi x < 2 \pi

0 < x < 1 4 o r 3 4 < x < 1 0<x<\dfrac{1}{4} or \dfrac{3}{4}<x<1

Required value = 0 + 1 + 4 + 3 + 4 + 1 = 13 0+1+4+3+4+1 = \boxed{13}

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