2 Roots and Sine!

Algebra Level 3

f ( x ) = x 3 2 a x 2 + ( b a ) x + a b x 2 \large f(x) = \frac {x^3-2ax^2+(b-a)x+ab}{x-2}

Function f ( x ) f(x) above, where a a and b b are integers, has two distinct real roots and f ( sin ( x ) ) = 0 f(\sin(x))=0 has one real solution for π < x < 2 π \pi<x<2\pi ,

Determine the product of the two roots.


The answer is -3.

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Yashas Ravi
Jun 22, 2019

Consider g ( x ) = x 3 ( 2 a ) x 2 + ( b a ) x + ( a b ) g(x)=x^3-(2a)x^2+(b-a)x+(ab) . If f ( x ) = 0 f(x)=0 , then g ( x ) = 0 g(x)=0 . If g ( x ) = 0 g(x)=0 , there can be either 1 1 solution or 3 3 solutions. However, since f ( x ) f(x) has 2 2 distinct roots, then ( x 2 ) (x-2) needs to be a factor of g ( x ) g(x) . By the Remainder Factor theorem, 2 3 2 a ( 2 2 ) + ( b a ) ( 2 ) + ( a b ) = 0 2^3-2a(2^2)+(b-a)(2)+(ab)=0 which simplifies to a b 10 a + 2 b + 8 = 0 ab-10a+2b+8=0 .

Since f ( sin ( x ) ) = 0 f(\sin(x))=0 has 1 1 solution in the given interval, then sin ( x ) \sin(x) can only take 1 1 distinct value in that interval. Since sin ( 1.5 π ) = 1 \sin(1.5π)=-1 is the only value that satisfies the condition, and 1 sin ( x ) 1 -1≤\sin(x)≤1 so even though ( sin ( x ) 2 ) (\sin(x)-2) is a factor, sin ( x ) = 2 \sin(x)=2 does not bring any real roots. As a result, ( x + 1 ) (x+1) is also a factor of g ( x ) g(x) . Again, by the Remainder Factor Theorem, ( 1 ) 3 2 a ( 1 ) 2 + ( b a ) ( 1 ) + ( a b ) = 0 (-1)^3-2a(-1)^2+(b-a)(-1)+(ab)=0 which simplifies to a b a b 1 = 0 ab-a-b-1=0 .

By solving the system regarding a a and b b by setting both equations to b and solving by substitution and the Quadratic Formula, a = 2 a=2 and b = 3 b=3 . This means that g ( x ) = x 3 4 x 2 + x + 6 g(x)=x^3-4x^2+x+6 . By Polynomial division, the quotient of g ( x ) g(x) and ( x 2 ) ( x + 1 ) (x-2)(x+1) is ( x 3 ) (x-3) . As a result, ( x 3 ) (x-3) is the remaining factor of g ( x ) g(x) . This means that ( x 3 ) (x-3) is also a factor of f ( x ) f(x) , making x = 3 x=3 and x = 1 x=-1 the 2 2 distinct roots. Finally, 1 3 = 3 -1*3=-3 which is the final answer.

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...