Roots of a polynomial mixed with a trigonometric function

[vcsharp2003]

Homework Statement

The question is as given in the image below and is part of a question bank for Calculus at high school (i.e. grade 12). The correct answer is supposed to be (C) two real roots.

Relevant Equations

None

When I look at the left hand side of the equation in above question then I can see that the highest degree of x would be 6 after the denominators are eliminated.

I know that a polynomial of degree n will have n roots, but this one is not a pure polynomial since there is also a trigonometric function.

I am confused on how to go about solving this problem. Any pointers would help. Probably something to do with functions chapter in Calculus.

 

[mathwonk]

the expression whose roots you want is an expression in y, which becomes, when the bottoms are multiplied out, a polynomial in y of degree 2. There is no trig function left. the trig function is only involved in determining the constants f(1), f(2), and f(3) in this polynomial. So there are at most 2 real roots y of that equation. i.e. first plug in x=1, x=2, and x=3 in the expression for f, and evaluate those constants. then you have an equation on the right, involving only y, and constants. i.e. if f(1) = a, and f(2) = b, and f(3) = c, you are asking about the number of roots of the quadratic equation:
(y-b)(y-c) + (y-a)(y-c) + (y-a)(y-b) = 0. you may need to use the fact that the value of 7sin(x) lies between -7 and 7, (inclusive), or you may need to estimate 7sin(1), 7sin(2), and 7sin(3) to get a handle on a, b, and c.

edit:
oops, I left out the coefficients 1,2,3. so I guess you get something like:
(y-b)(y-c) + 2(y-a)(y-c) + 3(y-a)(y-b) = 0. anyway, you have to do it yourself.

 

[WWGD]

Unless I'm missing something here, it seems a lot to expect students to be able to approximate the value of sinx near 0. Do they allow/expect approximations like sinx~x for x near 0? Though then we need to transform into radians.

 

[mathwonk]

I agree this is an unreasonable problem. But assuming x is already in radians, One can use a calculator to approximate sin(x) to the nearest integer, for x = 1,2,3, and then show that the discriminant of the quadratic polynomial is positive. Indeed it is over 1.5 million. hence there are two real roots. I dislike such problems, or at least such solutions. minimal thought, maximal tedious computation. Just approximating sin(x) as somewhere between -1 and 1, which does not require a calculator, does not suffice.

Of course one of you may well suggest a clever solution that proves me wrong in disliking this problem. E.g. I have not used calculus, and that is a hint here as to what should be used. But I think calculus tells you nothing about roots of quadratics that algebra does not already tell you.

 

[mathwonk]

Another odd thing about this problem is the suggestion that the polynomial could have only one real root. counted with appropriate multiplicity, a quadratic never has only one real root. i.e. many texts assume that when we say a polynomial has "two roots" we allow the possibility that they are equal, but whether that is meant here is not made clear.

 

[vcsharp2003]

the expression whose roots you want is an expression in y, which becomes, when the bottoms are multiplied out, a polynomial in y of degree 2. There is no trig function left. the trig function is only involved in determining the constants f(1), f(2), and f(3) in this polynomial. So there are at most 2 real roots y of that equation. i.e. first plug in x=1, x=2, and x=3 in the expression for f, and evaluate those constants. then you have an equation on the right, involving only y, and constants. i.e. if f(1) = a, and f(2) = b, and f(3) = c, you are asking about the number of roots of the quadratic equation:
(y-b)(y-c) + (y-a)(y-c) + (y-a)(y-b) = 0. you may need to use the fact that the value of 7sin(x) lies between -7 and 7, (inclusive), or you may need to estimate 7sin(1), 7sin(2), and 7sin(3) to get a handle on a, b, and c.

edit:
oops, I left out the coefficients 1,2,3. so I guess you get something like:
(y-b)(y-c) + 2(y-a)(y-c) + 3(y-a)(y-b) = 0. anyway, you have to do it yourself.

Thanks for your answer.

I was assuming that y needs to be substituted by f(x) resulting in an equation involving x, which was what was making it impossible to solve. The reason why I assumed this is because in Calculus we generally take it for granted that y=f(x).

 

[vcsharp2003]

Unless I'm missing something here, it seems a lot to expect students to be able to approximate the value of sinx near 0. Do they allow/expect approximations like sinx~x for x near 0? Though then we need to transform into radians.

The angle in these trigonometric ratios of sin 1, sin 2 and sin 3 must be treated as radians.

1 radian is about 57 degrees, 2 radians is about 114 degrees and 3 radians is about 171 degrees.
So, we cannot use the assumption that sin x = x since the angle x in these cases are nowhere close to 0 radians.

 

[vcsharp2003]

I agree this is an unreasonable problem. But assuming x is already in radians, One can use a calculator to approximate sin(x) to the nearest integer, for x = 1,2,3, and then show that the discriminant of the quadratic polynomial is positive. Indeed it is over 1.5 million. hence there are two real roots. I dislike such problems, or at least such solutions. minimal thought, maximal tedious computation. Just approximating sin(x) as somewhere between -1 and 1, which does not require a calculator, does not suffice.

Of course one of you may well suggest a clever solution that proves me wrong in disliking this problem. E.g. I have not used calculus, and that is a hint here as to what should be used. But I think calculus tells you nothing about roots of quadratics that algebra does not already tell you.

After understanding the question based on your post, I tried to go about it, but its far too complex since another requirement we have is that no calculator is allowed.

My analysis is as below.

Let $a=f(1)$, $b=f(2)$ and $c=f(3)$.
I can easily see that $a= 102 + 7\sin{1} \approx 102$, $b= 212 + 7\sin{2} \approx 212$ and $c= 336 + 7\sin{3} \approx 336$.

I end up with the following quadratic equation after simplification.
$$6y^2-(5a+4b+3c)y + (2ac + 3 ab+ bc) = 0$$
If I now let $m = 5a+4b+3c$ and $n= 2ac + 3 ab+ bc$, I end up with the following equation.
$$6y^2-my+n = 0$$.

Clearly we'll have two roots for the above equation whose nature will depend on the sign of the discriminant, which is $m^2 - 24n$. The problem arises at this stage since we need to decide the sign of this discriminant without using a calculator, but just manual calculations or some mathematical trick. Do you know of a simpler method to determine the sign other than substituting the approximate value of $a$, $b$ and $c$ into the discriminant?

 

[vcsharp2003]

I agree this is an unreasonable problem. But assuming x is already in radians, One can use a calculator to approximate sin(x) to the nearest integer, for x = 1,2,3, and then show that the discriminant of the quadratic polynomial is positive. Indeed it is over 1.5 million. hence there are two real roots. I dislike such problems, or at least such solutions. minimal thought, maximal tedious computation. Just approximating sin(x) as somewhere between -1 and 1, which does not require a calculator, does not suffice.

Of course one of you may well suggest a clever solution that proves me wrong in disliking this problem. E.g. I have not used calculus, and that is a hint here as to what should be used. But I think calculus tells you nothing about roots of quadratics that algebra does not already tell you.

I think the reason its under Calculus is because the first chapter in our Calculus course is Functions where the concept of $f(x)$ is explained. Here we need to use that concept since we need to know what $f(1)$,$f(2)$ and $f(3)$ mean.

 

[pasmith]

After understanding the question based on your post, I tried to go about it, but its far too complex since another requirement we have is that no calculator is allowed.

My analysis is as below.

Let $a=f(1)$, $b=f(2)$ and $c=f(3)$.
I can easily see that $a= 102 + 7\sin{1} \approx 102$, $b= 212 + 7\sin{2} \approx 212$ and $c= 336 + 7\sin{3} \approx 336$.

I end up with the following quadratic equation after simplification.
$$6y^2-(5a+4b+3c)y + (2ac + 3 ab+ bc) = 0$$
If I now let $m = 5a+4b+3c$ and $n= 2ac + 3 ab+ bc$, I end up with the following equation.
$$6y^2-my+n = 0$$.

Clearly we'll have two roots for the above equation whose nature will depend on the sign of the discriminant, which is $m^2 - 24n$. The problem arises at this stage since we need to decide the sign of this discriminant without using a calculator, but just manual calculations or some mathematical trick. Do you know of a simpler method to determine the sign other than substituting the approximate value of $a$, $b$ and $c$ into the discriminant?

It should be farily easy to calculate $$\begin{split} m^2 - 24n &= 25a^2 + 16b^2 + 9c^2 + (40 - 72)ab + (30 - 48)ac + (24 - 24)bc \\ &= 25a^2 + 16b^2 + 9c^2 - 32ab - 18ac \\ &= A^2 + B^2 + C^2 - \tfrac45 AB - \tfrac 65AC\end{split}$$ where $A = 5a$, $B = 4b$ and $C = 3c$. Now you can complete the square: $$\begin{split} m^2 - 24n &= (A - \tfrac25B - \tfrac35C)^2 + \tfrac{21}{25}B^2 + \tfrac{16}{25}C^2 - \tfrac{12}{25} BC \\ &= (A - \tfrac25B - \tfrac35C)^2 + \tfrac{1}{25}( 4C - \tfrac32 B )^2 + \tfrac34 B^2 \\ &\geq 0. \end{split}$$

Alternatively, and without doing any algebra at all, you could argue that $$g(y) = 1/(y - a) + 2/(y - b) + 3/(y - c)$$ for distinct $a$, $b$ and $c$ must cross the zero axis twice because its behaviour at each vertical asymptote is governed by $1/(y - y_0) + C$ which is negative for $y< y_0$ and positive for $y > y_0$.

 

[FactChecker]

I think the answer could be found by realizing that each term, $\frac {1}{y-f(n)}$, dominates the formula for $y$ near $f(n)$. Those terms at least force a certain number of zeros. It is not as clear to me how to show that there are not some more zeros. Maybe someone else has ideas about that.

 

[vcsharp2003]

It should be farily easy to calculate $$\begin{split} m^2 - 24n &= 25a^2 + 16b^2 + 9c^2 + (40 - 72)ab + (30 - 48)ac + (24 - 24)bc \\ &= 25a^2 + 16b^2 + 9c^2 - 32ab - 18ac \\ &= A^2 + B^2 + C^2 - \tfrac45 AB - \tfrac 65AC\end{split}$$ where $A = 5a$, $B = 4b$ and $C = 3c$. Now you can complete the square: $$\begin{split} m^2 - 24n &= (A - \tfrac25B - \tfrac35C)^2 + \tfrac{21}{25}B^2 + \tfrac{16}{25}C^2 - \tfrac{12}{25} BC \\ &= (A - \tfrac25B - \tfrac35C)^2 + \tfrac{1}{25}( 4C - \tfrac32 B )^2 + \tfrac34 B^2 \\ &\geq 0. \end{split}$$

Alternatively, and without doing any algebra at all, you could argue that $$g(y) = 1/(y - a) + 2/(y - b) + 3/(y - c)$$ for distinct $a$, $b$ and $c$ must cross the zero axis twice because its behaviour at each vertical asymptote is governed by $1/(y - y_0) + C$ which is negative for $y< y_0$ and positive for $y > y_0$.

Shouldn't $m^2-24n \gt 0$ rather than $m^2-24n \geq 0$, since the last term after completing the squares is $\frac {3}{4} B^2$ which will always be positive because $b \approx 212$?

 

[mathwonk]

The argument by pasmith in #10 is just beautiful, and exactly the kind of intelligent solution that justifies the problem! So my error was in multiplying out the bottoms and losing the tool of asymptotes. (still it really does not use calculus, or probably some people consider limits as calculus.).

Maybe a complete solution uses a bit more, to rule out further zeroes, as Factchecker points out. I.e. one could use the quadratic approach to argue there are at most 2 solutions, together with the asymptote argument that there are at least 2. Hence both solutions are "simple", i.e. not multiple.

Or one could actually use calculus to compute the slopes of the graph of the fractional equation in y, and since in fact its derivative is never zero (away from a,b,c), then arguing also about the sign of the three terms in the various intervals between a,b,c, that also shows there are exactly two distinct roots, and that the roots have multiplicity one each. but they key is using the original form of the equation in y, and not multiplying out. I.e. one could use calculus to graph the equation in y and eyeball the solutions.

Thank you pasmith, now I really like the solution, and thus grudgingly, also the problem.

Still I don't exactly love it, since the original equation in x was just a smokescreen designed to confuse. I.e. the only data needed from it is that a, b, c, are distinct. So as often happens, the problem is not designed to test knowledge of calculus, but cleverness, i.e. the ability ignore distractions and focus on essential information.

 

[pasmith]

A better argument: Set $g(y) = N(y)/D(y)$ where $$\begin{split} N(y) &= (y - b)(y - c) + 2(y - a)(y - c) + 3(y - a)(y - b) \\ D(y) &= (y- a)(y - b)(y - c)\end{split}$$ where $a = f(1) < b = f(2) < c = f(3)$. We have $$\begin{split} N(a) &= (a - b)(a - c) > 0 \\ N(b) &= 2(b- a)(b - c) < 0 \\ N(c) &= 3(c- a)(c - b) > 0\end{split}$$ so $N$ - which is a quadratic with positive $y^2$ coefficient - has a zero in $(a,b)$ and a zero in $(b, c)$. Since $D$ does not vanish at these points it follows that $g = N/D$ has zeroes at these two points and not elsewhere.

(This does rely on the specialization of the IVT to a quadratic with positive leading coefficient which falls below the horizontal axis somewhere, but I think this can be assumed, even in a section introducing the concept of function notation - if you know how to determine how many distinct real roots a quadratic has then you will know what its graph looks like in each of the possible cases.)

 

[mathwonk]

this is indeed a nice succinct way to combine the asymptote and quadratic arguments. again it completely avoids calculus, but uses limits in the form of IVT. It also implies of course the two roots are both simple.