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The original problem:

If $0\le a,b\le 3$ and the equation $$x^2+4+3\cos(ax+b)=2x$$ has at least one real solution, then find the value of $a+b$

$$$$

At first, on rearranging, I got the following expression: $$x^2-2x+(4+3\cos(ax+b))=0$$ I thought this was a quadratic in $x$, and thus from the quadratic formula(and that at least one real root exists), $D\ge 0$ ie $$4-4(4+\cos(ax+b))\ge 0$$ $$$$

However I'm not really sure about this. I've treated $\cos(ax+b)$ as a constant term even though the argument of the cosine includes $x$: the variable in which the quadratic expression is. $$$$Under these circumstances, is it correct to use $3\cos(ax+b)$ as a constant? If not, how could I use the quadratic formula to find values of $x$ satisfying $$x^2-2x+(4+3\cos(ax+b))=0$$

Many thanks in anticipation!

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  • $\begingroup$ Thanks, but how exactly do I proceed from here? $\endgroup$ – Better World Mar 15 '16 at 10:54
  • $\begingroup$ @Banach why did you delete comment $\endgroup$ – user5954246 Mar 15 '16 at 11:13
  • $\begingroup$ You missed to copy a $3$ and follow through to the conclusion $-12(1+\cos(ax+b))\ge0$. After that the result follows as in the answers. $\endgroup$ – LutzL Mar 15 '16 at 12:39
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Wih a little bit of manipulation we can rewrite your given equation as $$x^2 - 2x + 4 = -3\cos{\left( ax + b \right)}$$

Let $f(x) = x^2 - 2x + 4$. Differentiating to find the minimum, we get

$$f'(x) = 2x - 2 = 0 \implies x = 1$$ $$f''(x) = 2 > 0 \implies \text{minimum at } x = 1$$

The minimum value of LHS is thus $f(1) = 3$.

The RHS is a cosine whose value oscillates in the range $[-3,3]$. The maximum value of the RHS is thus $3$. So we can see that equality holds if and only if the LHS is minimum and RHS is maximum. We just saw that the LHS, $f(x)$, is minimal only at $x = 1$. Now the value of $x$ is fixed, so the value of the RHS depends only on $a$ and $b$.

Hence, we have that

$$f(1) = 3 = -3\cos(a\times 1 + b) \implies \cos(a+b) = -1$$

I leave it to you to complete the problem from here :)

On a side note: No you can't take $\cos(ax + b)$ as a constant since $x$ is not a constant :)

$a + b = (2n+1)\pi$ but $0 \le a+b \le 6$. So $n = 0$ and $a + b = \pi$

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Completing the square,

$$x^2-2x+1=(x-1)^2=-3(1+\cos(ax+b)).$$

The two expressions can only be equal if they are zero, i.e. $x=1$ and $\cos(ax+b)=-1$.

Hence,

$$a+b=\pi.$$

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  • $\begingroup$ What is completing the square? Is my solution incorrect? $\endgroup$ – Banach Tarski Mar 15 '16 at 12:28
  • $\begingroup$ @BanachTarski: you didn't provide a complete solution. $\endgroup$ – Yves Daoust Mar 15 '16 at 12:33
  • $\begingroup$ Why must they both be equal to $0$? The RHS can be anything from $-6$ to $6$, and the LHS covers values between $0$ and $6$, so all solutions may very well result in not equaling $0$, right? $\endgroup$ – Simply Beautiful Art Mar 20 '16 at 15:08
  • $\begingroup$ @SimpleArt wrong about the RHS. $\endgroup$ – Yves Daoust Mar 20 '16 at 16:13
  • $\begingroup$ I don't thank the downvoter. $\endgroup$ – Yves Daoust Mar 20 '16 at 16:15
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since the $3cos(ax+b)$ lies between $[-3,3]$

because $cos x$ has a max and min value as $1,-1$ (respectively) then $3cos x$ will have $3,-3$

i can write $$-3=<x^2-2x+4<=3$$

i am completing the square !

$$\implies -3=<(x-1)^2+3<=3$$

as you see that the term inside the square has a least value 0 and this equation is always greater than or equal to 3 then $(x-1)$ must be zero !! and $x=1$

and then substituting the value of x we get $cos(a+b)=-1$ $$\implies a+b=\arccos (-1)$$ $$a+b=\pi$$and that solved the problem!!

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  • $\begingroup$ What is completing the square? I don't see what I have missed $\endgroup$ – Banach Tarski Mar 15 '16 at 12:28
  • $\begingroup$ Completing the square means transforming $ax^2+bx+c=a(x+\frac{b}2a)^2+c-\frac{b^2}{4a}$ so that all $x$ terms are assembled in one complete square. $\endgroup$ – LutzL Mar 15 '16 at 12:34
  • $\begingroup$ Why is that necessary. I just used the quadratic as a function and found out that solution can only exist when the quadratuc function is minimum $\endgroup$ – Banach Tarski Mar 15 '16 at 12:38
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    $\begingroup$ Seems odd to me that you use \arccos but not \cos. $\endgroup$ – Kyle Kanos Mar 15 '16 at 12:54
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    $\begingroup$ @Banach Tarski, this problem can be solved both with and without calcilus. Completing the square is the main way you find the minimum of a quadratic without calculus. $\endgroup$ – Mark S. Mar 15 '16 at 17:56

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