RQ 1

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RQ - Random Question. This is the start of multiple upcoming entries consisting of random problems I created myself for fun. Some of them are just straightforward annoying, but I do think that some of the questions have a really cool solution and I want to put those out here. Have fun!

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Question

Let $f:\left[\frac{1}{2},1\right]\to [0,2]$ be a function such that $f(x)=4x^3-3x+1$ Let $g$ be the inverse function of $f$ . Then the value of $g\left(\frac{1}{\sqrt{2}}+1\right)$ is (use $\sqrt{3}=1.7,\sqrt{2}=1.4$ )

Hint

Trigonometry ftw

Answer

Let $x=\cos \theta$ . The question is practically done at this step. Regardless, lets complete it. Substituting this in $f$, we get

$f(x)=4{\cos }^3\theta -3\cos \theta +1$

$⟹f(x)=\cos 3\theta +1$

Therefore, in terms of the parameter $\theta$ , $f$ can be defined as $x=\cos \theta ,y=\cos 3\theta +1$ . Therefore, the parametric form of the inverse function would be

$(x=3\cos \theta +1,y=\cos \theta )$

$⟹g(x)=\cos \left(\frac{1}{3}\arccos (x-1)\right)$

Here it is worthy to note the domain and range of the functions involved so that to confirm the values of $\theta$ possible. Since $x\in [0.5,1]$ , $\theta \in [0,\pi /3]$.

For $\theta \in [0,\pi /3]$ , range of $f$ is $[0,2]$ and thus the codomain matches with the range now, which makes it invertible. Also notice that $\arccos (x-1)$ should take principal values i.e., $[0,\pi ]$.

Now, $g\left(\frac{1}{\sqrt{2}}+1\right)=\cos \left(\frac{1}{3}\arccos \left(\frac{1}{\sqrt{2}}\right)\right)$ . Also, from $\cos \theta =\sqrt{\frac{1+\cos 2\theta }{2}}$ , we have

$g\left(\frac{1}{\sqrt{2}}+1\right)=\cos \left(\frac{\pi }{12}\right)=\sqrt{\frac{2+\sqrt{3}}{4}}=\frac{\sqrt{3}+1}{2\sqrt{2}}$

$=\frac{2.7}{2.8}=0.964$

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Summary

Shortest entry so far. Nothing much to say. I really liked this trick because it somehow infuses the trigonometric domain with the normal functions and calculus domain.