Find the number of solutions of the equation $\sin^{-1}[x^2+\frac13]+\cos^{-1}[x^2-\frac23]=x^2$, for $x\in[-1,1]$

Question

The following question is taken from the practice set of JEE exam.

Find the number of solutions of the equation $\sin^{-1}[x^2+\frac13]+\cos^{-1}[x^2-\frac23]=x^2$, for $x\in[-1,1]$ and $[x]$ denotes the greatest integer less than or equal to $x$.

My attempt: $$-1\le[x^2+\frac13]\le1\\\implies -1\le x^2+\frac13\lt2\\\implies-\frac43\le x^2\lt\frac53\\\implies0\le x^2\lt\frac53$$

Also, $$-1\le[x^2-\frac23]\le1\\\implies -1\le x^2-\frac23\lt2\\\implies-\frac13\le x^2\lt\frac83\\\implies0\le x^2\lt\frac83$$

Taking intersection, I get the domain is $$0\le x^2\lt\frac53$$

Considering case-I, $$0\le x^2\lt\frac23\\\implies\frac13\le x^2+\frac13\lt1\\\implies[x^2+\frac13]=0\\\implies\sin^{-1}[x^2+\frac13]=0$$

Again, $$0\le x^2\lt\frac23\\\implies-\frac23\le x^2-\frac23\lt0\\\implies[x^2-\frac23]=-1\\\implies\cos^{-1}[x^2-\frac23]=\pi$$

Adding them, I get the output in case-I is $\pi$.

Considering case-II $$\frac23\le x^2\lt\frac53\\\implies1\le x^2+\frac13\lt2\\\implies[x^2+\frac13]=1\\\implies\sin^{-1}[x^2+\frac13]=\frac\pi2$$

Again, $$\frac23\le x^2\lt\frac53\\\implies0\le x^2-\frac23\lt1\\\implies[x^2-\frac23]=0\\\implies\cos^{-1}[x^2-\frac23]=\frac\pi2$$

Adding them, I get the output in case-II is $\pi$.

It implies that the range of LHS of the given equation is only $\pi$. And the right side is $x^2$ i.e. a parabola. So, they would intersect at two points. So, the answer to the posed question should be $2$. Is this correct?

Also, is there a way we could draw the graph for the two functions in LHS too to understand how exactly their sum is $\pi$? Thanks.

Answer 1

Your approach is correct. But it can be simplified. In general, consider the function; $$f(u)=\sin^{-1}[u]+\cos^{-1}([u]-1)$$ It is obvious that $[u]= 0,1$ for function to be defined. Now, if $[u]=1$, $f(u)=\pi$ and if $[u]=0$, $f(u)=\pi$. Hence, it is clear that the function takes a constant value in its domain. Now, if you let $$u=x^2+\frac 13$$ and solve for $0\leq u<2$, you will get your required domain. This is because $$[x^2-\frac 23]=[(x^2+ \frac 13)-1]=[x^2+\frac 13]-1$$

All that remains is to draw graph of $x^2$ and find out number of point of intersection.

Answer 2

I got the answer. When I got LHS=$\pi$, I need to see that it is true only for $0\le x^2\lt\frac53$. However, RHS, ie. $x^2=\pi$ is valid only for $x=\pm\sqrt\pi$. And $-\sqrt\frac53\gt-\sqrt\pi$ and $\sqrt\frac53\lt\sqrt\pi$. Thus, the graphs of LHS and RHS never intersect. Thus, the answer to the question zero number of solutions.