Let [ ] denote the greatest integer function.

Question

Let [   ] denote the greatest integer function. If the domain of the function $f(x)=\sin ^{-1}\left(\frac{x+[x]}{3}\right)$ is $[\alpha, \beta)$, then $\alpha^2+\beta^2$ is equal to:

Accepted Answer

For the function

$f(x)={\sin }^{-1}(\frac{x+[x]}{3}),$

the expression inside ${\sin }^{-1}$ must lie between $-1$ and $1$.

So, we need

$-1\le \frac{x+[x]}{3}\le 1$

Multiplying throughout by $3$,

$-3\le x+[x]\le 3$

Now let

$[x]=n$

Then by definition of greatest integer function,

$n\le xand so

$x+[x]=x+n$

Hence the condition becomes

$-3\le x+n\le 3$

But since $x\in [n,n+1)$, let us check possible integer values of $n$.

Case 1: $n=-2$
Then

$x\in [-2,-1)$

and

$x+[x]=x-2$

Since $x\in [-2,-1)$,

$x-2\in [-4,-3)$

This does not satisfy $-3\le x+[x]$ except at $x=-1$, but $-1
otin [-2,-1)$.

So no value from this interval is allowed.

Case 2: $n=-1$
Then

$x\in [-1,0)$

and

$x+[x]=x-1$

So

$x-1\in [-2,-1)$

This lies completely within $[-3,3]$.

Hence all $x\in [-1,0)$ are allowed.

Case 3: $n=0$
Then

$x\in [0,1)$

and

$x+[x]=x$

So

$x\in [0,1)$

This also lies within $[-3,3]$.

Hence all $x\in [0,1)$ are allowed.

Case 4: $n=1$
Then

$x\in [1,2)$

and

$x+[x]=x+1$

So

$x+1\in [2,3)$

This lies within $[-3,3]$.

Hence all $x\in [1,2)$ are allowed.

Case 5: $n=2$
Then

$x\in [2,3)$

and

$x+[x]=x+2$

So

$x+2\in [4,5)$

This exceeds $3$, so no value is allowed.

Now let us also check the boundary points:

At $x=-\frac{3}{2}$, actually $[x]=-2$, so this does not help.

Better to solve more carefully near the ends.

We want the full set of $x$ satisfying

$-3\le x+[x]\le 3$

Let $x\in [n,n+1)$, so $[x]=n$.

Then

$-3\le x+n\le 3$

which gives

$-3-n\le x\le 3-n$

Together with $n\le x$[n,n+1)\cap [-3-n,\, 3-n].$

Now check relevant integers:

For $n=-2$
$x\in [-2,-1)\cap [-1,5]={-1}$

but $-1$ is not included in $[-2,-1)$, so no solution.

For $n=-1$
$x\in [-1,0)\cap [-2,4]=[-1,0)$

For $n=0$
$x\in [0,1)\cap [-3,3]=[0,1)$

For $n=1$
$x\in [1,2)\cap [-4,2]=[1,2)$

For $n=2$
$x\in [2,3)\cap [-5,1]=\emptyset$

Thus domain is

$[-1,0)\cup [0,1)\cup [1,2)=[-1,2)$

So,

$\alpha =-1,\, \beta =2$

Therefore,

${\alpha }^{2}+{\beta }^{2}=(-1{)}^{2}+{2}^{2}=1+4=5$

Hence the correct answer is

$5$

So, Option B is correct.

Solutions · 1