Let [ ] denote the greatest integer function.

Question

Let [ ] denote the greatest integer function. If the domain of the function 
$f(x)=\cos ^{-1}\left(\frac{4 x+2[x]}{3}\right)$ is $[\alpha, \beta]$, then $12(\alpha+\beta)$ is equal to :

Accepted Answer

For the function

$f(x)={\cos }^{-1}(\frac{4x+2[x]}{3}),$

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

So, for domain:

$-1\le \frac{4x+2[x]}{3}\le 1$

Multiplying by $3$:

$-3\le 4x+2[x]\le 3$

Now let

$[x]=n$

where $n$ is an integer and

$n\le xThen the inequality becomes

$-3\le 4x+2n\le 3.$

Subtracting $2n$:

$-3-2n\le 4x\le 3-2n$

Dividing by $4$:

$\frac{-3-2n}{4}\le x\le \frac{3-2n}{4}$

But also,

$n\le xSo we find possible integer values of $n$ for which these intervals overlap.

Case 1: $n=-1$
Then

$[x]=-1,\, -1\le xand

$\frac{-3-2(-1)}{4}\le x\le \frac{3-2(-1)}{4}$

$\frac{-1}{4}\le x\le \frac{5}{4}$

Intersection with $[-1,0)$ gives

$-\frac{1}{4}\le xCase 2: $n=0$
Then

$0\le xand

$\frac{-3}{4}\le x\le \frac{3}{4}$

Intersection gives

$0\le x\le \frac{3}{4}.$

Case 3: $n=-2$
Then

$-2\le xand

$\frac{-3-2(-2)}{4}\le x\le \frac{3-2(-2)}{4}$

$\frac{1}{4}\le x\le \frac{7}{4}$

No intersection.

Case 4: $n=1$
Then

$1\le xand

$\frac{-5}{4}\le x\le \frac{1}{4}$

No intersection.

So the domain is

$[-\frac{1}{4},0)\cup [0,\frac{3}{4}]=[-\frac{1}{4},\frac{3}{4}].$

Thus,

$\alpha =-\frac{1}{4},\, \beta =\frac{3}{4}.$

Therefore,

$\alpha +\beta =\frac{1}{2}$

and

$12(\alpha +\beta )=12\cdot \frac{1}{2}=6.$

Hence the correct answer is

$

Solutions · 1