Let [cdot] denote the greatest integer functi | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The domain of $\sin^{-1}(u)$ is $u \in [-1, 1]$.Thus,$-1 \le \frac{x+[x]}{3} \le 1$,which implies $-3 \le x+[x] \le 3$.Case $1$: If $x \in [-2, -1

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(B) The domain of $\sin^{-1}(u)$ is $u \in [-1, 1]$.Thus,$-1 \le \frac{x+[x]}{3} \le 1$,which implies $-3 \le x+[x] \le 3$.Case $1$: If $x \in [-2, -1)$,then $[x] = -2$. So,$-3 \le x - 2 \le 3 \implies -1 \le x \le 5$. Since $x \in [-2, -1)$,the intersection is $[-1, -1)$ (empty).Case $2$: If $x \in [-1, 0)$,then $[x] = -1$. So,$-3 \le x - 1 \le 3 \implies -2 \le x \le 4$. The intersection with $[-1, 0)$ is $[-1, 0)$.Case $3$: If $x \in [0, 1)$,then $[x] = 0$. So,$-3 \le x \le 3$. The intersection with $[0, 1)$ is $[0, 1)$.Case $4$: If $x \in [1, 2)$,then $[x] = 1$. So,$-3 \le x + 1 \le 3 \implies -4 \le x \le 2$. The intersection with $[1, 2)$ is $[1, 2)$.Case $5$: If $x \in [2, 3)$,then $[x] = 2$. So,$-3 \le x + 2 \le 3 \implies -5 \le x \le 1$. The intersection with $[2, 3)$ is empty.Combining these,the domain is $[-1, 2)$.Thus,$\alpha = -1$ and $\beta = 2$.Therefore,$\alpha^2 + \beta^2 = (-1)^2 + (2)^2 = 1 + 4 = 5$.

Let $[\cdot]$ 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:

Similar Questions

The domain of the function defined by $f(x) = \cos^{-1} \sqrt{x-1}$ is

Domain of the function $f(x) = \frac{1}{\sqrt{\ln(\cot^{-1}x)}}$ is

The range of $\left|\cos ^{-1} x\right|$ is . . . . . . .

The domain of the function $f(x) = \sin^{-1}\left(\frac{3x^2+x-1}{(x-1)^2}\right) + \cos^{-1}\left(\frac{x-1}{x+1}\right)$ is:

Domain of function $f(x) = \sin^{-1}(5x)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module