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

Answer

(a): We know, $\frac{-\pi}{2} \leq \sin ^{-1} \sqrt{x-1} \leq \frac{\pi}{2}$
$\Rightarrow-1 \leq \sqrt{x-1} \leq 1 \Rightarrow 0 \leq x-1 \leq 1 \Rightarrow 1 \leq x \leq 2$
$\therefore \quad$ Domain of $f(x)$ is $[1,2]$.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Similar questions

If a line makes angle $\alpha,\beta$ and $\gamma$ with the axes respectively, then $\cos2\alpha+\cos2\beta+\cos2\gamma=$
  1. -2
  2. -1
  3. 1
  4. 2
The feasible region is :
A relation $R$ is defined on $N$. Which of the following is the reflexive relation?
The solution of the differential equation $(\text{x}^{2}+1)\frac{\text{dy}}{\text{dx}}+(\text{y}^{2}+1)=0$ is:  
  1. $\text{y}=2+\text{x}^{2}$
  2. $\text{y}=\frac{1+\text{x}}{1-\text{x}}$ 
  3. $\text{y}=\text{x}(\text{x}-1)$
  4. $\text{y}=\frac{1+\text{y}}{1-\text{y}}$ 
f : R → R given by $\text{f(x)}=\text{x}+\sqrt{\text{x}^2}$ is:
  1. Injective.
  2. Surjective.
  3. Bijective.
  4. None of these.
The function $f : A \rightarrow B$ defined by $f(x) = -x^2 + 6x- 8$ is a bijection if,
If a * b denote the bigger among a and b and if ab = (a * b) + 3, then 4.7 =
  1. 14
  2. 31
  3. 10
  4. 8
If $a, b, c$ are in $A.P.,$ then the determinant $\begin{vmatrix}\text{x}+2&\text{x}+3&\text{x}+2\text{a}\\\text{x}+3&\text{x}+4&\text{x}+2\text{b}\\\text{x}+4&\text{x}+5&\text{x}+2\text{c}\end{vmatrix}$
Which of the following transformation reduce the differential quation  into the form $\frac{\text{du}}{\text{dx}}+\text{P}(\text{x})\text{u}=\text{Q}(\text{x})$ into the from $\frac{\text{dz}}{\text{dx}}+\frac{\text{z}}{\text{x}}\log\text{z}=\frac{\text{z}}{\text{x}^{2}}(\log\text{z})^{2}$
  1. $\text{u}=\log\text{x}$
  2. $\text{u}=\text{e}^{\text{z}}$
  3. $\text{u}=(\log\text{z})^{-1}$
  4. $\text{u}=(\log\text{z})^{2}$ 
Evaluate: $\int\sqrt{1+\text{y}}^2.\text{2ydy:}$
  1. $\text{I}=\frac{2}{3}(1+\text{y}^2)^\frac{3}{2}+\text{c}$
  2. $\text{I}=\frac{2}{5}(1-\text{y}^2)\frac{3}{2}+\text{c}$
  3. $\text{I}=\frac{2}{3}(1-\text{y}^2)^\frac{3}{2}+\text{c}$
  4. None of these