Question

Find the domain of the following function

$f(x) = {sin}^{ - 1} \sqrt{x - 1}$
​

Answer 1

$\large\underline{\sf{Solution-}}$

Given function is

$\rm \: f(x) = {sin}^{ - 1} \sqrt{x - 1}$

We know,

$\boxed{\sf{ \:Domain \: of \: {sin}^{ - 1} x \: is \: x \: \in \: [ - 1,1] \: }}$

So, for the given function

$\rm \: \sqrt{x - 1} \: \in \: [ - 1, \: 1]$

$\rm \: - 1 \leqslant \sqrt{x - 1} \leqslant 1$

So, on squaring both sides, we get

$\rm \: 0 \leqslant x - 1 \leqslant 1$

On adding 1 in each term, we get

$\rm \: 0 + 1 \leqslant x - 1 + 1 \leqslant 1 + 1$

$\rm \: 1 \leqslant x \leqslant 2$

$\rm\implies \:Domain \: of \: {sin}^{ - 1} \sqrt{x - 1} \: is \: \: x \: \in \: [1, \: 2] \:$

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Additional Information

$\begin{gathered}\boxed{\begin{array}{c|c} \bf Function & \bf Range \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf y =  {sin}^{ - 1}(sinx) & \sf  x \:  \: if -\dfrac{\pi  }{2} \leqslant x \leqslant \dfrac{\pi  }{2}\\ \\ \sf y =  {cos}^{ - 1}(cosx) & \sf x \:  \: if \: 0 \leqslant y \leqslant \pi \\ \\ \sf y =  {tan}^{ - 1}(tanx) & \sf x \:  \: if \:  - \dfrac{\pi  }{2} < x < \dfrac{\pi  }{2}\\ \\ \sf y =  {cosec}^{ - 1}(cosecx) & \sf x \:  \: if \: x \:  \in \: \bigg[ - \dfrac{\pi}{2}, \: \dfrac{\pi  }{2}\bigg] -  \{0 \}\\ \\ \sf y =  {sec}^{ - 1}(secx) & \sf x \:  \: if \: x \:  \in \: [0, \: \pi] \:   -  \: \bigg\{\dfrac{\pi  }{2}\bigg\}\\ \\ \sf y =  {cot}^{ - 1}(cotx) & \sf x \:  \: if \:  \:  \in \: \bigg( -  \dfrac{\pi  }{2} , \dfrac{\pi  }{2}\bigg) -  \{0 \} \end{array}} \\ \end{gathered}$

$\begin{gathered}\boxed{\begin{array}{c|c} \bf Function & \bf Range \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf y = sinx & \sf   - 1 \leqslant y \leqslant 1\\ \\ \sf y = cosx & \sf  - 1 \leqslant y \leqslant 1 \\ \\ \sf y = tanx & \sf y \:  \in \: ( -  \infty , \infty )\\ \\ \sf y = cosec & \sf y \leqslant  - 1 \:  \: or \:  \: y \geqslant 1\\ \\ \sf y = secx & \sf y \leqslant  - 1 \:  \: or \:  \: y \geqslant 1\\ \\ \sf y = cotx & \sf y \:  \in \: ( -  \infty , \infty ) \end{array}} \\ \end{gathered}$

Answer 2

Answer:

the square root to be defined:

x−1≥0⟹x≥1 (1)

For inverse sine to be defined,

−1≤

x−1

≤1

As square is always non-negative:

0≤

x−1

≤1

⟹0≤x−1≤1

⟹1≤x≤2 (2)

Intersection of the two intervals (1) and (2) gives:

x∈[1,2]

Step-by-step explanation:

$hope \: it \: helps \: brainlist \: answer$