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The domain of the function by f(x) =

    \[\mathbf{si}{{\mathbf{n}}^{-\mathbf{1}}}~\sqrt{\left( \mathbf{x}\text{ }\text{ }\mathbf{1} \right)}\]

is (a)

    \[\left[ \mathbf{1},\text{ }\mathbf{2} \right]\]

(b)

    \[\left[ -\mathbf{1},\text{ }\mathbf{1} \right]\]

(c)

    \[\left[ \mathbf{0},\text{ }\mathbf{1} \right]\]

(d) none of these
CBSE | Class 12 | Exercise 1 | Inverse Trigonometric Functions | Maths | NCERT Exemplar
The domain of the function by f(x) =

    \[\mathbf{si}{{\mathbf{n}}^{-\mathbf{1}}}~\sqrt{\left( \mathbf{x}\text{ }\text{ }\mathbf{1} \right)}\]

is (a)

    \[\left[ \mathbf{1},\text{ }\mathbf{2} \right]\]

(b)

    \[\left[ -\mathbf{1},\text{ }\mathbf{1} \right]\]

(c)

    \[\left[ \mathbf{0},\text{ }\mathbf{1} \right]\]

(d) none of these

The correct option is  (a)

    \[\left[ \mathbf{1},\text{ }\mathbf{2} \right]\]

We know that,

    \[si{{n}^{-1}}~x\]

is defined for

    \[x\in \left[ -1,\text{ }1 \right]\]

So, f(x) =

    \[\mathbf{si}{{\mathbf{n}}^{-\mathbf{1}}}~\sqrt{\left( \mathbf{x}\text{ }\text{ }\mathbf{1} \right)}\]

is defined if

    \[<span class="ql-right-eqno"> (1) </span><span class="ql-left-eqno"> </span><img src="https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-8a0869979131ceddb0dd11dc8f006c30_l3.png" height="124" width="281" class="ql-img-displayed-equation quicklatex-auto-format" alt="\begin{align*} & \begin{array}{*{35}{l}} 0\text{ }\le \text{ }\sqrt{\left( x\text{ }\text{ }1 \right)}\text{ }\le \text{ }1 \\ 0\text{ }\le \text{ }x\text{ }\text{ }1\text{ }\le \text{ }1 \\ \end{array} \\ & 1\text{ }\le \text{ }x\text{ }\le \text{ }2 \\ \end{align*}" title="Rendered by QuickLaTeX.com"/>\]

Hence,

    \[x\in \left[ 1,\text{ }2 \right]\]

i

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