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Let \[\mathbf{f}:\text{ }\mathbf{R}\text{ }\to \text{ }\mathbf{R}\] be the function defined by \[\mathbf{f}\left( \mathbf{x} \right)\text{ }=\text{ }\mathbf{1}/\left( \mathbf{2}\text{ }\text{ }\mathbf{cos}\text{ }\mathbf{x} \right)\forall \mathbf{x}\in \mathbf{R}\]. Then, find the range of f.
Let \[\mathbf{f}:\text{ }\mathbf{R}\text{ }\to \text{ }\mathbf{R}\] be the function defined by \[\mathbf{f}\left( \mathbf{x} \right)\text{ }=\text{ }\mathbf{1}/\left( \mathbf{2}\text{ }\text{ }\mathbf{cos}\text{ }\mathbf{x} \right)\forall \mathbf{x}\in \mathbf{R}\]. Then, find the range of f.
CBSE | Class 12 | Exercise 1 | Maths | NCERT Exemplar | Relations and Functions
Let \[\mathbf{f}:\text{ }\mathbf{R}\text{ }\to \text{ }\mathbf{R}\] be the function defined by \[\mathbf{f}\left( \mathbf{x} \right)\text{ }=\text{ }\mathbf{1}/\left( \mathbf{2}\text{ }\text{ }\mathbf{cos}\text{ }\mathbf{x} \right)\forall \mathbf{x}\in \mathbf{R}\]. Then, find the range of f.

According to the question,

\[\begin{array}{*{35}{l}}

f\left( x \right)\text{ }=\text{ }1/\left( 2\text{ }\text{ }cos\text{ }x \right)\forall x\in R  \\

Let\text{ }y\text{ }=\text{ }1/\left( 2\text{ }\text{ }cos\text{ }x \right)  \\

2y\text{ }\text{ }ycos\text{ }x\text{ }=\text{ }1  \\

cos\text{ }x\text{ }=\text{ }\left( 2y\text{ }\text{ }1 \right)/\text{ }y  \\

cos\text{ }x\text{ }=\text{ }2\text{ }\text{ }1/y  \\

\end{array}\]

Now, we know that \[-1\text{ }\le \text{ }cos\text{ }x\text{ }\le \text{ }1\]

So,

\[\begin{array}{*{35}{l}}

-1\text{ }\le \text{ }2\text{ }\text{ }1/y\text{ }\le \text{ }1  \\

-3\text{ }\le \text{ }\text{ }1/y\text{ }\le \text{ }-1  \\

1\text{ }\le \text{ }\text{ }1/y\text{ }\le \text{ }3  \\

1/3\text{ }\le \text{ }y\text{ }\le \text{ }1  \\

\end{array}\]

Thus, the range of the given function is \[\left[ 1/3,\text{ }1 \right]\].

i

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