Let $f (x) = a + b {cos}^{- 1} x (b > 0)$ . If domain and range of f(x) are the same set then $(b - a)$ is equal to

1 - \frac{1}{π}

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\frac{2}{π} + 1

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1 - \frac{2}{π}

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Solution

The correct option is B $\frac{2}{π} + 1$
$f (x) = a + b {c o s}^{- 1} x$
Domain: $x \in [- 1, 1]$
Range: ${c o s}^{- 1} x \in [0, π] \Rightarrow a + b {c o s}^{- 1} x \in [a, a + b π]$
Therefore, $- 1 = a$
And $1 = a + b π \Rightarrow b = \frac{2}{π}$
Gives $b - a = \frac{2}{π} + 1$

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