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Q.
If range of $f(x)=\cos ^{-1}\left(x^2 \cdot \cos 1+\sin 1 \cdot \sqrt{1-x^4}\right)$ is $[a, b]$, then $(b-a)$ is equal to
Inverse Trigonometric Functions
Solution:
$f ( x )=\cos ^{-1}\left( x ^2 \cdot \cos 1+\sin 1 \cdot \sqrt{1- x ^4}\right)=\left|\cos ^{-1}\left( x ^2\right)-\cos ^{-1}(\cos 1)\right|=\left|\cos ^{-1}\left( x ^2\right)-1\right| \text { (even function) } $
$\text { Domian of } f ( x ) \text { is }[-1,1] $
$\text { Range of } f ( x ) \text { is }[0,1] $
$\therefore b - a =1$
Aliter: $f(x)=\begin{cases}1-\theta, & \theta \in[0,1] \\ \theta-1, & \theta \in\left[1, \frac{\pi}{2}\right]\end{cases}$
Range of $f(x)$ is $[0,1]$.