If $f(x) = cos^{-1} x + cos^{-1} \begin{Bmatrix}\frac{x}{2}+\frac{1}{2}\sqrt{3-3x^2}\end{Bmatrix}$, then f $(\frac{2}{3})$ equals
Answer & explanation
Correct answer: option 1
We have,
$cos^{-1}\begin{Bmatrix}xy + \sqrt{1-x^2}\sqrt{1-y^2}\end{Bmatrix}$
\(=\left\{\begin{matrix}cos^{-1}x - cos^{-1} y, if -1 ≤ x, y ≤1 \, and \, x ≤y \\ cos^{-1} y - cos^{-1} x, if -1 ≤y ≤ 0, 0 ≤ x ≤ 1 \, and \, y ≤ x \end{matrix}\right.\)
$∴ cos^{-1}\begin{Bmatrix}\frac{x}{2}+\frac{1}{2}\sqrt{3-3x^2}\end{Bmatrix}$
$= cos^{-1}\begin{Bmatrix}x× \frac{1}{2} + \sqrt{1-x^2}\sqrt{1-\frac{1}{4}}\end{Bmatrix}$
\(= \left\{\begin{matrix}cos^{-1} x - cos^{-1} \frac{1}{2}, if -1 ≤ x ≤\frac{1}{2}\\cos^{-1} \frac{1}{2} - cos^{-1}x, if \, \frac{1}{2} ≤ x ≤ 1 \end{matrix}\right.\)
\(∴ f(x) = \left\{\begin{matrix}2 cos^{-1}x - cos^{-1}\frac{1}{2}, if -1 ≤ x ≤ \frac{1}{2}\\cos^{-1}\frac{1}{2} \,\,\,\,\,\, , if \frac{1}{2} ≤ x ≤ 1 \end{matrix}\right.\)
$⇒ f(\frac{2}{3}) = cos^{-1}\frac{1}{2}=\frac{\pi}{3}$