CUET Preparation Today
Let f : [1/2, 1] → [-1, 1] is given by $f^{-1}(x) = 4x^3− 3x$ |
$\cos \left(\frac{1}{3} \cos ^{-1} x\right)$ $3 \cos \left(\cos ^{-1} x\right)$ $3 \sin ^{-1}(\sin x)$ $\sin \left(\frac{1}{3} \sin ^{-1} x\right)$ |
$\cos \left(\frac{1}{3} \cos ^{-1} x\right)$ |
$y=4x^3-3x$ let $x=\cos θ$ $y=4\cos^3 θ-3\cos θ$ $⇒θ\cos^{-1}(x)$ $y=\cos 3θ⇒\frac{1}{3}\cos^{-1}y=θ$ so $\cos\left(\frac{1}{3}\cos^{-1}y\right)=x$ $f^{-1}x=\cos \left(\frac{1}{3} \cos ^{-1} x\right)$ |
