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Question: If $f(x) = cos^{-1}x + cos^{-1}\{\frac{x}{2} + \frac{1}{2}\sqrt{3-3x^2}\}$ then:...

If f(x)=cos−1x+cos−1{x2+123−3x2}f(x) = cos^{-1}x + cos^{-1}\{\frac{x}{2} + \frac{1}{2}\sqrt{3-3x^2}\} then:

A

f(23)=π3f(\frac{2}{3}) = \frac{\pi}{3}

B

f(23)=2cos−123−π3f(\frac{2}{3}) = 2cos^{-1}\frac{2}{3} - \frac{\pi}{3}

C

f(13)=π3f(\frac{1}{3}) = \frac{\pi}{3}

D

f(13)=2cos−113−π3f(\frac{1}{3}) = 2cos^{-1}\frac{1}{3} - \frac{\pi}{3}

Answer

A, D

Explanation

Solution

The given function is f(x)=cos⁡−1x+cos⁡−1{x2+123−3x2}f(x) = \cos^{-1}x + \cos^{-1}\left\{\frac{x}{2} + \frac{1}{2}\sqrt{3-3x^2}\right\}. Let x=cos⁡θx = \cos\theta for θ∈[0,π]\theta \in [0, \pi]. Then 3−3x2=3(1−x2)=3sin⁡2θ=3∣sin⁡θ∣\sqrt{3-3x^2} = \sqrt{3(1-x^2)} = \sqrt{3\sin^2\theta} = \sqrt{3}|\sin\theta|. Since θ∈[0,π]\theta \in [0, \pi], sin⁡θ≥0\sin\theta \ge 0, so 3−3x2=3sin⁡θ\sqrt{3-3x^2} = \sqrt{3}\sin\theta.

Substitute x=cos⁡θx = \cos\theta into the expression for f(x)f(x): f(x)=cos⁡−1(cos⁡θ)+cos⁡−1{cos⁡θ2+3sin⁡θ2}f(x) = \cos^{-1}(\cos\theta) + \cos^{-1}\left\{\frac{\cos\theta}{2} + \frac{\sqrt{3}\sin\theta}{2}\right\} f(x)=θ+cos⁡−1{cos⁡θcos⁡π3+sin⁡θsin⁡π3}f(x) = \theta + \cos^{-1}\left\{\cos\theta \cos\frac{\pi}{3} + \sin\theta \sin\frac{\pi}{3}\right\} Using the identity cos⁡Acos⁡B+sin⁡Asin⁡B=cos⁡(A−B)\cos A \cos B + \sin A \sin B = \cos(A-B): f(x)=θ+cos⁡−1{cos⁡(θ−π3)}f(x) = \theta + \cos^{-1}\left\{\cos\left(\theta - \frac{\pi}{3}\right)\right\}

Now, we need to evaluate cos⁡−1{cos⁡(θ−π3)}\cos^{-1}\left\{\cos\left(\theta - \frac{\pi}{3}\right)\right\}. Since θ∈[0,π]\theta \in [0, \pi], we have θ−π3∈[−π3,2π3]\theta - \frac{\pi}{3} \in \left[-\frac{\pi}{3}, \frac{2\pi}{3}\right].

Case 1: θ−π3≥0  ⟹  θ≥π3\theta - \frac{\pi}{3} \ge 0 \implies \theta \ge \frac{\pi}{3}. This corresponds to cos⁡θ≤cos⁡(π3)=12\cos\theta \le \cos(\frac{\pi}{3}) = \frac{1}{2}. So, for x∈[−1,12]x \in [-1, \frac{1}{2}], we have θ≥π3\theta \ge \frac{\pi}{3}. In this case, cos⁡−1{cos⁡(θ−π3)}=θ−π3\cos^{-1}\left\{\cos\left(\theta - \frac{\pi}{3}\right)\right\} = \theta - \frac{\pi}{3}. Therefore, f(x)=θ+(θ−π3)=2θ−π3=2cos⁡−1x−π3f(x) = \theta + (\theta - \frac{\pi}{3}) = 2\theta - \frac{\pi}{3} = 2\cos^{-1}x - \frac{\pi}{3}.

Case 2: θ−π3<0  ⟹  θ<π3\theta - \frac{\pi}{3} < 0 \implies \theta < \frac{\pi}{3}. This corresponds to cos⁡θ>cos⁡(π3)=12\cos\theta > \cos(\frac{\pi}{3}) = \frac{1}{2}. So, for x∈(12,1]x \in (\frac{1}{2}, 1], we have θ<π3\theta < \frac{\pi}{3}. In this case, cos⁡−1{cos⁡(θ−π3)}=−(θ−π3)=π3−θ\cos^{-1}\left\{\cos\left(\theta - \frac{\pi}{3}\right)\right\} = -(\theta - \frac{\pi}{3}) = \frac{\pi}{3} - \theta. Therefore, f(x)=θ+(π3−θ)=π3f(x) = \theta + (\frac{\pi}{3} - \theta) = \frac{\pi}{3}.

Now, let's evaluate f(x)f(x) for the given options:

For x=23x = \frac{2}{3}: Since 23>12\frac{2}{3} > \frac{1}{2}, this falls under Case 2. So, f(23)=π3f\left(\frac{2}{3}\right) = \frac{\pi}{3}. This matches option (A).

For x=13x = \frac{1}{3}: Since 13≤12\frac{1}{3} \le \frac{1}{2}, this falls under Case 1. So, f(13)=2cos⁡−1(13)−π3f\left(\frac{1}{3}\right) = 2\cos^{-1}\left(\frac{1}{3}\right) - \frac{\pi}{3}. This matches option (D).