I am doing problems in the chapter Functions. I have enclosed images of the question and my solution for the problem.
Question: (Kindly ignore the question code - [16JM120379])
My Solution:
Clearly, there is no such option in the given question. So, did I go wrong anywhere? If yes, please tell where I went wrong. Or, is the question having a missing option?
Kindly guide me how to approach this problem.
You incorrectly identified the domain of $h(x)$. If we have that $$\cos{(x)}\in[-1,1/2]$$ Then there are infinitely many intervals in which $x$ can lie given by adding and subtracting $2\pi$ to the supplied interval. So you should instead have $$\text{dom }h=[\pi/3+2\pi k,5\pi/3+2\pi k]_{k\in\mathbb{Z}}$$ But you also made another mistake because you should then get $$x+\pi/3\in[2\pi/3+2\pi k,2\pi+2\pi k]_{k\in\mathbb{Z}}$$ where you added $\pi/3$ to $5\pi/3$ and somehow got $\pi$. One could continue to get $$\sin{(x+\pi/3)}\in[-1,\sqrt{3}/2]$$ $$\arccos{(\sin{(x+\pi/3)})}\in[\pi/6,\pi]$$ $$\exp{(\arccos{(\sin{(x+\pi/3)})})}\in\left[e^{\pi/6},e^\pi\right]$$