Entry Level Curve Sketching

Since $\ln \cos x = \ln \cos(-x)$ , and the only graph of an even function is A, the answer must hence be A.

More features of A resembling $y = \ln \cos x$ include:

1. At $x=0$ , $\ln \cos 0 = \ln 1 = 0$ . None of the other graphs pass through the origin.

2. On certain intervals, $\cos x$ is positive (namely $((2k-\frac12)\pi,(2k+\frac12)\pi)$ ). Since the domain of $\ln$ is $\mathbb{R}^+$ , $\ln \cos x$ is defined at exactly $x$ in those intervals. On the complement (namely $[(2k+\frac12)\pi,(2k+\frac32)\pi]$ ), $\cos x$ is nonpositive, and so $\ln \cos x$ is undefined at exactly $x$ in those intervals. A fits the bill.

3. Drawing a rough sketch of A's tangent's gradient, we get a graph closely resembling $-\tan x$ . Why? Why, because it is just that.

Since $\cos{x}$ is an even function, $y = \ln(\cos{x})$ is also an even function. Among the five answer options only graph A shows an even function. And since $\cos{x} < 0$ in the second and third quadrants ( $|x| > \left(2k+\frac{1}{2}\right)\pi$ ), $y$ only has real values in the first and last quadrants ( $|x| < \left(2k+\frac{1}{2}\right)\pi$ ), where $\cos{x}$ decreases from $1$ to $0$ and hence $y$ decreases from $0$ to $-\infty$ , as $|x|$ increased from $\left(2k +0\right) \pi$ to $\left(2k +\frac{1}{2} \right) \pi$ . Therefore, graph $\boxed{A}$ shows the function $y$ .

Easiest method :- Put x=0 , y will also be 0 , And there was only one graph touching origin

I differentiated it to get sinx/cosx which is tanx, and then plugged in 0 and the only graph with gradient 0 at x=0 is graph A

Since cos(x) is an even function (i-e cos(x)=cos(-x)) which implies that the left sub-graph must be a mirror image of right sub-graph which indicates that (A) must be the graph of ln(cos(x))