Find the acute angle of intersection of curves, $y=\left[\right|\sin x|+|\cos x\left|\right]$ and $x^2+y^2=5$, where [.] denotes the greatest integral function.

The given curves are $y = \left[\left|\sin x\right|+\left|\cos x\right|\right]$ and $x^2+y^2=5$.

Now, we know ${\left(\left|\sin x\right|+\left|\cos x\right|\right)}_{min} = 1$ (When either of them is zero).

And,

${\left(\left|\sin x\right|+\left|\cos x\right|\right)}_{max}= {(\sin x + \cos x)}_{max} ={\left(\sqrt{2}\sin \left(x+\frac{\mathrm{\pi }}{4}\right)\right)}_{max} =\sqrt{2}$.

 $$\begin{gathered} \Rightarrow (\left|\sin x\right|+\left|\cos x\right|)\in [1,\sqrt{2}] \\ \Rightarrow \left[\left|\sin x\right|+ \left|\cos x\right|\right] = 1, for x\in R \end{gathered}$$

This means that the first curve is simply the straight line $y = 1$ and ordinate of point of intersections is $1$.

$$\begin{gathered} \Rightarrow x^2 = 5 - y^2 = 4 \\ x = \pm 2 \end{gathered}$$

Hence, the points of intersections are $A(-2,1), B(2,1)$.

Now, on differentiating equation of second curve w.r.t $x$, we get,

$$\begin{gathered} 2x + 2y\frac{dy}{dx}=0 \\ \frac{dy}{dx}= -\frac{x}{y} \end{gathered}$$

For second curve, slope of tangent at $A = {\overline{)\frac{dy}{dx}}}_{(-2,1)}=2$ and that at $B = {\overline{)\frac{dy}{dx}}}_{(2,1)} = -2$.

Now, angle of intersections is the angle made by the line $y = 1$ and the tangents to second curve at $A, B$.

Since, $y = 1$ is parallel to $x-axis$, angle made by tangents with $x-axis$ and the line are equal.

$\Rightarrow \tan {\theta }_A = 2, \tan {\theta }_B= -2$.

Hence, the acute angle of intersections are both equal to ${\tan }^{-1}\left(2\right)$ and option $1$ is correct.