Question

The angle of intersection between the curves $y=\left[\left|\sin\,x\right|+\left|\cos\,x\right|\right]$ and $x^{2}+y^{2}=10,$ where $[x]$ denotes the greatest integer $= x,$ is

• A. $\tan^{-1}\, (3)$
• B. $\tan^{-1}\, (-3)$
• C. $\tan^{-1}\sqrt{3}$
• D. $\tan^{-1}\left(1/\sqrt{3}\right)$

Answer 1

Answer: (A)

$\tan^{-1}\, (3)$

Answer 2

Given, $y=[|\sin x|+|\cos x|]$ and $x^{2}+y^{2}=10$
We know that $(|\sin x|+|\cos x|) \in[1, \sqrt{2}]$
$\therefore y=1$
The point of intersection of given curve is $x^{2}+1^{2}=10$
$\Rightarrow x^{2}=9$
$\Rightarrow x=\pm 3$
$\therefore$ Point of intersection is $(\pm 3,1)$
Now, $x^{2}+y^{2}=10$
$\Rightarrow 2 x+2 y \frac{d y}{d x}=0$
$\Rightarrow \frac{d y}{d x}=-\frac{x}{y}$
At point $(-3,1)$
$\frac{d y}{d x}=\frac{3}{1}=3$
$\Rightarrow m_{1}=3$
Slope of line $y=1$ is $m_{2}=0$
$\therefore$ Angle between two curves is
$\tan \theta=\frac{m_{1}-m_{2}}{1+m_{1} \,m_{2}}=3$
$\Rightarrow \theta=\tan ^{-1}(3)$