Question

The angle of intersection of curves $y=[|sinx|+|cosx|]and{x}^2+y^2=5$, where [∙] denotes the greatest integer function, is

• A. $ta{n}^{-1}2$
• B. $ta{n}^{-1}\left(\frac{1}{2}\right)$
• C. $ta{n}^{-1}\sqrt{2}$
• D. $ta{n}^{-1}\frac{1}{\sqrt{2}}$

Answer 1

The correct option is A $ta{n}^{-1}2$

$\because 1\le |sinx|+|cosx|\le \sqrt{2}\therefore y=[|sinx|+|cosx|]=1\because -\sqrt{2}\le sinx+cosx\le \sqrt{2}|sinx|+|cosx|=\sqrt{\{|sinx|+|cosx|{\}}^2}=\sqrt{(1+|sin2x|)}\le 1$
Let P and Q be the points of intersection of given curves.
Now, solving y = 1 and $x^2+y^2=5$
$\therefore x^2+1=5\Rightarrow x=\pm 2$
$\therefore P\equiv (-2,1)$ and $Q\equiv (2,1)$
Clearly the slope of line y = 1 is zero
$\because x^2+y^2=5\therefore 2x+2y\frac{dy}{dx}=0\Rightarrow \frac{dy}{dx}=-\frac{x}{y}(\frac{dy}{dx}{)}_{(-2,1)}=2$ and $(\frac{dy}{dx}{)}_{(2,1)}=-2$
Thus, the angle of intersection is $ta{n}^{-1}\left(2\right)$ and $ta{n}^{-1}(-2)$