Question

The angle of intersection of curves. y = [|sin x| + |cos x|] and $x^2+y^2=5$ where [.] denotes greatest integral function is

• A. $\frac{\pi }{4}$
• B. $ta{n}^{-1}\left(\frac{1}{2}\right)$
• C. $ta{n}^1\left(2\right)$
• D. none of these

Answer 1

The correct option is C

$ta{n}^1\left(2\right)$

We know that. $1\le |sinx|+|cosx|\le \sqrt{2}$
$\Rightarrow y=[|sinx|+|cosx|]=1$
Let P and Q be the points of intersection of given curves.
Clearly, the given curves meet at points where y = 1 so, we get
$x^2+1=5\Rightarrow x=\pm 2$

Now, P(2, 1) and Q(-2, 1)
Differentiating $x^2+y^2=5$ w.r.t. x, we get 2x +
$2y\frac{dy}{dx}=0\Rightarrow \frac{dy}{dx}=-\frac{x}{y}{\left(\frac{dy}{dx}\right)}_{\left(2.1\right)}=-2and{\left(\frac{dy}{dx}\right)}_{(-2.1)}=2$
Clearly the slope of line y = 1 is zero and the slope of the tangents at P and Q are (-2) and (2) respectively.
Thus, the angle of intersection is $ta{n}^{-1}\left(2\right)$
Hence the correct option is c