The angle of intersection of curves. y = [|sin x| + |cos x|] and x2+y2=5 where [.] denotes greatest integral function is
tan1(2)
We know that. 1≤|sin x|+|cos x|≤√2
⇒y=[|sin x|+|cos x|]=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
x2+1=5⇒x=±2
Now, P(2, 1) and Q(-2, 1)
Differentiating x2+y2=5 w.r.t. x, we get 2x +
2ydydx=0⇒dydx=−xy(dydx)(2.1)=−2 and (dydx)(−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 tan−1(2)
Hence the correct option is c