The angle of intersection between the curves | vedclass

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(A) Given, $y=[|\sin x|+|\cos x|]$ and $x^{2}+y^{2}=10$.We know that $(|\sin x|+|\cos x|) \in [1, \sqrt{2}]$.Since $[x]$ is the greatest integer funct

Vedclass · Accepted Answer

$	an ^{-1} 3$

Vedclass · Answer

(A) Given, $y=[|\sin x|+|\cos x|]$ and $x^{2}+y^{2}=10$.We know that $(|\sin x|+|\cos x|) \in [1, \sqrt{2}]$.Since $[x]$ is the greatest integer function, $y = [|\sin x|+|\cos x|] = 1$.The point of intersection of the given curves is found by substituting $y=1$ into $x^{2}+y^{2}=10$:$x^{2}+1^{2}=10$ $\Rightarrow x^{2}=9$ $\Rightarrow x=\pm 3$.The points of intersection are $(3, 1)$ and $(-3, 1)$.For the circle $x^{2}+y^{2}=10$, differentiating with respect to $x$ gives $2x + 2y \frac{dy}{dx} = 0$, so $\frac{dy}{dx} = -\frac{x}{y}$.At point $(-3, 1)$, the slope $m_{1} = -(-3)/1 = 3$.The curve $y=1$ is a horizontal line, so its slope $m_{2} = 0$.The angle $	heta$ between the curves is given by $	an 	heta = |\frac{m_{1}-m_{2}}{1+m_{1}m_{2}}| = |\frac{3-0}{1+3(0)}| = 3$.Therefore, $	heta = 	an^{-1}(3)$.

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

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