Question

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

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

Answer 1

The correct option is A ${\tan }^{-1}3$

Given, $y=[\left|\sin x\right|+\left|\cos x\right|]$ and $x^2+y^2=10$
We know that $(\left|\sin x\right|+\left|\cos x\right|)\in [1,\sqrt{2}]$
$\therefore y=1$
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, $x^2+y^2=10$
$\Rightarrow 2x+2y\frac{dy}{dx}=0$
$\Rightarrow \frac{dy}{dx}=-\frac{x}{y}$
At point $(-3,1)$
$\frac{dy}{dx}=\frac{3}{1}=3\Rightarrow m_1=3$
Slope of line $y=1$ is $m_2=0$
$\therefore$ Angle between two curves is
$\tan \theta =\frac{m_1-m_2}{1+m_1{m}_2}=3$
$\Rightarrow \theta ={\tan }^{-1}\left(3\right)$