Question

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

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

Answer 1

The correct option is B

${\tan }^{-1}\left(\pm 3\right)$

Explanation for the correct option:

Find the angle of intersection of curves:

Step 1: Find the point of intersection of the curves.

Given,$y=\left[\left|\sin x\right|+\left|\cos x\right|\right]$

We know that $\left(\left|\sin x\right|+\left|\cos x\right|\right)\in \left[1,\sqrt{2}\right]$

$\therefore y=1\left[\because \left[x\right]\text{deonotes the greatest integer}\le x\right]$

Substitute $y=1$in the given curve is $x^2+y^2=10$,to find the point of intersection. So,

$$\begin{gathered} \Rightarrow x^2+1^2=10 \\ \Rightarrow x^2=9 \\ \Rightarrow x=\pm 3 \end{gathered}$$

Therefore the point of intersection is $(\pm 3,1)$

Step 2: Find the slope of curves at their point of intersection

Now, Differentiate the given curve $x^2+y^2=10$

$$\begin{gathered} \Rightarrow 2x+2y\frac{dy}{dx}=0 \\ \Rightarrow \frac{dy}{dx}=\frac{-x}{y} \\ \Rightarrow \frac{dy}{dx}=\pm \frac{3}{1}\left[\text{intersecting point at}(\pm 3,1)\right] \\ =\pm 3 \end{gathered}$$

So, slope of tangent to the curve $x^2+y^2=10$ at $(\pm 3,1)$ is $m_1=\pm 3$

and Slope of line $y=1$is $m_2=0$

Step4: Find the angle required angle.

The angle between two curves is

$$\begin{gathered} \tan \theta =\frac{m_2-m_1}{1+m_1{m}_2} \\ =\frac{0-\left(\pm 3\right)}{1+0\times \pm 3} \\ =\frac{\pm 3}{1} \\ =\pm 3 \end{gathered}$$

$\Rightarrow \theta ={\tan }^{-1}(\pm 3)$

Hence, option (B) is the correct answer.