Question

The angle of intersection between the curves $$y=[|\sin x|+|\cos x|]$$ and $x^2+y^2=10$ , where $x$ denote the greatest integer $\le x$ , is
A.${\tan }^{-1}3$
B. ${\tan }^{-1}(-3)$
C. ${\tan }^{-1}\sqrt{3}$
D. ${\tan }^{-1}(-{\displaystyle \frac{1}{\sqrt{3}}})$

Answer 1

Hint: We need to find the angle of intersection between the curves $$y=[|\sin x|+|\cos x|]$$ and $x^2+y^2=10$ . For this first find the range of $$y=[|\sin x|+|\cos x|]$$ . After that, we will be considering only the value of lower bound since $x$ denotes the greatest integer $\le x$ ,i.e. $y=1$ . After that, substitute that value in $x^2+y^2=10$ . Find its slope by differentiating. Then get the next slope by differentiating $y=1$ . Now use the equation $\tan \theta =\left|{\displaystyle \frac{m_2-m_1}{1+m_1{m}_2}}\right|$ to get the value of the angle.

Complete step by step answer:
We need to find the angle of intersection between the curves $$y=[|\sin x|+|\cos x|]$$ and $x^2+y^2=10$ .
Let us find the range of $$y=[|\sin x|+|\cos x|]$$ .
We know that the range of $|\sin x|$ is
$0\le |\sin x|\le 1$
And the range of $|\cos x|$ is
$0\le |\cos x|\le 1$
Therefore, range of $$y=[|\sin x|+|\cos x|]$$ can be found out as follows:
$$y=\sin x+\cos x$$where$x\in (0,{\displaystyle \frac{\pi }{2}})$ .
Now multiply and divide RHS by $\sqrt{2}$ . So the above equation becomes,
$$y=\sqrt{2}({\displaystyle \frac{1}{\sqrt{2}}}\sin x+{\displaystyle \frac{1}{\sqrt{2}}}\cos x)...(i)$$
Now, $\sin (x+{\displaystyle \frac{\pi }{4}})=\sin x\cos {\displaystyle \frac{\pi }{4}}+\cos x\sin {\displaystyle \frac{\pi }{4}}$
Solving, we get
$\sin (x+{\displaystyle \frac{\pi }{4}})={\displaystyle \frac{1}{\sqrt{2}}}\sin x+{\displaystyle \frac{1}{\sqrt{2}}}\cos x$
Therefore equation $(i)$ can be written as
$$y=\sqrt{2}\sin (x+{\displaystyle \frac{\pi }{4}})$$
We know that $\sin x$ ranges from $[-1,1]$ .
Therefore, $-1\le \sin (x+{\displaystyle \frac{\pi }{4}})\le 1$
Multiplying by $\sqrt{2}$ we get
$-\sqrt{2}\le \sqrt{2}\sin (x+{\displaystyle \frac{\pi }{4}})\le \sqrt{2}$
As $|\sin x|$ ranges from $0\le |\sin x|\le 1$ , comparing with the above one, we get
$$y=[|\sin x|+|\cos x|]=[1,\sqrt{2}]$$
It is given that $x$ denote the greatest integer $\le x$ . So we will consider the value $y=1$ .
Given that $x^2+y^2=10$ . Substituting the value of $y$ here, we get
$x^2+1=10\Rightarrow x^2=9$
$x=\pm 3$
Therefore, the intersection points are $q(3,1)$ and $p(-3,1)$ .

We need to find the slope of the tangent $(\pm 3,1)$ to $x^2+y^2=10$ .
Now differentiate $x^2+y^2=10$ with respect to $x$ . We will get
$2x+2y{\displaystyle \frac{dy}{dx}}=0$
$\Rightarrow x+y{\displaystyle \frac{dy}{dx}}=0$
$\Rightarrow {\displaystyle \frac{dy}{dx}}={\displaystyle \frac{-x}{y}}$
Now for the point $q(3,1)$ ,
${{\displaystyle \frac{dy}{dx}}|}_{p(3,1)}=-3$
For the point $p(-3,1)$ ,
${{\displaystyle \frac{dy}{dx}}|}_{p(-3,1)}=3$
Therefore, slope $m_1=\pm 3$ .
We have, $y=1$ .
Differentiating $y$ with respect to $x$ , we get
${{\displaystyle \frac{dy}{dx}}|}_p=0$
That is, the slope $m_2={{\displaystyle \frac{dy}{dx}}|}_p=0$ .
Now, to find the angle of intersection, we have
$\tan \theta =\left|{\displaystyle \frac{m_2-m_1}{1+m_1{m}_2}}\right|$
We will use in this case $m_2=-3$ as per the figure.
Substituting the value, we will get
$\tan \theta =\left|{\displaystyle \frac{0-(-3)}{1+0\times -3}}\right|=\left|3\right|=\pm 3$
Taking inverse of $\tan$ we will get the value of $\theta$ .
Therefore, $\theta ={\tan }^{-1}3$ and $\theta ={\tan }^{-1}(-3)$ .
Hence the correct options are A and B.

Note:
In this question, it is not necessary to write the steps to get the range of $$y=[|\sin x|+|\cos x|]$$ .
We know that when $\sin x$ increases $\cos x$ decreases. So the maximum value cannot be obtained.
We know that at $x={\displaystyle \frac{\pi }{4}}$ both $\sin x$ and $\cos x$ will be the same, i.e, ${\displaystyle \frac{1}{\sqrt{2}}}$ .
 So $$y=\sin x+\cos x={\displaystyle \frac{1}{\sqrt{2}}}+{\displaystyle \frac{1}{\sqrt{2}}}=\sqrt{2}$$ .
Therefore, the maximum value of $$y=[|\sin x|+|\cos x|]=\sqrt{2}$$ .
To find the minimum value, we know that minimum value of $\sin x=0$ and that of $\cos x=1$ .
Now $$y=[|\sin x|+|\cos x|]=0+1=1$$ .
Thus the range of $$y=[|\sin x|+|\cos x|]=[1,\sqrt{2}]$$ .