Question

If $\left[ x \right]$ denotes the greatest integer $\le x$, then the system of linear equations $\left[ \sin \theta \right]x+\left[ -\cos \theta \right]y=0$ and $\left[ \cot \theta \right]x+y=0$
(A) Has infinitely many solutions if $\theta \in \left( \dfrac{\pi }{2},\dfrac{2\pi }{3} \right)\cup \left( \pi ,\dfrac{7\pi }{6} \right)$.
(B) Has infinitely many solutions if $\theta \in \left( \dfrac{\pi }{2},\dfrac{2\pi }{3} \right)$ and has a unique solution if $\theta \in \left( \pi ,\dfrac{7\pi }{6} \right)$.
(C) Has a unique solution if $\theta \in \left( \dfrac{\pi }{2},\dfrac{2\pi }{3} \right)$ and have infinitely many solutions if $\theta \in \left( \pi ,\dfrac{7\pi }{6} \right)$.
(D) Has a unique solution if $\theta \in \left( \dfrac{\pi }{2},\dfrac{2\pi }{3} \right)\cup \left( \pi ,\dfrac{7\pi }{6} \right)$.

Answer 1

We start solving this question by first considering the intervals given in the options and then calculate the values of the step functions, $\left[ \sin \theta \right],\left[ -\cos \theta \right],\left[ \cot \theta \right]$ using the formula for step function $\left[ x \right]=n-1,\ \ \ x\in \left( n-1,n \right)$ and substitute them in the given system of equations to obtain the system of equations with integer coefficients. Then we find whether the system of equations has roots and if so the number of roots in both cases.

Complete step-by-step answer:
Let us start by considering the formula for the step function.
For any integer n,
$\left[ x \right]=n-1,\ \ \ x\in \left( n-1,n \right)$
First let us consider the given system of linear equations, $\left[ \sin \theta \right]x+\left[ -\cos \theta \right]y=0$ and $\left[ \cot \theta \right]x+y=0$.
Now consider the given intervals in the question, $\theta \in \left( \dfrac{\pi }{2},\dfrac{2\pi }{3} \right)$ and $\theta \in \left( \pi ,\dfrac{7\pi }{6} \right)$.
When $\theta \in \left( \dfrac{\pi }{2},\dfrac{2\pi }{3} \right)$, let us calculate the values of $\left[ \sin \theta \right],\left[ -\cos \theta \right],\left[ \cot \theta \right]$ and substitute them in the equations.
If $\theta \in \left( \dfrac{\pi }{2},\dfrac{2\pi }{3} \right)$ then
As $\cos \theta$ is decreasing when $\theta \in \left( \dfrac{\pi }{2},\dfrac{2\pi }{3} \right)$
$\begin{aligned} & \Rightarrow \cos \theta \in \left( \cos \dfrac{2\pi }{3},\cos \dfrac{\pi }{2} \right) \\\ & \Rightarrow \cos \theta \in \left( -\dfrac{1}{2},0 \right) \\\ & \Rightarrow -\cos \theta \in \left( 0,\dfrac{1}{2} \right) \\\ & \Rightarrow \left[ -\cos \theta \right]=0 \\\ \end{aligned}$
As $\sin \theta$ is decreasing in $\theta \in \left( \dfrac{\pi }{2},\dfrac{2\pi }{3} \right)$
$\begin{aligned} & \Rightarrow \sin \theta \in \left( \sin \dfrac{2\pi }{3},\sin \dfrac{\pi }{2} \right) \\\ & \Rightarrow \sin \theta \in \left( \dfrac{\sqrt{3}}{2},1 \right) \\\ & \Rightarrow \left[ \sin \theta \right]=0 \\\ \end{aligned}$
As $\cot \theta$ is decreasing in $\theta \in \left( \dfrac{\pi }{2},\dfrac{2\pi }{3} \right)$

& \Rightarrow \cot \theta \in \left( \cot \dfrac{2\pi }{3},\cot \dfrac{\pi }{2} \right) \\\ & \Rightarrow \cot \theta \in \left( -\dfrac{1}{\sqrt{3}},0 \right) \\\ & \Rightarrow \left[ \cot \theta \right]=-1 \\\ \end{aligned}$$ Now let us substitute these values in the given equations. Then the equations transform as, $\begin{aligned} & \left[ \sin \theta \right]x+\left[ -\cos \theta \right]y=0\Rightarrow \left( 0 \right)x+\left( 0 \right)y=0\Rightarrow 0=0 \\\ & \left[ \cot \theta \right]x+y=0\Rightarrow \left( -1 \right)x+y=0\Rightarrow y-x=0 \\\ \end{aligned}$ So, we have only one equation in this case, that is $y=x$. Therefore, a given system of equations has infinitely many solutions if $\theta \in \left( \dfrac{\pi }{2},\dfrac{2\pi }{3} \right)$. Now, let us consider the other interval, $\theta \in \left( \pi ,\dfrac{7\pi }{6} \right)$ and repeat the same procedure as above to this interval. If $\theta \in \left( \pi ,\dfrac{7\pi }{6} \right)$ then As $\cos \theta $ is increasing when $\theta \in \left( \pi ,\dfrac{7\pi }{6} \right)$ $\begin{aligned} & \Rightarrow \cos \theta \in \left( \cos \pi ,\cos \dfrac{7\pi }{6} \right) \\\ & \Rightarrow \cos \theta \in \left( -1,-\dfrac{\sqrt{3}}{2} \right) \\\ & \Rightarrow -\cos \theta \in \left( \dfrac{\sqrt{3}}{2},1 \right) \\\ & \Rightarrow \left[ -\cos \theta \right]=0 \\\ \end{aligned}$ As $\sin \theta $ is decreasing in $\theta \in \left( \pi ,\dfrac{7\pi }{6} \right)$ $\begin{aligned} & \Rightarrow \sin \theta \in \left( \sin \dfrac{7\pi }{6},\sin \pi \right) \\\ & \Rightarrow \sin \theta \in \left( -\dfrac{1}{2},0 \right) \\\ & \Rightarrow \left[ \sin \theta \right]=-1 \\\ \end{aligned}$ As $\cot \theta $ is decreasing in $\theta \in \left( \pi ,\dfrac{7\pi }{6} \right)$ $$\begin{aligned} & \Rightarrow \cot \theta \in \left( \cot \dfrac{2\pi }{3},\cot \dfrac{\pi }{2} \right) \\\ & \Rightarrow \cot \theta \in \left( -\dfrac{1}{\sqrt{3}},0 \right) \\\ & \Rightarrow \left[ \cot \theta \right]=-1 \\\ \end{aligned}$$ Now let us substitute these values in the given equations. Then the equations transform as, $\begin{aligned} & \left[ \sin \theta \right]x+\left[ -\cos \theta \right]y=0\Rightarrow \left( -1 \right)x+\left( 0 \right)y=0\Rightarrow -x=0\Rightarrow x=0 \\\ & \left[ \cot \theta \right]x+y=0\Rightarrow \left( -1 \right)x+y=0\Rightarrow y-x=0 \\\ \end{aligned}$ So, we have two equations in this case, that is $x=0$ and $y=x$. So, we get the solution for them as $\left( x,y \right)=\left( 0,0 \right)$. Therefore, a given system of equations has a unique solution if $\theta \in \left( \pi ,\dfrac{7\pi }{6} \right)$. So, the given system of equations has infinitely many solutions if $\theta \in \left( \dfrac{\pi }{2},\dfrac{2\pi }{3} \right)$ and has a unique solution if $\theta \in \left( \pi ,\dfrac{7\pi }{6} \right)$. **So, the correct answer is “Option B”.** **Note:** There is a possibility of one making a mistake while solving this problem by taking the formula for the function step of x wrongly as, If n is any integer, $\left[ x \right]=n\ \ if\ x\in \left( n-1,n \right)$ But it is wrong as the step function of x gives the greatest integer that is less than or equal to x.