Question

The vectors $\vec{a}(x)=(\cos x)\hat{i}+(\sin x)\hat{j}$ and $\vec{b}(x)=x\hat{i}+(\sin x)\hat{j}$ are collinear for

• A. unique value of $x,0<x<\frac{\pi }{6}$
• B. unique value of $x,\frac{\pi }{6}<x<\frac{\pi }{3}$
• C. no value of $x$
• D. infinitely many values of $x,0<x<\frac{\pi }{2}$

Answer 1

Answer: (B)

$\vec{a}(x)$ and $\vec{b}(x)$ are collinear if and only if $\cos x=x$
Now, let $f(x)=x-\cos x$
$\therefore f^{\prime }(x)=1+\sin x\Rightarrow f^{\prime }(x)\ge 0$
$\Rightarrow f(x)$ is an increasing function and hence $f(x)=0$ is true for a unique value of $x$.
As, $f\left(\frac{\pi }{6}\right)=\frac{\pi }{6}-\frac{\sqrt{3}}{2}<0$
and $f\left(\frac{\pi }{3}\right)=\frac{\pi }{3}-\frac{1}{2}>0$
So, using intermediate value theorem,
Thus, $\cos x=x$, for a unique value of $x,x\in \left(\frac{\pi }{6},\frac{\pi }{3}\right)$