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 $\frac{\pi }{6}<x<\frac{\pi }{3}$
• 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 $f(x)=x-\cos x$, then $f^{\prime }(x)=1+\sin x\ge 0$
$\Rightarrow f(x)$ is increasing and hence $f(x)=0$ for a unique value of $x$.
For $x\ge \frac{\pi }{3},f(x)>0$ and $x<\frac{\pi }{6},f(x)<0.$ Thus
$\cos x=x$, for a unique value of $x,x\in \left(\frac{\pi }{6},\frac{\pi }{3}\right)$