Question

If $f\left(x\right)=\text{sgn}({\sin }^{2}x-\sin x-1)$ has exactly four points of discontinuity for $x\in (0,n\pi ),n\in N,$ then

• A. Minimum value of $n$ is $5.$
• B. Maximum value of $n$ is $6.$
• C. There are exactly two possible values of $n$
• D. Minimum value of $n$ is $7$

Answer 1

The correct option is C There are exactly two possible values of $n$

$f\left(x\right)=\text{sgn}({\sin }^{2}x-\sin x-1)$ is discontinuous when ${\sin }^{2}x-\sin x-1=0\Rightarrow \sin x=\frac{1+\sqrt{5}}{2}\text{or}\sin x=\frac{1-\sqrt{5}}{2}\Rightarrow \sin x=\frac{1-\sqrt{5}}{2}(\because \sin x\in [-1,1\left]\right)$


Dotted line represents $y=\frac{1-\sqrt{5}}{2}$

For exactly four points of discontinuity, $n$ can take value $4$ or $5$ as shown in the diagram.