Question

For which of the following values of 'm' the line y = mx cuts the graph of y = sin–1(sin x) in exactly 5 points-

• A. 1/5
• B. -1/7
• C. -1/3
• D. 1/3

Answer 1

Answer: (A)

1/5

Answer 2

The graph of $y = \sin^{-1}(\sin x)$ is an odd, periodic, saw-tooth wave. The line $y = mx$ passes through the origin $(0,0)$. For exactly 5 intersection points, due to symmetry, there must be 1 point at the origin and 2 points for $x>0$ (and 2 for $x<0$).

For $x>0$, the graph of $y = \sin^{-1}(\sin x)$ consists of segments: $y=x$ for $x \in [0, \frac{\pi}{2}]$, $y=\pi-x$ for $x \in [\frac{\pi}{2}, \frac{3\pi}{2}]$, $y=x-2\pi$ for $x \in [\frac{3\pi}{2}, \frac{5\pi}{2}]$, etc.

Consider $m>0$. The line $y=mx$ intersects $y=\pi-x$ at $x_1 = \frac{\pi}{m+1}$. This is valid for $m \in (0,1]$. It intersects $y=x-2\pi$ at $x_2 = \frac{2\pi}{1-m}$. This is valid for $m \in (0, 1/5]$.

For exactly two intersections ($N=2$), the line must intersect the second segment ($y=x-2\pi$) at its endpoint $(\frac{5\pi}{2}, \frac{\pi}{2})$ and not intersect any further segments. This occurs when $m = \frac{\pi/2}{5\pi/2} = \frac{1}{5}$.

For $m=1/5$, $x_1 = \frac{5\pi}{6}$ and $x_2 = \frac{5\pi}{2}$. These are two distinct points for $x>0$.

Consider $m<0$. The line $y=mx$ intersects $y=\pi-x$ at $x_1 = \frac{\pi}{m+1}$ (valid for $m \in [-1/3, 0)$) and $y=x-2\pi$ at $x_2 = \frac{2\pi}{1-m}$ (valid for $m \in [-1/3, 0)$).

If $m=-1/3$, $x_1=x_2=\frac{3\pi}{2}$, resulting in $N=1$.

If $m=-1/7$, $x_1=\frac{7\pi}{6}$, $x_2=\frac{7\pi}{4}$, and the line also intersects $y=3\pi-x$ at $x_3=\frac{7\pi}{2}$, resulting in $N=3$.

For $N=2$ when $m<0$, $m$ must be in $(-\frac{1}{3}, -\frac{1}{7})$.

Comparing with typical multiple choice options, $m=1/5$ is the only value that fits.