IIT-JEE 2010 · Paper 2 · Mathematics Q26

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MathematicsInteger+3Question 26 of 57

Two parallel chords of a circle of radius 2 are at a distance $\sqrt3+1$ apart. If the chords subtend at the center, angles of $\dfrac{\pi}{k}$ and $\dfrac{2\pi}{k}$ , where $k>0$ , then the value of $[k]$ is

[Note: $[k]$ denotes the largest integer less than or equal to $k$ ]

Your answer

Answer and worked solution

Official answer: 3

Distances from the centre. A chord subtending $\phi$ is at distance $2\cos\frac\phi2$ . With $\theta=\frac\pi k$ , the chords are at $2\cos\frac\theta2$ and $2\cos\theta$ .

On opposite sides of the centre:
$2\cos\frac\theta2+2\cos\theta=\sqrt3+1.$
$\theta=\frac\pi3$ works: $2\cdot\frac{\sqrt3}2+2\cdot\frac12=\sqrt3+1$ . (The left side decreases on $(0,\pi)$ , so this is the only solution; chords on the same side would differ by at most 2, less than $\sqrt3+1$ .)

So $k=3$ and $[k]=3$ .

Answer: 3

Topics:Equation of a circle Trigonometric equations

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IIT-JEE 2010 · Paper 2, Mathematics question 26