JEE Advanced 2014 · Paper 2 · Mathematics Q57 · Dronacharya Exams

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MathematicsList match+3 / -1Question 57 of 60

Let $z_k=\cos\left(\frac{2k\pi}{10}\right)+i\sin\left(\frac{2k\pi}{10}\right)$ ; $k=1,2,\ldots,9$ .

List I

P
For each $z_k$ there exists a $z_j$ such that $z_k\cdot z_j$
Q
There exists a $k\in\{1,2,\ldots,9\}$ such that $z_1\cdot z=z_k$ has no solution $z$ in the set of complex numbers.
R
$\dfrac{|1-z_1||1-z_2|\cdots|1-z_9|}{10}$ equals
S
$1-\sum_{k=1}^{9}\cos\left(\frac{2k\pi}{10}\right)$ equals

List II

1
True
2
False
3
1
4
2

A

P → 1; Q → 2; R → 4; S → 3

B

P → 2; Q → 1; R → 3; S → 4

C

P → 1; Q → 2; R → 3; S → 4

D

P → 2; Q → 1; R → 4; S → 3

Answer and worked solution

Official answer: C

The $z_k$ are the tenth roots of unity other than 1.

P (printed as " $z_k\cdot z_j$ ", meaning $z_k\cdot z_j=1$ ): take $z_j=\overline{z_k}=z_{10-k}$ , which is in the set. True → 1.
Q: $z_1z=z_k$ always has the solution $z=z_k/z_1$ . The claim is False → 2.
R: $\prod_{k=1}^9(x-z_k)=1+x+\cdots+x^9$ ; at $x=1$ the product is 10. So the expression is $10/10=1$ → 3.
S: The sum of all ten roots is 0, so $\sum_{k=1}^9\cos\frac{2k\pi}{10}=-1$ and $1-(-1)=2$ → 4.

(C) P → 1, Q → 2, R → 3, S → 4

Topics:Roots of unity

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