JEE Advanced 2021 · Paper 1 · Mathematics Q4

← JEE Advanced 2021 · Paper 1

MathematicsSingle correct+3 / -1Question 42 of 57

Let $\theta_1, \theta_2, \ldots, \theta_{10}$ be positive valued angles (in radian) such that $\theta_1 + \theta_2 + \cdots + \theta_{10} = 2\pi$ . Define the complex numbers $z_1 = e^{i\theta_1}$ , $z_k = z_{k-1}e^{i\theta_k}$ for $k = 2, 3, \ldots, 10$ , where $i = \sqrt{-1}$ . Consider the statements $P$ and $Q$ given below:
$P: |z_2 - z_1| + |z_3 - z_2| + \cdots + |z_{10} - z_9| + |z_1 - z_{10}| \le 2\pi$
$Q: |z_2^2 - z_1^2| + |z_3^2 - z_2^2| + \cdots + |z_{10}^2 - z_9^2| + |z_1^2 - z_{10}^2| \le 4\pi$
Then,

P

is TRUE and

Q

is FALSE

Q

is TRUE and

P

is FALSE

both

P

and

Q

are TRUE

both

P

and

Q

are FALSE

Answer and worked solution

Official answer: C

The points. $z_k=e^{i(\theta_1+\cdots+\theta_k)}$ all lie on the unit circle, and $z_{10}=e^{2\pi i}=1$ .

Statement P. Each term is a chord:
$|z_{k+1}-z_k|=|e^{i\theta_{k+1}}-1|=2\sin\frac{\theta_{k+1}}2\le\theta_{k+1},$
and similarly $|z_1-z_{10}|\le\theta_1$ . Summing, the total is at most $\sum\theta_k=2\pi$ . So P is true.

Statement Q. Each term is
$|z_{k+1}^2-z_k^2|=|e^{2i\theta_{k+1}}-1|=2|\sin\theta_{k+1}|\le2\theta_{k+1}.$
Summing gives at most $4\pi$ . So Q is true.

(C)

Topics:Geometry in the complex plane Polar and Euler form

← Previous question Next question →

JEE Advanced 2021 · Paper 1, Mathematics question 4