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Try this problem from I.S.I. B.Stat Entrance Objective Problem based on Periodic Function.

## Periodic Function (B.Stat Objective Question )

If f(x) = $a_0+a_1cosx+a_2cos2x+….+a_ncosnx$ where $a_0,a_1,….,a_n$ are non zero real numbers and $a_n > |a_0|+|a_1|+….+|a_{n-1}|$, then number of roots of f(x)=0 in $0 \leq x \leq 2\pi$, is

• 0
• at least 2n
• 53361
• 5082

### Key Concepts

Periodic

Real Numbers

Inequality

But try the problem first…

Source

B.Stat Objective Problem 710

Challenges and Thrills of Pre-College Mathematics by University Press

## Try with Hints

First hint

f is periodic with period $2\pi$

here $0 <|\displaystyle\sum_{k=0}^{n-1}a_kcos(kx)| \leq \displaystyle\sum_{k=0}^{n-1}|a_k|<a_n, x\in$set of reals

Second Hint

for points $x_k=\frac{k\pi}{n}$ $1 \leq k \leq 2n$

$f(x_k)=a_n[cos(k\pi)+\theta], |\theta|<1 for 1 \leq k \leq 2n$

[ since cos $\theta$ is periodic and f(x) is expressed for every point x=$x_k$]

Final Step

f has at least 2n points in such a period interval where f has alternating sign.