If eix is a solution of the equation zn+p1zn-1+p2zn-2+……+pn=0, where pi are real i=1,2,3,………n,then pnsinnx+pn-1sinn-1x+…+p1sinx+1=

TS EAMCET · Maths · Complex Number

If $e^{i x}$ is a solution of the equation $z^{n} + p_{1} z^{n - 1} + p_{2} z^{n - 2} + \dots \dots + p_{n} = 0$ , where $p_{i}$ are real $(i = 1, 2, 3, \dots \dots \dots n)$ ,then $p_{n} \sin n x + p_{n - 1} \sin (n - 1) x + \dots + p_{1} \sin x + 1 =$

A $\cos (n + 1) x$
B $\sin (n (n + 1)) x$
C 0
D 1

Verified Solution

Answer & Solution

Correct Answer

(C) 0

Step-by-step Solution

Detailed explanation

Given that eix=cosx+isinx imaginary roots always occurs in pairs e-ix is also solution of zn+p1zn-1+p2zn-2+……+pn=0 i.e., pneixn+pn-1eixn-1+pn-2eixn-2+.....p1eix+1 = 0 ⇒pncosnx+isinnx+pn-1cosn-1x+isinn-1x+pn-2cosn-2x+isinn-2x+......p1cosx+isinx+1=0 By…

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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