Question

If $x^2+px+1$ is a factor of $2co{s}^2\theta x^3+2x+sin2\theta$, then

• A. $\theta =n\pi +\frac{\pi }{2},n\in I$
• B. $\theta =\frac{n\pi }{2},n\in I$
• C. $\theta =n\pi ,n\in I$
• D. $\theta =2n\pi ,n\in I$

Answer 1

The correct option is C $\theta =n\pi ,n\in I$

Let $2co{s}^2\theta x^3+2x+sin2\theta =(x^2+px+1)(ax+b)$
On expanding the RHS term, and comparing the coefficient of $x^3$ and the constant term with the LHS, we get
$a=2co{s}^2\theta$ and $b=sin2\theta$

$2co{s}^2\theta x^3+2x+sin2\theta =(x^2+px+1)(2co{s}^2\theta x+sin2\theta )$
On comparing coefficients of $x$ and $x^2,$
$2=psin2\theta +2co{s}^2\theta \Rightarrow 2si{n}^2\theta =psin2\theta orp=tan\theta ...\left(i\right)and0=sin2\theta +2pco{s}^2\theta \therefore p=-tan\theta ...\left(ii\right)$
From Eqs. $\left(i\right)$ and $\left(ii\right)$,
$tan\theta =-tan\theta 2tan\theta =0tan\theta =0\Rightarrow \theta =n\pi ,n\in I$