JEE maths#16

Since $\cos^2 \theta + \beta$ and $\sin^2 \theta + \beta$ are the roots of $x^2+4x+2$ , using Vieta's formula , we have:

$\begin{cases} \cos^2 \theta + \beta + \sin^2 \theta + \beta = - 4 &...(1) \\ (\cos^2 \theta + \beta)(\sin^2 \theta + \beta) = 2 &...(2) \end{cases}$

From equation (1):

$\begin{aligned} \cos^2 \theta + \beta + \sin^2 \theta + \beta & = - 4 \\ 1+2\beta &= - 4 \\ \implies \beta & = - \frac 52 \end{aligned}$

From equation (2):

$\begin{aligned} (\cos^2 \theta + \beta)(\sin^2 \theta + \beta) & = 2 \\ \cos^2 \theta \sin^2 \theta + \beta(\cos^2 \theta + \sin^2 \theta) + \beta^2 & = 2 \\ \frac 14 \sin^2 2 \theta + \beta + \beta^2 & = 2 \\ \frac 14 \sin^2 2 \theta - \frac 52 + \frac {25}4 & = 2 \\ \sin^2 2 \theta - 10 + 25 & = 8 \\ \implies \sin^2 2 \theta & = -7 & \small \color{#3D99F6} \text{where } \theta \text{ is complex.} \end{aligned}$

Also since $\cos^4 \theta + \alpha$ and $\sin^4 \theta + \alpha$ are the roots of $x^2+b(2x+1)$ , we have:

$\begin{aligned} \cos^4 \theta + \alpha + \sin^4 \theta + \alpha & = -2b \\ \cos^4 \theta + \sin^4 \theta + 2 \alpha & = -2b \\ (\cos^2 \theta + \sin^2 \theta)^2 - 2\cos^2 \theta \sin^2 \theta + 2\alpha & = -2b \\ 1^2 - \frac 12 \sin^2 2 \theta + 2\alpha & = -2b \\ 1 + \frac 72 + 2\alpha & = - 2b \\ 9 + 4\alpha & = -4b & ...(3) \end{aligned}$

$\begin{aligned} (\cos^4 \theta + \alpha)(\sin^4 \theta + \alpha) & = b \\ \cos^4 \theta \sin^4 \theta + \alpha (\cos^4 \theta + \sin^4 \theta) + \alpha^2 & = b \\ \frac 1{16} \sin^4 2\theta + \frac 92\alpha + \alpha^2 & = b \\ \frac {49}{16} + \frac 92\alpha + \alpha^2 & = b \\ 49 + 72 \alpha + 16\alpha^2 & = 16b & ...(4) \end{aligned}$

$\begin{aligned} 4(3)+(4): \quad 16\alpha^2 + 88\alpha + 85 & = 0 \\ (4\alpha + 5)(4\alpha + 17) & = 0 \end{aligned}$

$\begin{cases} \alpha = - \frac 54 & (1): & 9 + 4\left(-\frac 54\right) = -4b & \implies \color{#D61F06} b = -1 & \color{#D61F06} \text{negative, rejected.} \\ \alpha = - \frac {17}4 & (1): & 9 + 4\left(-\frac {17}4\right) = -4b & \implies \color{#3D99F6} b = \boxed{2} & \color{#3D99F6} \text{positive, accepted.} \end{cases}$

let the roots of the first eqn be $a_1, b_1$ and second one be $a_2,b_2$ . we see that $(a_1-b_1)^2=(\cos^4(\theta)-\sin^4(\theta))^2=((\cos^2(\theta)-\sin^2(\theta))(\cos^2(\theta)+\sin^2(\theta)))^2=(\cos^2(\theta)-\sin^2(\theta))^2=(a_2-b_2)^2$ we can write this as $(a_1+b_1)^2-4a_1 b_1=(a_2+b_2)^2-4a_2b_2$ using vieta's formula we have $(-2b)^2-4b=(-4)^2-4*2\to b^2-b-2=0\to b= \boxed{2},-1$