Let α_θ and β_θ be the distinct roots of 2 x^2+( θ) x-1=0, θ ∈(0,2 π) . If m and M are the minimum and the maximum values of α_θ^4+β_θ^4 , then 16(M+m) equals :

Correct answer: B. 25 aligned & 2 x^2+( θ) x-1=0 & α_θ+β_θ=(- θ)/(2) & α_θ · β_θ=(-1)/(2) & α_θ^2+β_θ^2=(α_θ+β_θ)^2-2 α_θ β_θ ( ^2 θ)/(4)+1 aligned aligned & α_θ^4+β_θ^4=(a_θ^2+β_θ^2)^2-2 α_θ^2 β_θ^2=(( ^2 θ)/(4)+1)^2-(2)/(4) & =(( ^2 θ)/(4)+1)^2-(1)/(2) aligned Maximum when θ=1 aligned & M=((1)/(4)+1)^2-(1)/(2) & M=(17)/(16) aligned Minimum when θ=0 aligned & m=1-(1)/(2)=(1)/(2) & 16(M+m)=16((17)/(16)+(1)/(2))=25 aligned

JEE Main 202522 JanuaryEvening Shift

Exam: JEE Main
Year: 2025
Date: 22 January
Shift: Evening Shift
Subject: Mathematics
Chapter: Quadratic Equation

JEE Main 2025 · 22 January · Evening Shift · Mathematics · Quadratic Equation

Let α_θ and β_θ be the distinct roots of 2 x^2+( θ) x-1=0, θ ∈(0,2...

Let

\alpha_\theta

and

\beta_\theta

be the distinct roots of

2 x^2+(\cos \theta) x-1=0, \theta \in(0,2 \pi)

. If m and M are the minimum and the maximum values of

\alpha_\theta^4+\beta_\theta^4

, then

16(M+m)

equals :

Options

Correct answer

B. 25

Step-by-step solution

\begin{aligned} & 2 x^2+(\cos \theta) x-1=0 \\ & \alpha_\theta+\beta_\theta=\frac{-\cos \theta}{2} \\ & \alpha_\theta \cdot \beta_\theta=\frac{-1}{2} \\ & \alpha_\theta^2+\beta_\theta^2=\left(\alpha_\theta+\beta_\theta\right)^2-2 \alpha_\theta \beta_\theta \frac{\cos ^2 \theta}{4}+1\end{aligned}

\begin{aligned} & \alpha_\theta^4+\beta_\theta^4=\left(a_\theta^2+\beta_\theta^2\right)^2-2 \alpha_\theta^2 \beta_\theta^2=\left(\frac{\cos ^2 \theta}{4}+1\right)^2-\frac{2}{4} \\ & =\left(\frac{\cos ^2 \theta}{4}+1\right)^2-\frac{1}{2} \end{aligned}

Maximum when

\cos \theta=1

\begin{aligned} & M=\left(\frac{1}{4}+1\right)^2-\frac{1}{2} \\ & M=\frac{17}{16} \end{aligned}

Minimum when

\cos \theta=0

\begin{aligned} & m=1-\frac{1}{2}=\frac{1}{2} \\ & 16(M+m)=16\left(\frac{17}{16}+\frac{1}{2}\right)=25 \end{aligned}

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