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If \(2 x^2+(\cos \theta) x-1=0, \theta \in[0,2 \pi]\) has roots \(\alpha\) and \(\beta\). Then the sum of maximum and mi…

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If \(2 x^2+(\cos \theta) x-1=0, \theta \in[0,2 \pi]\) has roots \(\alpha\) and \(\beta\). Then the sum of maximum and minimum value of \(\alpha^4+\beta^4\) is

[JEE Main 2025]

a

\(\frac{25}{16}\)

b

\(\frac{9}{16}\)

c

\(\frac{41}{16}\)

d

\(\frac{8}{17}\)

✓ Correct answer: a)

\(\frac{25}{16}\)

Explanation


\(\begin{aligned}
& 2 x^2+(\cos \theta) x-1=0 \\
& \alpha+\beta=\frac{-\cos \theta}{2}; \quad \alpha \beta=-\frac{1}{2} \\
& \alpha^2+\beta^2=(\alpha+\beta)^2-2 \alpha \beta =\frac{\cos ^2 \theta}{4}+1 \\
& \alpha^4+\beta^4=\left(\alpha^2+\beta^2\right)^2-2 \alpha^2 \beta^2=\left(\frac{\cos ^2 \theta}{4}+1\right)^2-\frac{2}{4} \\
& \alpha^4+\beta^4=\left(\frac{\cos ^2 \theta}{4}+1\right)^2-\frac{1}{2}
\end{aligned}
\)


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 \(\cos \theta=0\)

\(
m=1-\frac{1}{2}=\frac{1}{2}
\)

\(\left.\therefore \quad 16(M+m)=16 (\frac{17}{16}+\frac{1}{2}\right)=25\)

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