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Study Stuff for MCA Examinations - NIMCET
Study Stuff for MCA Examinations - NIMCET
Previous Year Question (PYQs)
The equation $\sin^4 x + \cos^4 x + \sin 2x + \alpha = 0$ is solvable for:
Solution
Simplify: $\sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x$$= 1 - \dfrac{1}{2}\sin^2(2x)$
So equation becomes:
$1 - \dfrac{1}{2}\sin^2(2x) + \sin(2x) + \alpha = 0$
Let $t = \sin(2x)$, where $t \in [-1,1]$
Expression becomes:
$1 + \alpha + t - \dfrac{t^2}{2} = 0$
Multiply by 2:
$t^2 - 2t - 2(1+\alpha) = 0$
For real $t$ in $[-1,1]$, discriminant must allow roots inside interval.
Solving gives the range:
${-\dfrac{3}{2} \le \alpha \le \dfrac{1}{2}}$
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