NIMCET 2009 Question Paper with solutions

Question : 28 of 120

Marks: +1, -0

The equation

\sin ^{4} x+\cos ^{4} x+\sin 2 x+\alpha=0

is solvable for

Solution:

\sin ^{4} x+\cos ^{4} x+\sin 2 x+α=0

⇒(\sin ^{2} x+\cos ^{2} x)^{2}-2 \sin ^{2} x \cos ^{2} x+\sin 2 x+α=0

⇒ \;\; 1-\;{(\sin 2 x)^{2}}/{2}+\sin 2 x+α=0

⇒ \;\; (\sin 2 x)^{2}-2 \sin 2 x-2(1+α)=0

⇒ \;\; \sin 2 x=\;{2 ± √{4+8(1+α)}}/{2}

=1 ± √{3+2 α}

So, if

|1 ± √{3+2 α}| ≤ 1 \;\; ( ∵ -1 ≤ \sin x ≤ 1)

-1 ≤ 1 ± √{3+2 α} ≤ 1

-2 ≤ ± √{3+2 α} ≤ 0

0 ≤ 3+2 α ≤ 4

-3 ≤ 2 α ≤ 1

-\;{3}/{2} ≤ α ≤ \;{1}/{2}

⇒ \;\; -\;{3}/{2} ≤ α ≤ \;{1}/{2}

Then, the given equation has solution.

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