Question

# The equation $$\sin ^{ 4 }{ x } -2\cos ^{ 2 }{ x } +{ a }^{ 2 }=0$$ is solvable if

A
3a3
B
2a2
C
1a1
D
None of these

Solution

## The correct option is D $$-\sqrt{2}\ge a\ge\sqrt{2}$$$$\sin ^{ 4 }{ x } -2\sin ^{ 2 }{ x } +{ a }^{ 2 }-2=0$$If $$\sin ^{ 2 }{ x } =y,$$ then the equation reduce to $$y^2+2y+a^2-2=0$$Now, $$\Delta =4-4\left( { a }^{ 2 }-2 \right) \ge 0\Rightarrow 3-{ a }^{ 2 }\ge 0\Rightarrow { a }^{ 2 }\le 3\Rightarrow y=\sin ^{ 2 }{ x } =-1\pm \sqrt { 3-{ a }^{ 2 } }$$    ...(1)Now, $$y=\sin ^{ 2 }{ x }$$ is positive$$\therefore 0\le y\le 1\Rightarrow 0\le -1\pm \sqrt { 3-{ a }^{ 2 } } \le 1\Rightarrow 1\le \pm \sqrt { 3-{ a }^{ 2 } } \le 2$$$$\Rightarrow 1\le 3-{ a }^{ 2 }\le 4\le -2\le -{ a }^{ 2 }\le 1\Rightarrow { a }^{ 2 }-2\le 0$$ and $$a^2+1\ge 0$$    ...(2)Using (1) and (2), we have$${ a }^{ 2 }-2\le 0\Rightarrow \left( a+\sqrt { 2 } \right) \left( a-\sqrt { 2 } \right) \le 0\Rightarrow -\sqrt { 2 } \le a\le \sqrt { 2 }$$Maths

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