CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

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


A
3a3
loader
B
2a2
loader
C
1a1
loader
D
None of these
loader

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

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image