If F(x) = \left[\begin{array}{ccc}\cos x & -\sin x & ...

CBSE Class 12 Math 2024 All Sets Solved Paper

Question : 17 of 20

Marks: +1, -0

F(x) = {[\table \cos x , -\sin x,0;\sin x,\cos x,0;0,0,1 ]}

and

[F(x)]^2 = F(kx)

, then the value of k is

Solution:

{[ {\F}(x)]^2= {\F}(kx)}

{{[\table \cos x , -\sin x , 0 ; \sin x , \cos x , 0 ; 0 , 0 , 1 ]}{[\table \cos x , -\sin x , 0 ; \sin x , \cos x , 0 ; 0 , 0 , 1 ]}={[\table \cos k x , -\sin k x , 0 ; \sin k x , \cos k x , 0 ; 0 , 0 , 1 ]}}

{{[\table \cos ^2 x-\sin ^2 x , -\cos x \sin x-\sin x \cos x , 0 ; \sin x \cos x+\cos x \sin x , -\sin ^2 x+\cos ^2 x , 0 ; 0 , 0 , 1 ]}={[\table \cos k x , -\sin k x , 0 ; \sin k x , \cos k x , 0 ; 0 , 0 , 1 ]}}

{{[\table \cos 2 x , -2 \sin x \cos x , 0 ; 2 \sin x \cos x , \cos 2 x , 0 ; 0 , 0 , 1 ]}={[\table \cos k x , -\sin k x , 0 ; \sin k x , \cos k x , 0 ; 0 , 0 , 1 ]}}

\text "(by using formula" \cos 2x = \cos^2 x - \sin^2 x )

{{[\table \cos 2 x , -\sin 2 x , 0 ; \sin 2 x , \cos 2 x , 0 ; 0 , 0 , 1 ]}={[\table \cos k x , -\sin k x , 0 ; \sin k x , \cos k x , 0 ; 0 , 0 , 1 ]}}

\text "(by using formula" \sin 2 x=2 \sin x \cos x )

The above matrix equation, we get

\cos 2x = \cos kx

and

\sin 2x = \sin kx

⇒ 2x = kx

⇒ k = 2

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