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  • Please,please help me - Matrices and Determinants - JEE Main

The number of distinct real roots of the

equation,in the

interval \left [ -\frac{\pi }{4} ,\frac{\pi }{4}\right ] is

 

  • Option 1)

    4

  • Option 2)

    3

  • Option 3)

    2

  • Option 4)

    1

 

Answers (1)

best_answer

As we leant in

Cramer's rule for solving system of linear equations -

When \Delta =0  and \Delta _{1}=\Delta _{2}=\Delta _{3}=0 ,

then  the system of equations has infinite solutions.

- wherein

a_{1}x+b_{1}y+c_{1}z=d_{1}

a_{2}x+b_{2}y+c_{2}z=d_{2}

a_{3}x+b_{3}y+c_{3}z=d_{3}

and 

\Delta =\begin{vmatrix} a_{1} &b_{1} &c_{1} \\ a_{2} & b_{2} &c_{2} \\ a_{3}&b _{3} & c_{3} \end{vmatrix}

\Delta _{1},\Delta _{2},\Delta _{3} are obtained by replacing column 1,2,3 of \Delta by \left ( d_{1},d_{2},d_{3} \right )  column

 

 \begin{vmatrix} \cos x & \sin x & \sin x \\ \sin x & \cos x & \sin x \\ \sin x & \sin x & \cos x \end{vmatrix}=0

C1\rightarrow C1+C2+C3

\begin{vmatrix} \cos x +2 \sin x & \sin x & \sin x \\ \cos x + 2\sin x & \cos x & \sin x\\ \cos x + 2\sin x & \sin x & \cos x \end{vmatrix} = 0

\Rightarrow (\cos x + 2\sin x )\begin{vmatrix} 1 & \sin x & \sin x \\ 1 & \cos x & \sin x\\ 1 & \sin x & \cos x \end{vmatrix} = 0

\Rightarrow (\cos x + 2\sin x )\begin{vmatrix} 0 & \sin x- \cos x & 0 \\ 0 & \cos x- \sin x & \sin x-cos x\\ 1 & \sin x & \cos x \end{vmatrix} = 0

\Rightarrow (\cos x + 2\sin x ) (\cos x - \sin x )^{2}

\\ \Rightarrow \tan x = -\frac{1}{2} \ and\ \tan x = 1 \\ so \ two \ solution \left [ \frac{-\pi }{4}, \frac{\pi }{4} \right ]

 


Option 1)

4

This option is incorrect.

Option 2)

3

This option is incorrect.

Option 3)

2

This option is correct.

Option 4)

1

This option is incorrect.

Posted by

divya.saini

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