Statement -1 : The system of linear equations x + ( , )y + ( , )z… - Vidyadip

MCQ

Statement $-1$ : The system of linear equations

$x + \left( {\sin \,\alpha } \right)y + \left( {\cos \,\alpha } \right)z = 0$

$x + \left( {\cos \,\alpha } \right)y + \left( {\sin \alpha } \right)z = 0$

$x - \left( {\sin \,\alpha } \right)y - \left( {\cos \alpha } \right)z = 0$

has a non-trivial solution for only one value of $\alpha $ lying in the interval $\left( {0\,,\,\frac{\pi }{2}} \right)$

Statement $-2$ : The equation in $\alpha $

$\left| {\begin{array}{*{20}{c}}
{\cos {\mkern 1mu} \alpha }&{\sin {\mkern 1mu} \alpha }&{\cos {\mkern 1mu} \alpha } \\
{\sin {\mkern 1mu} \alpha }&{\cos {\mkern 1mu} \alpha }&{\sin {\mkern 1mu} \alpha } \\
{\cos {\mkern 1mu} \alpha }&{ - \sin {\mkern 1mu} \alpha }&{ - \cos {\mkern 1mu} \alpha }
\end{array}} \right| = 0$

has only one solution lying in the interval $\left( {0\,,\,\frac{\pi }{2}} \right)$

A
Statement $- 1$ is true, Statement $-2$ is true,Statement $-2$ is not correct explantion for Statement $-1$
B
Statement $-1$ is true, Statement $-2$ is true,Statement $-2$ is a correct explantion for Statement $-1$
✓
Statement $- 1$ is true, Statement $-2$ is false
D
Statememt $-1$ is false, Statement $-2$ is true.

✓

Answer

Correct option: C.

Statement $- 1$ is true, Statement $-2$ is false

c
${\Delta _1} = \left| {\begin{array}{*{20}{c}}
1&{\sin \alpha }&{\cos \alpha }\\
1&{\cos \alpha }&{\sin \alpha }\\
1&{ - \sin \alpha }&{\cos \alpha }
\end{array}} \right|$

$ = \left| {\begin{array}{*{20}{c}}
0&{\sin \alpha - \cos \alpha }&{\cos \alpha - \sin \alpha }\\
0&{\cos \alpha + \sin \alpha }&{\sin \alpha - \cos \alpha }\\
1&{ - \sin \alpha }&{\cos \alpha }
\end{array}} \right|$

$ = {\left( {\sin \alpha - \cos \alpha } \right)^2} - \left( {{{\cos }^2}\alpha - {{\sin }^2}\alpha } \right)$

$ = {\sin ^2}\alpha - {\cos ^2}\alpha - 2\sin \alpha .\cos \alpha - {\cos ^2}\alpha + {\sin ^2}\alpha $

$ = 2{\sin ^2}\alpha - 2\sin \alpha .\cos \alpha $

$ = 2\sin \alpha \left( {\sin \alpha - \cos \alpha } \right)$

Now, $\sin \alpha - \cos \alpha = 0$ for only

$\alpha = \frac{\pi }{4}$ in $\left( {0,\frac{\pi }{2}} \right)$

${\Delta _1} = 2\left( {\sin \alpha } \right) \times 0 = 0$

Since value of $\sin \alpha $ is finite for $\alpha \in \left( {0,\frac{\pi }{2}} \right)$

Hence non- trivivial solution for only one value of $\alpha $ in $\left( {0,\frac{\pi }{2}} \right)$

$\left| {\begin{array}{*{20}{c}}
{\cos \alpha }&{\sin \alpha }&{\cos \alpha }\\
{\sin \alpha }&{\cos \alpha }&{\sin \alpha }\\
{\cos \alpha }&{ - \sin \alpha }&{ - \cos \alpha }
\end{array}} \right| = 0$

$ \Rightarrow \left| {\begin{array}{*{20}{c}}
0&{\sin \alpha }&{\cos \alpha }\\
0&{\cos \alpha }&{\sin \alpha }\\
{2\cos \alpha }&{ - \sin \alpha }&{ - \cos \alpha }
\end{array}} \right| = 0$

$ \Rightarrow 2\cos \alpha \left( {{{\sin }^2}\alpha - {{\cos }^2}\alpha } \right) = 0$

$\therefore \cos \alpha = 0$ or ${\sin ^2}\alpha - {\cos ^2}\alpha = 0$

But $\cos \alpha = 0$ not possible for any value of

$\alpha \in \left( {0,\frac{\pi }{2}} \right)$

$\therefore {\sin ^2}\alpha - {\cos ^2}\alpha = 0 \Rightarrow \sin \alpha = - \cos \alpha $

which is also not possible for any value of

$\alpha \in \left( {0,\frac{\pi }{2}} \right)$

Hence, there is no solution.

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