Question

If maximum and minimum values of the determinant $\left|\begin{array}{ccc}1+{\sin }^{2}x& {\cos }^{2}x& \sin 2x\\ {\sin }^{2}x& 1+{\cos }^{2}x& \sin 2x\\ {\sin }^{2}x& {\cos }^{2}x& 1+\sin 2x\end{array}\right|$ are $\alpha$ and $\beta$, then

• A. $\alpha +{\beta }^{99}=4$
• B. ${\alpha }^3-{\beta }^{17}=26$
• C. $({\alpha }^{2n}-{\beta }^{2n})$ is always an even integer for $n\in N$
• D. a triangle can be constructed having its sides as $\alpha -\beta ,\alpha +\beta$ and $\alpha +3\beta$

Answer 1

Answer: (A), (B), (C)

The correct options are
A ${\alpha }^3-{\beta }^{17}=26$
B $\alpha +{\beta }^{99}=4$
C $({\alpha }^{2n}-{\beta }^{2n})$ is always an even integer for $n\in N$

Given

$A=\left|\begin{array}{ccc}1+{\sin }^{2}x& {\cos }^{2}x& \sin 2x\\ {\sin }^{2}x& 1+{\cos }^{2}x& \sin 2x\\ {\sin }^{2}x& {\cos }^{2}x& 1+\sin 2x\end{array}\right|$

Apply $c_1\to c_1+c_2$ to get

$A=\left|\begin{array}{ccc}2& {\cos }^{2}x& \sin 2x\\ 2& 1+{\cos }^{2}x& \sin 2x\\ 1& {\cos }^{2}x& 1+\sin 2x\end{array}\right|$

Now apply $r_2\to r_2-r_1$ and $r_3\to r_3-r_1$

$A=\left|\begin{array}{ccc}2& {\cos }^{2}x& \sin 2x\\ 0& 1& 0\\ -1& 0& 1\end{array}\right|=2\sin 2x$

Since, the $\sin 2x$ has maximum value of $1$ and minimum value of $-1$

so $\alpha =3,\beta =1$

$\alpha -\beta =2,\alpha +\beta =4,\alpha =3,\beta =6$

Thus $(\alpha -\beta )+(\alpha +\beta )=\alpha +3\beta =6$

so $(\alpha -\beta ),(\alpha +\beta ),(\alpha +3\beta )$ cannot form a triangle

All other options are correct.