JEE MAIN - Mathematics (2019 - 12th January Evening Slot - No. 15)
If A = $$\left[ {\matrix{
1 & {\sin \theta } & 1 \cr
{ - \sin \theta } & 1 & {\sin \theta } \cr
{ - 1} & { - \sin \theta } & 1 \cr
} } \right]$$;
then for all $$\theta $$ $$ \in $$ $$\left( {{{3\pi } \over 4},{{5\pi } \over 4}} \right)$$, det (A) lies in the interval :
then for all $$\theta $$ $$ \in $$ $$\left( {{{3\pi } \over 4},{{5\pi } \over 4}} \right)$$, det (A) lies in the interval :
$$\left( {{3 \over 2},3} \right]$$
$$\left( {0,{3 \over 2}} \right]$$
$$\left[ {{5 \over 2},4} \right)$$
$$\left( {1,{5 \over 2}} \right]$$
Explanation
$$\left| A \right| = \left| {\matrix{
1 & {\sin \theta } & 1 \cr
{ - \sin \theta } & 1 & {\sin \theta } \cr
{ - 1} & { - \sin \theta } & 1 \cr
} } \right|$$
= 2(1 + sin2$$\theta $$)
$$\theta $$ $$ \in $$ $$\left( {{{3\pi } \over 4},{{5\pi } \over 4}} \right) \Rightarrow {1 \over {\sqrt 2 }} < \sin \theta < {1 \over {\sqrt 2 }}$$
$$ \Rightarrow $$ 0 $$ \le $$ sin2$$\theta $$ < $${1 \over 2}$$
$$ \therefore $$ $$\left| A \right| \in \left[ {2,3} \right)$$
= 2(1 + sin2$$\theta $$)
$$\theta $$ $$ \in $$ $$\left( {{{3\pi } \over 4},{{5\pi } \over 4}} \right) \Rightarrow {1 \over {\sqrt 2 }} < \sin \theta < {1 \over {\sqrt 2 }}$$
$$ \Rightarrow $$ 0 $$ \le $$ sin2$$\theta $$ < $${1 \over 2}$$
$$ \therefore $$ $$\left| A \right| \in \left[ {2,3} \right)$$
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