If the minimum and the maximum values of the function $f:\left[ {{\pi \over 4},{\pi \over 2}} \right] \to R$, defined by
$$f\left( \theta \right) = \left| {\matrix{
{ - {{\sin }^2}\theta } & { - 1 - {{\sin }^2}\theta } & 1 \cr
{ - {{\cos }^2}\theta } & { - 1 - {{\cos }^2}\theta } & 1 \cr
{12} & {10} & { - 2} \cr
} } \right|$$ are m and M respectively, then the ordered pair (m,M) is
equal to :
Solution
Given <br>$$f\left( \theta \right) = \left| {\matrix{
{ - {{\sin }^2}\theta } & { - 1 - {{\sin }^2}\theta } & 1 \cr
{ - {{\cos }^2}\theta } & { - 1 - {{\cos }^2}\theta } & 1 \cr
{12} & {10} & { - 2} \cr
} } \right|$$
<br><br>C<sub>1</sub> $\to$ C<sub>1</sub>
– C<sub>2</sub>
, C<sub>3</sub> $\to$ C<sub>3</sub>
+ C<sub>2</sub>
<br><br>= $$\left| {\matrix{
1 & { - 1 - {{\sin }^2}\theta } & { - {{\sin }^2}\theta } \cr
1 & { - 1 - {{\cos }^2}\theta } & { - {{\cos }^2}\theta } \cr
2 & {10} & 8 \cr
} } \right|$$
<br><br>C<sub>2</sub> $\to$ C<sub>2</sub>
– C<sub>3</sub>
<br><br>= $$\left| {\matrix{
1 & { - 1} & { - {{\sin }^2}\theta } \cr
1 & { - 1} & { - {{\cos }^2}\theta } \cr
2 & 2 & 8 \cr
} } \right|$$
<br><br>= 1(2cos<sup>2</sup>$\theta$ – 8) + (8 + 2cos<sup>2</sup>$\theta$) – 4sin<sup>2</sup>$\theta$
<br><br>= 4cos<sup>2</sup>$\theta$ - 4cos<sup>2</sup>$\theta$
<br><br>= 4 cos 2$\theta$
<br><br>$\theta$ $\in$ $\left[ {{\pi \over 4},{\pi \over 2}} \right]$
<br><br>$\Rightarrow$ 2$\theta$ $\in$ $\left[ {{\pi \over 2},{\pi }} \right]$
<br><br>$\Rightarrow$ cos 2$\theta$ $\in$ [-1, 0]
<br><br>$\Rightarrow$ 4cos 2$\theta$ $\in$ [-4, 0]
<br><br>$\Rightarrow$ $f\left( \theta \right)$ $\in$ [-4, 0]
<br><br>$\therefore$ (m, M) = (–4, 0)
About this question
Subject: Mathematics · Chapter: Matrices and Determinants · Topic: Types of Matrices and Operations
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