JEE MAIN - Mathematics (2020 - 5th September Morning Slot - No. 10)
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 :
$$f\left( \theta \right) = \left| {\matrix{ { - {{\sin }^2}\theta } & { - 1 - {{\sin }^2}\theta } & 1 \cr { - {{\cos }^2}\theta } & { - 1 - {{\cos }^2}\theta } & 1 \cr {12} & {10} & { - 2} \cr } } \right|$$ are m and M respectively, then the ordered pair (m,M) is equal to :
$$\left( {0,2\sqrt 2 } \right)$$
(-4, 0)
(-4, 4)
(0, 4)
Explanation
Given
$$f\left( \theta \right) = \left| {\matrix{ { - {{\sin }^2}\theta } & { - 1 - {{\sin }^2}\theta } & 1 \cr { - {{\cos }^2}\theta } & { - 1 - {{\cos }^2}\theta } & 1 \cr {12} & {10} & { - 2} \cr } } \right|$$
C1 $$ \to $$ C1 – C2 , C3 $$ \to $$ C3 + C2
= $$\left| {\matrix{ 1 & { - 1 - {{\sin }^2}\theta } & { - {{\sin }^2}\theta } \cr 1 & { - 1 - {{\cos }^2}\theta } & { - {{\cos }^2}\theta } \cr 2 & {10} & 8 \cr } } \right|$$
C2 $$ \to $$ C2 – C3
= $$\left| {\matrix{ 1 & { - 1} & { - {{\sin }^2}\theta } \cr 1 & { - 1} & { - {{\cos }^2}\theta } \cr 2 & 2 & 8 \cr } } \right|$$
= 1(2cos2$$\theta $$ – 8) + (8 + 2cos2$$\theta $$) – 4sin2$$\theta $$
= 4cos2$$\theta $$ - 4cos2$$\theta $$
= 4 cos 2$$\theta $$
$$\theta $$ $$ \in $$ $$\left[ {{\pi \over 4},{\pi \over 2}} \right]$$
$$ \Rightarrow $$ 2$$\theta $$ $$ \in $$ $$\left[ {{\pi \over 2},{\pi }} \right]$$
$$ \Rightarrow $$ cos 2$$\theta $$ $$ \in $$ [-1, 0]
$$ \Rightarrow $$ 4cos 2$$\theta $$ $$ \in $$ [-4, 0]
$$ \Rightarrow $$ $$f\left( \theta \right)$$ $$ \in $$ [-4, 0]
$$ \therefore $$ (m, M) = (–4, 0)
$$f\left( \theta \right) = \left| {\matrix{ { - {{\sin }^2}\theta } & { - 1 - {{\sin }^2}\theta } & 1 \cr { - {{\cos }^2}\theta } & { - 1 - {{\cos }^2}\theta } & 1 \cr {12} & {10} & { - 2} \cr } } \right|$$
C1 $$ \to $$ C1 – C2 , C3 $$ \to $$ C3 + C2
= $$\left| {\matrix{ 1 & { - 1 - {{\sin }^2}\theta } & { - {{\sin }^2}\theta } \cr 1 & { - 1 - {{\cos }^2}\theta } & { - {{\cos }^2}\theta } \cr 2 & {10} & 8 \cr } } \right|$$
C2 $$ \to $$ C2 – C3
= $$\left| {\matrix{ 1 & { - 1} & { - {{\sin }^2}\theta } \cr 1 & { - 1} & { - {{\cos }^2}\theta } \cr 2 & 2 & 8 \cr } } \right|$$
= 1(2cos2$$\theta $$ – 8) + (8 + 2cos2$$\theta $$) – 4sin2$$\theta $$
= 4cos2$$\theta $$ - 4cos2$$\theta $$
= 4 cos 2$$\theta $$
$$\theta $$ $$ \in $$ $$\left[ {{\pi \over 4},{\pi \over 2}} \right]$$
$$ \Rightarrow $$ 2$$\theta $$ $$ \in $$ $$\left[ {{\pi \over 2},{\pi }} \right]$$
$$ \Rightarrow $$ cos 2$$\theta $$ $$ \in $$ [-1, 0]
$$ \Rightarrow $$ 4cos 2$$\theta $$ $$ \in $$ [-4, 0]
$$ \Rightarrow $$ $$f\left( \theta \right)$$ $$ \in $$ [-4, 0]
$$ \therefore $$ (m, M) = (–4, 0)
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