JEE MAIN - Mathematics (2020 - 6th September Morning Slot - No. 18)
Let m and M be respectively the minimum and maximum values of
$$\left| {\matrix{ {{{\cos }^2}x} & {1 + {{\sin }^2}x} & {\sin 2x} \cr {1 + {{\cos }^2}x} & {{{\sin }^2}x} & {\sin 2x} \cr {{{\cos }^2}x} & {{{\sin }^2}x} & {1 + \sin 2x} \cr } } \right|$$
Then the ordered pair (m, M) is equal to :
$$\left| {\matrix{ {{{\cos }^2}x} & {1 + {{\sin }^2}x} & {\sin 2x} \cr {1 + {{\cos }^2}x} & {{{\sin }^2}x} & {\sin 2x} \cr {{{\cos }^2}x} & {{{\sin }^2}x} & {1 + \sin 2x} \cr } } \right|$$
Then the ordered pair (m, M) is equal to :
(–3, –1)
(–4, –1)
(1, 3)
(–3, 3)
Explanation
$$\left| {\matrix{
{{{\cos }^2}x} & {1 + {{\sin }^2}x} & {\sin 2x} \cr
{1 + {{\cos }^2}x} & {{{\sin }^2}x} & {\sin 2x} \cr
{{{\cos }^2}x} & {{{\sin }^2}x} & {1 + \sin 2x} \cr
} } \right|$$
R1 $$ \to $$ R1 – R2, R2 $$ \to $$ R2 – R3
$$\left| {\matrix{ { - 1} & 1 & 0 \cr 1 & 0 & { - 1} \cr {{{\cos }^2}x} & {{{\sin }^2}x} & {1 + \sin 2x} \cr } } \right|$$
= –1(sin2 x) – 1(1 + sin2x + cos2 x)
= - sin2x - 2
$$ \therefore $$ minimum value when sin2x = 1
m = - 2 - 1 = -3
$$ \therefore $$ Maximum value when sin2x = –1
M = -2 + 1 = -1
$$ \therefore $$ (m, M) = (–3, –1)
R1 $$ \to $$ R1 – R2, R2 $$ \to $$ R2 – R3
$$\left| {\matrix{ { - 1} & 1 & 0 \cr 1 & 0 & { - 1} \cr {{{\cos }^2}x} & {{{\sin }^2}x} & {1 + \sin 2x} \cr } } \right|$$
= –1(sin2 x) – 1(1 + sin2x + cos2 x)
= - sin2x - 2
$$ \therefore $$ minimum value when sin2x = 1
m = - 2 - 1 = -3
$$ \therefore $$ Maximum value when sin2x = –1
M = -2 + 1 = -1
$$ \therefore $$ (m, M) = (–3, –1)
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