JEE MAIN - Mathematics (2021 - 16th March Evening Shift - No. 17)
The maximum value of
$$f(x) = \left| {\matrix{ {{{\sin }^2}x} & {1 + {{\cos }^2}x} & {\cos 2x} \cr {1 + {{\sin }^2}x} & {{{\cos }^2}x} & {\cos 2x} \cr {{{\sin }^2}x} & {{{\cos }^2}x} & {\sin 2x} \cr } } \right|,x \in R$$ is :
$$f(x) = \left| {\matrix{ {{{\sin }^2}x} & {1 + {{\cos }^2}x} & {\cos 2x} \cr {1 + {{\sin }^2}x} & {{{\cos }^2}x} & {\cos 2x} \cr {{{\sin }^2}x} & {{{\cos }^2}x} & {\sin 2x} \cr } } \right|,x \in R$$ is :
$$\sqrt 5 $$
$${3 \over 4}$$
5
$$\sqrt 7 $$
Explanation
$$f(x) = \left| {\matrix{
{{{\sin }^2}x} & {1 + {{\cos }^2}x} & {\cos 2x} \cr
{1 + {{\sin }^2}x} & {{{\cos }^2}x} & {\cos 2x} \cr
{{{\sin }^2}x} & {{{\cos }^2}x} & {\sin 2x} \cr
} } \right|$$
$${C_1} \to {C_1} + {C_2}$$
= $$\left| {\matrix{ 2 & {1 + {{\cos }^2}x} & {\cos 2x} \cr 2 & {{{\cos }^2}x} & {\cos 2x} \cr 1 & {{{\cos }^2}x} & {\sin 2x} \cr } } \right|$$
$${R_1} \to {R_1} - {R_2}$$
= $$\left| {\matrix{ 0 & 1 & 0 \cr 2 & {{{\cos }^2}x} & {\cos 2x} \cr 1 & {{{\cos }^2}x} & {\sin 2x} \cr } } \right|$$
$$ = ( - 1)[2\sin 2x - \cos 2x] = \cos 2x - 2\sin 2x$$
We know, maximum value of acosx $$ \pm $$ bsinx
= $$\sqrt {{a^2} + {b^2}} $$
$$ \therefore $$ Here maximum value = $$\sqrt {{1^2} + {{\left( { - 2} \right)}^2}} $$$$ = \sqrt 5 $$
$${C_1} \to {C_1} + {C_2}$$
= $$\left| {\matrix{ 2 & {1 + {{\cos }^2}x} & {\cos 2x} \cr 2 & {{{\cos }^2}x} & {\cos 2x} \cr 1 & {{{\cos }^2}x} & {\sin 2x} \cr } } \right|$$
$${R_1} \to {R_1} - {R_2}$$
= $$\left| {\matrix{ 0 & 1 & 0 \cr 2 & {{{\cos }^2}x} & {\cos 2x} \cr 1 & {{{\cos }^2}x} & {\sin 2x} \cr } } \right|$$
$$ = ( - 1)[2\sin 2x - \cos 2x] = \cos 2x - 2\sin 2x$$
We know, maximum value of acosx $$ \pm $$ bsinx
= $$\sqrt {{a^2} + {b^2}} $$
$$ \therefore $$ Here maximum value = $$\sqrt {{1^2} + {{\left( { - 2} \right)}^2}} $$$$ = \sqrt 5 $$
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