JEE MAIN - Mathematics (2021 - 27th July Morning Shift - No. 21)
Let $$f(x) = \left| {\matrix{
{{{\sin }^2}x} & { - 2 + {{\cos }^2}x} & {\cos 2x} \cr
{2 + {{\sin }^2}x} & {{{\cos }^2}x} & {\cos 2x} \cr
{{{\sin }^2}x} & {{{\cos }^2}x} & {1 + \cos 2x} \cr
} } \right|,x \in [0,\pi ]$$. Then the maximum value of f(x) is equal to ______________.
Answer
6
Explanation
$$\left| {\matrix{
{ - 2} & { - 2} & 0 \cr
2 & 0 & { - 1} \cr
{{{\sin }^2}x} & {{{\cos }^2}x} & {1 + \cos 2x} \cr
} } \right|\left( \matrix{
{R_1} \to {R_1} - {R_2} \hfill \cr
\& \,{R_2} \to {R_2} - {R_3} \hfill \cr} \right)$$
= $$ - 2({\cos ^2}x) + 2(2 + 2\cos 2x + {\sin ^2}x)$$
= $$4 + 4\cos 2x - 2({\cos ^2}x - {\sin ^2}x)$$
$$ \therefore $$ $$f(x) = 4 + \underbrace {2\cos 2x}_{\max = 1}$$
$$ \Rightarrow $$ $$f{(x)_{\max }} = 4 + 2 = 6$$
= $$ - 2({\cos ^2}x) + 2(2 + 2\cos 2x + {\sin ^2}x)$$
= $$4 + 4\cos 2x - 2({\cos ^2}x - {\sin ^2}x)$$
$$ \therefore $$ $$f(x) = 4 + \underbrace {2\cos 2x}_{\max = 1}$$
$$ \Rightarrow $$ $$f{(x)_{\max }} = 4 + 2 = 6$$
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