JEE Main 2021 — Application Of Derivatives Question with Solution
From: JEE Main 2021 (Online) 16th March Evening Shift
Question
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 :
Choose an option
Show full solutionCorrect option: A
Correct answer
A
Step-by-step 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|
= \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|
= \left| {\matrix{ 0 & 1 & 0 \cr 2 & {{{\cos }^2}x} & {\cos 2x} \cr 1 & {{{\cos }^2}x} & {\sin 2x} \cr } } \right|
We know, maximum value of acosx bsinx
=
Here maximum value =
= \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|
= \left| {\matrix{ 0 & 1 & 0 \cr 2 & {{{\cos }^2}x} & {\cos 2x} \cr 1 & {{{\cos }^2}x} & {\sin 2x} \cr } } \right|
We know, maximum value of acosx bsinx
=
Here maximum value =
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This is a previous-year question from JEE Main 2021, covering the Application Of Derivatives chapter of Mathematics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.