F(x)=left| {begin{array}{*{20}{c}} {{{sin }^2}x}&{ - 2 + {{cos }^2}x}&{cos 2x} {2 + {{sin }^

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3 and 4 .Determinants and Matrices

hard

$$f(x)=\left| {\begin{array}{*{20}{c}} {{{\sin }^2}x}&{ - 2 + {{\cos }^2}x}&{\cos 2x} \\ {2 + {{\sin }^2}x}&{{{\cos }^2}x}&{\cos 2x} \\ {{{\sin }^2}x}&{{{\cos }^2}x}&{1 + \cos 2x} \end{array}} \right| ,x \in[0, \pi]$$

Then the maximum value of $f(x)$ is equal to $.....$

$6$

$7$

$8$

$9$

(JEE MAIN-2021)

Solution

$\left| {\begin{array}{*{20}{c}} { – 2}&{ – 2}&0 \\ 2&0&{ – 1} \\ {{{\sin }^2}x}&{{{\cos }^2}x}&{1 + \cos 2x} \end{array}} \right| (\mathrm{R}_{1} \rightarrow \mathrm{R}_{1}-\mathrm{R}_{2} \,and \,\mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{3})$

$-2\left(\cos ^{2} \mathrm{x}\right)+2\left(2+2 \cos 2 \mathrm{x}+\sin ^{2} \mathrm{x}\right)$

$4+4 \cos 2 \mathrm{x}-2\left(\cos ^{2} \mathrm{x}-\sin ^{2} \mathrm{x}\right)$

$\mathrm{f}(\mathrm{x})=4+\underbrace{2 \cos 2 \mathrm{x}}_{\max =1}$

$\mathrm{f}(\mathrm{x})_{\max }=4+2=6$

Standard 12

Mathematics

Using properties of determinants, prove that:

$\left|\begin{array}{lll}x & x^{2} & 1+p x^{3} \\ y & y^{2} & 1+p y^{3} \\ z & z^{2} & 1+p z^{3}\end{array}\right|=(1+p x y z)(x-y)(y-z)(z-x),$ where $p$ is any scalar.

hard

The value of the determinant $\left| {\,\begin{array}{*{20}{c}}{31}&{37}&{92}\\{31}&{58}&{71}\\{31}&{105}&{24}\end{array}\,} \right|$ is

easy

$\left| {\,\begin{array}{*{20}{c}}{a – b – c}&{2a}&{2a}\\{2b}&{b – c – a}&{2b}\\{2c}&{2c}&{c – a – b}\end{array}\,} \right| = $

hard

Show that $\left|\begin{array}{ccc}1+a & 1 & 1 \\ 1 & 1+b & 1 \\ 1 & 1 & 1+c\end{array}\right|=a b c\left(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=a b c+b c+c a+a b$

hard

If $a^2 + b^2 + c^2 = – 2$ and $f (x) = $ $\left| {\,\begin{array}{*{20}{c}}{1 + {a^2}x}&{(1 + {b^2})x}&{(1 + {c^2})x}\\{(1 + {a^2})x}&{1 + {b^2}x}&{(1 + {c^2})x}\\ {(1 + {a^2})x}&{(1 + {b^2})x}&{1 + {c^2}x}\end{array}\,} \right|$ then $f (x)$ is a polynomial of degree

normal

$$f(x)=\left| {\begin{array}{*{20}{c}} {{{\sin }^2}x}&{ - 2 + {{\cos }^2}x}&{\cos 2x} \\ {2 + {{\sin }^2}x}&{{{\cos }^2}x}&{\cos 2x} \\ {{{\sin }^2}x}&{{{\cos }^2}x}&{1 + \cos 2x} \end{array}} \right| ,x \in[0, \pi]$$ Then the maximum value of $f(x)$ is equal to $.....$

Solution

Similar Questions

Using properties of determinants, prove that: $\left|\begin{array}{lll}x & x^{2} & 1+p x^{3} \\ y & y^{2} & 1+p y^{3} \\ z & z^{2} & 1+p z^{3}\end{array}\right|=(1+p x y z)(x-y)(y-z)(z-x),$ where $p$ is any scalar.

The value of the determinant $\left| {\,\begin{array}{*{20}{c}}{31}&{37}&{92}\\{31}&{58}&{71}\\{31}&{105}&{24}\end{array}\,} \right|$ is

$\left| {\,\begin{array}{*{20}{c}}{a – b – c}&{2a}&{2a}\\{2b}&{b – c – a}&{2b}\\{2c}&{2c}&{c – a – b}\end{array}\,} \right| = $

Show that $\left|\begin{array}{ccc}1+a & 1 & 1 \\ 1 & 1+b & 1 \\ 1 & 1 & 1+c\end{array}\right|=a b c\left(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=a b c+b c+c a+a b$

If $a^2 + b^2 + c^2 = – 2$ and $f (x) = $ $\left| {\,\begin{array}{*{20}{c}}{1 + {a^2}x}&{(1 + {b^2})x}&{(1 + {c^2})x}\\{(1 + {a^2})x}&{1 + {b^2}x}&{(1 + {c^2})x}\\ {(1 + {a^2})x}&{(1 + {b^2})x}&{1 + {c^2}x}\end{array}\,} \right|$ then $f (x)$ is a polynomial of degree

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$$f(x)=\left| {\begin{array}{*{20}{c}} {{{\sin }^2}x}&{ - 2 + {{\cos }^2}x}&{\cos 2x} \\ {2 + {{\sin }^2}x}&{{{\cos }^2}x}&{\cos 2x} \\ {{{\sin }^2}x}&{{{\cos }^2}x}&{1 + \cos 2x} \end{array}} \right| ,x \in[0, \pi]$$

Then the maximum value of $f(x)$ is equal to $.....$

Using properties of determinants, prove that:

$\left|\begin{array}{lll}x & x^{2} & 1+p x^{3} \\ y & y^{2} & 1+p y^{3} \\ z & z^{2} & 1+p z^{3}\end{array}\right|=(1+p x y z)(x-y)(y-z)(z-x),$ where $p$ is any scalar.