Question

Let ∆(x) =

Answer 1

Step-by-step explanation:

Given Question is

$\sf \: If \: \triangle (x) = \begin{gathered}\sf \left | \begin{array}{ccc}1 + {sin}^{2}x & {cos}^{2}x &4sin2x\\ {sin}^{2}x &1 + {cos}^{2}x &4sin2x\\ {sin}^{2}x & {cos}^{2}x & 1 + 4sin2x\end{array}\right | \end{gathered}$

The the value of x for which above determinant is maximum, is equals to

$\sf \: \: \: \: (a) \: \: \dfrac{\pi}{2}$

$\sf \: \: \: \: (b) \: \: \dfrac{\pi}{6}$

$\sf \: \: \: \: (c) \: \: \dfrac{\pi}{3}$

$\sf \: \: \: \: (d) \: \: \dfrac{\pi}{4}$

$\red{\large\underline{\sf{Solution-}}}$

Given determinant is

$\sf \:\triangle (x) = \begin{gathered}\sf \left | \begin{array}{ccc}1 + {sin}^{2}x & {cos}^{2}x &4sin2x\\ {sin}^{2}x &1 + {cos}^{2}x &4sin2x\\ {sin}^{2}x & {cos}^{2}x & 1 + 4sin2x\end{array}\right | \end{gathered}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{\tt{ OP \: C_1 \: \to \: C_1 + C_2 + C_3 \: }}$

We get,

$\sf \:\triangle (x) = \begin{gathered}\sf \left | \begin{array}{ccc}1 + {sin}^{2}x + {cos}^{2}x + 4sin2x & {cos}^{2}x &4sin2x\\ 1 + {sin}^{2}x + {cos}^{2}x + 4sin2x &1 + {cos}^{2}x &4sin2x\\ 1 + {sin}^{2}x + {cos}^{2}x + 4sin2x & {cos}^{2}x & 1 + 4sin2x\end{array}\right | \end{gathered}$

$\sf \:\triangle (x) = 1 + {sin}^{2}x + {cos}^{2}x + 4sin2x\begin{gathered}\sf \left | \begin{array}{ccc}1 & {cos}^{2}x &4sin2x\\ 1 &1 + {cos}^{2}x &4sin2x\\ 1 & {cos}^{2}x & 1 + 4sin2x\end{array}\right | \end{gathered}$

$\boxed{\tt{ OP \: R_2 \: \to \: R_2 - R_1 \: }} \: \: and \: \: \boxed{\tt{ R_3 \: \to \: R_3 - R_1 \: }}$

We get,

$\sf \:\triangle (x) = 1 +1+ 4sin2x\begin{gathered}\sf \left | \begin{array}{ccc}1 & {cos}^{2}x &4sin2x\\ 0 &1 &0\\ 0 & 0 & 1\end{array}\right | \end{gathered}$

$\sf \:\triangle (x) = (2+ 4sin2x)(1 - 0)$

$\sf \:\triangle (x) = 2+ 4sin2x$

We know,

$\rm :\longmapsto\:\boxed{\rm{ sinx \: has \: maximum \: value \: 1 \: at \: x = \dfrac{\pi}{2} \: }}$

So,

$\rm \implies\:Maximum \: value \: of \: \triangle (x)\: occurs \: when \: 2x = \dfrac{\pi}{2}$

$\rm \implies\:Maximum \: value \: of \: \triangle (x)\: occurs \: when \: x = \dfrac{\pi}{4}$

So, option (d) is correct.

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More to Know :-

1. The determinant value remains unaltered if rows and columns are interchanged.

2. The determinant value is 0, if two rows or columns are identical.

3. The determinant value is multiplied by - 1, if successive rows or columns are interchanged.

4. The determinant value remains unaltered if rows or columns are added or subtracted.