Alexoooo137idn
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Let ∆(x) =
|1+ sin²x cos²x 4sin2x |
|sin²x 1+co²x 4sin2x |
|sin²x cos²x 1+4sin2x|

The value of x (0 ≤ x ≤ π/x) for which ∆(x) is maximum, is equal to
(a) π/2
(b) π/6
(c) π/3
(d) π/4​

Let X 1 Sinx Cosx 4sin2x Sinx 1cox 4sin2x Sinx Cosx 14sin2xThe Value Of X 0 X Πx For Which X Is Maximum Is Equal Toa Π2b Π6c Π3d Π4 class=

Sagot :

SalmonWorldTopperidn

Step-by-step explanation:

Given Question is

[tex] \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}[/tex]

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

[tex] \sf \: \: \: \: (a) \: \: \dfrac{\pi}{2} [/tex]

[tex] \sf \: \: \: \: (b) \: \: \dfrac{\pi}{6} [/tex]

[tex] \sf \: \: \: \: (c) \: \: \dfrac{\pi}{3} [/tex]

[tex] \sf \: \: \: \: (d) \: \: \dfrac{\pi}{4} [/tex]

[tex] \red{\large\underline{\sf{Solution-}}}[/tex]

Given determinant is

[tex] \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}[/tex]

[tex] \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{\tt{ OP \: C_1 \: \to \: C_1 + C_2 + C_3 \: }}[/tex]

We get,

[tex] \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}[/tex]

[tex] \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}[/tex]

[tex]\boxed{\tt{ OP \: R_2 \: \to \: R_2 - R_1 \: }} \: \: and \: \: \boxed{\tt{ R_3 \: \to \: R_3 - R_1 \: }}[/tex]

We get,

[tex] \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}[/tex]

[tex] \sf \:\triangle (x) = (2+ 4sin2x)(1 - 0)[/tex]

[tex] \sf \:\triangle (x) = 2+ 4sin2x[/tex]

We know,

[tex]\rm :\longmapsto\:\boxed{\rm{ sinx \: has \: maximum \: value \: 1 \: at \: x = \dfrac{\pi}{2} \: }}[/tex]

So,

[tex]\rm \implies\:Maximum \: value \: of \: \triangle (x)\: occurs \: when \: 2x = \dfrac{\pi}{2} [/tex]

[tex]\rm \implies\:Maximum \: value \: of \: \triangle (x)\: occurs \: when \: x = \dfrac{\pi}{4} [/tex]

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.

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