If - 1-x 3 3 -2 x 3 -2 -2 x and - 2-x 3 -2 x then

The correct option is A d/dx ( Δ _1 )=3Δ _2 Given, #160; Δ _ 1 = x amp; 3 amp; 3 2 amp; x amp; 3 2 amp; 2 amp; x ...

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Byju's Answer

Standard XII

Mathematics

Cofactor

If Δ 1=x ...

Question

If $Δ_{1} = ∣ ∣ ∣ ∣ \begin{matrix} x & 3 & 3 - 2 & x & 3 - 2 & - 2 & x \end{matrix} ∣ ∣ ∣ ∣$ and $Δ_{2} = ∣ ∣ ∣ \begin{matrix} x & 3 - 2 & x \end{matrix} ∣ ∣ ∣$ then

\frac{d}{d x} (Δ_{1}) = 3 Δ_{2}

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\frac{d}{d x} (Δ_{1}) = 3 {(Δ_{2})}_{2}^{2}

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Δ_{1} = \frac{d}{d x} (Δ_{2})

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Δ_{1} = 4 Δ_{2}^{2}

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Solution

The correct option is A $\frac{d}{d x} (Δ_{1}) = 3 Δ_{2}$
Given, $Δ_{1} = ∣ ∣ ∣ ∣ \begin{matrix} x & 3 & 3 - 2 & x & 3 - 2 & - 2 & x \end{matrix} ∣ ∣ ∣ ∣$

$\Rightarrow Δ_{1} = x^{3} + 18 x - 6$

$\frac{d}{d x} (Δ_{1}) = 3 x^{2} + 18$

$\Rightarrow \frac{d}{d x} (Δ_{1}) = 3 (x^{2} + 6)$
Now, $Δ_{2} = ∣ ∣ ∣ \begin{matrix} x & 3 - 2 & x \end{matrix} ∣ ∣ ∣$

$Δ_{2} = x^{2} + 6$
Hence, $\frac{d}{d x} (Δ_{1}) = 3 Δ_{2}$

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Similar questions

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and

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If Δ1=∣∣ ∣∣x33−2x3−2−2x∣∣ ∣∣ and Δ2=∣∣∣x3−2x∣∣∣ then

The correct option is A ddx(Δ1)=3Δ2Given, Δ1=∣∣ ∣∣x33−2x3−2−2x∣∣ ∣∣⇒Δ1=x3+18x−6ddx(Δ1)=3x2+18⇒ddx(Δ1)=3(x2+6)Now, Δ2=∣∣∣x3−2x∣∣∣Δ2=x2+6Hence, ddx(Δ1)=3Δ2

If $Δ_{1} = ∣ ∣ ∣ ∣ \begin{matrix} x & 3 & 3 - 2 & x & 3 - 2 & - 2 & x \end{matrix} ∣ ∣ ∣ ∣$ and $Δ_{2} = ∣ ∣ ∣ \begin{matrix} x & 3 - 2 & x \end{matrix} ∣ ∣ ∣$ then