If Delta x- left - matrix e - x -sin 2 x -tan x - 2ln 1-x -cos x -sin xcos x- 2 - e- x-1 -sin x- 2end matrix right - A - B x - Cx 2- Then B is equal toA- 0B- 1C- 2D- -1

The correct option is A 0Δ 'x= | e^x amp;2cos2x amp;2xsec^2x^2 ln(1+x) amp;cosx amp;sinx cosx^2 amp;e^x 1 amp;sinx^2 |+ | e^x ...

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Byju's Answer

Standard XII

Mathematics

Differentiation of a Determinant

If Delta x= l...

Question

If $Δ (x) = ∣ ∣ ∣ ∣ ∣ \begin{matrix} e^{x} & s i n 2 x & t a n x^{2} l n (1 + x) & c o s x & s i n x c o s x^{2} & e^{x} - 1 & s i n x^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣$ $= A + B x + C x^{2} + \dots$ Then B is equal to

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Solution

$Δ^{'} x =$
$∣ ∣ ∣ ∣ ∣ \begin{matrix} e^{x} & 2 c o s 2 x & 2 x s e c^{2} x^{2} l n (1 + x) & c o s x & s i n x c o s x^{2} & e^{x} - 1 & s i n x^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣ + ∣ ∣ ∣ ∣ ∣ \begin{matrix} e^{x} & s i n 2 x & t a n x^{2} \frac{1}{(1 + x)} & - s i n x & c o s x c o s x^{2} & e^{x} - 1 & s i n x^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣ + ∣ ∣ ∣ ∣ ∣ \begin{matrix} e^{x} & s i n 2 x & t a n x^{2} l n (x + x) & c o s x & s i n x - 2 x s i n x^{2} & e^{x} & 2 x c o s x^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣$
$= B + 2 C x + \dots$
$P u t x = 0, B = ∣ ∣ ∣ ∣ \begin{matrix} 1 & 2 & 0 0 & 1 & 0 1 & 0 & 0 \end{matrix} ∣ ∣ ∣ ∣ + ∣ ∣ ∣ ∣ \begin{matrix} 1 & 0 & 0 1 & 0 & - 1 1 & 0 & 0 \end{matrix} ∣ ∣ ∣ ∣ + ∣ ∣ ∣ ∣ \begin{matrix} 1 & 0 & 0 0 & 1 & 0 0 & 1 & 0 \end{matrix} ∣ ∣ ∣ ∣ = 0$

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If Δ(x)=∣∣ ∣ ∣∣exsin2xtanx2ln(1+x)cosxsinxcosx2ex−1sinx2∣∣ ∣ ∣∣ =A+Bx+Cx2+⋯ Then B is equal to

If $Δ (x) = ∣ ∣ ∣ ∣ ∣ \begin{matrix} e^{x} & s i n 2 x & t a n x^{2} l n (1 + x) & c o s x & s i n x c o s x^{2} & e^{x} - 1 & s i n x^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣$ $= A + B x + C x^{2} + \dots$ Then B is equal to