Question

Find the wronskian of the set {1 - x, 1 + x, 1 - 3x}.​

Answer 1

Answer:

Correct option is

A

−415

1(x+1x−1)=3x

⇒x+1x−1=3⇒x+1x−1=9

x−1=9x+9

−8x=10⇒x=4−5

f(3)=3(4−5)=4−15

Answer 2

Step-by-step explanation:

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

We know that,

$\rm :\longmapsto\:Wronkian \: of \: the \: set \: {\left \{f_{1}(x),f_{2}(x),f_{3}(x) \right \}} \: is \:$

$W{\left(f_{1},f_{2},f_{3} \right)}\left(x\right) = \left|\begin{array}{ccc}f_{1}\left(x\right) & f_{2}\left(x\right) & f_{3}\left(x\right)\\f_{1}^{\prime}\left(x\right) & f_{2}^{\prime}\left(x\right) & f_{3}^{\prime}\left(x\right)\\f_{1}^{\prime\prime}\left(x\right) & f_{2}^{\prime\prime}\left(x\right) & f_{3}^{\prime\prime}\left(x\right)\end{array}\right|$

Here,

$\rm :\longmapsto\:f_1(x) = 1 - x$

$\rm :\longmapsto\:f_2(x) = 1 + x$

$\rm :\longmapsto\:f_3(x) = 1 - 3x$

So, there respective differential coefficients are as follow

$\rm :\longmapsto\:f_1'(x) = 0 - 1 = - 1$

$\rm :\longmapsto\:f_2'(x) = 0 + 1 = 1$

$\rm :\longmapsto\:f_3'(x) = 0 - 3 = - 3$

Also,

$\rm :\longmapsto\:f_1''(x) = 0$

$\rm :\longmapsto\:f_2''(x) = 0$

$\rm :\longmapsto\:f_3''(x) = 0$

So, on substituting the values in

$W{\left(f_{1},f_{2},f_{3} \right)}\left(x\right) = \left|\begin{array}{ccc}f_{1}\left(x\right) & f_{2}\left(x\right) & f_{3}\left(x\right)\\f_{1}^{\prime}\left(x\right) & f_{2}^{\prime}\left(x\right) & f_{3}^{\prime}\left(x\right)\\f_{1}^{\prime\prime}\left(x\right) & f_{2}^{\prime\prime}\left(x\right) & f_{3}^{\prime\prime}\left(x\right)\end{array}\right|.$

We get

$\rm :\longmapsto\:W= \: \ \: \begin{gathered}\sf \left | \begin{array}{ccc}1 - x&1 + x&1 - 3x\\ - 1&1& - 3\\0&0&0\end{array}\right | \end{gathered}$

$\rm \: = \: 0$

[ If elements of any row all are 0, its determinant value is zero]

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