Question

If \[y=\dfrac{a+bx}{c+dx}\], where a, b, c and d are constants and \[\lambda {{y}_{1}}{{y}_{3}}=\mu y_{2}^{2}\], then value of \[{{\mu }^{{{\lambda }^{2}}}}\] must be?

Answer 1

Hint: \[{{y}_{1}}=\dfrac{dy}{dx};{{y}_{2}}=\dfrac{d}{dx}(\dfrac{dy}{dx})=\dfrac{d{{y}_{1}}}{dx}; {{y}_{3}}=\dfrac{d}{dx}(\dfrac{{{d}^{2}}y}{d{{x}^{2}}})=\dfrac{d{{y}_{2}}}{dx}\]. Apply these concepts to find the values of ${{y}_{1}}$ , ${{y}_{2}}$ and ${{y}_{3}}$ and substitute in given equation. By comparing equation we get ${\lambda}$ and ${\mu}$ values and then substitute in \[{{\mu }^{{{\lambda }^{2}}}}\] to get required answer.

Complete step-by-step answer:
We are given \[y=\dfrac{a+bx}{c+dx}\] . The given function is of the form \[y=\dfrac{f(x)}{g(x)}\] where , \[f(x)=a+bx\] and \[g(x)=c+dx\].
Now , first we will find the value of \[{{y}_{1}}\].
To find \[{{y}_{1}}\], we will differentiate \[y\] with respect to \[x\].
On differentiating \[y\] with respect to \[x\], we get ,
\[{{y}_{1}}=\dfrac{dy}{dx}=\dfrac{d}{dx}(\dfrac{f(x)}{g(x)})\]
For this we will use the quotient rule, which is given as
\[\dfrac{d}{dx}(\dfrac{f(x)}{g(x)})=\dfrac{g(x).{{f}^{'}}(x)-f(x).{{g}^{'}}(x)}{{{[g(x)]}^{2}}}\]
Here , \[{{f}^{'}}(x)=\dfrac{d}{dx}(a+bx)=b\] and \[g'(x)=\dfrac{d}{dx}(c+dx)=d\].
Putting these values in \[{{y}_{1}}\], we get,
\[{{y}_{1}}=\dfrac{(c+dx)b-(a+bx)d}{{{(c+dx)}^{2}}}\]
\[=\dfrac{bc+bdx-ad-bdx}{{{(c+dx)}^{2}}}\]
\[\Rightarrow {{y}_{1}}=\dfrac{bc-ad}{{{(c+dx)}^{2}}}\]
Now, we will find the value of \[{{y}_{2}}\].
To find \[{{y}_{2}}\] , we will differentiate \[{{y}_{1}}\] with respect to \[x\].
\[{{y}_{1}}\] is of the form \[\dfrac{k}{p{{(x)}^{n}}}\], where \[k=bc-ad\], \[p(x)=(c+dx)\]and \[n=2\]
On differentiating \[{{y}_{1}}\] with respect to\[x\], we get ,
\[{{y}_{2}}=\dfrac{-k.n}{p{{(x)}^{n+1}}}.{{p}^{'}}(x)\]
\[=\dfrac{2(ad-bc).d}{{{(c+dx)}^{3}}}\]
Now , we will find the value of \[{{y}_{3}}\].
To find \[{{y}_{3}}\] , we will differentiate \[{{y}_{2}}\] with respect to \[x\].
\[{{y}_{2}}\] is of the form \[\dfrac{c}{p{{(x)}^{m}}}\], where \[c=2d(ad-bc)\], \[p(x)=c+dx\] and \[m=3\]
On differentiating \[{{y}_{2}}\] with respect to \[x\], we get ,
So, \[{{y}_{3}}=\dfrac{-c.m}{p{{(x)}^{m+1}}}.{{p}^{'}}(x)\]
\[=\dfrac{-3(2d(ad-bc))}{{{(c+dx)}^{4}}}.d\]
\[=\dfrac{6{{d}^{2}}(bc-ad)}{{{(c+dx)}^{4}}}\]
Now , in the question, it is given that
\[\lambda {{y}_{1}}{{y}_{3}}=\mu y_{2}^{2}\]
Now , we will find the value of \[{{y}_{1}}{{y}_{3}}\].
To find the value of \[{{y}_{1}}{{y}_{3}}\], we will multiply the value of \[{{y}_{1}}\] with the value of \[{{y}_{3}}\].
So , \[{{y}_{1}}{{y}_{3}}=\dfrac{bc-ad}{{{(c+dx)}^{2}}}.\dfrac{6{{d}^{2}}(bc-ad)}{{{(c+dx)}^{4}}}\]
\[=\dfrac{(-(ad-bc))}{{{(c+dx)}^{2}}}.\dfrac{6{{d}^{2}}(-(ad-bc))}{{{(c+dx)}^{4}}}\]
\[=\dfrac{6{{d}^{2}}.{{(ad-bc)}^{2}}}{{{(c+dx)}^{6}}}\]
Now , we will find the value of \[y_{2}^{2}\] .
So, \[y_{2}^{2}={{[\dfrac{2d(ad-bc)}{{{(c+dx)}^{3}}}]}^{2}}\]
\[=\dfrac{4{{d}^{2}}{{(ad-bc)}^{2}}}{{{(c+dx)}^{6}}}\]
Now , comparing the values of \[{{y}_{1}}{{y}_{3}}\] and \[y_{2}^{2}\], we can conclude if we multiple \[{{y}_{1}}{{y}_{3}}\] by \[2\] and \[y_{2}^{2}\] by \[3\] , they can be equated as \[2\times {{y}_{1}}{{y}_{3}}=3y_{2}^{2}\]
i.e. \[2\times \dfrac{6{{d}^{2}}.{{(ad-bc)}^{2}}}{{{(c+dx)}^{6}}}=3\times \dfrac{4{{d}^{2}}{{(ad-bc)}^{2}}}{{{(c+dx)}^{6}}}\]
In the question , it is given \[\lambda {{y}_{1}}{{y}_{3}}=\mu y_{2}^{2}\].
Comparing with \[2{{y}_{1}}{{y}_{3}}=3y_{2}^{2}\] , we can see \[\lambda =2\] and \[\mu =3\]
So, \[{{\mu }^{{{\lambda }^{2}}}}={{3}^{{{2}^{2}}}}={{3}^{4}}=3\times 3\times 3\times 3=81\]

Note: \[{{y}_{1}}\] and \[{{y}_{2}}\] are composite functions , i.e of the form \[h(x)=a(b(x))\] and hence, the derivative will be of the form \[h'(x)=a'(b(x))\times b'(x)\].
So, \[\dfrac{d}{dx}({{y}_{1}})=\dfrac{kn}{p{{(x)}^{n+1}}}.{{p}^{'}}(x)\]
Here , most of the students make a mistake of not multiplying by \[{{p}^{'}}(x)\]. Such mistakes should be avoided as students can end up getting a wrong answer.