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.