Question

If $y=\frac{\left(a+b{x}^{{\displaystyle \frac{3}{2}}}\right)}{x^{{\displaystyle \frac{5}{4}}}}$and $y^{\text{'}}=0$ at $x=5$. Then the ratio of $a:b$is ?

• A. $5:2$
• B. $\surd 5:1$
• C. $1:2$
• D. $3:5$

Answer 1

The correct option is B

$\surd 5:1$

Explanation for the correct option:

Step 1: Simplifying the given equation:

$\begin{array}{rcl}y& =& \frac{\left(a+b{x}^{{\displaystyle \frac{3}{2}}}\right)}{x^{{\displaystyle \frac{5}{4}}}}\\ & =& \frac{a}{x^{{\displaystyle \frac{5}{4}}}}+\frac{b{x}^{{\displaystyle \frac{3}{2}}}}{x^{{\displaystyle \frac{5}{4}}}}\\ & =& a{x}^{\frac{-5}{4}}+b{x}^{\frac{1}{4}}\end{array}$

Step 2: Differentiate the above equation with respect $x$

$\frac{dy}{dx}=\left(\frac{-5}{4}\right)a{x}^{\frac{-9}{4}}+\left(\frac{1}{4}\right)b{x}^{\frac{-3}{4}}\left[\because \frac{d\left(x^n\right)}{dx}=n{x}^{n-1}\right]$

Step 3: Finding the ratio of $a$ and $b$:

At $x=5,y\text{'}=0$

$$\begin{gathered} \left(\frac{-5}{4}\right)a{\left(5\right)}^{\frac{-9}{4}}+\left(\frac{1}{4}\right)b{\left(5\right)}^{\frac{-3}{4}}=0 \\ \Rightarrow \left(\frac{-5}{4}\right)a{\left(5\right)}^{\frac{-9}{4}}=-\left(\frac{1}{4}\right)b{\left(5\right)}^{\frac{-3}{4}} \\ \Rightarrow \frac{a}{b}=\frac{1}{5}{\left(5\right)}^{\frac{-3}{4}+\left(\frac{9}{4}\right)} \\ ={\left(5\right)}^{\frac{3}{2}-1} \\ =5^{\frac{1}{2}} \\ =\sqrt{5} \end{gathered}$$

Hence, option (B) is correct option.