Question

If $f\left(x\right)={\left(x-7\right)}^2{\left(x-2\right)}^7,x\in \left[2,7\right]$ Then, the value of $\theta \in \left(2,7\right)$ such that $f\text{'}\left(\theta \right)=0$ is

• A. $\frac{49}{4}$
• B. $\frac{53}{9}$
• C. $\frac{53}{7}$
• D. $\frac{49}{9}$

Answer 1

The correct option is B

$\frac{53}{9}$

Explanation for the correct option:

Finding the value of $\theta$:

Given that,

$f\left(x\right)={\left(x-7\right)}^2{\left(x-2\right)}^7$

Differentiate the above equation with respect to $x$.

$\begin{array}{rcl}f\text{'}\left(x\right)& =& {\left(x-2\right)}^{7}2\left(x-7\right)+{\left(x-7\right)}^2\left(7{\left(x-2\right)}^6\right)\mathbf{[}\mathbf{\because }\frac{\mathbf{d}\left(\mathrm{uv}\right)}{\mathbf{dx}}\mathbf{=}\mathbf{vu}\mathbf{\text{'}}\mathbf{+}\mathbf{uv}\mathbf{\text{'}}\mathbf{,}\mathbf{}\frac{\mathbf{d}\left(x^n\right)}{\mathbf{dx}}\mathbf{=}{\mathbf{nx}}^{\mathbf{n}\mathbf{-}\mathbf{1}}\mathbf{]}\\ & =& 2\left(x-7\right){\left(x-2\right)}^7+7{\left(x-7\right)}^2{\left(x-2\right)}^6\end{array}$

Substitute $x=\theta$ in $f\text{'}\left(x\right)$,

$$\begin{gathered} f\text{'}\left(\theta \right)=0\left[\mathrm{given}\right] \\ \Rightarrow 2\left(\mathrm{\theta }-7\right){\left(\mathrm{\theta }-2\right)}^7+7{\left(\mathrm{\theta }-7\right)}^2{\left(\mathrm{\theta }-2\right)}^6=0 \\ \Rightarrow \left(\mathrm{\theta }-7\right){\left(\mathrm{\theta }-2\right)}^6\left[2\mathrm{\theta }-4+7\mathrm{\theta }-49\right]=0 \\ \Rightarrow \left(\mathrm{\theta }-7\right){\left(\mathrm{\theta }-2\right)}^6\left[9\mathrm{\theta }-53\right]=0 \\ \Rightarrow \mathrm{\theta }=7\mathrm{or}2\mathrm{or}\frac{53}{9} \end{gathered}$$

Since $\theta \in \left(2,7\right)$, so neglect $7$ and $2$.

Therefore, the value of $\theta =\frac{53}{9}$

Hence, the correct option is B.