Question

Let $f(x)={(x-7)}^2{(x-2)}^7{(x-2)}^7,x\in [2,7]$ . The value of $\theta \in (2,7)$ such that $f(\theta )=0$ is equal to

• A. $\frac{49}{4}$
• B. $\frac{53}{9}$
• C. $\frac{53}{7}$
• D. $\frac{49}{9}$
• E. $\frac{45}{7}$

Answer 1

Answer: (B)

Given, $f(x)={(x-7)}^2{(x-2)}^7$
$\Rightarrow$ $f(\theta )={(\theta -7)}^2{(\theta -2)}^7$
$\Rightarrow$ $f(\theta )=2(\theta -7){(\theta -2)}^7+7{(\theta -2)}^6{(\theta -7)}^2$ Put $f(\theta )=0$
$\Rightarrow$ $(\theta -7){(\theta -2)}^6[2(\theta -2)+7(\theta -7)]=0$
$\Rightarrow$ $9\theta =53\Rightarrow \theta =\frac{53}{9}$