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} $