Question

If $f:R\to R$ is a differentiable function and $f\left( 2 \right)=6$ then $\displaystyle \lim_{x \to 2}\int\limits_{6}^{f\left( x \right)}{\dfrac{2tdt}{\left( x-2 \right)}}$ is:
a. 0
b. 2 f(2)
c. 12 f’(2)
d. 24 f’(2)

Answer 1

In this question we have been given the multiple application of calculus used as definite integration and limit. So, first we will integrate and then apply the limit. When limit is applied it can be seen that it is in $\dfrac{0}{0}$ form. So, we will apply the L'Hospital rule and find the answer.

Complete step by step answer:
In the given question, we will use the concepts of calculus. As we can see, the value of the limit is asked. To solve this question, we should understand the concept of limits. In mathematics, a limit is a value that a function ‘approaches’ as the input ‘approaches’ some value. So, we have to solve the integration first. So, we get,
$\int\limits_{6}^{f\left( x \right)}{\dfrac{2tdt}{\left( x-2 \right)}}$
Here, the integration variable is dt, so (x-2) will be constant. So, we get,
$\dfrac{2}{\left( x-2 \right)}\int\limits_{6}^{f\left( x \right)}{tdt}$
We know that $\int{{{z}^{n}}dz=\dfrac{{{z}^{n+1}}}{n+1}}$. So, we can write,
$\dfrac{2}{\left( x-2 \right)}\left[ \dfrac{{{t}^{2}}}{2} \right]_{6}^{f\left( x \right)}$
On applying the limits, we get,

& \dfrac{2}{\left( x-2 \right)}\times \dfrac{1}{2}\left[ {{\left( f\left( x \right) \right)}^{2}}-{{\left( 6 \right)}^{2}} \right] \\\ & \dfrac{{{\left( f\left( x \right) \right)}^{2}}-36}{x-2} \\\ \end{aligned}$$ Now, we will apply the limit, so we get, $$\displaystyle \lim_{x \to 2}\dfrac{{{\left( f\left( x \right) \right)}^{2}}-36}{x-2}$$ When we will put x = 2, so we will get the answer in the form of $\dfrac{0}{0}$. So, we will apply the L’Hospital rule and differentiate the numerator and the denominator separately, so we get, $$\displaystyle \lim_{x \to 2}\dfrac{2f\left( x \right).f'\left( x \right)}{1}$$ Now, we will put x = 2, so we get, $\begin{aligned} & 2f\left( 2 \right)f'\left( 2 \right) \\\ & 2\times 6\times f'\left( 2 \right) \\\ & 12f'\left( 2 \right) \\\ \end{aligned}$ Therefore, we get the answer as $12f'\left( 2 \right)$. **So, the correct answer is “Option C”.** **Note:** In this question given, we need to understand that we cannot apply limits first and then integrate. Also, we need to understand the approach that (x-2) acts as a constant as the integration is in terms of the variable, ‘t’.