Question

Let $f:R\to R$ be a continuously differentiable function in which f(2) = 6 and $f'(2)=\dfrac{1}{48}$ .
If $\int\limits_{6}^{f(x)}{4{{t}^{3}}dt=(x-2)g(x),}$ then $\displaystyle \lim_{x \to 2}g(x)$ is equals to
(a)24
(b)36
(c)12
(d)18

Answer 1

To solve this question, first we will write the given expression again by keeping g(x) on only the left hand side. Then, we will check whether we get any indeterminate form on putting limits, if yes then we will use L’hospital rule to solve the $\dfrac{0}{0}$ case by using leibniz rule and after simplifying and putting limits we will get value of $\displaystyle \lim_{x \to 2}g(x)$.

Complete step-by-step answer:
Before we solve the question, let us see what is L’hopital rule and Leibniz rule.
If, we have expression $\displaystyle \lim_{x \to \infty}\dfrac{f(x)}{g(x)}$ , and if on putting limit we get indeterminate form $\dfrac{0}{0}$ or $\dfrac{\infty }{\infty }$, then we use L’hopital rule which states that we differentiate functions f(x) and g(x) unless and until we get out of indeterminate form that is $\displaystyle \lim_{x \to \infty}\dfrac{f(x)}{g(x)}=\displaystyle \lim_{x \to \infty}\dfrac{f'(x)}{g'(x)}$, $\displaystyle \lim_{x \to \infty}\dfrac{f(x)}{g(x)}=\displaystyle \lim_{x \to \infty}\dfrac{f''(x)}{g''(x)}$ and so on.
Leibniz rule state that $\dfrac{d}{dx}\int\limits_{u(x)}^{v(t)}{f(t)dt}=\dfrac{d}{dx}\left[ F(v(t))-F(u(x)) \right]$
Or, $\dfrac{d}{dx}\int\limits_{u(x)}^{v(t)}{f(t)dt}=F'(v(x))\dfrac{dv}{dx}-F'(u(x))\dfrac{du}{dx}$
Now, in question it is given that $\int\limits_{6}^{f(x)}{4{{t}^{3}}dt=(x-2)g(x),}$
So, we can write above equation by re – arranging as,
$g(x)=\dfrac{\int\limits_{6}^{f(x)}{4{{t}^{3}}dt}}{(x-2)}$
Now, we asked to find the value of g(x), when x tends to 2, that is $\displaystyle \lim_{x \to 2}g(x)$
So, applying limit we get
$\displaystyle \lim_{x \to 2}g(x)=\displaystyle \lim_{x \to 2}\dfrac{\int\limits_{6}^{f(x)}{4{{t}^{3}}dt}}{(x-2)}$
So, we can see that if we put a limit, then the denominator and numerator gets 0, so we get an indeterminate form of $\dfrac{0}{0}$ .
So, we need to use L’ hopital rule here, we get
$\displaystyle \lim_{x \to 2}g(x)=\displaystyle \lim_{x \to 2}\dfrac{\dfrac{d}{dx}\left( \int\limits_{6}^{f(x)}{4{{t}^{3}}dt} \right)}{\dfrac{d}{dx}(x-2)}$
Using Leibniz rule we get $\dfrac{d}{dx}\left( \int\limits_{6}^{f(x)}{4{{t}^{3}}dt} \right)=4{{[f(x)]}^{3}}.f'(x)-0$ and $\dfrac{d}{dx}(x-2)=1$, as we know that $\dfrac{d}{dx}({{x}^{n}})=n{{x}^{n-1}}$
So, $\displaystyle \lim_{x \to 2}g(x)=\displaystyle \lim_{x \to 2}4{{[f(x)]}^{3}}.f'(x)$
Or, $\displaystyle \lim_{x \to 2}g(x)=4\displaystyle \lim_{x \to 2}{{[f(x)]}^{3}}.\displaystyle \lim_{x \to 2}f'(x)$
$\displaystyle \lim_{x \to 2}g(x)=4\displaystyle \lim_{x \to 2}{{[f(2)]}^{3}}.\displaystyle \lim_{x \to 2}f'(2)$
Using values given in question which are f(2) = 6 and $f'(2)=\dfrac{1}{48}$ , we get
$\displaystyle \lim_{x \to 2}g(x)=4{{[6]}^{3}}.\dfrac{1}{48}$
On simplifying, we get
$\displaystyle \lim_{x \to 2}g(x)=4.216.\dfrac{1}{48}$
On solving, we get
$\displaystyle \lim_{x \to 2}g(x)=18$

So, the correct answer is “Option d”.

Note: To solve such questions one must know the application of L’hopital rule and most importantly Leibniz rule which stated that $\dfrac{d}{dx}\int\limits_{u(x)}^{v(t)}{f(t)dt}=F'(v(x))\dfrac{dv}{dx}-F'(u(x))\dfrac{du}{dx}$. Also, remember that $\dfrac{d}{dx}({{x}^{n}})=n{{x}^{n-1}}$. Try not to make calculation errors as the question is fully numerical based so any minor error will lead to wrong answers.