JEE Main 2019 — Mathematics Calculus

The given limit can be written as

$$l=\underset{x\rightarrow 2}{\mathrm{lim}}\frac{{\int }_{6}^{f(x)}2tdt}{(x-2)}$$l=x→2lim​(x−2)∫6f(x)​2tdt​

Applying L' Hospital’s rule i.e. if $l=\underset{x\rightarrow a}{\mathrm{lim}}\frac{g(x)}{h(x)}$l=x→alim​h(x)g(x)​ and \underset{x\rightarrow a}{\mathrm{lim}}g(x)\rightarrow 0&\underset{x\rightarrow a}{\mathrm{lim}}h(x)\rightarrow 0, then $l=\underset{x\rightarrow a}{\mathrm{lim}}\frac{{g}^{'}(x)}{{h}^{'}(x)}$l=x→alim​h′(x)g′(x)​

$$l=\underset{x\rightarrow 2}{lim}\frac{\frac{d}{dx}{\int }_{6}^{f(x)}2tdt}{1}$$l=x→2lim​1dxd​∫6f(x)​2tdt​

Now, applying Newton's Leibnitz rule i.e. $\displaystyle \frac{d}{dx}{\int }_{a}^{b}g(x)dx=g(b)\cdot {b}^{'}-g(a)\cdot {a}^{'},$dxd​∫ab​g(x)dx=g(b)⋅b′−g(a)⋅a′, we get

$$\Rightarrow l=\underset{x\rightarrow 2}{\mathrm{lim}}(2f(x)\cdot {f}^{'}(x)-0)$$⇒l=x→2lim​(2f(x)⋅f′(x)−0)

$$\Rightarrow l=2f(2)\cdot {f}^{'}(2)$$⇒l=2f(2)⋅f′(2)

Put the given value, to get

$$\Rightarrow l=2\times 6\times {f}^{'}(2)=12{f}^{'}(2).$$⇒l=2×6×f′(2)=12f′(2).