Let f:R\rightarrow R be a continuously differentiable function such that f(2)=6 and f'(2)=\frac{1}{48}. If \int_{6}^{f\left ( x \right )}4t^{3}dt=(x-2)g(x), then \lim_{x\rightarrow 2}g(x) is equal to : 

 

 

 

 

  • Option 1)

    36

  • Option 2)

    24

  • Option 3)

    12

  • Option 4)

    18

Answers (1)

f:R\rightarrow R             f(2)=6               and                f'(2)=\frac{1}{48}

\lim_{x\rightarrow 2}g(x)=\lim_{x\rightarrow 2}\frac{\int _6^{f(x)}4t^3dt}{x-2}

                  =\lim_{x\rightarrow 2}\frac{4\cdot f^{3}(x)\cdot f'(x)}{1}

                  =4f^{3}\left ( 2 \right )f'\left ( 2 \right )

                 =4\times \left ( 6 \right )^{3}\times \frac{1}{48}

                 =6\times 36\times \frac{1}{12}=18

 

  


Option 1)

36

Option 2)

24

Option 3)

12

Option 4)

18

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