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  • If graph of the continuous function passes through <img alt="\mathrm{(a, 0)}" src="ht

If graph of the continuous function \mathrm{y=f(x)} passes through \mathrm{(a, 0)} and graph of the function \mathrm{y=f^{\prime}(x)} passes through \mathrm{(a, b),(b>0), then \, \lim _{x \rightarrow a} \frac{\ln \left(1+6 f^2(x)-3 f(x)\right.}{3 f(x)}}

Option: 1

equal to 1
 


Option: 2

equal to b
 


Option: 3

equal to -1
 


Option: 4

does not exist


Answers (1)

best_answer

Since \mathrm{f(a)=0 \Rightarrow \lim _{x \rightarrow a} \ln \left(1+6 f^2(x)-3 f(x)\right)=0}

And  \mathrm{\lim _{x \rightarrow a} 3 f(x)=0}

Apply L's Hospital's rule

\mathrm{ L=\lim _{x \rightarrow a} \frac{12 f(x) f^{\prime}(x)-3 f^{\prime}(x)}{\left(1+6 f^2(x)-3 f(x)\right) 3 f^{\prime}(x)}=\lim _{x \rightarrow a} \frac{3 f^{\prime}(x)[4 f(x)-1]}{3 f^{\prime}(x)\left[1+6 f^2(x)-3 f(x)\right]}=-1 }

Posted by

manish painkra

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