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JEE Main & Advanced Mathematics Applications of Derivatives Question Bank Critical Thinking

  • question_answer
    Let \[f(x)\]and \[g(x)\]be two functions having finite non-zero 3rd order derivatives \[{f}'''(x)\]and \[{g}'''(x)\] for all, \[x\in R\]. If \[f(x)g(x)=1\]for all \[x\in R\], then \[\frac{{{f}'''}}{{{f}'}}-\frac{{{g}'''}}{{{g}'}}\]is equal to

    A) \[3\text{ }\left( \frac{{{f}''}}{g}-\frac{{{g}''}}{f} \right)\]

    B) \[3\text{ }\left( \frac{{{f}''}}{f}-\frac{{{g}''}}{g} \right)\]

    C) \[3\text{ }\left( \frac{g''}{g}-\frac{f''}{g} \right)\]

    D) \[3\text{ }\left( \frac{{{f}''}}{f}-\frac{{{g}''}}{f} \right)\]

    Correct Answer: B

    Solution :

    • We have \[f(x)g(x)=1\]           
    • Differentiating with respect to x, we get             
    • \[{f}'g+f{g}'=0\]                                  ?..(i)           
    • Differentiating (i) w.r.t. x, we get \[{f}''g+2{f}'{g}'+f{g}''=0\]  ?..(ii)           
    • Differentiating (ii) w.r.t. x, we get           
    • \[{f}'''g+{g}'''\,f+3{f}''{g}'+3{g}''{f}'=0\]           
    • Þ   \[\frac{{{f}'''}}{{{f}'}}({f}'g)+\frac{{{g}'''}}{{{g}'}}(f{g}')+\frac{3{f}''}{f}(f{g}')+\frac{3{g}''}{g}(g{f}')=0\]           
    • Þ  \[\left( \frac{{{f}'''}}{{{f}'}}+\frac{3{g}''}{g} \right)\,({f}'g)=-\left( \frac{{{g}'''}}{{{g}'}}+\frac{3{f}''}{f} \right)\,(f{g}')\]           
    • Þ \[-\left( \frac{{{f}'''}}{{{f}'}}+\frac{3{g}''}{g} \right)\,(f{g}')=-\left( \frac{{{g}'''}}{{{g}'}}+\frac{3{f}''}{g} \right)f{g}'\] ,  [using (i)]                   
    • Þ \[\frac{{{f}'''}}{{{f}'}}+\frac{3{g}''}{g}=\frac{{{g}'''}}{{{g}'}}+\frac{3{f}''}{f}\Rightarrow \frac{{{f}'''}}{{{f}'}}-\frac{{{g}'''}}{{{g}'}}=3\left( \frac{{{f}''}}{f}-\frac{{{g}''}}{g} \right)\].


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