JEE Main & Advanced JEE Main Paper (Held On 9 April 2014)

  • question_answer
    If \[f(x)={{\left( \frac{3}{5} \right)}^{x}}+{{\left( \frac{4}{5} \right)}^{x}}=-1,x\in R,\]then the equation f(x) =0 has :   [JEE Main Online Paper ( Held On 09 Apirl  2014  )

    A) no solution

    B) one solution

    C) two solutions

    D) more than two solutions

    Correct Answer: B

    Solution :

    \[f(x)={{\left( \frac{3}{5} \right)}^{x}}+{{\left( \frac{4}{5} \right)}^{x}}-1\] Put f(x) = 0 \[\Rightarrow \]\[0={{\left( \frac{3}{5} \right)}^{x}}+{{\left( \frac{4}{5} \right)}^{x}}-1\] \[\Rightarrow \]\[{{\left( \frac{3}{5} \right)}^{x}}+{{\left( \frac{4}{5} \right)}^{x}}=1\] \[\Rightarrow \]\[{{3}^{x}}+{{4}^{x}}={{5}^{x}}\] For\[x=1\] \[{{3}^{3}}+{{4}^{3}}=91<{{5}^{3}}\] Only for x = 2, equation (1) Satisfy So, only one solution (x = 2)


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