If $f\left( x \right) = {\left( {\frac{3}{5}} \right)^x} + {\left( {\frac{4}{5}} \right)^x} - 1$ , $x \in R$ , then the equation $f(x) = 0$ has

  • [JEE MAIN 2014]
  • A

    no solution

  • B

    one solution

  • C

    two solution

  • D

    more than two solutions

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  • [IIT 2001]

Let $\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}$ be defined as

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If $f({x_1}) - f({x_2}) = f\left( {\frac{{{x_1} - {x_2}}}{{1 - {x_1}{x_2}}}} \right)$ for ${x_1},{x_2} \in [ - 1,\,1]$, then $f(x)$ is

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The domain of definition of the function $f (x) = {\log _{\left[ {x + \frac{1}{x}} \right]}}|{x^2} - x - 6|+ ^{16-x}C_{2x-1} + ^{20-3x}P_{2x-5}$  is

Where $[x]$ denotes greatest integer function.