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
no solution
one solution
two solution
more than two solutions
Let $f(x) = (1 + {b^2}){x^2} + 2bx + 1$ and $m(b)$ the minimum value of $f(x)$ for a given $b$. As $b$ varies, the range of $m(b)$ is
Let $\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}$ be defined as
$f(x+y)+f(x-y)=2 f(x) f(y), f\left(\frac{1}{2}\right)=-1 .$ Then, the value of $\sum_{\mathrm{k}=1}^{20} \frac{1}{\sin (\mathrm{k}) \sin (\mathrm{k}+\mathrm{f}(\mathrm{k}))}$ is equal to:
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
Let $f : R -\{0,1\} \rightarrow R$ be a function such that $f(x)+f\left(\frac{1}{1-x}\right)=1+x$. Then $f(2)$ is equal to :
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.