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
The range of $f(x)=4 \sin ^{-1}\left(\frac{x^2}{x^2+1}\right)$ is
Let $f: R \rightarrow R$ be a continuous function such that $f\left(x^2\right)=f\left(x^3\right)$ for all $x \in R$. Consider the following statements.
$I.$ $f$ is an odd function.
$II.$ $f$ is an even function.
$III$. $f$ is differentiable everywhere. Then,
Given the function $f(x) = \frac{{{a^x} + {a^{ - x}}}}{2},\;(a > 2)$. Then $f(x + y) + f(x - y) = $
Show that the function $f: R_* \rightarrow R_*$ defined by $f(x)=\frac{1}{x}$ is one-one and onto, where $R_*$ is the set of all non-zero real numbers. Is the result true, if the domain $R_*$ is replaced by $N$ with co-domain being same as $R _*$ ?
Let $2{\sin ^2}x + 3\sin x - 2 > 0$ and ${x^2} - x - 2 < 0$ ($x$ is measured in radians). Then $x$ lies in the interval