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Let $f(x) = sin\,x,\,\,g(x) = x.$
Statement $1:$ $f(x)\, \le \,g\,(x)$ for $x$ in $(0,\infty )$
Statement $2:$ $f(x)\, \le \,1$ for $(x)$ in $(0,\infty )$ but $g(x)\,\to \infty$ as $x\,\to \infty$
Statement $1$ is true, Statement $2$ is false.
Statement $1$ is true, Statement $2$ is true,Statement $2$ is a correct explanation for Statement $1.$
Statement $1$ is true, Statement $2$ is true,Statement $2$ is not a correct explanation for Statement $1.$
Statement $1$ is false, Statement $2$ is true.
Solution
Let $f\left( x \right) = \sin \,x$ and $f\left( x \right) = \sin \,x$
Statement-$1$ : $f\left( x \right) \le gx\left( {\forall x} \right) \in \left( {0,\infty } \right)$
i.e, $\sin \,x \le x\forall x \in \left( {0,\infty } \right)$
which is true
Statement-$2$ : $f\left( x \right) \le 1\forall x \in \left( {0,\infty } \right)$
i.e., $\sin \,x \le 1\forall x \in \left( {0,\infty } \right)$
It is true and
$g\left( x \right) = x \to \infty $ as $x \to \infty $ also true.
