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Q. Let $ f : R \to R$ defined by
$f(x) = 2x^3 + 2x^2 + 300x + 5\, sin\, x$, then $f$ is

Relations and Functions - Part 2

Solution:

$f(x) = 2x^3 + 2x^2+ 300x + 5 \,sin\,x$
$f'(x) = 6x^2 + 4x + 300 + 5 \,cos\, x$
$= 2(3x^2 + 2x+ 147) + (6 + 5cos\, x)$
Since $-1 \le cos x \le 1 \forall x \in R$
$\therefore 6 + 5\,cos\,x > 0$
let $g(x) = 3x^2 + 2x +147$
Since $a = 3 > 0$ and $D = (2)^2 - 4 \times 3 \times 147 < 0$
$\therefore g(x) > 0$
$ f'(x) > 0$
$\therefore f(x)$ is an increasing function therefore said to be
$1 - 1$ function.
Now, $f(\infty) = - \infty$ and $f(\infty) = \infty$
$\therefore f(x)$ is continuous also.
$\therefore $ Range of $f =$ codomain of $f = R$.
$\therefore f$ is onto.