Let $f\left(x\right)=a{x}^3+b{x}^2+cx+d\sin x$. Find the condition that $f\left(x\right)$ is always one-one function.

Given: $f\left(x\right)=a{x}^3+b{x}^2+cx+d\sin x$

To find: The condition that $f\left(x\right)$ is always one-one function.

Step 1

Consider the given question,

$f\left(x\right)=a{x}^3+b{x}^2+cx+d\sin x$

By differentiating the given function, we get

$f^{\text{'}}\left(x\right)=3a{x}^2+2bx+c+d\cos$$x$

For one to one function, the differentiation should be greater than zero.

$\Rightarrow 3a{x}^2+2bx+c+d\cos x>0$

As $f^{\text{'}}\left(x\right)>0$

$Step2$

Consider the concept of quadratic equation, the discriminant should be less than zero,

$\Rightarrow D<0$

$\Rightarrow b^2-4ac<0$

$(2b{)}^2-4\times (3a)\times (cd)<0$

$4b^2-12a\left(cd\right)<0$

$Step3$

Consider the above equation,

$\Rightarrow b^2-3a\left(cd\right)<0$

$b^2<3a\left(cd\right)$

This is the condition $b^2<3a\left(cd\right)$ for $f\left(x\right)$ to be always one-one.

$f\left(x\right)=a{x}^3+b{x}^2+cx+d\sin x$, this is the condition $b^2<3a\left(cd\right)$ for $f\left(x\right)$ to be always one-one function.

Therefore, the correct answer is option $1.$