Question

$f\left(x\right)=x^3+3x^2+4x+b\sin x+c\cos x,\forall x\in R$ is a one-one function, then the value of $b^2+c^2$ is

• A. $\ge 1$
• B. $\ge 2$
• C. $\le 1$
• D. none of these

Answer 1

Answer: (C)

$\le 1$

Here, $f\left(x\right)=x^3+3x^2+4x+b\sin x+c\cos x$
$f^{\prime }\left(x\right)=3x^2+6x+4+b\cos x-c\sin x$
Now, for $f\left(x\right)$ to be one-one, the only possibility is
$f^{\prime }\left(x\right)\ge 0,\forall x\in R$
ie, $3x^2+6x+4+b\cos x-c\sin x\ge 0,\forall x\in R$
ie, $3x^2+6x+4\ge c\sin x-b\cos x,\forall x\in R$

As we are approximating lower bound for the function by a number instead of a function, hence we chose the number.
ie, $3x^2+6x+4\ge \sqrt{b^2+c^2},\forall x\in R$
ie, $\sqrt{b^2+c^2}\le 3(x^2+2x+1)+1,\forall x\in R$
$\sqrt{b^2+c^2}\le 3(x+1{)}^2+1,\forall x\in R$
$\sqrt{b^2+c^2}\le 1,\forall x\in R$
$\Rightarrow b^2+c^2\le 1,\forall x\in R$
Hence, (c) is the correct answer.