Question

Let $f\left(x\right)=x^2+\lambda x+\mu \cos x,\lambda$ is positive integer and $\mu$ is a real number.The number of ordered pairs $(\lambda ,\mu )$ for which $f\left(x\right)=0$ and $f\left(f\left(x\right)\right)=0$ have same set of real roots

Answer 1

Let $\alpha$ be the root of $f\left(x\right)=0\Rightarrow f\left(\alpha \right)=0$

$\Rightarrow f\left(f\left(\alpha \right)\right)=0$

$\Rightarrow f\left(0\right)=0\Rightarrow \mu =0$

$\therefore f\left(x\right)=x^2+\lambda x=0$

$\Rightarrow x(x+\lambda )=0$

$\Rightarrow x=0,-\lambda$

$f\left(f\left(x\right)\right)={(x^2+\lambda x)}^2+\lambda (x^2+\lambda x)$

$\Rightarrow (x^2+\lambda x)\{x^2+\lambda x+\lambda \}=0$

Now $f\left(x\right)=0,f\left(f\left(x\right)\right)=0$ have same roots iff

$x^2+\lambda x+\lambda =0$ has no real roots

$\Rightarrow {\lambda }^2-4\lambda <0$

$\Rightarrow \lambda (\lambda -4)<0$

$\Rightarrow 0<\lambda <4$

Since $\lambda$ is a positive integer $\lambda =\mathrm{1.2.3}$

$\therefore (\lambda ,\mu )\approx (1,0),(2,0),(3,0)$

$\therefore 3$ pairs are possible.