Mathematics - Functions Question with Solution | TestHub

Let $f(x)=x^2+\lambda x+\mu \cos x$. Let $\alpha$ be a root of $f(x)=0$, which implies $f(\alpha )=0$.

This also implies $f(f(\alpha ))=f(0)=0$, because $\alpha$ is also a root of $f(f(x))=0$.

Now, $f(0)=\mu =0$.

So, we have:

$$f(x)=x^2+\lambda x=0\Rightarrow x=0,x=-\lambda$$

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

The roots will be $x=0$ and $x=-\lambda$. If $x^2+\lambda x+\lambda =0$ has no real roots, then ${\lambda }^2-4\lambda <0$.

This implies $0<\lambda <4$, so $\lambda =1,2,3$.

However, $\lambda =0$ also satisfies the condition. Thus, there are 4 different pairs $(\lambda ,\mu )$ that exist: $(0,0),(1,0),(2,0),(3,0)$.