Question

The function $f(x)=x^3+\lambda x^2+5x+sin2x$ will be an invertible function if $\lambda$ belongs to

• A. (-\infty,-3)
• B. (-3,3)
• C. (3,+\infty)
• D. (-3,8)

Answer 1

Answer: (B)

2

Answer 2

For $f(x)$ to be invertible, it must be strictly monotonic. Since $f(x)$ is a polynomial of degree 3 plus $\sin(2x)$, its derivative $f'(x)$ will determine its monotonicity.

$f'(x) = 3x^2 + 2\lambda x + 5 + 2\cos2x$.

As $x \to \pm\infty$, $3x^2 \to \infty$, so $f'(x)$ will eventually be positive. Thus, $f(x)$ must be strictly increasing, which means $f'(x) \ge 0$ for all $x \in \mathbb{R}$.

Let $g(x) = 3x^2 + 2\lambda x + 5$. We need $g(x) + 2\cos2x \ge 0$.

Since $2\cos2x \ge -2$, a sufficient condition for $f'(x) \ge 0$ is $g(x) \ge 2$ for all $x$.

The minimum value of $g(x)$ occurs at $x = -\frac{2\lambda}{2 \cdot 3} = -\frac{\lambda}{3}$.

Minimum value of $g(x) = 3(-\frac{\lambda}{3})^2 + 2\lambda(-\frac{\lambda}{3}) + 5 = \frac{\lambda^2}{3} - \frac{2\lambda^2}{3} + 5 = 5 - \frac{\lambda^2}{3}$.

We require $5 - \frac{\lambda^2}{3} \ge 2$, which simplifies to $3 \ge \frac{\lambda^2}{3}$, or $\lambda^2 \le 9$.

This gives $-3 \le \lambda \le 3$.

For $\lambda \in [-3, 3]$, we have $3x^2 + 2\lambda x + 5 \ge 2$.

Therefore, $f'(x) = (3x^2 + 2\lambda x + 5) + 2\cos2x \ge 2 + 2\cos2x$.

Since $2+2\cos2x \ge 0$, we have $f'(x) \ge 0$ for all $x \in \mathbb{R}$.

For $f(x)$ to be invertible, $f'(x)$ must not be zero on any interval.

$f'(x)=0$ only if $3x^2+2\lambda x+5=2$ and $2\cos2x=-2$, i.e., $3x^2+2\lambda x+3=0$ and $\cos2x=-1$.

The quadratic $3x^2+2\lambda x+3=0$ has real roots only if its discriminant $D=(2\lambda)^2-4(3)(3) = 4\lambda^2-36 \ge 0$, meaning $\lambda^2 \ge 9$.

This condition contradicts $\lambda^2 \le 9$, unless $\lambda^2 = 9$, i.e., $\lambda = \pm 3$.

If $\lambda=3$, $3x^2+6x+3=0 \implies 3(x+1)^2=0 \implies x=-1$. At $x=-1$, $\cos(2(-1))=\cos(2) \ne -1$. So $f'(-1) \ne 0$.

If $\lambda=-3$, $3x^2-6x+3=0 \implies 3(x-1)^2=0 \implies x=1$. At $x=1$, $\cos(2(1))=\cos(2) \ne -1$. So $f'(1) \ne 0$.

Thus, for $\lambda \in [-3, 3]$, $f'(x)$ is always strictly greater than zero.

Therefore, $f(x)$ is strictly increasing for $\lambda \in [-3, 3]$, making it invertible.

Among the given options, $(-3,3)$ is the correct choice.