Question

Let $f(x) = \frac{x^4 - \lambda x^3 - 3x^2 + 3\lambda x}{x-\lambda}$. If range of f(x) is the set of intire real number then the true set in which $\lambda$ lies is

• A. [-2, 2]
• B. [0, 4]
• C. (1, 3)
• D. None of them

Answer 1

Answer: (A)

[-2, 2]

Answer 2

The given function is $f(x) = \frac{x^4 - \lambda x^3 - 3x^2 + 3\lambda x}{x-\lambda}$. The function is defined for $x \neq \lambda$.

We can factor the numerator: $x^4 - \lambda x^3 - 3x^2 + 3\lambda x = x^3(x - \lambda) - 3x(x - \lambda) = (x^3 - 3x)(x - \lambda)$.

So, for $x \neq \lambda$, we have $f(x) = \frac{(x^3 - 3x)(x - \lambda)}{x - \lambda} = x^3 - 3x$. Let $g(x) = x^3 - 3x$. Then $f(x) = g(x)$ for all $x \neq \lambda$.

The range of $f(x)$ is the set of values $g(x)$ takes for $x \in \mathbb{R} \setminus \{\lambda\}$. The function $g(x) = x^3 - 3x$ is a cubic polynomial. The range of $g(x)$ for $x \in \mathbb{R}$ is $\mathbb{R}$. The range of $f(x)$ is the set $\{g(x) \mid x \in \mathbb{R}, x \neq \lambda\}$. This set is equal to the range of $g(x)$ for $x \in \mathbb{R}$, which is $\mathbb{R}$, minus the value $g(\lambda)$, if $g(\lambda)$ is only attained by $g(x)$ at $x = \lambda$. If $g(\lambda)$ is attained by $g(x)$ at some value $x_0 \neq \lambda$, then $g(\lambda)$ is in the range of $f(x)$ because $f(x_0) = g(x_0) = g(\lambda)$ and $x_0 \neq \lambda$.

The range of $f(x)$ is the entire set of real numbers $\mathbb{R}$ if and only if the value $g(\lambda)$ is in the range of $f(x)$. This happens if and only if there exists some $x_0 \neq \lambda$ such that $f(x_0) = g(\lambda)$. Since $f(x_0) = g(x_0)$ for $x_0 \neq \lambda$, this condition becomes: there exists $x_0 \neq \lambda$ such that $g(x_0) = g(\lambda)$.

We need to find the values of $\lambda$ such that the equation $g(x) = g(\lambda)$ has at least one solution $x \neq \lambda$. $x^3 - 3x = \lambda^3 - 3\lambda$ $x^3 - \lambda^3 - 3x + 3\lambda = 0$ $(x - \lambda)(x^2 + x\lambda + \lambda^2) - 3(x - \lambda) = 0$ $(x - \lambda)(x^2 + x\lambda + \lambda^2 - 3) = 0$ The solutions are $x = \lambda$ or $x^2 + \lambda x + \lambda^2 - 3 = 0$.

The equation $g(x) = g(\lambda)$ always has the solution $x = \lambda$. For there to be a solution $x \neq \lambda$, the quadratic equation $x^2 + \lambda x + \lambda^2 - 3 = 0$ must have at least one real root. Let $h(x) = x^2 + \lambda x + \lambda^2 - 3$. The discriminant of this quadratic is $D = \lambda^2 - 4(1)(\lambda^2 - 3) = \lambda^2 - 4\lambda^2 + 12 = -3\lambda^2 + 12$. For the quadratic equation to have real roots, we must have $D \ge 0$. $-3\lambda^2 + 12 \ge 0$ $12 \ge 3\lambda^2$ $4 \ge \lambda^2$ $\lambda^2 \le 4$ $-2 \le \lambda \le 2$.

Therefore, the final answer is $\boxed{[-2, 2]}$.