Question

Let $f(x)$ be a cubic polynomial defined by $f(x)=\frac{x^{3}}{3}+(a-3)x^{2}+x-13$. Then, the sum of all possible value(s) of 'a' for which $f(x)$ has negative point of local minimum in the interval $[1, 5]$, is

Answer 1

1

Answer 2

To find the local extrema of $f(x)$, we first compute its derivative $f'(x)$:

$f(x)=\frac{x^{3}}{3}+(a-3)x^{2}+x-13$

$f'(x) = \frac{d}{dx}\left(\frac{x^{3}}{3}+(a-3)x^{2}+x-13\right) = x^2 + 2(a-3)x + 1$

For local extrema, we set $f'(x) = 0$:

$x^2 + 2(a-3)x + 1 = 0$

This is a quadratic equation. Let its roots be $x_1$ and $x_2$. For real roots, the discriminant $D$ must be non-negative:

$D = (2(a-3))^2 - 4(1)(1) = 4(a-3)^2 - 4 = 4((a-3)^2 - 1)$

For real roots, $D \ge 0 \implies (a-3)^2 - 1 \ge 0 \implies (a-3)^2 \ge 1$.

This implies $|a-3| \ge 1$, so $a-3 \ge 1$ or $a-3 \le -1$.

Thus, $a \ge 4$ or $a \le 2$.

Now, let's determine which root corresponds to a local minimum. We use the second derivative test:

$f''(x) = \frac{d}{dx}(x^2 + 2(a-3)x + 1) = 2x + 2(a-3)$

The roots of $f'(x)=0$ are $x = \frac{-2(a-3) \pm \sqrt{4((a-3)^2 - 1)}}{2} = -(a-3) \pm \sqrt{(a-3)^2 - 1}$.

Let $x_1 = -(a-3) - \sqrt{(a-3)^2 - 1}$ and $x_2 = -(a-3) + \sqrt{(a-3)^2 - 1}$. Note that $x_1 < x_2$.

Evaluate $f''(x)$ at these roots:

For $x_1$: $f''(x_1) = 2x_1 + 2(a-3) = 2(-(a-3) - \sqrt{(a-3)^2 - 1}) + 2(a-3) = -2\sqrt{(a-3)^2 - 1}$.

For $x_2$: $f''(x_2) = 2x_2 + 2(a-3) = 2(-(a-3) + \sqrt{(a-3)^2 - 1}) + 2(a-3) = 2\sqrt{(a-3)^2 - 1}$.

For a local minimum, $f''(x) > 0$.

If $D > 0$, then $\sqrt{(a-3)^2 - 1} > 0$. In this case, $f''(x_1) < 0$ (local maximum) and $f''(x_2) > 0$ (local minimum).

Thus, the point of local minimum is $x_{min} = x_2 = -(a-3) + \sqrt{(a-3)^2 - 1}$.

If $D = 0$, i.e., $a=4$ or $a=2$, then $f''(x_1)=f''(x_2)=0$. In this case, $f'(x)$ is a perfect square, e.g., for $a=2$, $f'(x)=(x-1)^2$. At $x=1$, $f'(1)=0, f''(1)=0, f'''(1)=2 \ne 0$. This indicates an inflection point, not a local extremum. So, for a local minimum to exist, we must have $D>0$, which means $a < 2$ or $a > 4$.

We need the local minimum to be in the interval $[1, 5]$, i.e., $1 \le x_{min} \le 5$.

Case 1: $a > 4$.

Let $k = a-3$. Then $k > 1$.

$x_{min} = -k + \sqrt{k^2 - 1}$.

Since $k > 1$, $k^2 - 1 < k^2$, so $\sqrt{k^2 - 1} < k$.

Therefore, $x_{min} = -k + \sqrt{k^2 - 1} < 0$.

This means $x_{min}$ is negative. This contradicts the condition $x_{min} \in [1, 5]$.

So, there are no solutions for $a > 4$.

Case 2: $a < 2$.

Let $k = a-3$. Then $k < -1$.

$x_{min} = -k + \sqrt{k^2 - 1}$.

Since $k < -1$, let $k = -m$ where $m > 1$.

$x_{min} = -(-m) + \sqrt{(-m)^2 - 1} = m + \sqrt{m^2 - 1}$.

Since $m > 1$, $m^2 - 1 > 0$.

So $x_{min} = m + \sqrt{m^2 - 1} > 1 + \sqrt{1^2 - 1} = 1$.

Thus, the condition $x_{min} \ge 1$ is always satisfied for $a < 2$.

Now we need to satisfy $x_{min} \le 5$:

$m + \sqrt{m^2 - 1} \le 5$.

$\sqrt{m^2 - 1} \le 5 - m$.

For this inequality to hold, we must have $5 - m \ge 0$, so $m \le 5$.

Since we are in the case $m > 1$, we have $1 < m \le 5$.

Square both sides of $\sqrt{m^2 - 1} \le 5 - m$:

$m^2 - 1 \le (5 - m)^2$

$m^2 - 1 \le 25 - 10m + m^2$

$-1 \le 25 - 10m$

$10m \le 26$

$m \le 2.6$.

Combining $1 < m \le 5$ and $m \le 2.6$, we get $1 < m \le 2.6$.

Recall that $m = -(a-3) = 3-a$.

So, $1 < 3-a \le 2.6$.

$1 < 3-a \implies a < 2$.

$3-a \le 2.6 \implies 0.4 \le a$.

Combining these, the range for $a$ is $0.4 \le a < 2$.

Next, we need to satisfy the condition that $f(x_{min}) < 0$ (the point of local minimum is negative, referring to the function value).

We know that $x_{min}$ is a root of $x^2 + 2(a-3)x + 1 = 0$, so $2(a-3) = -\frac{x_{min}^2+1}{x_{min}}$.

Substitute $a-3$ into $f(x_{min})$:

$f(x_{min}) = \frac{x_{min}^3}{3}+(a-3)x_{min}^2+x_{min}-13$

$f(x_{min}) = \frac{x_{min}^3}{3} + \left(-\frac{x_{min}^2+1}{2x_{min}}\right)x_{min}^2 + x_{min} - 13$

$f(x_{min}) = \frac{x_{min}^3}{3} - \frac{x_{min}(x_{min}^2+1)}{2} + x_{min} - 13$

$f(x_{min}) = \frac{2x_{min}^3 - 3x_{min}^3 - 3x_{min}}{6} + x_{min} - 13$

$f(x_{min}) = \frac{-x_{min}^3 - 3x_{min} + 6x_{min}}{6} - 13$

$f(x_{min}) = \frac{-x_{min}^3 + 3x_{min}}{6} - 13 = \frac{x_{min}(3-x_{min}^2)}{6} - 13$.

We need $f(x_{min}) < 0$:

$\frac{x_{min}(3-x_{min}^2)}{6} - 13 < 0$

$\frac{x_{min}(3-x_{min}^2)}{6} < 13$

$x_{min}(3-x_{min}^2) < 78$.

Let $g(x) = x(3-x^2) = 3x - x^3$. We need to check if $g(x_{min}) < 78$ for $x_{min} \in [1, 5]$.

To find the range of $g(x)$ on $[1, 5]$, we find $g'(x)$:

$g'(x) = 3 - 3x^2 = 3(1-x^2)$.

For $x \in [1, 5]$, $x^2 \ge 1$, so $1-x^2 \le 0$. Thus $g'(x) \le 0$, which means $g(x)$ is a decreasing function on $[1, 5]$.

The maximum value of $g(x)$ in this interval occurs at $x=1$:

$g(1) = 1(3-1^2) = 2$.

The minimum value of $g(x)$ in this interval occurs at $x=5$:

$g(5) = 5(3-5^2) = 5(3-25) = 5(-22) = -110$.

So, for $x_{min} \in [1, 5]$, we have $-110 \le g(x_{min}) \le 2$.

Since $g(x_{min}) \le 2$, the condition $g(x_{min}) < 78$ is always satisfied.

Therefore, $f(x_{min}) < 0$ is always true for $x_{min} \in [1, 5]$.

The only constraint on $a$ is $0.4 \le a < 2$.

The question asks for the "sum of all possible value(s) of 'a'". If 'a' is a real number, there are infinitely many values in this interval, and their sum is undefined. This phrasing typically implies that 'a' must be an integer, or specific discrete values are expected. In the context of JEE/NEET, if not explicitly stated, integer values are often implied for "sum of values" questions.

Assuming 'a' is an integer:

The integers in the interval $[0.4, 2)$ are $a=1$.

Let's verify $a=1$:

If $a=1$, then $a-3 = -2$.

$x_{min} = -(-2) + \sqrt{(-2)^2 - 1} = 2 + \sqrt{4-1} = 2 + \sqrt{3}$.

$1 < 2+\sqrt{3} \approx 2+1.732 = 3.732 \le 5$. This satisfies $x_{min} \in [1, 5]$.

Also, $f(2+\sqrt{3}) = \frac{(2+\sqrt{3})(3-(2+\sqrt{3})^2)}{6} - 13$.

$(2+\sqrt{3})^2 = 4+3+4\sqrt{3} = 7+4\sqrt{3}$.

$3-(7+4\sqrt{3}) = -4-4\sqrt{3}$.

$f(2+\sqrt{3}) = \frac{(2+\sqrt{3})(-4-4\sqrt{3})}{6} - 13 = \frac{-4(2+\sqrt{3})(1+\sqrt{3})}{6} - 13$

$= \frac{-2(2+2\sqrt{3}+\sqrt{3}+3)}{3} - 13 = \frac{-2(5+3\sqrt{3})}{3} - 13$

$= \frac{-10-6\sqrt{3}}{3} - 13 = -\frac{10}{3} - 2\sqrt{3} - 13 = -\frac{49}{3} - 2\sqrt{3}$.

Since $\frac{49}{3} \approx 16.33$ and $2\sqrt{3} \approx 3.46$, $f(x_{min})$ is clearly negative.

Thus, $a=1$ is the only possible integer value for 'a'.

The sum of all possible value(s) of 'a' is 1.

The final answer is $\boxed{1}$.

Explanation of the solution:

1. Find Critical Points: Calculate the first derivative $f'(x)$ and set it to zero to find the critical points. $f'(x) = x^2 + 2(a-3)x + 1 = 0$.
2. Determine Local Minimum: Calculate the second derivative $f''(x) = 2x + 2(a-3)$. The local minimum occurs at the root $x_{min}$ where $f''(x_{min}) > 0$. This root is $x_{min} = -(a-3) + \sqrt{(a-3)^2 - 1}$.
3. Conditions for Existence of Local Minimum: For real roots and a local minimum to exist (not an inflection point), the discriminant $D = 4((a-3)^2 - 1)$ must be strictly positive, i.e., $D > 0$. This implies $|a-3| > 1$, which means $a < 2$ or $a > 4$.
4. Condition on $x_{min}$ interval: The local minimum must lie in the interval $[1, 5]$.
   • If $a > 4$, then $x_{min} < 0$, which contradicts $x_{min} \in [1, 5]$. So, no solutions for $a > 4$.
   • If $a < 2$, let $m = 3-a$. Then $m > 1$. The local minimum $x_{min} = m + \sqrt{m^2-1}$. We need $1 \le m + \sqrt{m^2-1} \le 5$. The condition $m > 1$ ensures $m + \sqrt{m^2-1} > 1$. The condition $m + \sqrt{m^2-1} \le 5$ leads to $m \le 2.6$. Substituting back $m=3-a$, we get $0.4 \le a < 2$.
5. Condition on $f(x_{min})$: The problem states "negative point of local minimum", which implies $f(x_{min}) < 0$. We expressed $f(x_{min})$ in terms of $x_{min}$ as $f(x_{min}) = \frac{x_{min}(3-x_{min}^2)}{6} - 13$. For $x_{min} \in [1, 5]$, the function $g(x_{min}) = x_{min}(3-x_{min}^2)$ ranges from $-110$ (at $x_{min}=5$) to $2$ (at $x_{min}=1$). Since $g(x_{min}) \le 2$, then $\frac{g(x_{min})}{6} - 13 \le \frac{2}{6} - 13 = \frac{1}{3} - 13 = -\frac{38}{3}$, which is always less than 0. Thus, $f(x_{min}) < 0$ is always satisfied for $x_{min} \in [1, 5]$.
6. Find 'a' values: The overall condition for 'a' is $0.4 \le a < 2$. Assuming 'a' is an integer, the only possible value is $a=1$.
7. Sum of 'a' values: The sum is 1.