Question

Let a cubic polynomial be $f(x) = \frac{x^3}{3} - x^2 + ax + 2$. If all the values of $a$ for which $f(x)$ has a positive point of local maxima also satisfy the inequality $3x^2 - (b+1)x + b(b-2) < 0$ and range of values of $b$ is $[p, q]$, then find the value of $(p-q)$ _______

Answer 1

-1

Answer 2

To solve this problem, we need to follow these steps:

1. Determine the conditions on 'a' for which $f(x)$ has a positive point of local maxima.
2. Use this range of 'a' to find the range of 'b' that satisfies the given inequality.
3. Calculate the value of $(p-q)$.

Step 1: Analyze the cubic polynomial $f(x)$

The given polynomial is $f(x) = \frac{x^3}{3} - x^2 + ax + 2$.

1. Find the first derivative: $f'(x) = \frac{d}{dx}(\frac{x^3}{3} - x^2 + ax + 2) = x^2 - 2x + a$.

2. Find critical points: Set $f'(x) = 0$. $x^2 - 2x + a = 0$. For real critical points to exist, the discriminant of this quadratic equation must be non-negative: $D = (-2)^2 - 4(1)(a) = 4 - 4a$. So, $4 - 4a \ge 0 \implies 4 \ge 4a \implies a \le 1$.

3. Find the second derivative: $f''(x) = \frac{d}{dx}(x^2 - 2x + a) = 2x - 2$.

4. Determine the point of local maxima: The roots of $x^2 - 2x + a = 0$ are $x = \frac{-(-2) \pm \sqrt{4-4a}}{2} = \frac{2 \pm 2\sqrt{1-a}}{2} = 1 \pm \sqrt{1-a}$. Let $x_1 = 1 - \sqrt{1-a}$ and $x_2 = 1 + \sqrt{1-a}$. Note that $x_1 < x_2$.

   To identify a local maximum, we use the second derivative test:

   • $f''(x_1) = 2(1 - \sqrt{1-a}) - 2 = -2\sqrt{1-a}$.
   • $f''(x_2) = 2(1 + \sqrt{1-a}) - 2 = 2\sqrt{1-a}$.

   For a local maximum, $f''(x) < 0$. This occurs at $x_1 = 1 - \sqrt{1-a}$ because $-2\sqrt{1-a}$ is negative (or zero). For a strict local maximum, we need $f''(x_1) < 0$, which implies $-2\sqrt{1-a} < 0$, so $\sqrt{1-a} > 0$. This means $1-a > 0 \implies a < 1$. If $a=1$, then $f'(x) = (x-1)^2$, $f''(1)=0$, and $f'''(1)=2 \ne 0$. This indicates an inflection point, not a local extremum. So, for a local maximum to exist, we must have $a < 1$.

5. Apply the condition "positive point of local maxima": The point of local maxima is $x_{max} = 1 - \sqrt{1-a}$. We are given that this point must be positive: $1 - \sqrt{1-a} > 0$ $1 > \sqrt{1-a}$ Since both sides are positive, we can square both sides: $1^2 > (\sqrt{1-a})^2$ $1 > 1-a$ $0 > -a \implies a > 0$.

   Combining the conditions $a < 1$ and $a > 0$, the range of values for $a$ is $0 < a < 1$.

Step 2: Find the range of 'b'

The problem states that all values of $a$ for which $f(x)$ has a positive point of local maxima (i.e., $a \in (0,1)$) also satisfy the inequality $3x^2 - (b+1)x + b(b-2) < 0$. The most plausible interpretation is that the inequality must hold for $x=a$ for all $a \in (0,1)$. So, let $g(a) = 3a^2 - (b+1)a + b(b-2)$. We need $g(a) < 0$ for all $a \in (0,1)$.

Since $g(a)$ is a quadratic in $a$ with a positive leading coefficient (3 > 0), its parabola opens upwards. For $g(a)$ to be negative over an interval $(0,1)$, the roots of $g(a)=0$ must lie on either side of this interval. That is, $a=0$ and $a=1$ must lie between the roots. This implies two conditions:

1. $g(0) < 0$
2. $g(1) < 0$

Let's apply these conditions:

1. $g(0) = 3(0)^2 - (b+1)(0) + b(b-2) = b(b-2)$. We need $b(b-2) < 0$. This implies $0 < b < 2$.

2. $g(1) = 3(1)^2 - (b+1)(1) + b(b-2) = 3 - b - 1 + b^2 - 2b = b^2 - 3b + 2$. We need $b^2 - 3b + 2 < 0$. Factoring the quadratic: $(b-1)(b-2) < 0$. This implies $1 < b < 2$.

For both conditions to be satisfied, we take the intersection of $0 < b < 2$ and $1 < b < 2$. The intersection is $1 < b < 2$.

We should also check that the vertex of the parabola $g(a)$ lies within the interval $(0,1)$. The vertex is at $a_v = -\frac{-(b+1)}{2(3)} = \frac{b+1}{6}$. Since $1 < b < 2$, we have $2 < b+1 < 3$, so $\frac{2}{6} < \frac{b+1}{6} < \frac{3}{6}$. This means $\frac{1}{3} < a_v < \frac{1}{2}$. Since $a_v \in (0,1)$, the minimum value of $g(a)$ for $a \in (0,1)$ is $g(a_v)$. For $g(a) < 0$ for all $a \in (0,1)$, we must have $g(a_v) < 0$. $g(a_v) = \frac{11b^2 - 26b - 1}{12}$. We need $\frac{11b^2 - 26b - 1}{12} < 0 \implies 11b^2 - 26b - 1 < 0$. The roots of $11b^2 - 26b - 1 = 0$ are $b = \frac{26 \pm \sqrt{26^2 - 4(11)(-1)}}{2(11)} = \frac{26 \pm \sqrt{676 + 44}}{22} = \frac{26 \pm \sqrt{720}}{22} = \frac{26 \pm 12\sqrt{5}}{22} = \frac{13 \pm 6\sqrt{5}}{11}$. Numerically, $\frac{13 - 6\sqrt{5}}{11} \approx \frac{13 - 6(2.236)}{11} = \frac{13 - 13.416}{11} \approx -0.037$. And $\frac{13 + 6\sqrt{5}}{11} \approx \frac{13 + 13.416}{11} \approx 2.401$. So, $11b^2 - 26b - 1 < 0$ implies approximately $-0.037 < b < 2.401$.

The range of $b$ must satisfy all conditions: $1 < b < 2$ AND $-0.037 < b < 2.401$. The intersection of these ranges is $1 < b < 2$. So, the range of values of $b$ is $(1, 2)$.

Step 3: Find the value of $(p-q)$

The problem states that the range of values of $b$ is $[p, q]$. However, our result is an open interval $(1, 2)$. This usually implies that the question meant to ask for the range of $b$ to be an open interval, or that $p$ and $q$ are the infimum and supremum of the range. Assuming $p=1$ and $q=2$, then $(p-q) = (1-2) = -1$. If the question intended for a closed interval, there might be a subtle point missed, but based on the standard interpretation of "for all values in an open interval", the boundaries are excluded. Given the options for JEE/NEET, usually, such questions imply finding the interval $(p,q)$ and then calculating $p-q$.

Let's assume the question means $p=1$ and $q=2$. Then $(p-q) = 1 - 2 = -1$.

Final check: If $b=1$, $g(a) = 3a^2 - 2a - 1 = (3a+1)(a-1)$. For $a \in (0,1)$, $3a+1 > 1$ and $a-1 < 0$, so $g(a) < 0$. This works. If $b=2$, $g(a) = 3a^2 - 3a + 0 = 3a(a-1)$. For $a \in (0,1)$, $3a > 0$ and $a-1 < 0$, so $g(a) < 0$. This works. The boundary values $p=1$ and $q=2$ are indeed part of the valid range. So the range is $[1,2]$.

If $b=1$, $b(b-2) = 1(1-2) = -1$. $b^2-3b+2 = 1-3+2 = 0$. So $g(1)=0$. If $b=2$, $b(b-2) = 2(2-2) = 0$. $b^2-3b+2 = 4-6+2 = 0$. So $g(0)=0$. If $g(0) \le 0$ and $g(1) \le 0$ are the conditions, then the range for $b$ would be $[1,2]$. $b(b-2) \le 0 \implies 0 \le b \le 2$. $(b-1)(b-2) \le 0 \implies 1 \le b \le 2$. The intersection is $1 \le b \le 2$. And $g(a_v) = \frac{11b^2 - 26b - 1}{12} < 0$ still holds for $b \in [1,2]$ because $11(1)^2 - 26(1) - 1 = 11-26-1 = -16 < 0$ and $11(2)^2 - 26(2) - 1 = 44-52-1 = -9 < 0$. So the range is indeed $[1,2]$.

Thus, $p=1$ and $q=2$. The value of $(p-q) = 1 - 2 = -1$.

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