Q.Find two positive numbers whose sum is $15$ and the sum of whose squares is minimum.

We are given a discrete frequency distribution with values $x_i$ and frequencies $f_i$. The table is:

Notice that $x_i = 3$ appears twice in the table (with frequencies 5 and 1), so we should combine them: total frequency for $x=3$ is $5+1=6$. The corrected distribution is:

Now we compute the variance step by step.

1. Find total frequency $N$

   $N = \sum f_i = 3 + 6 + 3 = 12$.

2. Compute $\sum f_i x_i$ (the sum of all values)

   $\sum f_i x_i = (3 \times 1) + (6 \times 3) + (3 \times 5) = 3 + 18 + 15 = 36$.

3. Compute the mean $\bar{x}$

   $\bar{x} = \frac{\sum f_i x_i}{N} = \frac{36}{12} = 3$.

4. Compute $\sum f_i x_i^2$ (the sum of squares)

   $\sum f_i x_i^2 = (3 \times 1^2) + (6 \times 3^2) + (3 \times 5^2) = (3 \times 1) + (6 \times 9) + (3 \times 25) = 3 + 54 + 75 = 132$. …

We are given the parabola $y^2 = 4x$ and the point $P(-1, 4)$. We need the point $Q(h, k)$ on the parabola that is closest to $P$, then compute $h^2 + k^2 + d^2$, where $d = PQ$.

Concept and Intuition

The distance between two points is minimized when the squared distance is minimized (since square root is monotonic). For a point on the parabola, we can write its coordinates in terms of a parameter. The squared distance becomes a function of that parameter; we find its minimum by differentiation. Then we compute the required sum directly — no need to find $d$ separately, because $d^2$ is already part of the expression we minimize.

Step-by-step solution

1. Parameterize the parabola

   The parabola $y^2 = 4x$ can be written as $x = t^2$, $y = 2t$ (standard parametric form).

   So any point on it is $Q(t) = (t^2, 2t)$.

2. Write the squared distance from $P(-1, 4)$

because $x_Q - x_P = t^2 - (-1) = t^2 + 1$ and $y_Q - y_P = 2t - 4$.

3. Expand and simplify

4. Minimize by differentiation

Set $D'(t) = 0$:

5. Solve the cubic

   Try $t = 1$: $1 + 3 - 4 = 0$. So $t = 1$ is a root.

   Factor: $(t - 1)(t^2 + t + 4) = 0$.

   The quadratic $t^2 + t + 4 = 0$ has discriminant $1 - 16 = -15 < 0$, so no real roots.

   Hence the only real critical point is $t = 1$.

6. Verify it’s a minimum …

The key insight here is that the two series are not arbitrary arithmetic progressions — they are the natural sequences of odd and even numbers. The sum of the first $n$ odd numbers has a famously neat closed form: it’s simply $n^2$. Similarly, the sum of the first $m$ even numbers is $m(m+1)$. Once you recognise these patterns, the problem reduces to solving two simple equations.

Let’s work through it step by step.

1. The odd-number series

   The sequence $1, 3, 5, \ldots, l_1$ is just the first $n$ odd numbers, where $l_1$ is the $n$-th odd number. The $n$-th odd number is $2n-1$, so $l_1 = 2n-1$.

   The sum of the first $n$ odd numbers is $n^2$. We are told this sum is 1521.

   So $n^2 = 1521$. Taking the positive square root (since $n$ is a positive integer), $n = \sqrt{1521} = 39$.

   Therefore $l_1 = 2(39) - 1 = 78 - 1 = 77$.

2. The even-number series

   The sequence $2, 4, 6, \ldots, l_2$ is the first $m$ even numbers, where $l_2$ is the $m$-th even number. The $m$-th even number is $2m$, so $l_2 = 2m$.

   The sum of the first $m$ even numbers is $2 + 4 + 6 + \cdots + 2m = 2(1 + 2 + 3 + \cdots + m) = 2 \cdot \frac{m(m+1)}{2} = m(m+1)$.

   We are told this sum is 1722. So $m(m+1) = 1722$.

3. Solving for $m$ …

Factor by grouping:

The five roots are $-4, -3, -2, 2, 3$. With the ordering $\alpha < \beta < \gamma < 2 < \epsilon$: …

We are told that the sum of two roots of the quartic $x^4 - x^3 - 16x^2 + 4x + 48 = 0$ is zero. That means there exist two roots, say $\alpha$ and $\beta$, such that $\alpha + \beta = 0$, i.e., $\beta = -\alpha$. This is a powerful clue: it suggests the polynomial has a factor of the form $(x - \alpha)(x + \alpha) = x^2 - \alpha^2$. So the quartic can be factored into two quadratics, one of which is $x^2 - a$ for some $a > 0$ (since $\alpha^2 = a$). Our job is to find that factorization, then find all roots, and finally compute the sum of their fourth powers.

1. Set up the factorization Since two roots are opposites, the quartic must be divisible by $x^2 - k$ for some $k = \alpha^2$. Let’s write:

where the second quadratic has coefficients to be determined.

2. Expand and match coefficients Expanding:

Compare with the given polynomial:

We get the system:

3. Solve for $k$, $p$, $q$ From $p = -1$ and $-k p = 4$, we have $-k(-1) = k = 4$. So $k = 4$. Then $q - k = -16$ gives $q - 4 = -16 \Rightarrow q = -12$. Check $-k q = -4(-12) = 48$, which matches. So the factorization is:

4. Find all four roots
   • From $x^2 - 4 = 0$: $x = \pm 2$. …

Concept & Intuition

We need two different measures of spread: variance (average squared distance from the mean) and mean deviation from the median (average absolute distance from the median). The trick is to compute each carefully — variance uses the mean, mean deviation uses the median — and then subtract. A common mistake is using the wrong denominator (n vs n−1) or confusing median with mean.

Step-by-step solution

1. Find the mean

   Data: 2, 3, 5, 8, 12.

   Sum = 2 + 3 + 5 + 8 + 12 = 30.

   Number of values, n = 5.

   Mean, $\bar{x} = \frac{30}{5} = 6$.

2. Compute the variance $\sigma^2$ (population variance, denominator n)

   Deviations from mean:

   • $2 - 6 = -4$ → square = 16
   • $3 - 6 = -3$ → square = 9
   • $5 - 6 = -1$ → square = 1
   • $8 - 6 = 2$ → square = 4
   • $12 - 6 = 6$ → square = 36 Sum of squared deviations = 16 + 9 + 1 + 4 + 36 = 66. $\sigma^2 = \frac{66}{5} = 13.2$.

3. Find the median

   Sorted data: 2, 3, 5, 8, 12.

   Since n = 5 (odd), median is the middle value: 5.

4. Compute the mean deviation from the median (M) …

Concept and Intuition

We have an ellipse centered at the origin, with the major axis along the X‑axis and the minor axis along the Y‑axis. Its standard equation is

The foci are at $S = (-c, 0)$ and $S' = (c, 0)$, where $c^2 = a^2 - b^2$.

One end of the minor axis is $B = (0, b)$.

The condition $|S'SB| = \frac{\pi}{6}$ is a bit unusual: it’s an angle, not a length. The notation $|S'SB|$ means the angle $\angle S'SB$ — that is, the angle at vertex $S$ formed by points $S'$, $S$, and $B$. So we have $\angle S'SB = 30^\circ$.

This gives a geometric relation between $a$, $b$, and $c$. Then the point $(2\sqrt{3}, 1)$ lies on the ellipse, giving a second equation. Solving these yields $a^2$ and $b^2$, and the problem asks for the sum of the squares of the lengths of the major and minor axes: $(2a)^2 + (2b)^2 = 4(a^2 + b^2)$.

Step‑by‑step solution

1. Set up the geometry of the angle condition Points: $S' = (c, 0)$, $S = (-c, 0)$, $B = (0, b)$. The angle $\angle S'SB$ is at $S$, between vectors $\overrightarrow{SS'} = (2c, 0)$ and $\overrightarrow{SB} = (c, b)$. The cosine of the angle is given by the dot product:

Since $\cos 30^\circ = \frac{\sqrt{3}}{2}$, we have:

2. Simplify the relation Square both sides:

So $c = \sqrt{3}\,b$.

Recall $c^2 = a^2 - b^2$, hence:

Thus $a = 2b$.

3. Use the given point on the ellipse The point $(2\sqrt{3}, 1)$ satisfies: $$\frac{(2\sqrt{3})^2}{a^2} + \frac{1^2}{b^2} = 1 …$$

The four given lines form a square because they come in two pairs of parallel lines:

• $x+y=2$ and $x+y=6$ are parallel (slope $-1$).
• $x-y=-1$ and $x-y=-5$ are parallel (slope $1$).

These two families are perpendicular (slopes $-1$ and $1$), so they indeed form a square. The inscribed circle (incircle) touches all four sides — its center is the intersection of the diagonals, i.e., the center of the square, and its radius is half the distance between each pair of parallel lines.

1. Find the center of the square. The center lies at the intersection of the lines that are midway between each pair of parallel sides.
   • Midline between $x+y=2$ and $x+y=6$: average constant $= \frac{2+6}{2}=4$, so $x+y=4$.
   • Midline between $x-y=-1$ and $x-y=-5$: average constant $= \frac{-1+(-5)}{2} = -3$, so $x-y=-3$. Solve:

Adding: $2x = 1 \Rightarrow x = \frac12$. Then $y = 4 - x = \frac72$.

So the center is $C\left(\frac12, \frac72\right)$.

2. Find the radius of the inscribed circle. The distance between the two parallel lines $x+y=2$ and $x+y=6$ is

This is the side length of the square. The incircle’s radius is half the side length:

3. Write the equation of the circle. Center $(h,k) = \left(\frac12, \frac72\right)$, radius $r = \sqrt{2}$. Equation:

Expand:

Combine constants: $\frac14 + \frac{49}{4} = \frac{50}{4} = \frac{25}{2}$.

So:

The problem asks for $\text{Var}(X) + (E[X])^2$. A useful identity simplifies this:

so

That means we don’t need to compute the variance separately — just find $E[X^2]$ directly. But first, we must determine $K$ from the probability distribution.

1. Find $K$ using $\sum P(X=r) = 1$ The values of $r$ are $-2, -1, 0, 1, 2, 3$.

So:

Compute the squares: $4 + 1 + 0 + 1 + 4 + 9 = 19$

Hence $K \cdot 19 = 1$, so

2. Compute $E[X]$ (the mean)

Sum $r^3$ for $r = -2, -1, 0, 1, 2, 3$:

$(-8) + (-1) + 0 + 1 + 8 + 27 = 27$

So

3. Compute $E[X^2]$ …