CBSE Class 10 Maths Chapter 2 Polynomials Model Questions - 4 Marks
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CBSE Class 10 Maths Chapter 2 Polynomials Model Questions - 4 Marks
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SECTION D — Case-Based/Source-Based Integrated Questions
[4 Marks Each]
Q1. Case Study: The Gateway Arch (Structural Architecture)
An architect designs a grand entrance monument for an international exhibition. The shape of the main structural arch resembles a mathematical curve known as a parabola, opening downwards. The path of the archway can be modeled by the quadratic polynomial $p(x) = -x^2 + 6x + 16$, where $x$ represents the horizontal distance from the left pillar base (in meters) and $p(x)$ represents the height of the arch at that point.
1 MarksSub-question (a): Find the number of points where the arch touches the ground level.
Answer:
1. Ground level corresponds to the height $p(x) = 0$.
2. Set the quadratic polynomial equal to zero: $-x^2 + 6x + 16 = 0$, or $x^2 - 6x - 16 = 0$.
4. Since the discriminant is positive ($100 > 0$), the quadratic equation has two distinct real roots, meaning there are two distinct points where the arch touches the ground.
Result: 2 points
1 MarksSub-question (b): Determine the horizontal coordinates of the two base pillars by finding the zeroes of $p(x)$.
Answer:
1. Set the polynomial to zero: $-x^2 + 6x + 16 = 0$.
4. Solve for $x$: $x = 8$ or $x = -2$. (Since distance cannot be negative, the left base is at $x = -2$ relative to the reference frame or solved mathematically, yielding zeroes at $-2$ and $8$).
Result: $x = -2$ and $x = 8$ (or base points at $-2\text{ m}$ and $8\text{ m}$)
2 MarksSub-question (c): Calculate the maximum height of the archway and find the coordinate point where this peak occurs.
Answer:
1. For a downward-opening parabola given by $p(x) = ax^2 + bx + c$, the maximum value occurs at the vertex $x = -\frac{b}{2a}$.
2. Here, $a = -1$ and $b = 6$. Substitute these values: $x = -\frac{6}{2(-1)} = 3$.
3. Substitute $x = 3$ back into the polynomial to find the maximum height: $p(3) = -(3)^2 + 6(3) + 16 = -9 + 18 + 16 = 25$.
Result: The maximum height is $25\text{ meters}$, and the peak coordinate is $(3, 25)$.
Q2. Case Study: The Path of a Roller Coaster (Amusement Park Engineering)
An engineer models a steep drop and climb sequence on a new roller coaster track using a coordinate plane. The track profile for this segment corresponds to the shape of a parabola opening upwards. The track height relative to safety beams is represented by the polynomial $f(x) = x^2 - 4x - 5$.
1 MarksSub-question (a): What type of polynomial is represented by the track profile, and what is its maximum number of real zeroes?
Answer:
1. The track profile is given by the polynomial $f(x) = x^2 - 4x - 5$, which has a degree of 2.
2. A polynomial of degree 2 is known as a quadratic polynomial.
3. The maximum number of real zeroes for a polynomial of degree $n$ is $n$. Therefore, a quadratic polynomial has a maximum of 2 real zeroes.
Result: Quadratic polynomial; Maximum of 2 real zeroes.
1 MarksSub-question (b): Find the zeroes of the polynomial $f(x)$ by using the factorization method.
Answer:
1. Set the polynomial equal to zero to find the zeroes: $x^2 - 4x - 5 = 0$.
2. Factor the quadratic expression by splitting the middle term: find two numbers that multiply to $-5$ and add up to $-4$. These numbers are $-5$ and $+1$.
5. Set each factor to zero: $x + 1 = 0 \implies x = -1$, and $x - 5 = 0 \implies x = 5$.
Result: The zeroes of the polynomial are $x = -1$ and $x = 5$.
2 MarksSub-question (c): Verify the relationship between the zeroes and the coefficients by calculating the sum and product of the zeroes directly from the polynomial structure.
Answer:
1. Identify the coefficients from the given polynomial $f(x) = x^2 - 4x - 5$ in the standard form $ax^2 + bx + c$:
$a = 1$, $b = -4$, $c = -5$.
2. From part (b), the zeroes are $\alpha = -1$ and $\beta = 5$.
3. Verify Sum of Zeroes:
• Direct sum from zeroes: $\alpha + \beta = (-1) + 5 = 4$.
• Formula $-\frac{b}{a}$: $-\frac{-4}{1} = 4$.
• Since $4 = 4$, the sum of zeroes is verified.
4. Verify Product of Zeroes:
• Direct product from zeroes: $\alpha \cdot \beta = (-1) \cdot (5) = -5$.
• Formula $\frac{c}{a}$: $\frac{-5}{1} = -5$.
• Since $-5 = -5$, the product of zeroes is verified.
Result: Relationship between zeroes and coefficients successfully verified.
Q3. Case Study: Throwing a Basketball (Sports Trajectories)
During a practice drill, a basketball player releases a ball toward the hoop. The flight path of the basketball travels along a parabolic arc modeled by a quadratic expression. Let $\alpha$ and $\beta$ be the real zeroes of this polynomial. It is observed that the sum of the zeroes of this trajectory is $5$ and the product of the zeroes is $4$.
1 MarksSub-question (a): Formulate the quadratic polynomial $p(x)$ representing the basketball's path, assuming the leading coefficient $a = -1$.
Answer:
1. A general quadratic polynomial given its sum of zeroes ($S$) and product of zeroes ($P$) is expressed as $p(x) = a(x^2 - Sx + P)$.
2. From the problem, the sum of zeroes $S = 5$ and the product of zeroes $P = 4$.
3. Substitute the given leading coefficient $a = -1$, sum ($5$), and product ($4$) into the formula:
$p(x) = -1(x^2 - 5x + 4)$
4. Expand the expression:
$p(x) = -x^2 + 5x - 4$
Result: $p(x) = -x^2 + 5x - 4$
1 MarksSub-question (b): Find the individual values of the zeroes $\alpha$ and $\beta$.
Answer:
1. From the previous part, the polynomial is $p(x) = -x^2 + 5x - 4$. To find the zeroes, set $p(x) = 0$:
$-x^2 + 5x - 4 = 0 \implies x^2 - 5x + 4 = 0$.
2. Factor the quadratic expression by splitting the middle term: find two numbers that multiply to $4$ and add up to $-5$. These numbers are $-1$ and $-4$.
4. Set each factor to zero:
$x - 1 = 0 \implies x = 1$
$x - 4 = 0 \implies x = 4$
Result: The individual values of the zeroes are $\alpha = 1$ and $\beta = 4$ (or vice versa).
2 MarksSub-question (c): If the path equation shifts due to a different release angle such that the new polynomial is $g(x) = x^2 - 5x + 6$, calculate the value of $\alpha^2 + \beta^2$ for the new zeroes.
Answer:
1. Identify the coefficients from the new polynomial $g(x) = x^2 - 5x + 6$ in the standard form $ax^2 + bx + c$:
$a = 1$, $b = -5$, $c = 6$.
2. Find the sum and product of the new zeroes ($\alpha$ and $\beta$) using relations:
• Sum of zeroes: $\alpha + \beta = -\frac{b}{a} = -\frac{-5}{1} = 5$.
• Product of zeroes: $\alpha \cdot \beta = \frac{c}{a} = \frac{6}{1} = 6$.
3. Use the algebraic identity for $\alpha^2 + \beta^2$:
$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$
4. Substitute the values of the sum and product into the identity:
$\alpha^2 + \beta^2 = (5)^2 - 2(6) = 25 - 12 = 13$.
Result: $\alpha^2 + \beta^2 = 13$
Q4. Case Study: Highway Overpass Tunnel (Civil Engineering)
A concrete overpass tunnel is built across a national highway. The inner cross-sectional boundary of the tunnel is engineered as a parabolic arch opening downwards. The shape is mathematically defined by the polynomial $p(x) = -x^2 + 2x + 8$.
1 MarksSub-question (a): State whether the value of the coefficient of $x^2$ is positive or negative for a parabola that opens downwards.
Answer:
1. In a standard quadratic polynomial $p(x) = ax^2 + bx + c$, the sign of the leading coefficient $a$ (the coefficient of $x^2$) determines the direction the parabola opens.
2. When $a$ is positive ($a > 0$), the parabola opens upwards.
3. When $a$ is negative ($a < 0$), the parabola opens downwards.
Result: Negative
1 MarksSub-question (b): Find the width of the tunnel at its base on the ground level.
Answer:
1. Ground level corresponds to the height $p(x) = 0$.
2. Set the polynomial equal to zero: $-x^2 + 2x + 8 = 0$, or $x^2 - 2x - 8 = 0$.
3. Factor the quadratic equation by splitting the middle term: find two numbers that multiply to $-8$ and add up to $-2$. These numbers are $-4$ and $+2$.
5. Set each factor to zero to find the roots (base points): $x = -2$ and $x = 4$.
6. The width of the tunnel at the base is the distance between these two points: $\text{Width} = |4 - (-2)| = 6$ meters.
Result: 6 meters
2 MarksSub-question (c): If a new design update replaces the tunnel model with a polynomial whose zeroes are the negatives of the zeroes of $p(x)$, form this new quadratic polynomial.
Answer:
1. From part (b), the zeroes of the original polynomial $p(x) = -x^2 + 2x + 8$ are $x_1 = -2$ and $x_2 = 4$.
2. The zeroes of the new polynomial are the negatives of these zeroes:
$\alpha = -(-2) = 2$
$\beta = -(4) = -4$
3. Calculate the sum ($S$) and product ($P$) of these new zeroes:
• Sum ($S$) = $2 + (-4) = -2$
• Product ($P$) = $2 \cdot (-4) = -8$
4. Assuming the standard leading coefficient $a = 1$ (or keeping the same shape scaling where applicable, typically $a=1$ for a monic polynomial), form the new quadratic polynomial using $g(x) = x^2 - Sx + P$:
$g(x) = x^2 - (-2)x + (-8)$
$g(x) = x^2 + 2x - 8$
Result: $g(x) = x^2 + 2x - 8$ (or any non-zero scalar multiple of it)
Q5. Case Study: Corporate Sales Revenue (Financial Tracking)
A manufacturing company tracks its monthly sales revenue performance over a seasonal production cycle. The financial analysts discover that the net profit trend matches a quadratic trajectory. The zeroes of this profit polynomial are $\alpha = 3$ and $\beta = -2$.
1 MarksSub-question (a): Write the standard expression used to construct a quadratic polynomial when its zeroes are given.
Answer:
1. If $\alpha$ and $\beta$ are the zeroes of a quadratic polynomial, the polynomial can be expressed in terms of its sum of zeroes ($S = \alpha + \beta$) and product of zeroes ($P = \alpha\beta$).
2. The standard expression is:
$p(x) = k(x^2 - Sx + P)$
or expanded as:
$p(x) = k(x^2 - (\alpha + \beta)x + \alpha\beta)$
where $k$ is a non-zero real number (scaling factor).
1 MarksSub-question (b): Form the specific quadratic polynomial representing this net profit model.
Answer:
1. Given the zeroes $\alpha = 3$ and $\beta = -2$, calculate their sum ($S$) and product ($P$):
• Sum ($S$) = $3 + (-2) = 1$
• Product ($P$) = $3 \cdot (-2) = -6$
2. Substitute these values into the standard quadratic polynomial formula $p(x) = k(x^2 - Sx + P)$, choosing $k = 1$ for the standard monic polynomial:
$p(x) = x^2 - (1)x + (-6)$
$p(x) = x^2 - x - 6$
Result: $p(x) = x^2 - x - 6$ (or any non-zero real multiple of it)
2 MarksSub-question (c): Find a new polynomial whose zeroes are $2\alpha$ and $2\beta$.
Answer:
1. From the given information, the original zeroes are $\alpha = 3$ and $\beta = -2$.
2. The new zeroes are given as $2\alpha$ and $2\beta$:
• New zero 1 = $2(3) = 6$
• New zero 2 = $2(-2) = -4$
3. Calculate the sum ($S'$) and product ($P'$) of these new zeroes:
• Sum ($S'$) = $6 + (-4) = 2$
• Product ($P'$) = $6 \cdot (-4) = -24$
4. Form the new quadratic polynomial using the standard form $g(x) = k(x^2 - S'x + P')$ with $k = 1$:
$g(x) = x^2 - 2x - 24$
Result: $g(x) = x^2 - 2x - 24$ (or any non-zero scalar multiple of it)
Q6. Case Study: Olympic Javelin Throw (Projectile Dynamics)
An athlete throws a javelin during an Olympic field event. The height of the tip of the javelin above the ground at any horizontal displacement $x$ is dictated by a specific polynomial function. The curve crosses the horizontal reference axis at points representing its roots. The product of these roots for the javelin's equation $2x^2 - 8x + k$ is exactly $3$.
1 MarksSub-question (a): Find the numerical value of the constant term $k$.
Answer:
1. Identify the coefficients from the given quadratic polynomial $2x^2 - 8x + k$ in the standard form $ax^2 + bx + c$:
$a = 2$, $b = -8$, $c = k$.
2. Use the relation for the product of the zeroes (roots) of a quadratic polynomial, which is given by $\frac{c}{a}$.
3. According to the problem, the product of the roots is $3$:
$\frac{c}{a} = 3 \implies \frac{k}{2} = 3$
4. Solve for $k$:
$k = 3 \cdot 2 = 6$
Result: $k = 6$
1 MarksSub-question (b): Write down the complete updated polynomial after substituting the value of $k$.
Answer:
1. From part (a), we found that the constant term $k = 6$.
2. Substitute $k = 6$ into the original polynomial expression $2x^2 - 8x + k$:
$p(x) = 2x^2 - 8x + 6$
Result: $p(x) = 2x^2 - 8x + 6$
2 MarksSub-question (c): Find the exact zeroes of this updated polynomial and verify their sum against the coefficient ratio.
Answer:
1. The updated polynomial from part (b) is $p(x) = 2x^2 - 8x + 6$. To find the zeroes, set $p(x) = 0$ and simplify by dividing the entire equation by 2:
$x^2 - 4x + 3 = 0$
2. Factor the quadratic expression by splitting the middle term: find two numbers that multiply to $3$ and add up to $-4$. These numbers are $-1$ and $-3$.
4. Set each factor to zero to find the exact zeroes: $\alpha = 1$ and $\beta = 3$.
5. Verify Sum of Zeroes:
• Direct sum from zeroes: $\alpha + \beta = 1 + 3 = 4$.
• Formula $-\frac{b}{a}$ using original coefficients ($a=2, b=-8$): $-\frac{-8}{2} = \frac{8}{2} = 4$.
• Since $4 = 4$, the sum of zeroes is successfully verified.
Result: Zeroes are $x = 1$ and $x = 3$; sum verified as $4$.
Q7. Case Study:Suspension Bridge Cables (Infrastructure Design)
The main load-bearing cable of a suspension bridge hangs between two high concrete towers. The curve formed by the heavy cable under a uniform load forms a symmetric parabola. The algebraic curve can be traced by the polynomial $f(x) = 3x^2 - 12x + 9$.
a) Find the value of $x$ where the cable reaches its lowest point relative to the deck by analyzing the symmetry axis $x = -b/(2a)$. (1 Mark)
1 MarksSub-question (a): Find the numerical value of the constant term $k$.
Answer:
1. Identify the coefficients from the polynomial $f(x) = 3x^2 - 12x + 9$ in standard form $ax^2 + bx + c$:
$a = 3$, $b = -12$, $c = 9$.
2. Use the axis of symmetry formula for a parabola to find the lowest point (vertex):
$x = -\frac{b}{2a}$
3. Substitute the values of $a$ and $b$ into the formula:
$x = -\frac{-12}{2(3)}$
$x = \frac{12}{6} = 2$
Result: The cable reaches its lowest point at $x = 2$.
b) Find the points where the cable coordinates evaluate to zero on this coordinate system. (1 Mark)
1 MarksSub-question (b): Write down the complete updated polynomial after substituting the value of $k$.
Answer:
1. The points where the cable coordinates evaluate to zero refer to the zeroes (or roots) of the polynomial where $f(x) = 0$.
2. Set the polynomial equal to zero:
$3x^2 - 12x + 9 = 0$
3. Divide the entire equation by $3$ to simplify:
$x^2 - 4x + 3 = 0$
4. Factor the quadratic equation by splitting the middle term:
$x^2 - 3x - x + 3 = 0$
$x(x - 3) - 1(x - 3) = 0$
$(x - 1)(x - 3) = 0$
5. Solve for $x$:
$x = 1$ or $x = 3$
Result: The points where the cable coordinates evaluate to zero are $(1, 0)$ and $(3, 0)$.
c) Find the value of $\frac{1}{\alpha} + \frac{1}{\beta}$, where $\alpha$ and $\beta$ are the zeroes of the cable polynomial. (2 Marks)
2 MarksSub-question (c): Find the exact zeroes of this updated polynomial and verify their sum against the coefficient ratio.
Answer:
1. Identify the coefficients from the polynomial $f(x) = 3x^2 - 12x + 9$ in standard form $ax^2 + bx + c$:
$a = 3$, $b = -12$, $c = 9$.
2. Find the sum and product of the zeroes ($\alpha$ and $\beta$) using coefficient relationships:
• Sum ($\alpha + \beta$) = $-\frac{b}{a} = -\frac{-12}{3} = \frac{12}{3} = 4$
• Product ($\alpha\beta$) = $\frac{c}{a} = \frac{9}{3} = 3$
3. Express the target expression in terms of the sum and product of zeroes:
$\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta}$
4. Substitute the values of the sum ($4$) and product ($3$) into the expression:
$\frac{1}{\alpha} + \frac{1}{\beta} = \frac{4}{3}$
Result: The value of $\frac{1}{\alpha} + \frac{1}{\beta}$ is $\frac{4}{3}$.
Q8. Case Study: Satellite Dish Receiver (Telecommunications)
A telecommunications engineer designs a parabolic satellite dish to focus incoming signals onto a central receiver unit. The cross-section curve of the dish is modeled by a quadratic polynomial $p(x) = x^2 - kx + 12$. The difference between the two zeroes of this polynomial is given as $\alpha - \beta = 1$.
1 MarksSub-question (a): Express $(\alpha - \beta)^2$ explicitly in terms of $(\alpha + \beta)$ and $\alpha\beta$.
Answer:
1. Start with the algebraic expansion of $(\alpha + \beta)^2$:
$(\alpha + \beta)^2 = \alpha^2 + 2\alpha\beta + \beta^2$
2. Rearrange the terms to isolate $\alpha^2 + \beta^2$:
$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$
3. Now consider the square of the difference, $(\alpha - \beta)^2$:
$(\alpha - \beta)^2 = \alpha^2 - 2\alpha\beta + \beta^2$
1 MarksSub-question (b): Substitute the known properties of coefficients from $p(x)$ to find the value of $k$.
Answer:
1. Identify the coefficients from the polynomial $p(x) = x^2 - kx + 12$ in standard form $ax^2 + bx + c$:
$a = 1$, $b = -k$, $c = 12$.
2. Find the sum and product of the zeroes in terms of $k$ using relationship formulas:
• Sum ($\alpha + \beta$) = $-\frac{b}{a} = -\frac{-k}{1} = k$
• Product ($\alpha\beta$) = $\frac{c}{a} = \frac{12}{1} = 12$
3. From part (a), we have the identity: $(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta$.
4. Given that $\alpha - \beta = 1$, substitute the known values into the identity:
$(1)^2 = (k)^2 - 4(12)$
$1 = k^2 - 48$
5. Solve for $k^2$ and $k$:
$k^2 = 1 + 48 = 49$
$k = \pm\sqrt{49} = \pm 7$
Result: $k = 7$ or $k = -7$
2 MarksSub-question (c): Find the actual zeroes of the polynomial using the calculated value of $k$.
Answer:
1. From part (b), we found that $k = \pm 7$. We can test both cases to find the zeroes:
2. Case 1: If $k = 7$, the polynomial becomes $p(x) = x^2 - 7x + 12$.
• Set $p(x) = 0$: $x^2 - 7x + 12 = 0$
• Factor by splitting the middle term (numbers multiplying to $12$ and adding to $-7$ are $-3$ and $-4$):
$(x - 3)(x - 4) = 0$
• Zeroes are $x = 3$ and $x = 4$. (Note: Their difference is $4 - 3 = 1$, which matches the given condition).
3. Case 2: If $k = -7$, the polynomial becomes $p(x) = x^2 + 7x + 12$.
• Set $p(x) = 0$: $x^2 + 7x + 12 = 0$
• Factor (numbers multiplying to $12$ and adding to $7$ are $3$ and $4$):
$(x + 3)(x + 4) = 0$
• Zeroes are $x = -3$ and $x = -4$. (Note: Their difference is $|-3 - (-4)| = 1$, which also matches the condition).
Result:
• If $k = 7$, the zeroes are $3$ and $4$.
• If $k = -7$, the zeroes are $-3$ and $-4$.
Q9. Case Study: Undersea Research Submersible (Oceanographic Dives)
An undersea research drone dives from the surface of the ocean, levels out to collect seabed core samples, and then ascends back up to the research ship. The depth profile relative to the time axis follows a clean parabola. The polynomial tracking this dive path is $p(x) = x^2 - 7x + 10$.
1 MarksSub-question (a): At what time coordinates does the drone cross the ocean surface baseline (where depth equals zero)?
Answer:
1. Crossing the ocean surface baseline means the depth polynomial $p(x) = 0$.
2. Set the polynomial equal to zero: $x^2 - 7x + 10 = 0$.
3. Factor the quadratic expression by splitting the middle term: find two numbers that multiply to $10$ and add up to $-7$. These numbers are $-2$ and $-5$.
5. Set each factor to zero to find the time coordinates: $x = 2$ and $x = 5$.
Result: $x = 2$ and $x = 5$ units of time
1 MarksSub-question (b): What is the product of the zeroes of this dive path polynomial?
Answer:
1. Identify the coefficients from the polynomial $p(x) = x^2 - 7x + 10$ in standard form $ax^2 + bx + c$:
$a = 1$, $b = -7$, $c = 10$.
2. Use the relation for the product of the zeroes of a quadratic polynomial, which is given by $\frac{c}{a}$：
$\text{Product} = \frac{c}{a} = \frac{10}{1} = 10$.
3. Alternatively, using the zeroes found in part (a) ($2$ and $5$):
$2 \cdot 5 = 10$.
Result: $10$
2 MarksSub-question (c): Form a new quadratic polynomial whose zeroes are $\frac{1}{\alpha}$ and $\frac{1}{\beta}$.
Answer:
1. From the original polynomial $p(x) = x^2 - 7x + 10$, the sum and product of the zeroes ($\alpha$ and $\beta$) are:
• Sum ($\alpha + \beta$) = $7$
• Product ($\alpha\beta$) = $10$
2. Let the new zeroes be $\alpha' = \frac{1}{\alpha}$ and $\beta' = \frac{1}{\beta}$.
3. Calculate the sum of the new zeroes ($S'$):
$S' = \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{7}{10}$
4. Calculate the product of the new zeroes ($P'$):
$P' = \left(\frac{1}{\alpha}\right) \cdot \left(\frac{1}{\beta}\right) = \frac{1}{\alpha\beta} = \frac{1}{10}$
5. Form the new quadratic polynomial using $g(x) = k(x^2 - S'x + P')$. Choosing $k = 10$ to clear the fractions gives a clean integer-coefficient polynomial:
$g(x) = 10 \left(x^2 - \frac{7}{10}x + \frac{1}{10}\right)$
$g(x) = 10x^2 - 7x + 1$
Result: $10x^2 - 7x + 1$ (or any non-zero scalar multiple, such as $x^2 - \frac{7}{10}x + \frac{1}{10}$)
Q10. Case Study: High-Diver Trajectory (Aquatic Sports)
A platform diver leaps upward and outward from a 10-meter tower before falling into the pool. The flight curve is a downward-facing parabola. The quadratic polynomial representing this motion is $f(x) = -x^2 + 5x + 6$.
1 MarksSub-question (a): Find the total horizontal distance traveled by the diver when they hit the water surface level ($f(x) = 0$).
Answer:
1. Hitting the water surface level corresponds to the height function being zero: $f(x) = 0$.
2. Set the polynomial equal to zero: $-x^2 + 5x + 6 = 0$, or multiply by $-1$ to simplify: $x^2 - 5x - 6 = 0$.
3. Factor the quadratic equation by splitting the middle term: find two numbers that multiply to $-6$ and add up to $-5$. These numbers are $-6$ and $+1$.
3. Physical Representation: The y-intercept represents the initial height of the diver above the water level when the horizontal distance $x = 0$ (i.e., at the starting edge of the diving platform).
Result: y-intercept is $6$, representing the initial height of $6$ meters from which the dive begins (or the vertical offset of the trajectory model at launch).
2 MarksSub-question (c): Find the value of the algebraic expression $\alpha^3 + \beta^3$ using the zeroes of $f(x)$.
Answer:
1. From part (a), the zeroes of the polynomial $f(x) = -x^2 + 5x + 6$ are $\alpha = 6$ and $\beta = -1$ (or vice versa).
2. Calculate the sum ($S$) and product ($P$) of the zeroes using coefficient relationships ($a = -1, b = 5, c = 6$):
• Sum ($\alpha + \beta$) = $-\frac{b}{a} = -\frac{5}{-1} = 5$
• Product ($\alpha\beta$) = $\frac{c}{a} = \frac{6}{-1} = -6$
3. Use the algebraic identity for the sum of cubes:
$\alpha^3 + \beta^3 = (\alpha + \beta)(\alpha^2 - \alpha\beta + \beta^2)$
or alternatively:
$\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta)$
4. Substitute the values of sum ($5$) and product ($-6$) into the identity:
$\alpha^3 + \beta^3 = (5)^3 - 3(-6)(5)$
$\alpha^3 + \beta^3 = 125 + 90 = 215$
5. Verification using actual zeroes ($6$ and $-1$):
$6^3 + (-1)^3 = 216 - 1 = 215$
Result: $215$
Q11. Case Study: Architectural Fountain Jets (Fluid Mechanics)
A decorative public fountain shoots water droplets through a nozzle array. The stream forms an aesthetic liquid arch. The trajectory corresponds to the polynomial $p(x) = -2x^2 + 8x$.
1 MarksSub-question (a): Factorize the polynomial to find the coordinates of the water nozzle and the landing basin pool.
Answer:
1. Start with the given polynomial: $p(x) = -2x^2 + 8x$.
2. Factor out the common terms ($-2x$):
$p(x) = -2x(x - 4)$
3. Set $p(x) = 0$ to find where the water stream meets the ground level (nozzle launch point and landing basin pool):
$-2x(x - 4) = 0$
4. Solve for the roots:
• $x = 0$ (the water nozzle position at the origin)
• $x - 4 = 0 \implies x = 4$ (the landing basin pool position)
Result: Coordinates/points are $x = 0$ (nozzle) and $x = 4$ (landing basin pool).
1 MarksSub-question (b): What is the degree of this polynomial, and what name is given to its graphical shape?
Answer:
1. Examine the highest power of the variable $x$ in the polynomial $p(x) = -2x^2 + 8x$. The highest exponent is $2$, which makes it a quadratic polynomial.
2. Degree: $2$
3. Graphical Shape: The graph of a second-degree (quadratic) polynomial is called a parabola (specifically, opening downwards since the leading coefficient is negative).
Result: Degree is $2$; graphical shape is a parabola.
2 MarksSub-question (c): Find the coordinates of the highest apex point reached by the water stream.
Answer:
1. The trajectory polynomial is $p(x) = -2x^2 + 8x$. For a downward-facing parabola ($ax^2 + bx + c$ where $a = -2, b = 8, c = 0$), the vertex (highest apex point) occurs at the axis of symmetry, given by the formula:
$x = -\frac{b}{2a}$
2. Substitute the values of $a$ and $b$:
$x = -\frac{8}{2(-2)} = -\frac{8}{-4} = 2$
3. To find the corresponding height (y-coordinate) at this horizontal distance, evaluate the polynomial at $x = 2$:
$p(2) = -2(2)^2 + 8(2)$
$p(2) = -2(4) + 16 = -8 + 16 = 8$
4. Alternatively, by symmetry, the apex lies midway between the roots $x = 0$ and $x = 4$, which is $\frac{0 + 4}{2} = 2$.
Result: The coordinates of the highest apex point are $(2, 8)$.
Q12. Case Study: Microeconomic Cost Structures (Business Production)
A manufacturing shop finds that its marginal cost layout per unit batch is tied to the total count of batches produced per shift. The cost curve is a quadratic polynomial. One zero of this polynomial $p(x) = 4x^2 - 8kx - 9$ is equal in magnitude but opposite in sign to the other zero.
1 MarksSub-question (a): What is the value of the sum of the zeroes when one zero is the negative of the other?
Answer:
1. Let the two zeroes of the quadratic polynomial be $\alpha$ and $\beta$.
2. According to the problem, one zero is equal in magnitude but opposite in sign to the other, which means $\beta = -\alpha$.
3. Calculate the sum of the zeroes:
$\text{Sum} = \alpha + \beta = \alpha + (-\alpha) = 0$
Result: $0$
1 MarksSub-question (b): Calculate the numerical value of the parameter $k$ based on this condition.
Answer:
1. Identify the coefficients from the polynomial $p(x) = 4x^2 - 8kx - 9$ in standard form $ax^2 + bx + c$:
$a = 4$, $b = -8k$, $c = -9$.
2. Use the relation for the sum of the zeroes of a quadratic polynomial: $\text{Sum} = -\frac{b}{a}$.
3. From part (a), we know that the sum of the zeroes is equal to $0$ because they are additive inverses of each other ($\beta = -\alpha$).
4. Set up the equation using the coefficient ratio:
$-\frac{-8k}{4} = 0$
$\frac{8k}{4} = 0$
$2k = 0$
5. Solve for $k$:
$k = 0$
Result: $k = 0$
2 MarksSub-question (c): Write the final simplified polynomial and find its real roots.
Answer:
1. From part (b), we found that the parameter $k = 0$.
2. Substitute $k = 0$ into the original polynomial expression $p(x) = 4x^2 - 8kx - 9$ to get the final simplified polynomial:
$p(x) = 4x^2 - 8(0)x - 9$
$p(x) = 4x^2 - 9$
3. To find the real roots (zeroes), set $p(x) = 0$:
$4x^2 - 9 = 0$
4. Solve for $x$ by factoring as a difference of squares $(2x)^2 - (3)^2 = 0$ or by isolating $x^2$:
$4x^2 = 9$
$x^2 = \frac{9}{4}$
$x = \pm\sqrt{\frac{9}{4}} = \pm\frac{3}{2}$
5. Alternatively, using factors: $(2x - 3)(2x + 3) = 0 \implies x = \frac{3}{2}, -\frac{3}{2}$. (Notice that these roots are equal in magnitude and opposite in sign, which verifies our earlier condition).
Result: Simplified polynomial is $p(x) = 4x^2 - 9$; real roots are $x = \frac{3}{2}$ and $x = -\frac{3}{2}$ (or $1.5$ and $-1.5$).
Q13. Case Study: Skatepark Half-Pipe Ramp (Recreational Engineering)
A city constructs a concrete skatepark with a symmetrical half-pipe ramp facility. The profile line of the ramp is a parabola opening upwards. The polynomial tracking the curve is $f(x) = x^2 - 6x + 8$.
1 MarksSub-question (a): Find the horizontal points where the ramp profile matches the intermediate coping platform baseline ($f(x) = 0$).
Answer:
1. Set the polynomial equal to zero to find the baseline intersection points: $x^2 - 6x + 8 = 0$.
2. Factor the quadratic expression by splitting the middle term: find two numbers that multiply to $8$ and add up to $-6$. These numbers are $-2$ and $-4$.
4. Solve for the horizontal points: $x = 2$ and $x = 4$.
Result: $x = 2$ and $x = 4$ units of horizontal distance
1 MarksSub-question (b): Find the value of $\alpha^2\beta + \alpha\beta^2$ for this system.
Answer:
1. Identify the coefficients from the polynomial $f(x) = x^2 - 6x + 8$ in standard form $ax^2 + bx + c$:
$a = 1$, $b = -6$, $c = 8$.
2. Find the sum and product of the zeroes ($\alpha$ and $\beta$) using coefficient relationships:
• Sum ($\alpha + \beta$) = $-\frac{b}{a} = -\frac{-6}{1} = 6$
• Product ($\alpha\beta$) = $\frac{c}{a} = \frac{8}{1} = 8$
3. Factor the given expression $\alpha^2\beta + \alpha\beta^2$ by taking out the common term $\alpha\beta$:
$\alpha^2\beta + \alpha\beta^2 = \alpha\beta(\alpha + \beta)$
4. Substitute the values of the product and sum into the factored expression:
$\alpha\beta(\alpha + \beta) = (8)(6) = 48$
5. Verification using the actual zeroes from part (a) ($2$ and $4$):
$(2)^2(4) + (2)(4)^2 = (4)(4) + (2)(16) = 16 + 32 = 48$
Result: $48$
2 MarksSub-question (c): Form a new quadratic polynomial whose roots are $(\alpha + 1)$ and $(\beta + 1)$.
Answer:
1. From previous parts, for the polynomial $f(x) = x^2 - 6x + 8$, the sum and product of the original zeroes ($\alpha$ and $\beta$) are:
• Sum ($\alpha + \beta$) = $6$
• Product ($\alpha\beta$) = $8$
2. Let the new zeroes be $\alpha' = \alpha + 1$ and $\beta' = \beta + 1$.
3. Calculate the sum of the new zeroes ($S'$):
$S' = (\alpha + 1) + (\beta + 1) = (\alpha + \beta) + 2 = 6 + 2 = 8$
4. Calculate the product of the new zeroes ($P'$):
$P' = (\alpha + 1)(\beta + 1) = \alpha\beta + \alpha + \beta + 1 = 8 + 6 + 1 = 15$
5. Form the new quadratic polynomial using the standard formula $g(x) = x^2 - S'x + P'$:
$g(x) = x^2 - 8x + 15$
Q14. Case Study: Agricultural Drone Coverage (Precision Farming)
An autonomous automated agricultural drone flies a pre-programmed path to spray fertilizer over a fruit orchard. The flight path coordinate map shows a parabolic turn represented by $p(x) = x^2 - 9x + 20$.
1 MarksSub-question (a): Determine the zeroes of the drone path polynomial.
Answer:
1. Set the polynomial equal to zero to find the zeroes: $x^2 - 9x + 20 = 0$.
2. Factor the quadratic expression by splitting the middle term: find two numbers that multiply to $20$ and add up to $-9$. These numbers are $-4$ and $-5$.
1 MarksSub-question (b): Find the value of the sum of the roots directly from the coefficients.
Answer:
1. Identify the coefficients from the polynomial $p(x) = x^2 - 9x + 20$ in standard form $ax^2 + bx + c$:
$a = 1$, $b = -9$, $c = 20$.
2. Use the standard formula for the sum of the roots of a quadratic polynomial: $\text{Sum} = -\frac{b}{a}$.
3. Substitute the values of $a$ and $b$ into the formula:
$\text{Sum} = -\frac{-9}{1} = 9$
4. Verification using the roots found in part (a) ($4$ and $5$):
$4 + 5 = 9$
Result: $9$
2 MarksSub-question (c): Calculate the numerical value of the expression $\alpha^2 + \beta^2$ for this flight layout.
Answer:
1. From previous parts, for the polynomial $p(x) = x^2 - 9x + 20$, the sum and product of the zeroes ($\alpha$ and $\beta$) are:
• Sum ($\alpha + \beta$) = $9$
• Product ($\alpha\beta$) = $20$
2. Use the algebraic identity relating the sum of squares to the sum and product of zeroes:
$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$
3. Substitute the values of the sum ($9$) and product ($20$) into the identity:
$\alpha^2 + \beta^2 = (9)^2 - 2(20)$
$\alpha^2 + \beta^2 = 81 - 40$
$\alpha^2 + \beta^2 = 41$
4. Verification using the actual zeroes from part (a) ($4$ and $5$):
$4^2 + 5^2 = 16 + 25 = 41$
Result: $41$
Q15. Case Study: Solar Concentrator Vane (Green Energy Systems)
A solar heating trough relies on a parabolic mirror sheet to focus sunlight onto a central oil pipe line. The mirror shape curve is governed by the polynomial $g(x) = 2x^2 - 5x + 2$.
1 MarksSub-question (a): Solve the polynomial to find its zeroes.
Answer:
1. Set the polynomial equal to zero to find its zeroes: $2x^2 - 5x + 2 = 0$.
2. Factor the quadratic expression by splitting the middle term: find two numbers that multiply to $(2 \times 2 = 4)$ and add up to $-5$. These numbers are $-1$ and $-4$.
4. Solve for the zeroes:
• $2x - 1 = 0 \implies x = \frac{1}{2}$ (or $0.5$)
• $x - 2 = 0 \implies x = 2$
Result: $x = \frac{1}{2}$ and $x = 2$ units of distance
1 MarksSub-question (b): Write the value of the product of the zeroes using the formula $\frac{c}{a}$.
Answer:
1. Identify the coefficients from the polynomial $g(x) = 2x^2 - 5x + 2$ in standard form $ax^2 + bx + c$:
$a = 2$, $b = -5$, $c = 2$.
2. Use the standard formula for the product of the roots of a quadratic polynomial: $\text{Product} = \frac{c}{a}$.
3. Substitute the values of $a$ and $c$ into the formula:
$\text{Product} = \frac{2}{2} = 1$
4. Verification using the zeroes found in part (a) ($\frac{1}{2}$ and $2$):
$\frac{1}{2} \times 2 = 1$
Result: $1$
2 MarksSub-question (c): Find the numerical value of $\frac{\alpha}{\beta} + \frac{\beta}{\alpha}$ for the system.
Answer:
1. Identify the coefficients from the polynomial $g(x) = 2x^2 - 5x + 2$ in standard form $ax^2 + bx + c$:
$a = 2$, $b = -5$, $c = 2$.
2. Find the sum and product of the zeroes ($\alpha$ and $\beta$) using coefficient relationships:
• Sum ($\alpha + \beta$) = $-\frac{b}{a} = -\frac{-5}{2} = \frac{5}{2}$
• Product ($\alpha\beta$) = $\frac{c}{a} = \frac{2}{2} = 1$
3. Simplify the given expression by finding a common denominator:
$\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2 + \beta^2}{\alpha\beta}$
4. Rewrite the numerator $\alpha^2 + \beta^2$ in terms of $(\alpha + \beta)$ and $\alpha\beta$ using algebraic identities:
$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$
So, $\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{(\alpha + \beta)^2 - 2\alpha\beta}{\alpha\beta}$
5. Substitute the values of the sum ($\frac{5}{2}$) and product ($1$) into the expression:
$\frac{\left(\frac{5}{2}\right)^2 - 2(1)}{1} = \frac{\frac{25}{4} - 2}{1} = \frac{25}{4} - \frac{8}{4} = \frac{17}{4} = 4.25$
6. Verification using the actual zeroes from part (a) ($\frac{1}{2}$ and $2$):
$\frac{\frac{1}{2}}{2} + \frac{2}{\frac{1}{2}} = \frac{1}{4} + 4 = 0.25 + 4 = 4.25$
Result: $\frac{17}{4}$ (or $4.25$)
Q16. Case Study: Railway Tunnel Clearance (Transportation Networks)
A mountain rail line passes through a semi-parabolic tunnel cut. The structural archway line configuration is given by the expression $f(x) = -x^2 + 10x - 21$.
1 MarksSub-question (a): Identify the points on the ground line where the tunnel edges are located.
Answer:
1. Set the polynomial equal to zero to find the tunnel edges on the ground line: $-x^2 + 10x - 21 = 0$.
2. Multiply the entire equation by $-1$ to make the leading coefficient positive:
$x^2 - 10x + 21 = 0$
3. Factor the quadratic expression by finding two numbers that multiply to $21$ and add up to $-10$. These numbers are $-3$ and $-7$.
5. Solve for the tunnel edge points: $x = 3$ and $x = 7$.
Result: $x = 3$ and $x = 7$ units of distance
1 MarksSub-question (b): What is the axis of symmetry equation for this railway cut?
Answer:
1. Identify the coefficients from the polynomial $f(x) = -x^2 + 10x - 21$ in standard form $ax^2 + bx + c$:
$a = -1$, $b = 10$, $c = -21$.
2. The axis of symmetry for a parabola is given by the vertical line formula:
$x = -\frac{b}{2a}$
3. Substitute the values of $a$ and $b$ into the formula:
$x = -\frac{10}{2(-1)} = -\frac{10}{-2} = 5$
4. Alternatively, use the midpoint formula between the two roots ($x = 3$ and $x = 7$) found in part (a):
$x = \frac{3 + 7}{2} = \frac{10}{2} = 5$
Result: $x = 5$
2 MarksSub-question (c): If an engineer modifies the path so that the new roots are reciprocals of the original roots, determine the new polynomial.
Answer:
1. From previous parts, for the polynomial $f(x) = -x^2 + 10x - 21$, let the original roots be $\alpha$ and $\beta$. The sum and product of these roots are:
• Sum ($\alpha + \beta$) = $-\frac{b}{a} = -\frac{10}{-1} = 10$
• Product ($\alpha\beta$) = $\frac{c}{a} = \frac{-21}{-1} = 21$
2. Let the new roots be $\alpha' = \frac{1}{\alpha}$ and $\beta' = \frac{1}{\beta}$.
3. Calculate the sum of the new roots ($S'$):
$S' = \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{10}{21}$
4. Calculate the product of the new roots ($P'$):
$P' = \left(\frac{1}{\alpha}\right)\left(\frac{1}{\beta}\right) = \frac{1}{\alpha\beta} = \frac{1}{21}$
5. Form the new quadratic polynomial using the standard formula $g(x) = x^2 - S'x + P'$:
$g(x) = x^2 - \frac{10}{21}x + \frac{1}{21}$
6. Multiplying through by $21$ to clear the fractions gives the equivalent polynomial expression:
$21x^2 - 10x + 1$
Result: $21x^2 - 10x + 1$ (or any non-zero scalar multiple, such as $x^2 - \frac{10}{21}x + \frac{1}{21}$)
Q17. Case Study: Ballistic Missile Test Flight (Defense Lab Simulations)
A defense laboratory monitors a short-range test missile trajectory. The coordinate trace maps a clean downward parabola. The mathematical model for the height profile is $p(x) = -x^2 + 12x - 32$.
1 MarksSub-question (a): Find the launch point and the terminal impact point from the zeroes of $p(x)$.
Answer:
1. Set the polynomial equal to zero to find the launch and impact points on the ground line: $-x^2 + 12x - 32 = 0$.
2. Multiply the entire equation by $-1$ to make the leading coefficient positive:
$x^2 - 12x + 32 = 0$
3. Factor the quadratic expression by finding two numbers that multiply to $32$ and add up to $-12$. These numbers are $-4$ and $-8$.
Result: Launch point at $x = 4$ and terminal impact point at $x = 8$ units of distance
1 MarksSub-question (b): Find the horizontal coordinate value where the missile reaches its highest point.
Answer:
1. Identify the coefficients from the polynomial $p(x) = -x^2 + 12x - 32$ in standard form $ax^2 + bx + c$:
$a = -1$, $b = 12$, $c = -32$.
2. The highest point (vertex) of a downward-opening parabola occurs at the axis of symmetry, given by the formula:
$x = -\frac{b}{2a}$
3. Substitute the values of $a$ and $b$ into the formula:
$x = -\frac{12}{2(-1)} = -\frac{12}{-2} = 6$
4. Alternatively, use the midpoint formula between the launch and impact points ($x = 4$ and $x = 8$) found in part (a):
$x = \frac{4 + 8}{2} = \frac{12}{2} = 6$
Result: $x = 6$ units of horizontal distance
2 MarksSub-question (c): Evaluate the value of $(\alpha - \beta)^2$ for this rocket vector.
Answer:
1. From previous parts, for the polynomial $p(x) = -x^2 + 12x - 32$, the sum and product of the zeroes ($\alpha$ and $\beta$) are:
• Sum ($\alpha + \beta$) = $-\frac{b}{a} = -\frac{12}{-1} = 12$
• Product ($\alpha\beta$) = $\frac{c}{a} = \frac{-32}{-1} = 32$
2. Use the algebraic identity linking the square of the difference of roots to their sum and product:
$(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta$
3. Substitute the values of the sum ($12$) and product ($32$) into the identity:
$(\alpha - \beta)^2 = (12)^2 - 4(32)$
$(\alpha - \beta)^2 = 144 - 128$
$(\alpha - \beta)^2 = 16$
4. Verification using the actual zeroes from part (a) ($4$ and $8$):
$(4 - 8)^2 = (-4)^2 = 16$
Result: $16$
Q18. Case Study: Aircraft Parabolic Flight (Zero-Gravity Maneuvers)
An aerospace training jet flies in a parabolic arc to simulate a zero-gravity environment for astronauts. The microgravity vector follows the equation $f(x) = x^2 - 14x + 45$.
1 MarksSub-question (a): Factorize this equation to discover the entry and exit time markings of the zero-gravity maneuver.
Answer:
1. Set the polynomial equal to zero to find the entry and exit points: $x^2 - 14x + 45 = 0$.
2. Factor the quadratic expression by finding two numbers that multiply to $45$ and add up to $-14$. These numbers are $-5$ and $-9$.
4. Solve for the time markings: $x = 5$ and $x = 9$.
Result: Entry time at $x = 5$ and exit time at $x = 9$ units of time
1 MarksSub-question (b): State the sum of the zeroes of $f(x)$.
Answer:
1. Identify the coefficients from the polynomial $f(x) = x^2 - 14x + 45$ in standard form $ax^2 + bx + c$:
$a = 1$, $b = -14$, $c = 45$.
2. Use the standard formula for the sum of the roots of a quadratic polynomial: $\text{Sum} = -\frac{b}{a}$.
3. Substitute the values of $a$ and $b$ into the formula:
$\text{Sum} = -\frac{-14}{1} = 14$
4. Verification using the roots found in part (a) ($5$ and $9$):
$5 + 9 = 14$
Result: $14$
2 MarksSub-question (c): Compute the value of $\frac{1}{\alpha^2} + \frac{1}{\beta^2}$ for this aerospace profile.
Answer:
1. From previous parts, for the polynomial $f(x) = x^2 - 14x + 45$, the sum and product of the zeroes ($\alpha$ and $\beta$) are:
• Sum ($\alpha + \beta$) = $14$
• Product ($\alpha\beta$) = $45$
2. Simplify the given expression by finding a common denominator:
$\frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{\alpha^2 + \beta^2}{(\alpha\beta)^2}$
3. Rewrite the numerator $\alpha^2 + \beta^2$ using the sum and product of zeroes:
$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$
So, $\frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{(\alpha + \beta)^2 - 2\alpha\beta}{(\alpha\beta)^2}$
4. Substitute the values of the sum ($14$) and product ($45$) into the expression:
$\frac{(14)^2 - 2(45)}{(45)^2} = \frac{196 - 90}{2025} = \frac{106}{2025}$
5. Verification using the actual zeroes from part (a) ($5$ and $9$):
$\frac{1}{5^2} + \frac{1}{9^2} = \frac{1}{25} + \frac{1}{81} = \frac{81 + 25}{2025} = \frac{106}{2025}$
Result: $\frac{106}{2025}$
Q19. Case Study: Architectural Dome Section (Auditorium Design)
An interior designer drafts the ceiling layout for a premium music auditorium. The structural dome section is a parabola opening downwards, given by the polynomial $p(x) = -x^2 + x + 12$.
1 MarksSub-question (a): Find the real zeroes of this architectural dome section.
Answer:
1. Set the polynomial equal to zero to find the real zeroes: $-x^2 + x + 12 = 0$.
2. Multiply the entire equation by $-1$ to make the leading coefficient positive:
$x^2 - x - 12 = 0$
3. Factor the quadratic expression by finding two numbers that multiply to $-12$ and add up to $-1$. These numbers are $-4$ and $3$.
1 MarksSub-question (b): State the product of the zeroes of $p(x)$.
Answer:
1. Identify the coefficients from the polynomial $p(x) = -x^2 + x + 12$ in standard form $ax^2 + bx + c$:
$a = -1$, $b = 1$, $c = 12$.
2. Use the standard formula for the product of the roots of a quadratic polynomial: $\text{Product} = \frac{c}{a}$.
3. Substitute the values of $a$ and $c$ into the formula:
$\text{Product} = \frac{12}{-1} = -12$
4. Verification using the zeroes found in part (a) ($4$ and $-3$):
$4 \times (-3) = -12$
Result: $-12$
2 MarksSub-question (c): Find a quadratic polynomial whose zeroes are $\alpha^2$ and $\beta^2$.
Answer:
1. From previous parts, for the polynomial $p(x) = -x^2 + x + 12$, the sum and product of the original zeroes ($\alpha$ and $\beta$) are:
• Sum ($\alpha + \beta$) = $-\frac{b}{a} = -\frac{1}{-1} = 1$
• Product ($\alpha\beta$) = $\frac{c}{a} = \frac{12}{-1} = -12$
2. Let the new zeroes be $\alpha' = \alpha^2$ and $\beta' = \beta^2$.
3. Calculate the sum of the new zeroes ($S'$):
$S' = \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$
$S' = (1)^2 - 2(-12) = 1 + 24 = 25$
4. Calculate the product of the new zeroes ($P'$):
$P' = (\alpha^2)(\beta^2) = (\alpha\beta)^2 = (-12)^2 = 144$
5. Form the new quadratic polynomial using the standard formula $g(x) = x^2 - S'x + P'$:
$g(x) = x^2 - 25x + 144$
6. Verification using the actual zeroes from part (a) ($4$ and $-3$):
New zeroes are $4^2 = 16$ and $(-3)^2 = 9$.
Sum = $16 + 9 = 25$, Product = $16 \times 9 = 144$.
$g(x) = x^2 - 25x + 144$
Q20. Case Study: Logistics Supply Conveyor Belt (Industrial Automation)
A automated factory uses a curved conveyor assembly to transfer sorted parcels between different floor levels. The alignment follows a parabolic drop and curve modeled by the polynomial $f(x) = x^2 - 8x + 15$.
1 MarksSub-question (a): Find the zeroes of the sorting line path.
Answer:
1. Set the polynomial equal to zero to find the zeroes of the path: $x^2 - 8x + 15 = 0$.
2. Factor the quadratic expression by finding two numbers that multiply to $15$ and add up to $-8$. These numbers are $-3$ and $-5$.
1 MarksSub-question (b): State the value of the coefficients $a, b,$ and $c$ for this polynomial.
Answer:
1. Compare the given polynomial $f(x) = x^2 - 8x + 15$ with the standard quadratic form $ax^2 + bx + c$.
2. Identify each coefficient by matching terms:
• $a$ is the coefficient of $x^2$: $a = 1$
• $b$ is the coefficient of $x$: $b = -8$
• $c$ is the constant term: $c = 15$
Result: $a = 1$, $b = -8$, $c = 15$
2 MarksSub-question (c): Evaluate the value of $\frac{\alpha^2}{\beta} + \frac{\beta^2}{\alpha}$ using the properties of roots.
Answer:
1. From previous parts, for the polynomial $f(x) = x^2 - 8x + 15$, the sum and product of the zeroes ($\alpha$ and $\beta$) are:
• Sum ($\alpha + \beta$) = $-\frac{b}{a} = -\frac{-8}{1} = 8$
• Product ($\alpha\beta$) = $\frac{c}{a} = \frac{15}{1} = 15$
2. Simplify the given expression by finding a common denominator:
$\frac{\alpha^2}{\beta} + \frac{\beta^2}{\alpha} = \frac{\alpha^3 + \beta^3}{\alpha\beta}$
3. Use the algebraic identity for the sum of cubes in the numerator: $\alpha^3 + \beta^3 = (\alpha + \beta)(\alpha^2 - \alpha\beta + \beta^2)$
Alternatively, rewrite using $(\alpha + \beta)$ and $\alpha\beta$:
$\alpha^3 + \beta^3 = (\alpha + \beta)[(\alpha + \beta)^2 - 3\alpha\beta]$
4. Substitute the sum ($8$) and product ($15$) into the numerator:
$\alpha^3 + \beta^3 = (8)[(8)^2 - 3(15)] = 8[64 - 45] = 8[19] = 152$
5. Divide the result by the product ($\alpha\beta = 15$):
$\frac{\alpha^3 + \beta^3}{\alpha\beta} = \frac{152}{15}$
6. Verification using the actual zeroes from part (a) ($3$ and $5$):
$\frac{3^2}{5} + \frac{5^2}{3} = \frac{9}{5} + \frac{25}{3} = \frac{27 + 125}{15} = \frac{152}{15}$
Result: $\frac{152}{15}$ (or $10.133\dots$)
Academic Repository Disclaimer & எச்சரிக்கை அறிவிப்பு :
English: These model solutions are structured to global academic standards for self-assessment and practice purposes. Students are advised to verify with official textbooks for board examinations.
தமிழ்: இந்த மாதிரி கணிதத் தீர்வுகள் மாணவர்களின் சுய கற்றல் மற்றும் பயிற்சித் திறனுக்காக உலகளாவிய கல்வித் தரத்தில் உருவாக்கப்பட்டுள்ளன. தேர்வுகளுக்குத் தயாராகும் போது உங்கள் அதிகாரப்பூர்வ பாடப்புத்தகத்துடன் சரிபார்க்கவும்.
