Week 4 - Graded Assignment 4

Course: Jan 2026 - Mathematics I

Question 1

Consider the line ($L_{x}$) and parabola ($P_{x}$) as shown in below figure.

Abstract Solution (Strategy)

1. Common Factor Identification: If the line $L_x$ is a factor of the parabola $P_x$, then $P_x/L_x$ will be a linear function (straight line).
2. Roots Comparison: If $L_x$ has a root at $x=r_1$, and $P_x$ has roots at $x=r_1$ and $x=r_2$, the quotient $Q(x) = a(x-r_2)$ for $x \neq r_1$.
3. Discontinuity: Note the removable discontinuity (hole) at $x=r_1$. ^maths-w4-q1-strategy

Procedure

• Parabola $P_x$ has roots at $x = -2$ and $x = 2$ (from graph).
• Line $L_x$ has a root at $x = -2$ (passes through $(-2, 0)$).
• $P_x = a(x+2)(x-2)$. $L_x = b(x+2)$.
• $P_x / L_x = \frac{a}{b}(x-2)$.
• This represents a straight line with an $x$-intercept at $x = 2$.
• The graph in the accepted answer shows exactly this linear behavior.

Question 2

• The $f(x)$ cuts the X-axis at 3, 5 and 4.
• The $g(x)$ cuts the X-axis at 2, -6 and -1.
• If $x \in (-6, 2)$, then $g(x)$ is positive.
• $f(x)$ is negative when $x \in [-4,3] \cup (3, \infty)$.
• $g(x)$ is positive when $x \in (-\infty,-6) \cup (-1, -2)$.

1. Factorization: Use the candidate roots to find one factor, then perform division to find the remaining quadratic.
2. Sign Scheme: Determine intervals where the function is positive/negative by plotting roots on a number line. ^maths-w4-q2-strategy

Procedure

• For $g(x)$: $g(2) = 2^3 + 5(2^2) - 8(2) - 12 = 8 + 20 - 16 - 12 = 0$.
  • $(x-2)$ is a factor. Dividing $g(x)$ by $(x-2)$ gives $x^2 + 7x + 6 = (x+6)(x+1)$.
  • Roots: $2, -6, -1$.
• For $f(x)$: $f(3) = 3^3 - 4(3^2) - 17(3) + 60 = 27 - 36 - 51 + 60 = 0$.
  • $(x-3)$ is a factor. Dividing $f(x)$ by $(x-3)$ gives $x^2 - x - 20 = (x-5)(x+4)$.
  • Roots: $3, 5, -4$.
• Statement "The $g(x)$ cuts the X-axis at 2, -6 and -1" is correct.

Question 3

• $a(x-2)^2 (x^2 +7x +12), a > 0$
• $a(x^4+4x^{3}-7x^2-22x+24), a > 0$
• $a(x-2)^2 (x^2 +2x -8), a > 0$
• $a(x^4 - 5x^3 - 7x^2 -50x - 24), a > 0$

1. Identify Roots: Given roots are $2, -3, -4$. For a degree 4 polynomial, there must be a 4th root or one root must have multiplicity 2.
2. Sign Analysis: $f(x) > 0$ for $x \in (-1, 1)$ means at $x=0$, $f(0) > 0$.
3. Check Options: Multiply the factors to see which one matches the sign behavior. ^maths-w4-q3-strategy

Procedure

• Option 2 expansion: $f(x) = a(x-1)(x-2)(x+3)(x+4)$.
• Checking signs:
  • If $x \in (1, 2)$, say $1.5$: $(0.5)(-0.5)(4.5)(5.5) = (\text{negative})$. Correct.
  • If $x \in (-1, 1)$, say $0$: $(-1)(-2)(3)(4) = 24$. Correct ($a > 0$).
• Therefore, roots are $1, 2, -3, -4$.

Question 4

• The function $f(x)$ has exactly 7 turning points.
• The function $f(x)$ has exactly 6 points where the slope is 0.
• The function $f(x)$ is neither even nor odd function.
• In the interval $x \in (3,5)$, $f(x)$ is increasing first and then decreasing.
• The function $f(x)$ is negative when $x \in (-1,2)$.

1. Critical Points: Slope is 0 at turning points and at roots with multiplicity $> 1$.
2. Multiplicity Logic: Even multiplicity ($x=2, x=-1$) means the curve touches the axis (tangential). Odd multiplicity means it crosses.
3. Interval Tests: Evaluate sign of $f(x)$ in given intervals to determine behavior. ^maths-w4-q4-strategy

Procedure

• Degree of $f(x)$ is $2+1+2+1+1 = 7$. Max turning points $= 6$.
• $f(x)$ has roots at $2, 3, -1, -4, 5$.
• Roots with multiplicity $>1$: $x=2$ (mult 2), $x=-1$ (mult 2). Slope is 0 at these points.
• Excluding these, there are 4 other turning points between the distinct roots. Total $= 6$ points where slope is 0.
• For $x \in (-1, 2)$, test $x=0$: $f(0) = \frac{-1}{300}(-2)^2(-3)(1)^2(4)(-5) = \frac{-1}{300}(4)(-3)(4)(-5) = \frac{-240}{300} = \text{negative}$. Correct.

Question 5

• $4x^4 + 14x^3 + 8x^2 -17x - 14$
• $4x^4 + 14x^3 -6x^2 -19x - 34$
• $4x^4 + 2x^3 + 8x^2 -17x - 14$
• $4x^4 + 2x^3 + 8x^2 -18x - 34$

1. Find M(x): Calculate $Q(2)$, use the point-slope form $y - y_1 = m(x - x_1)$.
2. Substitution: Form the expression $P(x) + M(x)Q(x)$.
3. Expansion & Collection: Multiply polynomials and combine like terms. ^maths-w4-q5-strategy

Procedure

• $Q(2) = 2^3 + 2(2^2) - 6 = 8 + 8 - 6 = 10$. Point: $(2, 10)$, Slope: $3$.
• $M(x): y - 10 = 3(x - 2) \implies y = 3x + 4$.
• $M(x)Q(x) = (3x + 4)(x^3 + 2x^2 - 6) = 3x^4 + 6x^3 - 18x + 4x^3 + 8x^2 - 24 = 3x^4 + 10x^3 + 8x^2 - 18x - 24$.
• $P(x) + M(x)Q(x) = (x^4 + 4x^3 + x + 10) + (3x^4 + 10x^3 + 8x^2 - 18x - 24) = 4x^4 + 14x^3 + 8x^2 - 17x - 14$.

Question 6

• Multiplicity of -1 and 1 must be the same.
• $p(x)$ is increasing in the interval $(3, \infty )$.
• The total number of local minima is 3.
• The number of turning points is 5.

1. Turning Points Identification: Count every peak (local maximum) and valley (local minimum).
2. Multiplicity from Graph: Observe if the graph crosses the X-axis (odd multiplicity) or touches and turns (even multiplicity).
3. Interval Analysis: Check the slope of the curve in the given interval to determine if it is increasing or decreasing. ^maths-w4-q6-strategy

Procedure

• Turning Points: From the graph, we see peaks/valleys at approx $x=-4, -2, 0, 2, 4$? No, looking at the graph, there are 2 hills and 3 valleys. Total = 5 turning points. Correct.
• Local Minima: There are 3 distinct "valleys" at the bottom. Correct.
• Interval $(3, \infty)$: To the right of $x=3$, the graph moves upwards. Correct.
• Multiplicity: At both $x=-1$ and $x=1$, the graph crosses the axis in a similar way (linear-like crossing). Multiplicity for both is odd (likely 1). Correct.

Question 7

• $q(x) \longrightarrow \infty$ as $x \longrightarrow \infty$.
• $p(x) \longrightarrow - \infty$ as $x \longrightarrow \infty$.
• $p(x)$ has at most 4 turning points.
• The quotient obtained while dividing $q(x)$ by $p(x)$ is a constant.

1. End Behavior: Lead term $-x^5$. As $x \to \infty$, high power dominates with a negative sign.
2. Turning Points: For a polynomial of degree $n$, the maximum number of turning points is $n-1$.
3. Division Quotient: When dividing two polynomials of the same degree, the quotient is the ratio of their leading coefficients. ^maths-w4-q7-strategy

Procedure

• End Behavior: $\lim_{x \to \infty} (-x^5) = -\infty$. Statement 2 is True.
• Turning Points: Degree is $5$. Max turning points $= 4$. Statement 3 is True.
• Division: Both $p(x)$ and $q(x)$ are $5^{th}$ degree. Quotient = $(-1)/(-1) = 1$ (a constant). Statement 4 is True.

Question 8

• The degree of $p(x)+q(x)$ is 3.
• The degree of $p(x)r(x)$ is 5.
• When $p(x)$ divides $r(x)$ then obtained remainder is a linear function.
• When $p(x)$ divides $r(x)$ then obtained remainder is a quadratic function.

1. Degree of Sum: $\max(\text{deg } p, \text{deg } q)$.
2. Degree of Product: $\text{deg } p + \text{deg } q$.
3. Remainder Theorem: If degree of divisor is $D$, the degree of remainder is at most $D-1$. ^maths-w4-q8-strategy

Procedure

• $\text{deg}(p+q) = \max(2, 1) = 2$. Statement 1 is False.
• $\text{deg}(p \cdot r) = 2 + 3 = 5$. Statement 2 is True.
• Remainder: $p(x)$ is degree 2. Any remainder when dividing by $p(x)$ must be degree 1 (linear) or 0 (constant). Statement 3 is True.

Question 9

• There are 5 distinct roots in $s(x)$.
• There are 3 turning points in $s(x)$.
• Multiplicity of the root 1 is 2 in $s(x)$.
• Multiplicity of the root 3 is 1 in $s(x)$.

Abstract Solution (Strategy)

1. Factor All Polynomials: Express $p, q, r$ in fully factored form.
2. Combine Factors: Form $s(x)$ by multiplying the factored forms.
3. Count Multiplicities: Sum the exponents of each distinct factor.

Procedure

• $p(x) = (x-6)(x+1)$.
• $q(x) = (x+1)$.
• $r(x) = 2x(x^2 - 2x - 3) = 2x(x-3)(x+1)$.
• $s(x) = (x-6)(x+1) \cdot (x+1) \cdot 2x(x-3)(x+1) = 2x(x-6)(x-3)(x+1)^3$.
• Roots: $0, 6, 3$ (multiplicity 1 each) and $-1$ (multiplicity 3).
• Multiplicity of root $3$ is indeed 1. Correct.

Question 10

1. Condition: $M(n) \ge 40$.
2. Inequality Setup: $- \frac{n^2}{1000} (n^3 - 15n^2 + 50n) + 40 \ge 40$.
3. Sign Flip: Subtract 40 and divide by negative constant, which reverses the inequality.
4. Interval Solution: Solve the polynomial inequality $n^{2}(n^{3}-15n^{2}+50n) \le 0$. ^maths-w4-q10-strategy

Procedure

• $n^2 [n(n-5)(n-10)] \le 0$.
• $n^3 (n-5)(n-10) \le 0$.
• Since $n \in \{1, 2, ..., 12\}$, we analyze the sign of $f(n) = n^3(n-5)(n-10)$ on a number line.
• $f(n) \le 0$ when $n \in [5, 10]$ (where roots are $0, 5, 10$).
• Integers in range $\{1..12\}$ that satisfy this: $5, 6, 7, 8, 9, 10$.
• Total $= 6$ tests.

Question 11

Abstract Solution (Strategy)

1. Factor Analysis: $(x-5)^2$ means there is a root at $x=5$ with multiplicity 2 (touches the X-axis).
2. Positive Roots: "Exactly two distinct positive roots". Since $x=5$ is one (double), there must be exactly one more distinct root $x > 0$.
3. Degree & Roots: "Odd degree" and "at least three distinct real roots".
4. Intercept: $p(0) \neq 0$ means it doesn't pass through the origin.

Procedure

• Option 1 shows:
  • Touches at $x=5$ (multiplicity 2).
  • Crosses at another positive point. (Total 2 distinct positive roots).
  • Crosses at a negative point. (Total 3 distinct real roots).
  • Doesn't go through origin.
  • End behavior (one side up, one side down) indicates odd degree.
• This perfectly matches all constraints.

Question 12

Abstract Solution (Strategy)

1. Factorization: Rewrite all quadratic terms into linear factors.
2. Simplification: Cancel out common factors from numerator and denominator.
3. Graphing: Identify roots of the remaining polynomial and check its end behavior.

Procedure

• $p(x) = (x+5)(x-3)(x-2)(x+2)$.
• $q(x) = (x-2)(x+2)$.
• $r(x) = p(x)/q(x) = (x+5)(x-3) = x^2 + 2x - 15$.
• This is a parabola opening upwards with roots at $x = -5$ and $x = 3$.
• Option 1 shows this parabola.

Question 13

                         

Abstract Solution (Strategy)

1. Model Property: At points $x=1, 3, 5, 7$, the predicted value $\hat{y}$ is exactly $c$ because the product term becomes zero.
2. SSE Minimization: For a set of observed values $y_i$ at these points, the constant $c$ that minimizes $\sum (y_i - c)^2$ is simply the arithmetic mean of the $y_i$ values. ^sse-minimization-logic

Procedure

• Observed $y$ values at $x \in \{1, 3, 5, 7\}$ from Table-1: $3.2, 2.8, 3.8, 3.8$.
• Mean $= (3.2 + 2.8 + 3.8 + 3.8) / 4 = 13.6 / 4 = 3.4$.
• Therefore, $c = 3.4$ minimizes SSE.

Question 14

• $(4x^2 + 12x + 3)$
• $(4x^2 + 12x + 5)$
• $(2x^2 + 6x + 5)$
• $(2x^2 + 3x + 1)$

Abstract Solution (Strategy)

1. Division: Volume $=$ Length $\times$ Width $\times$ Height. First divide $V(x)$ by $L(x)$ to get the Area of the other two dimensions ($W \times h$).
2. Factorization: Factor the resulting quadratic into two linear terms $W(x)$ and $h(x)$.
3. Identification: Identify "Base Area" (typically Length $\times$ Width).

Procedure

• $V(x) / (x+1) = (4x^3 + 16x^2 + 17x + 5) / (x+1) = 4x^2 + 12x + 5$ (via synthetic division).
• Factor $4x^2 + 12x + 5$: find numbers that sum to 12 and multiply to 20 $\implies 10, 2$.
• $4x^2 + 10x + 2x + 5 = 2x(2x+5) + 1(2x+5) = (2x+1)(2x+5)$.
• Since $h(x) > W(x)$ for $x>0$, $h(x) = 2x+5$ and $W(x) = 2x+1$.
• Base Area ($L \times W$) $= (x+1)(2x+1) = 2x^2 + 3x + 1$.

Question 15

Abstract Solution (Strategy)

1. Equation Setup: Set $S(t) = 5000$.
2. Solve for t: Simplify the equation and find the positive integer root for $t$.

Procedure

• $t^3 - 2t^2 - 24t + 5000 = 5000 \implies t^3 - 2t^2 - 24t = 0$.
• $t(t^2 - 2t - 24) = 0$.
• $t(t - 6)(t + 4) = 0$.
• For $t > 0$, the only valid solution is $t = 6$ months.

Question 16

            

                                                          Fig - 1

             
             

                                                        Fig - 2

• Fig - 1 and Fig - 2 are polynomials of $x$.
• The polynomial of $x that represents the Fig - 1 have minimum 6 Zeros
• The polynomial of $x that represent the Fig - 2 have minimum 3 Zeros
• Fig - 2 represents the function of $x$.
• Fig - 2 is not a polynomial of $x$.
• The polynomial of $x$ that represents the Fig - 1 have 3 distinct Zeros

Abstract Solution (Strategy)

1. Vertical Line Test: If any vertical line intersects the curve more than once, it is not a function of $x$.
2. Polynomial Continuity: Polynomials are smooth, continuous, and defined everywhere. Sharp corners or cusps (like in Fig 2) indicate absolute values or piecewise functions, not simple polynomials.
3. Roots vs Maxima: Distinct zeros are where it crosses the axis. Minimum degree is related to turning points ($n-1$).

Procedure

• Fig 1:
  • Crosses the axis at 3 points. (3 distinct zeros).
  • Has 5 turning points. Polynomial degree is at least $5+1=6$. (Minimum 6 complex zeros). Correct.
• Fig 2:
  • Passes Vertical Line Test $\implies$ is a Function. Correct.
  • Has sharp corners (v-shapes). Simple polynomials cannot have sharp corners. Correct.

Question 17

• The roller coaster will first go up and then go down in the interval $(-5,-1)$.
• The roller coaster will first go down and then go up in the interval $(10,20)$.
• The roller coaster will first go up and then go down in the interval $(-1,-2)$.
• The roller coaster will first go up and then go down in the interval $(2,5)$.

Abstract Solution (Strategy)

1. Analyze Zeroes: Roots are segments where the graph crosses or touches.
2. Multiplicity behavior:
   • $(t+5)^2$: touches at $-5$ (even).
   • $(t+1)$: crosses at $-1$ (odd).
   • $(2-t)^3$: crosses at $2$ (odd, cubic flattening).
   • $(t-5)$: crosses at $5$ (odd).
3. Cubic Term: $(-0.01t^3...)$ factor also has roots. Testing values in intervals is safest.

Procedure

• Term 1: $(-0.01t^3 + 0.35t^2 - 3.5t + 10)$. At $t=10$, it's zero? $-10 + 35 - 35 + 10 = 0$. So $x=10$ is a root.
• Interval $(-5, -1)$: Touches at $-5$, crosses at $-1$. Between these, it must peak. Correct.
• Interval $(2, 5)$: Crosses at both points. Between them it peaks. Correct.

Question 18

• 6
• 7
• 5
• 8

Abstract Solution (Strategy)

1. Degree Calculation: Sum all the powers in the factored expression.
2. Turning Points Rule: Polynomial of degree $n$ has exactly $n-1$ points where the derivative is zero (slope 0).

Procedure

• Term 1 (Cubic): Degree 3.
• $(t+5)^2$: Degree 2.
• $(t-5)$: Degree 1.
• $(t+1)$: Degree 1.
• $(2-t)^3$: Degree 3.
• Total Degree $n = 3 + 2 + 1 + 1 + 3 = 10$.
• Turning Points (stationary points) $= 10 - 1 = 9$?
• Wait, the question asks for "turning points" (peaks/valleys). If roots have even multiplicity or are inflection points, they count as slope-0 points but not necessarily turning points.
• Stationary points = 9. But $x=2$ is an inflection point (degree 3). It isn't a "turning point".
• Stationary points $= 9 - 1 = 8$.

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