Week 7 - Graded Assignment 7

Course: Jan 2026 - Mathematics I

Week 7 - Graded Assignment 7

Last Submitted: You have last submitted on: 2026-04-01, 03:33 IST

Question 1

• i) $\rightarrow$ d) $\rightarrow$ 2)
• i) $\rightarrow$ a) $\rightarrow$ 2)
• ii) $\rightarrow$ c) $\rightarrow$ 1)
• ii) $\rightarrow$ c) $\rightarrow$ 4)
• iii) $\rightarrow$ d) $\rightarrow$ 2)
• iii) $\rightarrow$ a) $\rightarrow$ 3)
• iv) $\rightarrow$ b) $\rightarrow$ 1)
• iv) $\rightarrow$ b) $\rightarrow$ 3)

Solution

Abstract Solution (Strategy)

1. [Visual Recognition]: Identify graph types based on curvature, intercepts, and asymptotes.
2. [Decision rule]: Match linear through origin to $y=mx$, quadratic to $y=ax^2$, etc.

Procedure

• Step 1 – (i): Quadratic shape with vertex. Matches (d) and Graph 2.
• Step 2 – (ii): Straight line through origin. Matches (c) and Graph 4.
• Step 3 – (iii): Logarithmic growth (asymptote at $x=0$). Matches (a) and Graph 3.
• Step 4 – (iv): Exponential curve. Matches (b) and Graph 1.
• Result: (i)-d-2, (ii)-c-4, (iii)-a-3, (iv)-b-1.

Question 2

• $f(x) \geq g(x)$ for all $x \geq x_0$.
• There exists a point $x_1 > x_0$ such that $f(x_1) = g(x_1)$.
• $g(x) \geq f(x)$ for all $x \geq x_0$.
• $g(x) \geq f(x)$ for all $x_0 \geq x$.
• $f(x) \geq g(x)$ for all $x_0 \geq x$.

Solution

Abstract Solution (Strategy)

1. [Monotonicity Comparison]: If $f$ climbs and $g$ falls, they can only meet once. Past that point, $f$ is higher. Before that point, $g$ is higher.
2. [Decision rule]: Visualize the relative positions of the curves before and after $x_0$.

Procedure

• Step 1 – Right Side: For $x > x_0$, $f(x) > f(x_0)$ and $g(x) < g(x_0)$. Since $f(x_0)=g(x_0)$, we have $f(x) > g(x)$.
• Step 2 – Left Side: For $x < x_0$, $f(x) < f(x_0)$ and $g(x) > g(x_0)$. Since $f(x_0)=g(x_0)$, we have $g(x) > f(x)$.
• Result: $f \ge g$ for $x \ge x_0$; $g \ge f$ for $x \le x_0$.

Question 3

Solution

Abstract Solution (Strategy)

1. [Limit of Difference]: Combine the two rational expressions into one common denominator.
2. [Decision rule]: The limit of a rational function as $n \to \infty$ is the ratio of the coefficients of the highest power of $n$.

Procedure

• Step 1 – Combine: $a_n = \frac{12n^2(n+5) - (4n^2+23)(3n+27)}{(3n+27)(n+5)}$.
• Step 2 – Expand Numerator: $(12n^3 + 60n^2) - (12n^3 + 108n^2 + 69n + 621) = -48n^2 - 69n - 621$.
• Step 3 – Expand Denominator: $3n^2 + 15n + 27n + 135 = 3n^2 + 42n + 135$.
• Step 4 – Limit: $\lim_{n \to \infty} \frac{-48n^2}{3n^2} = -16$.
• Result: -16

Question 4

Solution

Abstract Solution (Strategy)

1. [Differentiability & Tangent Existence]: A function has a unique tangent at a point if it is differentiable at that point (the derivative exists and is unique).
2. [Formula]: $f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h}$.
3. [Decision rule]: Visually, look for points where the curve is "smooth" and well-defined. Sharp corners (cusps) or jumps (discontinuities) at (0,0) imply no unique tangent.

Procedure

• Step 1 – Graph Analysis: Examine the provided images.
• Step 2 – Count Smoothness: Identify curves that pass through (0,0) with a defined, continuous slope.
• Step 3 – Conclusion: 2 of the 4 provided curves exhibit standard differentiable properties at the origin origin.
• Result: 2

Question 5

Solution

Abstract Solution (Strategy)

1. [Substitution]: Use small-x approximations: $\log(1+16/n) \approx 16/n$ and $e^{9/n}-1 \approx 9/n$.
2. [Decision rule]: Recognize the Stirling-style limit $\frac{\sqrt[n]{n!}}{n} \to \frac{1}{e}$.

Procedure

• Step 1 – First Term: $\log(1 + 16/n) \approx 16/n$ for large $n$.
• Step 2 – Second Term: $(e^{9/n}-1) \approx 9/n$. The $(2\pi n)^{1/n}$ term goes to $1$ as $n \to \infty$.
• Step 3 – Sum: The bracket becomes $\approx \frac{16}{n} - \frac{9}{n} = \frac{7}{n}$.
• Step 4 – Combine: $e \cdot \sqrt[n]{n!} \cdot \frac{7}{n} \implies 7 \cdot e \left( \frac{\sqrt[n]{n!}}{n} \right)$.
• Step 5 – Limit: Since $\frac{\sqrt[n]{n!}}{n} \to 1/e$, the overall expression becomes $7 \cdot e \cdot (1/e) = 7$.
• Result: 7

Question 6

Solution

Abstract Solution (Strategy)

1. [Series Summation]: The numerator is the sum of $n-1$ arithmetic progression terms.
2. [Formula]: Sum of first $k$ odds $= k^2$.

Procedure

• Step 1 – Formulate: The sequence is $3, 5, ..., (2n-1)$. This is the sum of the first $n$ odd numbers $(1, 3, 5, ..., 2n-1)$ minus the first odd number ($1$).
• Step 2 – Sum: Numerator $= n^2 - 1$.
• Step 3 – Limit: $\lim_{n \to \infty} \frac{n^2-1}{n^2} = \lim_{n \to \infty} \left( 1 - \frac{1}{n^2} \right)$.
• Step 4 – Evaluate: $1 - 0 = 1$.
• Result: 1

Question 7

Solution

Abstract Solution (Strategy)

1. [Floor Function Limits (LHL/RHL)]: The floor function $\lfloor x \rfloor$ is discontinuous at integers. Its limit depends strictly on the direction of approach (Left-Hand Limit vs Right-Hand Limit).
2. [Formula]: $\lim_{x \to a^+} \lfloor x \rfloor = a$ (for integer $a$) and $\lim_{x \to a^-} \lfloor x \rfloor = a - 1$.
3. [Decision rule]: Evaluate the directional limits independently before performing arithmetic operations.

Procedure

• Step 1 – Right-Hand Limit (21): $\lim_{x \to 21^+} \lfloor x \rfloor$. Since $x$ is slightly above 21 (e.g., 21.001), the floor is 21.
• Step 2 – Left-Hand Limit (4): $\lim_{x \to 4^-} \lfloor x \rfloor. Since$x$ is slightly below 4 (e.g., 3.999), the floor is 3.
• Step 3 – Composite Calculation: $5(21) - 3(3) = 105 - 9 = 96$.
• Result: 96

Question 8

• Error in estimation by Algorithm 1: $\displaystyle a_n= \frac{n^2+5n}{6n^2+1}$.
• Error in estimation by Algorithm 2: $\displaystyle b_n=\frac{1}{8}+(-1)^n\frac{1}{n}$
• Error in estimation by Algorithm 3: $\displaystyle c_n=\frac{e^n+4}{7e^n}$

• Error in estimation by Algorithm 2 will be 0.500.
• Error in estimation by Algorithm 2 will give the minimum error.
• Error in estimation by Algorithm 2 will give the maximum error.
• Both Algorithm 1 and Algorithm 2 will give the same error and that will be the maximum.
• Error in estimation by Algorithm 1 will be 0.166 approximately.

Solution

Abstract Solution (Strategy)

1. [Sequence Convergence & Limits]: Evaluate the limit of each sequence as $n \to \infty$.
2. [Formula]: $a_n \to \frac{A}{B}$, $b_n \to C \pm 0$, $c_n \to \frac{Ae^n}{Be^n}$.
3. [Decision rule]: Find theoretical terminal error for each algorithm by evaluating its $n o \infty$ behavior.

Procedure

• Step 1 – Alg 1 Limit: $a_n = \frac{n^2+5n}{6n^2+1} o \frac{1}{6} \approx 0.166$.
• Step 2 – Alg 2 Limit: $b_n = \frac{1}{8} + \frac{(-1)^n}{n}$. As $n o \infty$, $\frac{1}{n} o 0$. Limit evaluates to $\frac{1}{8} = 0.125$.
• Step 3 – Alg 3 Limit: $c_n = \frac{e^n+4}{7e^n} = \frac{1}{7} + \frac{4}{7e^n} o \frac{1}{7} \approx 0.142$.
• Step 4 – Compare: Max error is Alg 1 (0.166). Min error is Alg 2 (0.125).
• Result: Error by Algorithm 2 is minimum, Algorithm 1 will be 0.166.

Question 9

• New algorithm error is less than Algorithm 1.
• Algorithm 2 error is less than New algorithm.
• New algorithm error is less than Algorithm 3.
• New algorithm error cannot be compared with Algorithm 3.

Solution

Abstract Solution (Strategy)

1. [Limit Difference]: The limit of the difference is the difference of the limits.
2. [Decision rule]: Compare $L_d = L_a - L_b$ with $L_a$ and $L_c$.

Procedure

• Step 1 – Vals: From previous: $L_a = 1/6$, $L_b = 1/8$, $L_c = 1/7$.
• Step 2 – Diff: $L_d = 1/6 - 1/8 = \frac{4-3}{24} = 1/24 \approx 0.041$.
• Step 3 – Compare: $1/24$ is smaller than $1/6$, $1/8$, and $1/7$.
• Result: The new error ($0.041$) is lower than any individual algorithm error.

Question 10

Solution

Abstract Solution (Strategy)

1. [Standard Limit Form]: Transform the expression to match $\frac{e^x-1}{x} \to 1$ as $x \to 0$.
2. [Decision rule]: Let $x = 1/n$. As $n \to \infty, x \to 0$.

Procedure

• Step 1 – Factor: $c_n' = n (e^{1/8n} - 1)$.
• Step 2 – Substitute: Let $h = 1/n$. $c_n' = \frac{e^{h/8} - 1}{h}$.
• Step 3 – Align Coefficient: Multiply and divide by $8$. $\frac{1}{8} \left( \frac{e^{h/8} - 1}{h/8} \right)$.
• Step 4 – Solve: As $n \to \infty, h \to 0$, so $\frac{e^{h/8}-1}{h/8} \to 1$.
• Result: $\frac{1}{8} \times 1 = 0.125$.