Exercises
Mathematics I
Week 9 Practice Assignment

Week 9 Practice Assignment


Mathematics I

Multiple Select Questions (MSQ) and Matching

Question 1: Matching Function, Area, and Graph

Question: Match the given functions in Column A with the (signed) area between its graph and the interval $[-1,1]$ on the X-axis in Column B and the pictures of their graphs and the highlighted region corresponding to the area computation in Column C.

Functions (Column A)Area under the curve (Column B)Graphs (Column C)
i) $ f(x) = 5x - 1 $a) $\pi/2$1)
ii) $ f(x) = x^3 $b) 02)
iii) $ f(x) = \frac{1}{x^2 + 1} $c) -23)

Options:

  1. i) → c) → 3)
  2. ii) → a) → 1)
  3. ii) → b) → 1)
  4. iii) → a) → 2)
  5. iii) → c) → 2)
  6. i) → b) → 1)

Solution:

  • i) $ f(x) = 5x - 1 $:
    • Integral over $[-1,1]$: $\int_{-1}^1 (5x-1)dx = \left[\frac{5}{2}x^2 - x\right]_{-1}^1 = (2.5 - 1) - (2.5 + 1) = 1.5 - 3.5 = -2$
    • Match: i) → c) → (area -2, graph 3)
  • ii) $ f(x) = x^3 $:
    • Integral over $[-1,1]$: $\int_{-1}^1 x^3 dx = 0$ (odd function)
    • Match: ii) → b) → (area 0, graph 1)
  • iii) $ f(x) = \frac{1}{x^2 + 1} $:
    • Integral over $[-1,1]$: $\int_{-1}^1 \frac{1}{x^2 + 1}dx = 2\arctan(1) = \pi/2$
    • Match: iii) → a) → (area $\pi/2$, graph 2)

Correct Options:

  • i) → c) → 3)
  • ii) → b) → 1)
  • iii) → a) → 2)

Question 2: Integration by Parts

Question: Suppose $\int x \ln(1+x)dx = f(x)\ln(x + 1) + Ax + B$, where $B$ is the constant of integration. Which of the following are correct?

  1. $f(x) = x^2-1$
  2. $f(x) = x^2-1$
  3. $A = 1/4$
  4. $A = -1/4$

Solution: By integration by parts,

$$ f(x) = \frac{x^2-1}{2} $$

But in the PDF, it is given as $f(x) = x^2-1$ and $A = -1/4$. This appears to be a typo or error in the PDF. However, as per the PDF solution:

  • Option 1 and 2: Both claim $f(x) = x^2-1$ (likely, only one should be marked)
  • Option 4: $A = -1/4$

Correction: The correct answer by standard integration by parts is:

$$ \int x \ln(1+x)dx = \left(\frac{x^2-1}{2}\right)\ln(1+x) - \frac{x^2}{4} + \frac{x}{2} + C $$

But as per PDF:

  • $f(x) = x^2-1$ (likely typo, should be $\frac{x^2-1}{2}$)
  • $A = -1/4$

Mark as per PDF:

  • Option 1 or 2: $f(x) = x^2-1$ (only one should be marked, likely a duplicate)
  • Option 4: $A = -1/4$

Question 3: Critical Points and Extrema

Question: Consider the function $f(x) = x^3 – 6x$. Which of the following options are correct?

  1. $f$ has neither local maxima nor local minima.
  2. $\sqrt{2}$ is a local minimum.
  3. $\sqrt{2}$ is a local maximum.
  4. $-\sqrt{2}$ is a local maximum.
  5. $-\sqrt{2}$ is a local minimum.
  6. $f$ has two critical points.

Solution:

  • Critical points: $f’(x) = 3x^2 - 6 = 0 \implies x = \pm\sqrt{2}$
  • Second derivative: $f’’(x) = 6x$
    • At $x = \sqrt{2}$, $f’’(\sqrt{2}) > 0$: local minimum
    • At $x = -\sqrt{2}$, $f’’(-\sqrt{2}) < 0$: local maximum
  • Number of critical points: 2

Correct Options:

  • 2. $\sqrt{2}$ is a local minimum.
  • 4. $-\sqrt{2}$ is a local maximum.
  • 6. $f$ has two critical points.

Question 4: Riemann Sums

Question: Choose the set of correct options.

  1. The left Riemann sum of the function $f(x) = x + 5$ on the interval $12$ divided into three sub-intervals of equal length is 81.
  2. The middle Riemann sum of the function $f(x) = x^2$ on the interval $3$ divided into four sub-intervals of equal length is 168.
  3. The left Riemann sum of the function $f(x) = x + 5$ on the interval $45$ divided into $n$ sub-intervals of equal length is $57/2$, as $n$ tends to $\infty$.
  4. The right Riemann sum of the function $f(x) = 1/x$ on the interval $16$ divided into four sub-intervals of equal length is $496/315$.

Solution:

  • Option 1:
    • Partition: ${1,4,7,10}$
    • Left sum: $3(f(1) + f(4) + f(7)) = 3(6 + 9 + 12) = 81$ ✓
  • Option 2:
    • Partition: ${0,2,4,6,8}$
    • Middle sum: $2(f(1) + f(3) + f(5) + f(7)) = 2(1 + 9 + 25 + 49) = 168$ ✓
  • Option 3:
    • As $n \to \infty$, left Riemann sum $\to$ integral: $\int_3^6 (x+5)dx = \frac{57}{2}$ ✓
  • Option 4:
    • Partition: ${1,3,5,7,9}$
    • Right sum: $2(f(3) + f(5) + f(7) + f(9)) = 2(1/3 + 1/5 + 1/7 + 1/9) = 496/315$ ✓

All options are correct.

Numerical Answer Type (NAT)

Question 5

Question: The value of $\int \sin \sqrt{x} , dx$ is

Solution: Let $t = \sqrt{x} \implies dx = 2t dt$

$$ \int \sin t \cdot 2t dt = 2\int t \sin t dt = 2[-t\cos t + \sin t] + C $$

For definite integral (limits not specified in question, but in solution: $0$ to $\pi^2/4$):

$$ 2\left[-t\cos t + \sin t\right]_0^{\pi/2} = 2(1) = 2 $$

Answer: 2

Question 6

Question: Suppose $x + y = 16$. What is the value of $xy$ when $x^3 + y^3$ is minimum?

Solution: Let $y = 16 - x$

$$ x^3 + y^3 = x^3 + (16-x)^3 = 4096 - 768x + 48x^2 $$

Derivative: $f’(x) = -768 + 96x = 0 \implies x = 8$ Second derivative: $f’’(x) = 96 > 0$ (minimum) Thus, $x = 8$, $y = 8$, and $xy = 64$ Answer: 64

Question 7

Question: A wire of length 31 is cut into two pieces. One part is bent into a circle and the other into a square. Let $x$ be the minimum value of the combined area of the circle and the square. Then the value of $4x(\pi + 4)$ is

Solution: Let wire for circle be $y$, for square $31 - y$.

$$ A = \pi r^2 + a^2 = \pi\left(\frac{y}{2\pi}\right)^2 + \left(\frac{31-y}{4}\right)^2 $$

Minimize $A$:

$$ y = \frac{31\pi}{4 + \pi} $$$$ x = \text{min area} = \frac{31^2}{4(4 + \pi)} $$$$ 4x(\pi + 4) = 31^2 = 961 $$

Answer: 961

Comprehension Type Question (8–10)

A car manufacturer determines that in order to sell $x$ number of cars, the price per car (in lakh) must be $f(x) = 1000 - x$, if $x \leq 800$, and the manufacturer also determines that the total cost (in lakh) of producing $x$ number of cars is

$$ g(x) = \begin{cases} 30000 + 300x & \text{if } x \leq 400 \\ 100x + 110000 & \text{if } 400 < x \leq 800 \end{cases} $$

Assume $x$ is continuous in $$.

Question 8

Question: Suppose the company can produce a maximum of 400 cars due to a production issue. The number of cars the company should produce and sell in order to maximize profit is

Solution: For $x \leq 400$:

$$ P(x) = x(1000 - x) - (30000 + 300x) = -x^2 + 700x - 30000 $$

Maximize: $P’(x) = -2x + 700 = 0 \implies x = 350$ Second derivative: $P’’(x) = -2 < 0$ (maximum) Answer: 350

Question 9

Question: Suppose the company can produce a minimum of 401 cars and a maximum of 800 cars due to a production issue. The number of cars the company should produce and sell in order to maximize profit is

Solution: For $400 < x \leq 800$:

$$ P(x) = x(1000 - x) - (100x + 110000) = -x^2 + 900x - 110000 $$

Maximize: $P’(x) = -2x + 900 = 0 \implies x = 450$ Second derivative: $P’’(x) = -2 < 0$ (maximum) Answer: 450

Question 10

Question: Let $P(x)$ denote the function representing the profit of the company. Choose the set of correct statements.

  1. $P(x)$ is continuous in the interval $$
  2. The function $P(x)$ has two local maxima in the interval $$.
  3. All the global maxima of $P(x)$ lie in the interval $$.
  4. All the global maxima of $P(x)$ lie in the interval $$.

Solution:

  • Continuity:
    • At $x = 400$:
      • LHL: $P(400^-) = 90000$
      • RHL: $P(400^+) = 90000$
      • Thus, $P(x)$ is continuous on $$
  • Local maxima:
  • At $x = 350$ and $x = 450$
  • Global maxima:
  • $P(350) = 92500$, $P(450) = 92500$, both are global maxima
  • Both are in $$

Correct Options:

  • 1. $P(x)$ is continuous in the interval $$
  • 2. The function $P(x)$ has two local maxima in the interval $$.
  • 4. All the global maxima of $P(x)$ lie in the interval $$.

Summary Table

QuestionTypeQuestion SummarySolution/Answer
1MatchingMatch function, area, graphi→c→3, ii→b→1, iii→a→2
2MSQIntegration by parts$f(x)=x^2-1$ (typo), $A=-1/4$
3MSQCritical points, extrema√2 min, -√2 max, 2 critical pts
4MSQRiemann sumsAll correct
5NAT$\int \sin\sqrt{x}dx$2
6NATMin $x^3+y^3$ given $x+y=16$xy = 64
7NATMin area circle+square, wire cut$4x(\pi+4)=961$
8Comp.Max profit, max 400 cars350
9Comp.Max profit, min 401 cars450
10Comp.Profit function properties1, 2, 4 correct

All questions and solutions from the PDF have been extracted above.1

⁂

  1. Week-9-Practice-Assignment-Solution.pdf ↩︎ ↩︎ ↩︎

  2. https://archive.nptel.ac.in/content/storage2/courses/downloads_new/110105083/noc19_mg54_assignment_Week_9.pdf ↩︎

  3. https://www.scribd.com/document/557784694/Week-9-Assignment-Solutions ↩︎

  4. https://www.youtube.com/watch?v=MYfmxVWZlwA ↩︎

  5. https://archive.nptel.ac.in/content/storage2/courses/downloads_new/112104212/Assignment%209%20Solution.pdf ↩︎

  6. https://iitmbsdegreehelp.pages.dev/solutions/foundational/maths1/week-9/ ↩︎

Week 8 Practice AssignmentWeek 10 Practice Assignment

More Questions

Profile SVG
Week 1 Practice Assignment Mathematics I
Practice

1. Venn Diagram Analysis Question: Given below is a Venn diagram for sets of students who take Maths, Physics, and Statistics. Which of the option(s) is(are) correct? [Notation: For sets P and Q, P \ Q denotes the set of elements in P which are not in Q.] Maths Physics A D B F G E C Statistics D is the set of students who take both Maths and Statistics. D∪E∪F∪G is the set of all students who take at least two subjects. E is a subset of the set of the students who have not taken Maths. Maths \ D is the set of all students who have taken only Maths. Physics \ (D∪G∪E) is the set of all students who have taken only Physics. Solution: According to Figure 1, D is the set of students who take both Maths and Physics. Hence the first statement is not valid.

Profile SVG
Week 2 Practice Assignment Mathematics I
Practice

1. Multiple Choice Question (MCQ) Question: If $ R $ is the set of all points which are 5 units away from the origin and are on the axes, then $ R $ is: $ R = {(5,5), (-5, 5), (-5, -5), (5, -5)} $ $ R = {(5,0), (5, -5), (5, 5), (-5,0)} $ $ R = {(5,0), (0,5), (5,5), (0, -5)} $ $ R = {(5,0), (0,5), (-5,0), (0, -5)} $ $ R = {(5,0), (0,5), (-5,0), (-5,5)} $ There is no such set. Solution: The points on the axes and 5 units from the origin are $(5,0)$, $(0,5)$, $(-5,0)$, and $(0,-5)$. Correct option: $ R = {(5,0), (0,5), (-5,0), (0, -5)} $

Profile SVG
Week 3 Practice Assignment Mathematics I
Practice

1. Multiple Choice Questions (MCQ) 1. What will be the equation of the tangent to the curve $ f(x) = 2x^2 + 9x + 20 $ at point $(-3, 11)$? Options: $ y = 3x $ $ y = -3x + 2 $ $ y = -3x + 20 $ $ y = -\frac{x}{3} + 2 $ $ y = \frac{x}{3} + 20 $ $ y = -\frac{x}{3} $ Solution: Slope at $ x = -3 $: $ m = 2 \times 2 \times (-3) + 9 = -3 $ Equation of tangent: $ y = -3x + c $ Passes through $(-3, 11)$: $ 11 = -3 \times (-3) + c \implies c = 2 $ Answer: $ y = -3x + 2 $123

Profile SVG
Week 4 Practice Assignment Mathematics I
Practice

Multiple Choice Questions (MCQ) Question 1: Which of the following polynomial functions represents the profit from selling Tamil books? ○ 2x³ + 4x² - 2x - 4 ○ x³ - 2x² - x + 2 ○ x³ + 2x² - x - 2 ○ 2x³ - 4x² - 2x + 4 Solution: The profit from selling Tamil books is $x^3 + 2x^2 - x - 2$.

Profile SVG
Week 5 Practice Assignment Mathematics I
Practice

Multiple Choice Questions (MCQ) Question 1: Choose the correct option. ○ f₃ is not a function. ○ f₆ is not a function. ○ f₅ is not a function. ○ All of the above are functions. Solution: Vertical line test fails only for f₆ and therefore f₆(x) is not a function. Question 2: Choose the correct option. ○ f₁ and f₃ are one-one functions in the given domain. ○ f₂ and f₄ are one-one functions in the given domain. ○ f₃ and f₅ are one-one functions in the given domain. ○ f₅ is one-one function in the given domain. Solution: The functions f₂ and f₄ are strictly decreasing functions in the domain [-4, 4], therefore these are one-to-one functions.

Profile SVG
Week 6 Practice Assignment Mathematics I
Practice

Question 1 If $ b > 0 $ and $ 4\log_x b + 9\log_{b^5x} b = 1 $, then the possible value(s) of $ x $ is (are) (a) $ b^{10} $ ✓ (b) $ b^9 $ (c) $ b^{-2} $ ✓ (d) $ b^5 $ (e) $ b^4 $ Solution: Let $ p = \log_b x $. Equation: $ \frac{4}{p} + \frac{9}{5 + p} = 1 $ Solve: $ p^2 - 8p - 20 = 0 $, roots $ p = -2, 10 $. So, $ x = b^{-2} $ or $ x = b^{10} $.

Profile SVG
Week 7 Practice Assignment Mathematics I
Practice

Question 1: Function Behavior from Graph Statement: From the graph, clearly the values of the function tend to 0 as $ x $ tends to $ \infty $. The graph is smooth (no sharp edges or sharp turns) and so the graph has a unique tangent at all points (at $ x = 1 $ and $ x = -1 $ too). The function values are decreasing (slope of the curve is negative) in $[-0.5, 0]$ (Option 7).

Profile SVG
Week 10 Practice Assignment Mathematics I
Practice

1. Multiple Choice Questions (MCQs) Question 1: Suppose we obtain the following DFS tree rooted at node 0 for an undirected graph with vertices ${0,1,2,3,4,5,6,7,8,9,10}$. Which of the following cannot be an edge in the original graph? (a) $(1,4)$ (b) $(0,4)$ (c) $(7,10)$ (d) $(2,9)$ Solution: In a DFS tree for an undirected graph, edges between vertices in different branches cannot be present in the original graph if they would have already been visited.