case study questions on application of derivatives

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Case Study on Application Of Derivatives Class 12 Maths PDF

The passage-based questions are commonly known as case study questions. Students looking for Case Study on Application Of Derivatives Class 12 Maths can use this page to download the PDF file.

The case study questions on Application Of Derivatives are based on the CBSE Class 12 Maths Syllabus, and therefore, referring to the Application Of Derivatives case study questions enable students to gain the appropriate knowledge and prepare better for the Class 12 Maths board examination. Continue reading to know how should students answer it and why it is essential to solve it, etc.

Case Study on Application Of Derivatives Class 12 Maths with Solutions in PDF

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The case study on Application Of Derivatives Class 12 Maths with solutions in PDF helps students tackle questions that appear confusing or difficult to answer. The answers to the Application Of Derivatives case study questions are very easy to grasp from the PDF - download links are given on this page.

Why Solve Application Of Derivatives Case Study Questions on Class 12 Maths?

There are three major reasons why one should solve Application Of Derivatives case study questions on Class 12 Maths - all those major reasons are discussed below:

To Prepare for the Board Examination: For many years CBSE board is asking case-based questions to the Class 12 Maths students, therefore, it is important to solve Application Of Derivatives Case study questions as it will help better prepare for the Class 12 board exam preparation.
Develop Problem-Solving Skills: Class 12 Maths Application Of Derivatives case study questions require students to analyze a given situation, identify the key issues, and apply relevant concepts to find out a solution. This can help CBSE Class 12 students develop their problem-solving skills, which are essential for success in any profession rather than Class 12 board exam preparation.
Understand Real-Life Applications: Several Application Of Derivatives Class 12 Maths Case Study questions are linked with real-life applications, therefore, solving them enables students to gain the theoretical knowledge of Application Of Derivatives as well as real-life implications of those learnings too.

How to Answer Case Study Questions on Application Of Derivatives?

Students can choose their own way to answer Case Study on Application Of Derivatives Class 12 Maths, however, we believe following these three steps would help a lot in answering Class 12 Maths Application Of Derivatives Case Study questions.

Read Question Properly: Many make mistakes in the first step which is not reading the questions properly, therefore, it is important to read the question properly and answer questions accordingly.
Highlight Important Points Discussed in the Clause: While reading the paragraph, highlight the important points discussed as it will help you save your time and answer Application Of Derivatives questions quickly.
Go Through Each Question One-By-One: Ideally, going through each question gradually is advised so, that a sync between each question and the answer can be maintained. When you are solving Application Of Derivatives Class 12 Maths case study questions make sure you are approaching each question in a step-wise manner.

What to Know to Solve Case Study Questions on Class 12 Application Of Derivatives?

A few essential things to know to solve Case Study Questions on Class 12 Application Of Derivatives are -

Basic Formulas of Application Of Derivatives: One of the most important things to know to solve Case Study Questions on Class 12 Application Of Derivatives is to learn about the basic formulas or revise them before solving the case-based questions on Application Of Derivatives.
To Think Analytically: Analytical thinkers have the ability to detect patterns and that is why it is an essential skill to learn to solve the CBSE Class 12 Maths Application Of Derivatives case study questions.
Strong Command of Calculations: Another important thing to do is to build a strong command of calculations especially, mental Maths calculations.

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CBSE Case Study Questions for Class 12 Maths Applications of Derivatives Free PDF

Mere Bacchon, you must practice the CBSE Case Study Questions for Class 12 Maths Applications of Derivatives in order to fully complete your preparation . They are very very important from exam point of view. These tricky Case Study Based Questions can act as a villain in your heroic exams! I have made sure the questions (along with the solutions) prepare you fully for the upcoming exams. To download the latest CBSE Case Study Questions , just click ‘ Download PDF ’.

CBSE Case Study Questions for Class 12 Maths Applications of Derivatives PDF

Checkout our case study questions for other chapters.

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Chapter 5 Continuity and Differentiability Case Study Questions
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Chapter 8 Applications of Integrals Case Study Questions

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Class 12 Maths: Case Study of Chapter 6 Applications of Derivatives PDF Download

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In Class 12 Boards there will be Case studies and Passage Based Questions will be asked, So practice these types of questions. Study Rate is always there to help you. Free PDF Download of CBSE Class 12 Mathematics Chapter 6 Applications of Derivatives Case Study and Passage Based Questions with Answers were Prepared Based on Latest Exam Pattern. Students can solve NCERT Class 12 Maths Applications of Derivatives to know their preparation level.

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In CBSE Class 12 Maths Paper, There will be a few questions based on case studies and passage-based as well. In that, a paragraph will be given, and then the MCQ questions based on it will be asked.

Applications of Derivatives Case Study Questions With answers

Here, we have provided case-based/passage-based questions for Class 12 Mathematics Chapter 6 Applications of Derivatives

Case Study/Passage-Based Questions

(i) To construct a garden using 200 ft of fencing, we need to maximize its

Answer: (b) area

(ii) If x denote the length of side of garden perpendicular to brick wall and y denote the length, of side parallel to brick wall, then find the relation representing total amount of fencing wire.

Answer: (c) y+2x=200

(iii) Area of the garden as a function of x, say A(x), can be represented as

Answer: (c) 200x – 2×2

(iv) Maximum value of A(x) occurs at x equals

Answer: (a) 50 ft

(v) Maximum area of garden will be

Answer: (c) 5000sq.ft

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Class 12th Maths - Application of Derivatives Case Study Questions and Answers 2022 - 2023

By QB365 on 08 Sep, 2022

QB365 provides a detailed and simple solution for every Possible Case Study Questions in Class 12 Maths Subject - Application of Derivatives, CBSE. It will help Students to get more practice questions, Students can Practice these question papers in addition to score best marks.

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Application of derivatives case study questions with answer key.

12th Standard CBSE

Final Semester - June 2015

(ii) Volume of the open box formed by folding up the cutting corner can be expressed as

(iii) The values of x for which \(\begin{equation} \frac{d V}{d x}=0 \end{equation}\) ,are

(iv) Megha is interested in maximising the volume of the box. So, what should be the side of the square to be cut off so that the volume of the box is maximum?

(v) The maximum value of the volume is

(ii) If x denote the length of side of garden perpendicular to brick wall and y denote the length, of side parallel to brick wall, then find the relation representing total amount of fencing wire.

(iii) Area of the garden as a function of x, say A(x), can be represented as

(iv) Maximum value of A(x) occurs at x equals

(v) Maximum area of garden will be

(ii) Revenue R as a function of x can be represented as

(iii) Find the number of days after 1stJuly, when Shyams father attain maximum revenue.

(iv) On which day should Shyam's father harvest the onions to maximise his revenue?

(v) Maximum revenue is equal to

An owner of an electric bi~e rental company have determined that if they charge customers Rs. x per day to rent a bike, where 50 Rs. x Rs. 200, then number of bikes (n), they rent per day can be shown by linear function n(x) = 2000 - 10x. If they charge Rs. 50 per day or less, they will rent all their bikes. If they charge Rs. 200 or more per day, they will not rent any bike. Based on the above information, answer the following questions.

(ii) If R(x) denote the revenue, then maximum value of R(x) occur when x equals

(iii) At x = 260, the revenue collected by the company is

(iv) The number of bikes rented per day, if x = 105 is

(v) Maximum revenue collected by company is

(ii) If x represent the number of apartments which are not rented, then the profit expressed as a function of x is

(iii) If P = 10500, then N =

(iv) If P = 11,000, then the profit is

(ii) The range of x is

(iii) The value of xfor which revenue is maximum, is

(iv) When the revenue is maximum, the price of the ticket is

(v) How any spectators should be present to maximize the revenue?

(ii) The radius that will minimize the cost of the material to manufacture the tin can is

(iii) The height thatt will minimize the cost of the material to manufacture the tin can is

(iv) If the cost of material used to manufacture the tin can is Rs.100/m 2 and \(\sqrt[3]{\frac{1500}{\pi}} \approx 7.8\) then minimum cost is approximately

(v) To minimize the cost of the material used to manufacture the tin can, we need to minimize the

(ii) The relation between a and b is given by

(iii) Area of poster in terms of b is given by

(iv) The value of b, so that area of the poster is minimized, is

(v) The value of a, so that area of the poster is minimized, is

(i) In order to make a least expensive water tank, Nitin need to minimize its

(ii) Total cost of tank as a function of h can' be' represented as

(iii) Range of h is

(iv) Value of h at which c(h) is minimum, is

(v) The cost ofleast expensive tank is

(ii) If sum of the surface areas of box and ball are given to be constant k 2 , then x is equal to

(iii) The radius of the ball, when S is minimum, is

(iv) Relation between length of the box and radius of the ball can be represented as

(v) Minimum value of S is

(ii) If C(x) denote the maintenance cost function, then maximum value of C(x) occur at x =

(iii) The maximum value of C(x) would be

(iv) The number of apartments, that the complex should have in order to minimize the maintenance cost, is

(v) If the minimum maintenance cost is attain, then the maintenance cost for each apartment would be

(ii) If x is the length of the outer rectangle, then area of inner rectangle in terms of x is

(iii) Find the range of x.

(iv) If area of inner rectangle is m~imum, then x is equal to

(v) If area of inner rectangle is maximum, then length and breadth of this rectangle are respectively

(ii) If magazine company increases Rs.500 as annual charges, then R is equal to

(iv) What amount of increase in annual charges will bring maximum revenue?

(ii) The maximum value of x can not be

(iii) The rainimum value of x can not be

(iv) If l(x) denote the combined light intensity, then lex) will be minimum when x =

(v) The darkest spot between the two lights is

(ii) The relation between x and y is

(iii) The outer surface area of tank will be minimum when depth of tank is equal to

(iv) The cost of material will be least when width of tank is equal to

(v) If cost of aluminium sheet is Rs.360/m 2 , then the minimum cost for the construction of tank will be

(ii) Distance (say D) between Arun and Manita will be

(iii) For which real value(s) of x, first derivative of D 2 w.r.t. 'x' will Vanish?

(iv) Find the position of Arun when Manita will hit the paper hall.

(v) The minimum value of D is

(ii) The area (A) of green grass, in terms of x, is given by

(iii) The maximum value of A is

(iv) The value oflength of rectangle, when A is maximum, is

(v) The area of gravelling path is

(ii) The length PQ is

(iii) Let there be a quantity S such that S = Rp 2 + RQ 2 , then S is given by

(iv) Find the value of x for which value of S is minimum.

(v) For minimum value of S, find the value of PR and RQ

(ii) The area (A) of the window can be given by

(iii) Rohan is interested in maximizing the area of the whole window, for this to happen, the value of x should be

(iv) Maximum area of the window is

(v) For maximum value of A, the breadth of rectangular part of the window is

(ii) The area (A) of the rectangular region, as a function of x, can be expressed as

(iii) School's manager is interested in maximising the area of floor 'A' for this to be happen, the value of x should be

(iv) The value of y, for which the area of floor is maximum is

(v) Maximum area of floor is

*****************************************

Application of derivatives case study questions with answer key answer keys.

(i) (b) : Since, side of square is of length 20 cm therefore \(\begin{equation} x \in(0,10) \end{equation}\) . ( ii) (a) : Clearly, height of open box = x cm Length of open box = 20 - 2x and width of open box = 20 - 2x \(\therefore\) Volume (V) of the open box = x x (20 - 2x) x (20 - 2x (iii) (d) : We have, V = x(20 - 2X) 2 \(\begin{equation} \therefore \frac{d V}{d x}=x \cdot 2(20-2 x)(-2)+(20-2 x)^{2} \end{equation}\) = (20 - 2x)( -4x + 20 - 2x) = (20 - 2x)(20 - 6x) Now, \(\begin{equation} \frac{d V}{d x}=0 \Rightarrow 20-2 x=0 \text { or } 20-6 x=0 \end{equation}\) \(\begin{equation} \Rightarrow x=10 \text { or } \frac{10}{3} \end{equation}\) (iv) (c) : We have, V = x(20 - 2X) 2 and \(\begin{equation} \frac{d V}{d x}=(20-2 x)(20-6 x) \end{equation}\) \(\begin{equation} \Rightarrow \frac{d^{2} V}{d x^{2}}=(20-2 x)(-6)+(20-6 x)(-2) \end{equation}\) = (-2)[60 - 6x + 20 - 6x] = (-2)[80 - 12x] = 24x - 160 For \(\begin{equation} x=\frac{10}{3}, \frac{d^{2} V}{d x^{2}}<0 \end{equation}\) and for \(\begin{equation} x=10, \frac{d^{2} V}{d x^{2}}>0 \end{equation}\) So, volume will be maximum when \(\begin{equation} x=\frac{10}{3} \end{equation}\) . (v) (d) : We have, V = x(20 - 2x) 2 ,which will be maximum when \(\begin{equation} x=\frac{10}{3} \end{equation}\) . \(\begin{equation} \therefore \ \text { Maximum volume }=\frac{10}{3}\left(20-2 \times \frac{10}{3}\right)^{2} \end{equation}\) \(\begin{equation} =\frac{10}{3} \times \frac{40}{3} \times \frac{40}{3}=\frac{16000}{27} \mathrm{~cm}^{3} \end{equation}\)

(i) (b) : To create a garden using 200 ft fencing, we need to maximise its area. (ii) (c) : Required relation is given by 2x + y = 200. (iii) (c) : Area of garden as a function of x can be represented as \(A(x)=x \cdot y=x(200-2 x)=200 x-2 x^{2}\) (iv) (a) : \(\begin{equation} A(x)=200 x-2 x^{2} \Rightarrow A^{\prime}(x)=200-4 x \end{equation}\) For the area to be maximum A'(x) = 0 \(\begin{equation} \Rightarrow 200-4 x=0 \Rightarrow x=50 \mathrm{ft} \end{equation}\) (v) (c) : Maximum-area of the garden = 200(50) - 2(50)2 = 10000 - 5000 = 5000 sq. ft

(i) (a) : Let x be the number of extra days after 1 st July. \(\therefore\) Price = Rs.(300 - 3Xx) = Rs.(300 - 3x) Quantity = 80 quintals + x(1 quintal per day) = (80 + x) quintals (ii) (b) : R(x) = Quantity x Price = (80 + x) (300 - 3x) = 24000 - 240x + 300x -3x 2 = 24000 + 60x - 3x 2 (iii) (a) : We have, R(x) = 24000 + 60x - 3x 2 \(\begin{equation} \Rightarrow R^{\prime}(x)=60-6 x \Rightarrow R^{\prime \prime}(x)=-6 \end{equation}\) For R(x) to be maximum, R'(x) = 0 and R"(x) < 0 \(\begin{equation} \Rightarrow 60-6 x=0 \Rightarrow x=10 \end{equation}\) (iv) (a) : Shyams father will attain maximum revenue after 10 days. So, he should harvest the onions after 10 days of 1 st July i.e., on 11 th July. (v) (c) : Maximum revenue is collected by Shyams father when x = 10 \(\therefore\) Maximum revenue = R(10) = 24000 + 60(10) - 3(10)2 = 24000 + 600 - 300 = 24300

(i) (a) : Let x be the charges per bike per day and n be the number of bikes rented per day. R(x) = n x x = (2000 - lOx) x = -10x 2 + 2000x (ii) (b) : We have, R(x) = 2000x - 10x 2 \(\Rightarrow R^{\prime}(x)=2000-20 x\) For R(x) to be maximum or minimum, R'(x) = 0 \(\Rightarrow 2000-20 x=0 \Rightarrow x=100\) Also, \(R^{\prime \prime}(x)=-20<0\) Thus, R(x) is maximum at x = 100 (iii) (c) : If company charge ~ 200 or more, they will not rent any bike. Therefore, revenue collected by him will be zero. (iv) (c) : If x = 105, number of bikes rented per day is given by n = 2000 - 10 x 105 = 950 (v) (d) : At x = 100, R(x) is maximum \(\therefore\) Maximum revenue = R(100) = -10(100) 2 + 2000(100) = Rs. 1,00,000

(i) (c) : If P is the rent price per apartment and N is the number of rented apartment, the profit is given by NP - 500 N = N(P- 500) [ \(\therefore\) Rs. 500/month is the maintenance charges for each occupied unit] (ii) (c) : Now, if x be the number of non-rented apartments, then N= 50 - x and P = 10000 + 250 x Thus, profit = N(P- 500) = (50 - x ) (10000 + 250x - 500 = (50 - x) (9500 + 250 x) = 250(50 - x) (38 + x) (iii) (b) : Clearly, if P = 10500, then \(10500=10000+250 x \Rightarrow x=2 \Rightarrow N=48\) (iv) (a) : Also, if P = 11000, then \(11000=10000+250 x \Rightarrow x=4\) and so profit (v) (b) : We have, P(x) = 250(50 - x) (38 + x) Now, P'(x) = 250[50 - x - (38 + x)] = 250[12 - 2x] For maxima/minima, put P'(x) = 0 \(\Rightarrow 12-2 x=0 \Rightarrow x=6\) Thus, price per apartment is, P = 10000 + 1500 = 11500 Hence, the rent that maximizes the profit is Rs. 11500.

(i) (a) : Let p be the price per ticket and x be the number of tickets sold. Then, revenue function \(R(x)=p \times x=\left(15-\frac{x}{3000}\right) x\) \(=15 x-\frac{x^{2}}{3000}\) (ii) (c) : Since, more than 36000 tickets cannot be sold. So, range of x is [0, 36000]. (iii) (c) : We have, \(R(x)=15 x-\frac{x^{2}}{3000}\) \(\Rightarrow R^{\prime}(x)=15-\frac{x}{1500}\) For maxima/minima, put R'(x) = 0 \(\Rightarrow \quad 15-\frac{x}{1500}=0 \Rightarrow x=22500\) Also, \(R^{\prime \prime}(x)=-\frac{1}{1500}<0\) (iv) (d) : Maximum revenue will be at x = 22500 \(\therefore \text { Price of a ticket }=15-\frac{22500}{3000}=15-7.5=Rs.7.5\) (v) (d) : Number of spectators will be equal to number of tickets sold. \(\therefore\) Required number of spectators = 22500

(i) (d) : Given, r cm is the radius and h cm is the height of required cylindrical can Given that, volume = 3 l= 3000 cm 3 \(\left(\because 1 l=1000 \mathrm{~cm}^{3}\right)\) \(\Rightarrow \pi r^{2} h=3000 \Rightarrow h=\frac{3000}{\pi r^{2}}\) Now, the surface area, as a function of r is given by \(S(r)=2 \pi r^{2}+2 \pi r h=2 \pi r^{2}+2 \pi r\left(\frac{3000}{\pi r^{2}}\right)\) \(=2 \pi r^{2}+\frac{6000}{r}\) (ii) (c) : Now, \(S(r)=2 \pi r^{2}+\frac{6000}{r}\) \(\Rightarrow S^{\prime}(r)=4 \pi r-\frac{6000}{r^{2}}\) To find criti£al points, put S'(r) = 0 \(\Rightarrow \frac{4 \pi r^{3}-6000}{r^{2}}=0\) \(\Rightarrow r^{3}=\frac{6000}{4 \pi} \Rightarrow r=\left(\frac{1500}{\pi}\right)^{1 / 3}\) Also, \(\left.S^{\prime \prime}(r)\right|_{r=} \sqrt[3]{\frac{1500}{\pi}}=4 \pi+\frac{12000 \times \pi}{1500}\) \(=4 \pi+8 \pi=12 \pi>0\) Thus, the critical point is the point of minima. (iii) (b) : The cost of material for the tin can is minimized when \(r=\sqrt[3]{\frac{1500}{\pi}} \mathrm{cm}\) and the height is \(\frac{3000}{\pi\left(\sqrt[3]{\frac{1500}{\pi}}\right)^{2}}=2 \sqrt[3]{\frac{1500}{\pi}} \mathrm{cm} .\) . (iv) (a) : We have,minimum surface area = \(\frac{2 \pi r^{3}+6000}{r}\) . \(=\frac{2 \pi \cdot \frac{1500}{\pi}+6000}{\sqrt[3]{\frac{1500}{\pi}}}=\frac{9000}{7.8}=1153.84 \mathrm{~cm}^{2}\) Cost of 1 m 2 material = Rs.100 \(\therefore \ \text { Cost of } 1 \mathrm{~cm}^{2} \text { material }=Rs. \frac{1}{100}\) \(\therefore \ \text { Minimum cost }=Rs. \frac{1153.84}{100}=Rs. 11.538\) (v) (c) : To minimize the cost we need to minimize the total surface area.

(i) (d) : In order to make least expensive water tank, Nitin need to minimize its cost. (ii) (d) : Let 1ft be the length and h ft be the height of the tank. Since breadth is equal to 5 ft. (Given) \(\therefore\) Two sides will be 5h sq. feet and two sides will be 1h sq. feet. So, the total area of the sides is (10 h + 2 1h)ft 2 Cost of the sides is Rs.10 per sq. foot. So, the cost to build the sides is (10h + 21h) x 10 = Rs.(100h + 20lh) Also, cost of base = (5l) x 20 = Rs. 100 l \(\therefore\) Total cost of the tank in Rs. is given by c = 100 h + 20 I h + 100 l Since, volume of tank = 80ft 3 \(\therefore \quad 5 l h=80 \mathrm{ft}^{3} \quad \therefore l=\frac{80}{5 h}=\frac{16}{h}\) \(\therefore \quad c(h)=100 h+20\left(\frac{16}{h}\right) h+100\left(\frac{16}{h}\right)\) \(=100 h+320+\frac{1600}{h}\) (iii) (b) : Since, all side lengths must be positive \(\therefore \quad h>0\) and \(\frac{16}{h}>0\) Since, \(\frac{16}{h}>0, \text { whenever } h>0\) \(\therefore \text { Range of } h \text { is }(0, \infty)\) (iv) (a) : To minimize cost, \(\frac{d c}{d h}=0\) \(\Rightarrow \quad 100-\frac{1600}{h^{2}}=0\) \(\Rightarrow 100 h^{2}=1600 \Rightarrow h^{2}=16 \Rightarrow h=\pm 4\) \(\Rightarrow h=4\) [ \(\therefore\) height can not be negative] (v) (c) : Cost of least expensive tank is given by \(c(4)=400+320+\frac{1600}{4}\) = 720 + 400 = Rs. 1120

(i) (c) : Let Sbe the sum of volume of parallelopiped and sphere, then \(S=x(2 x)\left(\frac{x}{3}\right)+\frac{4}{3} \pi r^{3}=\frac{2 x^{3}}{3}+\frac{4}{3} \pi r^{3}\) ...(i) (ii) (a) : Since, sum of surface area of box and sphere is given to be constant k 2 . \(\therefore \quad 2\left(x \times 2 x+2 x \times \frac{x}{3}+\frac{x}{3} \times x\right)+4 \pi r^{2}=k^{2}\) \(\Rightarrow 6 x^{2}+4 \pi r^{2}=k^{2}\) \(\Rightarrow x^{2}=\frac{k^{2}-4 \pi r^{2}}{6} \Rightarrow x=\sqrt{\frac{k^{2}-4 \pi r^{2}}{6}}\) ...(2) (iii) (b) : From (1) and (2), we get \(S=\frac{2}{3}\left(\frac{k^{2}-4 \pi r^{2}}{6}\right)^{3 / 2}+\frac{4}{3} \pi r^{3}\) \(=\frac{2}{3 \times 6 \sqrt{6}}\left(k^{2}-4 \pi r^{2}\right)^{3 / 2}+\frac{4}{3} \pi r^{3}\) \(\Rightarrow \quad \frac{d S}{d r}=\frac{1}{9 \sqrt{6}} \frac{3}{2}\left(k^{2}-4 \pi r^{2}\right)^{1 / 2}(-8 \pi r)+4 \pi r^{2}\) \(=4 \pi r\left[r-\frac{1}{3 \sqrt{6}} \sqrt{k^{2}-4 \pi r^{2}}\right]\) For maximum /minimum \(\frac{d S}{d r}=0\) \(\Rightarrow \frac{-4 \pi r}{3 \sqrt{6}} \sqrt{k^{2}-4 \pi r^{2}}=-4 \pi r^{2}\) \(\Rightarrow k^{2}-4 \pi r^{2}=54 r^{2}\) \(\Rightarrow r^{2}=\frac{k^{2}}{54+4 \pi} \Rightarrow r=\sqrt{\frac{k^{2}}{54+4 \pi}}\) ...(3) (iv) (d) : Since, \(x^{2}=\frac{k^{2}-4 \pi r^{2}}{6}=\frac{1}{6}\left[k^{2}-4 \pi\left(\frac{k^{2}}{54+4 \pi}\right)\right]\) [From (2) and (3)] \(=\frac{9 k^{2}}{54+4 \pi}=9\left(\frac{k^{2}}{54+4 \pi}\right)=9 r^{2}=(3 r)^{2}\) \(\Rightarrow x=3 r\) (v) (c) : Minimum value of S is given by \(\frac{2}{3}(3 r)^{3}+\frac{4}{3} \pi r^{3}\) \(=18 r^{3}+\frac{4}{3} \pi r^{3}=\left(18+\frac{4}{3} \pi\right) r^{3}\) \(=\left(18+\frac{4}{3} \pi\right)\left(\frac{k^{2}}{54+4 \pi}\right)^{3 / 2}\) [Using (3)] \(=\frac{1}{3} \frac{k^{3}}{(54+4 \pi)^{1 / 2}}\)

(i) (c) : Let C(x) be the maintenance cost function, then C(x) = 5000000 + 160x - 0.04x 2 (ii) (b) : We have, C(x) = 5000000 + 160x - 0.04x 2 Now, C(x) = 160 - 0.08x For maxima/minima, put C'(x) = 0 \(\Rightarrow\) 160 = 0.08x \(\Rightarrow\) x = 2000 (iii) (b) : Clearly, from the given condition we can see that we only want critical points that are in the interval [0,4500]. Now, we have C(0) = 5000000 C(2000) = 5160000 and C(4500) = 4910000 \(\therefore\) Maximum value of C(x)would be Rs.5160000 (iv) (a) : The complex must have 4500 apartments to minimise the maintenance cost. (v) (a) : The minimum maintenance cost for each apartment woud be Rs.1091.11

(i) (b) : In order to paint in the maximum area, Kyra needs to maximize the area of inner rectangle. (ii) (c) : Let x be the length and y be the breadth of outer rectangle. \(\therefore\) Length of inner rectangle = x - 1 and breadth of inner rectangle = y - 1.5 \(\therefore A(x)=(x-1)(y-1.5)\) \([\because x y=24 \text { (given) }]\) \(=(x-1)\left(\frac{24}{x}-1.5\right)\) (iii) (b) : Dimensions of rectangle (outer/inner) should be positive. \(\therefore \ x-1>0\) and \(\frac{24}{x}-1.5>0\) \(\Rightarrow x>1\) and \(x<16\) (iv) (c) : We have, \(A(x)=(x-1)\left(\frac{24}{x}-1.5\right)\) and \(A^{\prime \prime}(x)=\frac{-48}{x^{3}}\) For A(x) to be maximum or minimum, A'(x) = 0 \(\Rightarrow -1.5+\frac{24}{x^{2}}=0 \Rightarrow x^{2}=16 \Rightarrow x=\pm 4\) \(\therefore \ x=4\) [Since, length can't be negative] Also, \(A^{\prime \prime}(4)=\frac{-48}{4^{3}}<0\) Thus, at x = 4, area is maximum. (v) (a) : If area of inner rectangle is maximum, then Length of inner rectangle = x-1 = 4 - 1 = 3 ft And breadth of inner rectangle = \(y-1.5=\frac{24}{x}-1.5\) \(=\frac{24}{4}-1.5=6-1.5=4.5 \mathrm{ft}\)

(i) (d) : If x be the amount of increase in annual charges, then number of subscriber reduces to 5000 - x. \(\therefore\) Revenue, R(x) = (3000 + x) (5000 - x) \(=15000000+2000 x-x^{2}, 0 (ii) (a) : Clearly, at x = 500 R(500) = 15000000 + 2000(500) - (500) 2 = 15000000 + 1000000 - 250000 = Rs.15750000 (iii) (c) : Since, 15000000 + 2000x - x 2 = 15640000 (Given) \(\Rightarrow x^{2}-2000 x+640000=0\) \(\Rightarrow \quad x^{2}-1600 x-400 x+640000=0\) \(\Rightarrow x(x-1600)-400(x-1600)=0 \Rightarrow x=400,1600\) (iv) (a) : \(\frac{d R}{d x}=2000-2 x\) and \(\frac{d^{2} R}{d x^{2}}=-2<0\) For maximum revenue, \(\frac{d R}{d x}=0 \Rightarrow x=1000\) \(\therefore\) Required amount = Rs. 1000 (v) (b) : Maximum revenue = R(1000) = (3000 + 1000) (5000 - 1000) = 4000 x 4000 = ~ 16000000

(i) (c) : Since, the distance is x feet from the stronger light, therefore the distance from the weaker light will be 600 - x. So, the combined light intensity from both lamp posts is given by \(\frac{1000}{x^{2}}+\frac{125}{(600-x)^{2}}\) . (ii) (c) : Since, the person is in between the lamp posts, therefore x will lie in the interval (0, 600). So, maximum value of x can't be 600. (iii) (a) : Since, \(0 ,therefore minimum value of x can't be 0. (iv) (b) : We have , \(I(x)=\frac{1000}{x^{2}}+\frac{125}{(600-x)^{2}}\) \(\Rightarrow I^{\prime}(x)=\frac{-2000}{x^{3}}+\frac{250}{(600-x)^{3}}\) and \(\Rightarrow I^{\prime \prime}(x)=\frac{6000}{x^{4}}+\frac{750}{(600-x)^{4}}\) For maxima/minima, I'(x) = 0 \(\Rightarrow \frac{2000}{x^{3}}=\frac{250}{(600-x)^{3}} \Rightarrow 8(600-x)^{3}=x^{3}\) Taking cube root on both sides, we get \(2(600-x)=x \Rightarrow 1200=3 x \Rightarrow x=400\) Thus, I(x) is minimum when you are at 400 feet from the strong intensity lamp post. (v) (a) : Since, I(x) is minimum when x = 400 feet,therefore the darkest spot between the two light is at a distance of 400 feet from stronger lamp post, i.e., at a distance of 600 - 400 = 200 feet from the weaker lamp post.

(i) (c) : Length, AB = 2x Breadth, BC = 2y Also, radius, OA = 10 \(\therefore\) AC = 20 In \(\Delta A B C, A B+B C^{2}=A C^{2}\) \(\Rightarrow (2 x)^{2}+(2 y)^{2}=(20)^{2}\) \(\Rightarrow x^{2}+y^{2}=100\) (ii) (b) : Area of green grass = Area of rectangular part \(\therefore \ A=2 x \cdot 2 y\) [ \(\therefore\) Area of rectangle = length x breadth] \(=4 x y=4 x \sqrt{100-x^{2}}\) \(\left[\because x^{2}+y^{2}=100\right]\) (iii) (b) : We have, \(A=4 x \sqrt{100-x^{2}}\) \(\frac{d A}{d x}=\frac{4 x(-2 x)}{2 \sqrt{100-x^{2}}}+\sqrt{100-x^{2}} \cdot 4\) \(=\frac{-4 x^{2}+4\left(100-x^{2}\right)}{\sqrt{100-x^{2}}}\) For maximum value, \(\frac{d A}{d x}=0\) \(\Rightarrow-4 x^{2}+400-4 x^{2}=0\) \(\Rightarrow-8 x^{2}+400=0\) \(\Rightarrow x^{2}=50 \Rightarrow x=5 \sqrt{2}\) At \(x=5 \sqrt{2}\) \(A=4 x \sqrt{100-x^{2}}\) \(=4 \times 5 \sqrt{2} \cdot \sqrt{100-50}=4 \times 5 \sqrt{2} \times 5 \sqrt{2}=200 \mathrm{~m}^{2}\) (iv) (a) : Length of rectangle for which A is maximum \(=2 \times 5 \sqrt{2}=10 \sqrt{2}\) (v) (b) : Area of gravelling path = \(\pi(10)^{2}-200\) \(=100(\pi-2) \mathrm{m}^{2}\)

(i) (b) : Given, perimeter of window = 10 m \(\therefore\) x + y + y + perimeter of semicircle = 10 \(\Rightarrow x+2 y+\pi \frac{x}{2}=10\) (ii) (b) : \(A=x \cdot y+\frac{1}{2} \pi\left(\frac{x}{2}\right)^{2}\) \(=x\left(5-\frac{x}{2}-\frac{\pi x}{4}\right)+\frac{1}{2} \frac{\pi x^{2}}{4}\left[\because \text { From }(\mathrm{i}), y=5-\frac{x}{2}-\frac{\pi x}{4}\right]\) \(=5 x-\frac{x^{2}}{2}-\frac{\pi x^{2}}{4}+\frac{\pi x^{2}}{8}=5 x-\frac{x^{2}}{2}-\frac{\pi x^{2}}{8}\) (iii) (c) : We have, \(A=5 x-\frac{x^{2}}{2}-\frac{\pi x^{2}}{8}\) \(\Rightarrow \quad \frac{d A}{d x}=5-x-\frac{\pi x}{4}\) Now, \(\frac{d A}{d x}=0 \Rightarrow 5=x+\frac{\pi x}{4}\) \(\Rightarrow x(4+\pi)=20 \Rightarrow x=\frac{20}{4+\pi}\) \(\left[\text { Clearly, } \frac{d^{2} A}{d x^{2}}<0 \text { at } x=\frac{20}{4+\pi}\right]\) (iv) (d) : At \(x=\frac{20}{4+\pi}\) \(A=5\left(\frac{20}{4+\pi}\right)-\left(\frac{20}{4+\pi}\right)^{2} \frac{1}{2}-\frac{\pi}{8}\left(\frac{20}{4+\pi}\right)^{2}\) \(=\frac{100}{4+\pi}-\frac{200}{(4+\pi)^{2}}-\frac{50 \pi}{(4+\pi)^{2}}\) \(=\frac{(4+\pi)(100)-200-50 \pi}{(4+\pi)^{2}}=\frac{400+100 \pi-200-50 \pi}{(4+\pi)^{2}}\) \(=\frac{200+50 \pi}{(4+\pi)^{2}}=\frac{50(4+\pi)}{(4+\pi)^{2}}=\frac{50}{4+\pi}\) (v) (a) : We have, \(y=5-\frac{x}{2}-\frac{\pi x}{4}=5-x\left(\frac{1}{2}+\frac{\pi}{4}\right)\) \(=5-x\left(\frac{2+\pi}{4}\right)=5-\left(\frac{20}{4+\pi}\right)\left(\frac{2+\pi}{4}\right)\) \(=5-5 \frac{(2+\pi)}{4+\pi}=\frac{20+5 \pi-10-5 \pi}{4+\pi}=\frac{10}{4+\pi}\)

(i) (c) : Perimeter of floor = 2(length + breadth) \(\Rightarrow P=2(x+y)\) (ii) (c) : Area, A = length x breadth \(\Rightarrow A=x y\) Since, P = 2(x + y) \(\Rightarrow \frac{P-2 x}{2}=y\) \(\therefore \quad A=x\left(\frac{P-2 x}{2}\right) \Rightarrow A=\frac{P x-2 x^{2}}{2}\) (iii) (d) : We have, \(A=\frac{1}{2}\left(P x-2 x^{2}\right)\) \(\frac{d A}{d x}=\frac{1}{2}(P-4 x)=0\) \(\Rightarrow P-4 x=0 \Rightarrow x=\frac{P}{4}\) Clearly, at \(x=\frac{P}{4}, \frac{d^{2} A}{d x^{2}}=-2<0\) \(\therefore\) Area is maximum at \(x=\frac{P}{4}\) (iv) (c) : We have , \(y=\frac{P-2 x}{2}=\frac{P}{2}-\frac{P}{4}=\frac{P}{4}\) (v) (a) : We have, \(A=x y=\frac{P}{4} \cdot \frac{P}{4}=\frac{P^{2}}{16}\)

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Class 12 CBSE A.O.D ML AGGARWAL Case Study

Please select, case-study-1 a political party placed an order to a screen printer for the printing of party's slogan on a rectangular cloth sheets. a margin of 2.5 cm along length and 1 cm along width of cloth was left. the pictorial view of the slogan in shown below: if the total area of cloth is 640 cm2, based on the above information, answer the following questions: (i) the length and the breadth (in cm) of the rectangular part on which the slogan is printed respectively are (a)x-2.5, y-1 (b) x-5,y-2 (c) x-2,y-5 (d) x-5,y-1 (ii) the relation between x and y is (a) (x-2.5)(y-1)=640 (b) (x-5)(y-2)=640 (c) xy=640 (d) (x-2)(y-5)= (iii) area 'a' of cloth on which slogan matter was written is given by (a) 650-2x-3200/x (b) 650+2x+ 3200/x (c) 600-2x-3200/x (d) 600 +2x + 3200/x (iv) what is the value of x for which the printing area is maximum (a) 16 cm (b) 40 cm (c) 14 cm (d) 11 cm (v) what is the value of 'a' when printing of slogan is done on maximum area (a) 640 cm² (b) 418 cm² (c) 490 cm² (d) none of these.

Case-study-2 Saloni has a piece of tin rectangular in shape as shown in the diagram given below. She is going to cut squares from each corners and fold up the sides to form an open box. Based on the above information, answer the following questions: (i) If x is the side of square cut off from each corner of the sheet, then what is the length of open box? (a) 45-x (b) 45-2x (c) 45 +2x (d) none of these (ii) What is width of open box? (a) 24-x (b) 24+ x (c) 24-2x (d) none of these (iii) Volume 'V' of open box is given by (a) V = (45-x)(24-x) x (b) V=(45+x)(24+x) x (c) V=(45+2x) (24+2x) x (d) V=(45-2x)(24-2x) x (iv) What should be the side of square to be cut off so that the volume of box is maximum? (a) 18 cm (b) 5 cm (c) both (a) and (b) (d) none of these (v) The maximum value of V (in cm³) is (a) 5400 (b) 4200 (c) 3600 (d) 2450

Case-study-3 A company is planning to launch a new product and decides to pack the new product in closed right circular cylindrical cans of volume 432 πcm³. The cans are to be made from tin sheet. The company tried different options. Based on the above information, answer the following questions: (i) If 7 cm is the radius of the base of the cylinder and h cm is height, then (a) rh=216 (b) r(r+h)=216 (c) rh² = 432 (d) r 2 h=432 (ii) If S cm² is the surface area of the closed cylindrical can, then (a) S= 2π (r 2 +432/r) (b) S= π (r 2 +864/r) (c) S= π (r 2 +432/r) (d) S= 432π/r (iii) For S to be minimum r is equal to (a) 3 cm (b) 6 cm (c) 8 cm (d) 12 cm (iv) Minimum surface area of cylindrical can is (a) 54 πcm 2 (b) 108 πcm² (c) 216 πcm² (d) none of these (v) The relation between the radius and the height of cylindrical area is (a) height is equal to radius of base (b) height is equal to twice the radius of base (c) radius is equal to twice the height (d) radius is two-third of the height.

Case-study-4 A factory owner wants to construct a tank with rectangular base and rectangular sides, open at the top, so that its depth is 2 m and capacity is 8 m³. The building of the tank costs ₹250 per square metre for the base and ₹180 per square metre for the sides. Based on the above information, answer the following questions: (i) If the length and the breadth of the rectangular base of the tank are x metres and mimiy metres respectively, then the relation between x and y is (a)x+y=4 (b) xy=4 (c) xy=8 (d) xy+x+y=4 (ii) The cost of construction of the sides of the tank is (a) ₹180 (x+y) (b) ₹360 (x+y) (c) ₹720 (x + y) (d) ₹1120 (x+y) (iii) If C (in ₹) is the cost of construction of the tank, then C as a function of x is (d) C= 1120 + 720 (x+4/x) (b) C=720+1120 (x+4/x) (c) C=1120+360 (x+4/x) (d) C=1120+180 (x+4/x) (iv) The cost of construction of the tank is least when the value of x is (a) 1 (b) 3 /2 (c) 2 (d) 3 (v) The least cost of construction of the tank is (a) ₹2000 (b) ₹3000 (c)₹3600 (d) ₹4000

Case-study-5 A carpenter has a wire of length 28 m. He wants to cut into two pieces, one of the two pieces is to be made into a square and other into a circle. Based on the above information, answer the following questions: (i)If x metres wire is used in making a square, then what is the expression of combined area A? (ii) What is the length of radius for minimum combined area? (a) 28/π+4 (b) 112/π+4 (c) 14/π+4 (d) none of these (iii) What is the length of circular part? (a) 28/π+4 (b) 14/π+4 (c) 14π/π+4 (d) 28π/π+4 (iv) What is the length of square part? (a) 112/π+4 (b) 112π/π+4 (c) 14π/π+4 (d) 28/π+4 (v) The minimum combined area of square and circle is (a) 196/π+4 m 2 (b) 196/(π+4) 2 m 2 (c) 196/π-4 m 2 (d) none of these

Case-study-6 A firm has the cost function C(x)=x 3 /3-7x 2 + 111x +50 and demand function x = 100-p. Based on the above information, answer the following questions: (i) The total revenue function is (a) R(x) = x²-100x (b) R(x) = 100x-x 2 (c) R(x) = 100-x (d) none of these (ii) The total profit function is (a)-x 3 /3 +6x²-11x-50 (b)-x 3 /3-6x²-11x+50 (c)x 3 /3+6x²-11x+50 (d) -x 3 /3-6x²-11x+50 (iii) The value of x for which profit is maximum is (a) 8 (b) 9 (c) 10 (d) 11 (iv) The maximum profit is (a) ₹133.11 (b) ₹113.31 (c) ₹111.33 (d) ₹133.11 (b) The marginal revenue when x = 10 units is (a) 900 (b) 80 (c) 90 (d) 800

Case-study-7 The average cost function associated with producing and marketing x units of an item is given by 50 AC=2x-11 + 50/x Based on the above information, answer the following questions: (i) The total cost function is (a) C(x)=2x²-11x+50 (b) C(x)=2- 50/x 2 (c) C(x)=2x- 50/x (d) C(x)=2x²+11x-50 (ii) The marginal cost function is (a) MC=4x+ 11 (b) MC=4x-11 (c) MC = 2 + 50/x 2 (d) MC= 100/x 3 (iii) The marginal cost when x = 4 units is (a) ₹5 (b) ₹27 (c) ₹18 (d) ₹13 (iv) The range of values of x for which AC is increasing is (a) x (b) x>-5 (c) x (d) x>5 (v) The range of values of x for which total cost is decreasing is (a) x ≤ 11/4 (b) x ≥ 11/4 (c) x (d) x>-11/4

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Case Study Questions for Class 12 Maths Chapter 6 Application of Derivatives

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[PDF] Download Case Study Questions for Class 12 Maths Chapter 6 Application of Derivatives

Here we are providing case study questions for class 12 maths. In this article, we are sharing Class 12 Maths Chapter 6 Application of Derivatives case study questions. All case study questions of class 12 maths are solved so that students can check their solutions after attempting questions.

Case Study Questions for Class 12 Maths

Case study questions are a type of question that is commonly used in academic and professional settings to evaluate a person’s ability to analyze, interpret, and solve problems based on a given scenario or case study.

Typically, a case study question presents a real-world situation or problem that requires the individual to apply their knowledge and skills to identify the issues, consider various solutions, and recommend a course of action.

Case study questions are designed to test critical thinking skills, problem-solving abilities, and the capacity to work through complex and ambiguous situations.

Preparing for case study questions involves developing a deep understanding of the subject matter, being able to analyze and synthesize information quickly, and being able to communicate ideas clearly and effectively.

Importance of Solving Case Study Questions for Class 12 Maths

Solving case study questions for Class 12 Maths is extremely important as it provides students with an opportunity to apply the mathematical concepts they have learned to real-world scenarios. These questions present a situation or problem that requires students to use their problem-solving skills and critical thinking abilities to arrive at a solution.

The importance of solving case study questions for Class 12 Maths can be summarized as follows:

Enhances problem-solving skills: Case study questions challenge students to think beyond textbook examples and apply their knowledge to real-world situations. This enhances their problem-solving skills and helps them develop a deeper understanding of the mathematical concepts.
Improves critical thinking abilities: Case study questions require students to analyze and evaluate information, and draw conclusions based on their understanding of the situation. This helps them develop their critical thinking abilities, which are essential for success in many areas of life.
Helps in retaining concepts: Solving case study questions helps students retain the concepts they have learned for a longer period of time. This is because they are more likely to remember the concepts when they have applied them to a real-world situation.
Better preparation for exams: Many competitive exams, including the Class 12 Maths board exam, contain case study questions. Solving these questions helps students become familiar with the format of the questions and the skills required to solve them, which can improve their performance in exams.

In conclusion, solving case study questions for Class 12 Maths is important as it helps students develop problem-solving and critical thinking skills, retain concepts better, and prepare for exams.

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Unit 7: Applications of derivatives

About this unit.

Differential calculus is all about instantaneous rate of change. Let's see how the tools we've developed are applied in order to solve real-world word problems.

Rates of change in applied contexts

Applied rate of change: forgetfulness (Opens a modal)
Analyzing problems involving rates of change in applied contexts (Opens a modal)
Marginal cost & differential calculus (Opens a modal)
Rates of change in other applied contexts (non-motion problems) Get 3 of 4 questions to level up!

Related rates intro

Related rates intro (Opens a modal)
Analyzing problems involving related rates (Opens a modal)
Analyzing related rates problems: expressions (Opens a modal)
Analyzing related rates problems: equations (Pythagoras) (Opens a modal)
Analyzing related rates problems: equations (trig) (Opens a modal)
Analyzing related rates problems: expressions Get 3 of 4 questions to level up!
Analyzing related rates problems: equations Get 3 of 4 questions to level up!

Optimization

Optimization: sum of squares (Opens a modal)
Optimization: box volume (Part 1) (Opens a modal)
Optimization: box volume (Part 2) (Opens a modal)
Optimization: profit (Opens a modal)
Optimization: cost of materials (Opens a modal)
Optimization: area of triangle & square (Part 1) (Opens a modal)
Optimization: area of triangle & square (Part 2) (Opens a modal)
Optimization problem: extreme normaline to y=x² (Opens a modal)
Optimization Get 3 of 4 questions to level up!

Motion problems

Motion problems: when a particle is speeding up (Opens a modal)
Motion problems: finding the maximum acceleration (Opens a modal)
Motion problems (differential calc) Get 3 of 4 questions to level up!

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Chapter 6 Class 12 Application of Derivatives

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Learn Chapter 6 Application of Derivatives (AOD) of Class 12 free with solutions of all NCERT Questions for Maths Boards

We learned Derivatives in the last chapter, in Chapter 5 Class 12. In this Chapter we will learn the applications of those derivatives.

The topics in the chapter include

Finding rate of change
Checking if a function is increasing or decreasing in an interval
Checking if a function is increasing or decreasing in whole domain
Finding if function is strictly increasing or decreasing in an interval
Finding equation (and slope) of tangent and normal using derivatives
Finding approximate value of numbers and functions
Finding minimum and maximum values from graph of a function
Definition of - Maxima, Minima, Absolute Maxima, Absolute Minima, Point of Inflexion
Finding local maxima, local minima using first derivative test
Finding local maxima, local minima using second derivative test
Finding maximum and minimum values in closed interval ( Finding Absolute Maxima and Absolute Minima )
Statement questions of maxima and minima where we form equations and solve

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Mathematics
Class 12 Maths Case...

Class 12 Maths Case Study Questions

Table of Contents

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Class 12 Maths question paper will have 1-2 Case Study Questions. These questions will carry 5 MCQs and students will attempt any four of them. As all of these are only MCQs, it is easy to score good marks with a little practice. Class 12 Maths Case Study Questions are available on the myCBSEguide App and Student Dashboard .

Why Case Studies in CBSE Syllabus?

CBSE has introduced case study questions in the CBSE curriculum recently. The purpose was to make students ready to face real-life challenges with the knowledge acquired in their classrooms. It means, there was a need to connect theories with practicals. Whatsoever the students are learning, they must know how to apply it in their day-to-day life. That’s why CBSE is emphasizing case studies and competency-based education .

Case Study Questions in Maths

Let’s have a look over the class 12 Mathematics sample question paper issued by CBSE, New Delhi. Question numbers 17 and 18 are case study questions.

Focus on concepts

If you go through each MCQ there, you will find that the theme/case study is common but the questions are based on different concepts related to the theme. It means, that if you have done ample practice on the various concepts, you can solve all these MCQs in minutes.

Easy Questions with a Practical Approach

The difficulty level of the questions is average or say easy in some cases. On the other hand, you get four options to choose from. So, you get two levels of support to get full marks with very little effort.

Practice Questions Regularly

Most of the time we feel that it’s easy and neglect it. But in the end, we have to pay for this negligence. This may happen here too. Although it’s easy to score good marks on the case study questions if you don’t practice such questions, you may lose your marks. So, we suggest students should practice at least 30-40 such questions before writing the board exam.

12 Maths Case-Based Questions

We are giving you some examples of case study questions here. We have arranged hundreds of such questions chapter-wise on the myCBSEguide App. It is the complete guide for CBSE students. You can download the myCBSEguide App and get more case study questions there.

Case Study Question – 1

A is a diagonal matrix
A is a scalar matrix
A is a zero matrix
A is a square matrix
If A and B are two matrices such that AB = B and BA = A, then B 2 is equal to

Case Study Question – 2

4(x 3 – 24x 2 + 144x)
4(x 3 – 34x 2 + 244x)
x 3 – 24x 2 + 144x
4x 3 – 24x 2 + 144x
Local maxima at x = c 1
Local minima at x = c 1
Neither maxima nor minima at x = c 1
None of these

Case Study Questions Matrices -1

Answer Key:

Case Study Questions Matrices – 2

Read the case study carefully and answer any four out of the following questions: Once a mathematics teacher drew a triangle ABC on the blackboard. Now he asked Jose,” If I increase AB by 11 cm and decrease the side BC by 11 cm, then what type of triangle it would be?” Jose said, “It will become an equilateral triangle.”

Again teacher asked Suraj,” If I multiply the side AB by 4 then what will be the relation of this with side AC?” Suraj said it will be 10 cm more than the three times AC.

Find the sides of the triangle using the matrix method and answer the following questions:

(a) 3 × 3

Case Study Questions Determinants – 01

DETERMINANTS: A determinant is a square array of numbers (written within a pair of vertical lines) that represents a certain sum of products. We can solve a system of equations using determinants, but it becomes very tedious for large systems. We will only do 2 × 2 and 3 × 3 systems using determinants. Using the properties of determinants solve the problem given below and answer the questions that follow:

Three shopkeepers Ram Lal, Shyam Lal, and Ghansham are using polythene bags, handmade bags (prepared by prisoners), and newspaper envelopes as carrying bags. It is found that the shopkeepers Ram Lal, Shyam Lal, and Ghansham are using (20,30,40), (30,40,20), and (40,20,30) polythene bags, handmade bags, and newspapers envelopes respectively. The shopkeeper’s Ram Lal, Shyam Lal, and Ghansham spent ₹250, ₹270, and ₹200 on these carry bags respectively.

(b) Shyam Lal
(a) Ram Lal

Case Study Questions Determinants – 02

Case study questions application of derivatives.

R(x) = -x 2 + 200x + 150000
R(x) = x 2 – 200x – 140000
R(x) = 200x 2 + x + 150000
R(x) = -x 2 + 100 x + 100000
R'(x) > 0
R'(x) < 0
R”(x) = 0
(a) -x 2 + 200x + 150000
(a) R'(x) = 0
(c) 257, -63

Case Study Questions Vector Algebra

tan−1⁡(5/12)
tan−1⁡(12/3)
(b) 130 m/s
(a) tan−1⁡(5/12)
(b) 170 m/s

12 Maths Exam pattern

Question Paper Design of CBSE class 12 maths is as below. It clearly shows that 20% weightage will be given to HOTS questions. Whereas 55% of questions will be easy to solve.

No. chapter-wise weightage. Care to be taken to cover all the chapters
Suitable internal variations may be made for generating various templates keeping the overall weightage to different forms of questions and typology of questions the same.

Choice(s): There will be no overall choice in the question paper. However, 33% of internal choices will be given in all the sections

12 Maths Prescribed Books

Mathematics Part I – Textbook for Class XII, NCERT Publication
Mathematics Part II – Textbook for Class XII, NCERT Publication
Mathematics Exemplar Problem for Class XII, Published by NCERT
Mathematics Lab Manual class XII, published by NCERT

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12th Class Mathematics Applications of Derivatives Question Bank

Done case based (mcqs) - derivatives total questions - 40.

A) 1 done clear

B) -1 done clear

C) 0 done clear

D) 2 done clear

question_answer 2) L.H.D. of \[f\left( x \right)\] at \[x=1\] is

question_answer 3) \[f\left( x \right)\] is non-differentiable at

A) x = 1 done clear

B) x = 2 done clear

C) x = 3 done clear

D) x = 4 done clear

question_answer 4) Find the value of\[f'\left( 2 \right)\]

B) 2 done clear

C) 3 done clear

D) -1 done clear

question_answer 5) The value of f''(-1) is

A) 2 done clear

B) 1 done clear

C) -2 done clear

A) \[\frac{1}{\sqrt{2}}\] done clear

B) \[\sqrt{2}\] done clear

C) 1 done clear

D) 0 done clear

question_answer 7) The derivative of \[{{\sin }^{-1}}\left( \frac{2x}{1+{{x}^{2}}} \right)\]with respect to \[{{\cos }^{-1}}\left( \frac{1-{{x}^{2}}}{1+{{x}^{2}}} \right)\] is

A) - 1 done clear

C) 2 done clear

D) 4 done clear

question_answer 8) The derivative of \[{{e}^{{{x}^{3}}}}\]with respect to log x is

A) \[{{e}^{{{x}^{3}}}}\] done clear

B) \[3{{x}^{2}}\,2{{e}^{{{x}^{3}}}}\] done clear

C) \[3{{x}^{3}}{{e}^{{{x}^{3}}}}\] done clear

D) \[3{{x}^{2}}{{e}^{{{x}^{3}}}}+3x\] done clear

question_answer 9) The derivative of \[{{\cos }^{-1}}\left( 2x-1 \right)\]w.r.t. \[{{\cos }^{-1}}x\]is

B) \[\frac{-1}{2\sqrt{1-{{x}^{2}}}}\] done clear

C) \[\frac{2}{x}\] done clear

D) \[1-{{x}^{2}}\] done clear

question_answer 10) If \[y=\frac{1}{4}\,{{u}^{4}}\] and \[u=\frac{2}{3}\,{{x}^{3}}+5\], then \[\frac{dy}{dx}=\]

A) \[\frac{2}{27}{{x}^{2}}{{\left( 2{{x}^{3}}+15 \right)}^{3}}\] done clear

B) \[\frac{2}{27}{{x}^{2}}{{\left( 2{{x}^{3}}+15 \right)}^{3}}\] done clear

C) \[\frac{2}{27}x{{\left( 2{{x}^{3}}+5 \right)}^{3}}\] done clear

D) \[\frac{2}{27}{{\left( 2{{x}^{3}}+15 \right)}^{3}}\] done clear

question_answer 12) \[\frac{d}{dx}\left\{ gof\left( x \right) \right\}=\]

question_answer 13) R.H.D. of \[gof\left( x \right)\]at x = 0 is

A) 0 done clear

C) -1 done clear

question_answer 14) L.H.D. of \[gof\left( x \right)\]at x = 0 is

C) - 1 done clear

question_answer 15) The value of \[f'\left( x \right)\]at \[x=\frac{\pi }{4}\] is

A) 1/9 done clear

B) \[1/\sqrt{2}\] done clear

C) 1/2 done clear

D) not defined done clear

A) \[f\left( x \right)\] is differentiable and continuous done clear

B) \[f\left( x \right)\] is neither continuous nor differentiable done clear

C) \[f\left( x \right)\] is continuous but not differentiable done clear

D) none of these done clear

question_answer 17) If \[f\left( x \right)=\left| x-1 \right|,\,\,x\in R\], then at x=1

A) \[f\left( x \right)\]is not continuous done clear

B) \[f\left( x \right)\] is continuous but not differentiable done clear

C) \[f\left( x \right)\] is continuous and differentiable done clear

question_answer 18) \[f\left( x \right)={{x}^{3}}\] is

A) continuous but not differentiable at x = 3 done clear

B) continuous and differentiable at x = 3 done clear

C) neither continuous nor differentiable at x = 3 done clear

question_answer 19) If \[f\left( x \right)=\left[ sin\text{ }x \right]\], then which of the following is true?

A) \[f\left( x \right)\]is continuous and differentiable at x = 0. done clear

B) \[f\left( x \right)\]is discontinuous at x = 0 done clear

C) \[f\left( x \right)\]is discontinuous at x = 0 but not differentiable. done clear

D) \[f\left( x \right)\] is differentiable but not continuous at \[x=\pi /2\]. done clear

question_answer 20) If\[f\left( x \right)={{\sin }^{-1}}x,\,-1\,\le \,x\,\le \,\,1\], then

A) \[f\left( x \right)\] is both continuous and differentiable done clear

B) \[f\left( x \right)\] is neither continuous nor differentiable. done clear

C) \[f\left( x \right)\] is continuous but not differentiable. done clear

D) None of these. done clear

question_answer 22) If \[u={{x}^{2}}+{{y}^{2}}\] and \[x=s+3t\], \[y=2s-t\], then \[\frac{{{d}^{2}}u}{d{{s}^{2}}}\] is equal to

A) 12 done clear

B) 32 done clear

C) 36 done clear

D) 10 done clear

question_answer 23) \[f\left( x \right)=2\,\,\log \,\,\sin \,x\], then \[f''\left( x \right)\] is equal to

A) \[2\,\,\cos e{{c}^{3}}\,x\] done clear

B) \[2\,{{\cot }^{2}}x-4{{x}^{2}}\,\cos e{{c}^{2}}\,{{x}^{2}}\] done clear

C) \[2x\,\cot \,{{x}^{2}}\] done clear

D) \[-2\,\cos e{{c}^{2}}\,x\] done clear

question_answer 24) If \[f\left( x \right)={{e}^{x}}\,\sin \,x\], then \[f'''\left( x \right)=\]

A) \[2{{e}^{x}}\left( \sin \,x+\cos \,x \right)\] done clear

B) \[2{{e}^{x}}\left( \cos x-\sin x \right)\] done clear

C) \[2{{e}^{x}}\left( \sin \,x-\cos \,x \right)\] done clear

D) \[2{{e}^{x}}\,\cos x\] done clear

question_answer 25) If \[{{y}^{2}}=a{{x}^{2}}+bx+c\], then \[\frac{d}{dx}\left( {{y}^{3}}{{y}_{2}} \right)=\]

C) \[\frac{4ac-{{b}^{2}}}{{{a}^{2}}}\] done clear

A) \[{{x}^{x}}\left( 1+\log \,x \right)\] done clear

B) \[{{x}^{x}}\left( 1-\log \,x \right)\] done clear

C) \[-{{x}^{x}}\left( 1+\log \,x \right)\] done clear

D) \[{{x}^{x}}\,\log \,x\] done clear

question_answer 27) Differentiate \[{{x}^{x}}+{{a}^{x}}+{{x}^{a}}+{{a}^{a}}\] w.r.t. x

A) \[\left( 1+\log \,x \right)+\left( {{a}^{x}}\,\log \,a+a{{x}^{a-1}} \right)\] done clear

B) \[{{x}^{x}}\left( 1+\log \,x \right)+\log \,a+a{{x}^{a-1}}\] done clear

C) \[{{x}^{x}}\left( 1+\log \,x \right)+{{x}^{a}}\,\operatorname{logx}+a{{x}^{a-1}}\] done clear

D) \[{{x}^{x}}\left( 1+\log \,x \right)+{{a}^{x}}\,\log \,a\,+\,a{{x}^{a-1}}\] done clear

question_answer 28) If \[x={{e}^{x/y}}\], then find \[\frac{dy}{dx}\].

A) \[-\frac{\left( x+y \right)}{x\,\log \,x}\] done clear

B) \[-\frac{\left( x-y \right)}{x\,\log \,x}\] done clear

C) \[\frac{\left( x+y \right)}{x\,\log \,x}\] done clear

D) \[\frac{\left( x-y \right)}{x\,\log \,x}\] done clear

question_answer 29) If \[y={{\left( 2-x \right)}^{3}}\,{{\left( 3+2x \right)}^{5}}\], then find \[\frac{dy}{dx}\].

A) \[{{\left( 2-x \right)}^{3}}{{\left( 3+2x \right)}^{5}}\left[ \frac{15}{3+2x}-\frac{8}{2-x} \right]\] done clear

B) \[{{\left( 2-x \right)}^{3}}{{\left( 3+2x \right)}^{5}}\left[ \frac{15}{3+2x}+\frac{3}{2-x} \right]\] done clear

C) \[{{\left( 2-x \right)}^{3}}{{\left( 3+2x \right)}^{5}}\left[ \frac{10}{3+2x}-\frac{3}{2-x} \right]\] done clear

D) \[{{\left( 2-x \right)}^{3}}{{\left( 3+2x \right)}^{5}}\left[ \frac{10}{3+2x}+\frac{3}{2-x} \right]\] done clear

question_answer 30) If \[y={{x}^{x}}\,\,{{e}^{\left( 2x+5 \right)}}\] , then \[\frac{dy}{dx}\]is

A) \[{{x}^{x}}{{e}^{\left( 2x+5 \right)}}\,\left( 2+\log \,\,x \right)\] done clear

B) \[{{x}^{x}}{{e}^{\left( 2x+5 \right)}}\,\left( 3+2\,\log \,x \right)\] done clear

C) \[{{x}^{x}}\,{{e}^{\left( 2x+5 \right)}}\,\,\left( 2+3\,\log \,x \right)\] done clear

D) \[{{x}^{x}}\,{{e}^{\left( 2x+5 \right)}}\,\left( 3+\log \,x \right)\] done clear

A) \[\frac{-\sin \sqrt{x}}{2\sqrt{x}}\] done clear

B) \[\frac{\sin \sqrt{x}}{-\sin \sqrt{x}}\] done clear

C) \[sin\sqrt{x}\] done clear

D) \[-\sin \sqrt{x}\] done clear

question_answer 32) \[{{7}^{x+\frac{1}{x}}}\]

A) \[\left( \frac{{{x}^{2}}-1}{{{x}^{2}}} \right)\centerdot {{7}^{x+\frac{1}{x}}}\centerdot \log 7\] done clear

B) \[\left( \frac{{{x}^{2}}+1}{{{x}^{2}}} \right)\centerdot {{7}^{x+\frac{1}{x}}}\centerdot \log 7\] done clear

C) \[\left( \frac{{{x}^{2}}-1}{{{x}^{2}}} \right)\centerdot {{7}^{x-\frac{1}{x}}}\centerdot \log 7\] done clear

D) \[\left( \frac{{{x}^{2}}+1}{{{x}^{2}}} \right)\centerdot {{7}^{x+\frac{1}{x}}}\centerdot \log 7\] done clear

question_answer 33) \[\sqrt{\frac{1-\cos x}{1+\cos \,x}}\]

A) \[\frac{1}{2}{{\sec }^{2}}\frac{x}{2}\] done clear

B) \[-\frac{1}{2}{{\sec }^{2}}\frac{x}{2}\] done clear

C) \[{{\sec }^{2}}\frac{x}{2}\] done clear

D) \[-{{\sec }^{2}}\frac{x}{2}\] done clear

question_answer 34) \[\frac{1}{b}{{\tan }^{-1}}\left( \frac{x}{b} \right)+\frac{1}{a}{{\tan }^{-1}}\left( \frac{x}{a} \right)\]

A) \[\frac{-1}{{{x}^{2}}+{{b}^{2}}}+\frac{1}{{{x}^{2}}+{{a}^{2}}}\] done clear

B) \[\frac{1}{{{x}^{2}}+{{b}^{2}}}+\frac{1}{{{x}^{2}}+{{a}^{2}}}\] done clear

C) \[\frac{1}{{{x}^{2}}+{{b}^{2}}}-\frac{1}{{{x}^{2}}+{{a}^{2}}}\] done clear

question_answer 35) \[{{\sec }^{-1}}x+\cos e{{c}^{-1}}\frac{x}{\sqrt{{{x}^{2}}-1}}\]

A) \[\frac{2}{\sqrt{{{x}^{2}}-1}}\] done clear

B) \[\frac{-2}{\sqrt{{{x}^{2}}-1}}\] done clear

C) \[\frac{1}{\left| x \right|\,\sqrt{{{x}^{2}}-1}}\] done clear

D) \[\frac{2}{\left| x \right|\,\sqrt{{{x}^{2}}-1}}\] done clear

A) \[\frac{\left( 3{{x}^{2}}+2xy+{{y}^{2}} \right)}{{{x}^{2}}+2xy+3{{y}^{2}}}\] done clear

B) \[\frac{-\left( 3{{x}^{2}}+2xy+{{y}^{2}} \right)}{{{x}^{2}}+2xy+3{{y}^{2}}}\] done clear

C) \[\frac{\left( 3{{x}^{2}}+2xy-{{y}^{2}} \right)}{{{x}^{2}}-2xy+3{{y}^{2}}}\] done clear

D) \[\frac{3{{x}^{2}}+xy+{{y}^{2}}}{{{x}^{2}}+xy+3{{y}^{2}}}\] done clear

question_answer 37) \[{{x}^{y}}={{e}^{x-y}}\]

A) \[\frac{x-y}{\left( 1+\log \,x \right)}\] done clear

B) \[\frac{x+y}{\left( 1+\log \,x \right)}\] done clear

C) \[\frac{x-y}{x\left( 1+\log x \right)}\] done clear

D) \[\frac{x+y}{x\left( 1+\log \,x \right)}\] done clear

question_answer 38) \[{{e}^{\sin \,y}}=xy\]

A) \[\frac{-y}{x\,\left( y\,\cos y-1 \right)}\] done clear

B) \[\frac{y}{v\,\cos \,y-1}\] done clear

C) \[\frac{y}{y\,cos\,y+1}\] done clear

D) \[\frac{y}{x\left( y\,\cos y-1 \right)}\] done clear

question_answer 39) \[{{\sin }^{2}}x+{{\cos }^{2}}y=1\]

A) \[\frac{\sin \,2y}{\sin \,2x}\] done clear

B) \[-\frac{\sin \,2x}{\sin \,2y}\] done clear

C) \[-\frac{\sin \,2y}{\sin \,2x}\] done clear

D) \[\frac{\sin \,2y}{\sin \,2x}\] done clear

question_answer 40) \[y={{\left( \sqrt{x} \right)}^{{{\sqrt{x}}^{\sqrt{x}....\infty }}}}\]

A) \[\frac{-{{y}^{2}}}{x\left( 2-y\,\log \,x \right)}\] done clear

B) \[\frac{{{y}^{2}}}{2+y\,\log \,x}\] done clear

C) \[\frac{{{y}^{2}}}{x\left( 2+y\,\log \,x \right)}\] done clear

D) \[\frac{{{y}^{2}}}{x\left( 2-y\,\log \,x \right)}\] done clear

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IIT JEE Study Material

JEE Main Maths Applications of Derivatives Previous Year Questions With Solutions

JEE main previous year solved questions on Applications of Derivatives give students the opportunity to learn and solve questions in a more effective manner. In calculus, we use derivative to determine the maximum and minimum values of particular functions and many more. There are many important applications of derivatives. In this section, we will solve the problems based on the concepts such as maxima and minima of a function, approximation, Rolle’s theorem and intermediate value theorem, monotonicity, tangent and normal, subtangent and the subnormal, shortest distance between two curves etc. Collection of chapter wise JEE previous year questions with solutions is very useful study material for the JEE aspirants to check their preparation level. Students can expect about 1-2 questions from this chapter in the JEE examination. Click on the below link to get your PDF.

JEE Main Applications of Derivatives Past Year Questions With Solutions

A) Increasing

B) Decreasing

D) None of these

f(x) is an odd function.

f(x) is an increasing function.

Question 2: If f(x) = 2x 3 – 3x 2 – 12x + 5 and x belongs to [-2, 4], then the maximum value of function is at the following value of x

Therefore, the maximum value of function is 37 at x = 4.

Hence option D is the answer.

Question 3: The ratio of the height of cone of maximum volume inscribed in a sphere to its radius is

Diameter of the sphere = 2r

The radius of the cone is x and height is y.

Volume of cone is given by

Put \(\begin{array}{l}y=\frac{4}{3}r\end{array} \) \(\begin{array}{l}\frac{{{d}^{2}}V}{d{{y}^{2}}}=\frac{1}{3}\pi \,\left( 4r-6\times \frac{4}{3}r \right)\end{array} \) = negative value

Question 4: If ab = 2a + 3b, a > 0, b > 0 then the minimum value of ab is

When a = 6, b = 4;

(ab)min. = 6 × 4 = 24.

Question 5: If y = cot -1 (cos 2x) 1/2, then find the value of dy/dx at x = π/6.

Question 6: If f(x + y) = f(x).f(y) for all x and y and f(5) = 2, f'(0) = 3, then find f'(5).

Let x = 5, y = 0,

⇒ f(5+0) = f(5) f(0)

⇒ f(5) = f(5) f(0)

⇒ f(0) = 1

Question 7: If xe xy = y + sin 2 x, then at x = 0, dy/dx =

When x = 0, we get y = 0

Differentiating both sides with respect to x, we get;

Putting x = 0, y = 0, we get;

Question 8: If y = f(2x – 1)/(x 2 + 1) and f ‘(x) = sin x 2 then find dy/dx.

Let t = (2x – 1)/(x 2 + 1) then

Question 9: If x = sec θ – cos θ and y = sec n θ – cos n θ, then which of the following option is correct?

Option (A) is correct.

Question 10: If y 2 = p(x) is a polynomial of degree three, then 2(d/dx)(y 3 . d 2 y/dx 2 ) is

Question 11: At what points of the curve y = (2/3)x 3 + (1/2)x 2 , tangent makes the equal angle with axis?

Now tangent makes equal angle with axis

y = 45 0 and y = -45 0

and when x = -1,

Therefore, the required points are (1/2, 5/24) and (-1, -1/6).

Question 12: The function \(\begin{array}{l}f(x)=\int\limits_{-1}^{x}{t({{e}^{t}}-1)(t-1){{(t-2)}^{3}}{{(t-3)}^{5}}}dt\end{array} \) has a local minimum at x =

For local minima, slope, i.e. f′ (x) should change the sign from −ve to +ve.

If x = 0 + h; f′ (x) = (+) (+) (−) (−1) (−1) = −ve

Hence at x = 0, neither maxima or minima.

If x = 1 − h; f′ (x) = (+) (+) (−) (−1) (−1) = −ve

If x = 1 + h; f′ (x) = (+) (+) (+) (−1) (−1) = +ve

Hence at x = 1, there is a local minima.

If x = 2 − h; f′ (x) = (+) (+1) (+) (−) (−) = +ve

If x = 2 + h; f′ (x) = (+) (+) (+) (+) (−1) = −ve

Hence at x = 2, there is a local maxima.

If x = 3 − h; f′ (x) = (+) (+) (+) (+) (−) = −ve

If x = 3 + h; f′ (x) = (+) (+) (+) (+) (+) = +ve

Hence at x = 3, there is a local minima.

Question 13: Check whether the function \(\begin{array}{l}f(x)=\frac{\text{ln}(\pi +x)}{\text{ln}(e+x)}\end{array} \) is increasing or decreasing.

Hence, f (x) is decreasing in [0, ∞).

Question 14: Let \(\begin{array}{l}f(x)=\left\{\begin{Bmatrix} {x}^{\alpha}lnx, &x>0 \\ 0,& x=0 \end{Bmatrix}\right.\end{array} \) Rolle’s theorem is applicable to f for x ∈ [0, 1], if α =

For Rolle’s theorem to be applicable to f, for x ∈ [0, 1] we should have

(i) f (1) = f (0),

From (i), f (1) = 0, which is true.

From (ii), \(\begin{array}{l}0=f(0)=f({{0}_{+}})=\underset{x\to {{0}_{+}}}{\mathop{\lim }}\,{{x}^{\alpha }}\ln x\end{array} \) which is true only for positive values of α > 0, thus option (4) is correct.

Question 15: The distance travelled s (in metre) by a particle in t seconds is given by, s = t 3 + 2t 2 + t. What is the speed of the particle after 1 second?

Speed of the particle after 1 second is given by

Application of Derivatives Important JEE Main Questions 1

Application of Derivatives Important JEE Main Questions 2

JEE Advanced Maths Method of Differentiation Previous Year Questions with Solutions

Applications of Derivatives – Top 10 Most Important and Expected JEE Questions

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Man or bear? Hypothetical question sparks conversation about women's safety

Women explain why they would feel safer encountering a bear in the forest than a man they didn't know. the hypothetical has sparked a broader discussion about why women fear men..

If you were alone in the woods, would you rather encounter a bear or a man? Answers to that hypothetical question have sparked a debate about why the vast majority say they would feel more comfortable choosing a bear.

The topic has been hotly discussed for weeks as men and women chimed in with their thoughts all over social media.

Screenshot HQ , a TikTok account, started the conversation, asking a group of women whether they would rather run into a man they didn't know or a bear in the forest. Out of the seven women interviewed for the piece, only one picked a man.

"Bear. Man is scary," one of the women responds.

A number of women echoed the responses given in the original video, writing in the comments that they, too, would pick a bear over a man. The hypothetical has people split, with some expressing their sadness over the state of the world and others cracking jokes. Some men were flabbergasted.

Here's what we know.

A bear is the safer choice, no doubt about it, many say

There were a lot of responses, more than 65,000, under the original post. Many wrote that they understood why the women would choose a bear.

"No one’s gonna ask me if I led the bear on or give me a pamphlet on bear attack prevention tips," @celestiallystunning wrote.

@Brennduhh wrote: "When I die leave my body in the woods, the wolves will be gentler than any man."

"I know a bear's intentions," another woman wrote. "I don't know a man's intentions. no matter how nice they are."

Other TikTok users took it one step further, posing the hypothetical question to loved ones. Meredith Steele, who goes by @babiesofsteele , asked her husband last week whether he would rather have their daughter encounter a bear or a man in the woods. Her husband said he "didn't like either option" but said he was leaning toward the bear.

"Maybe it's a friendly bear," he says.

Diana, another TikTok user , asked her sister-in-law what she would choose and was left speechless.

"I asked her the question, you know, just for giggles. She was like, 'You know, I would rather it be a bear because if the bear attacks me, and I make it out of the woods, everybody’s gonna believe me and have sympathy for me," she said. "But if a man attacks me and I make it out, I’m gonna spend my whole life trying to get people to believe me and have sympathy for me.'"

Bear vs. man debate stirs the pot, woman and some men at odds

The hypothetical has caused some tension, with some women arguing that men will never truly understand what it's like to be a woman or the inherent dangers at play.

Social media users answered this question for themselves, producing memes, spoken word poetry and skits in the days and weeks since.

So, what would you choose?

Noncovalent interactions between quinoxalines and protoporphyrinogen oxidase (PPO): a computational case study for herbicidal applications

Original Paper
Published: 08 May 2024

Cite this article

Melek Hajji ORCID: orcid.org/0000-0002-6145-2858 2 nAff1 ,
Nadeem Abad 3 , 4 ,
Meriem Dallel 5 ,
Hanan Al-Ghulikah 6 ,
Mohamed El Hafi 4 ,
Taha Guerfel 2 ,
Joel T. Mague 7 ,
El Mokhtar Essassi 4 &
Youssef Ramli 3 , 8

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Quinoxaline-based compounds show promising inhibition of protoporphyrinogen oxidase (PPO), a key enzyme in chlorophyll production and a prime target for herbicide development. However, their precise molecular interactions remain largely unexplored. This study investigates 1-propyl-3-phenylquinoxalin-2(1H)-one (Qnz), a novel quinoxaline derivative designed and synthesized as a representative system. We examined its interaction, at the atomic level, with the active site of Nicotiana tabacum PPO (NtPPO) using an integrated molecular docking and DFT-based quantum mechanics cluster approach. The nature and strength of observed noncovalent interactions were theoretically evaluated using quantum theory of atoms-in-molecules (QTAIM), independent gradient model (IGM), and natural bond orbital (NBO) methods. Our analysis revealed a fascinating set of unconventional interactions that contribute to the stability of the Qnz–NtPPO complex, namely weak hydrogen bonding, homopolar dihydrogen interactions, π –stacking, and carbonyl–carbonyl interactions. Interestingly, the study uncovered the unexpected role of several less common amino acids—including Gly178, Ser235, and nonpolar aliphatic leucines—in facilitating molecular recognition. Moreover, the employed computational approaches have proven to be a powerful tool for analyzing interactions within the binding pocket.

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Acknowledgements

The authors extend their appreciation to Princess Nourah bint Abdulrahman University Researchers Supporting Project number (PNURSP2024R95), Princess Nourah bint Abdulrahman University, Riyadh, Saudi Arabia. The author, Melek Hajji, acknowledges the invaluable support and resources provided by Assalam International University-Sirte, throughout this research.

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Melek Hajji

Present address: Department of Chemistry, College of Pharmacy, Assalam International University-Sirte, Sirte, Libya

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Research Unit: Electrochemistry, Materials and Environment, University of Kairouan, 3100, Kairouan, Tunisia

Melek Hajji & Taha Guerfel

Laboratory of Medicinal Chemistry, Drug Sciences Research Center, Faculty of Medicine and Pharmacy, Mohammed V University in Rabat, Rabat, Morocco

Nadeem Abad & Youssef Ramli

Laboratoire de Chimie Organique Heterocyclique, Faculté Des Sciences, Université Mohammed V Rabat, Rabat, Morocco

Nadeem Abad, Mohamed El Hafi & El Mokhtar Essassi

Laboratory of Human Genome and Multifactorial Diseases (LR12ES07), Faculty of Pharmacy of Monastir, University of Monastir, Monastir, Tunisia

Meriem Dallel

Department of Chemistry, College of Science, Princess Nourah Bint Abdulrahman University, P.O. Box 84428, 11671, Riyadh, Saudi Arabia

Hanan Al-Ghulikah

Department of Chemistry, Tulane University, New Orleans, LA, 70118, USA

Joel T. Mague

Mohammed VI Center for Research and Innovation (CM6), 10000, Rabat, Morocco

Youssef Ramli

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Hajji, M., Abad, N., Dallel, M. et al. Noncovalent interactions between quinoxalines and protoporphyrinogen oxidase (PPO): a computational case study for herbicidal applications. Chem. Pap. (2024). https://doi.org/10.1007/s11696-024-03485-4

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