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Case Study on Application Of Derivatives Class 12 Maths PDF
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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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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.
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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.
- Chapter 4 Determinants Case Study Questions
- Chapter 5 Continuity and Differentiability Case Study Questions
- Chapter 7 Integrals Case Study Questions
- 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
Hope the information shed above regarding Case Study and Passage Based Questions for Class 12 Maths Chapter 6 Applications of Derivatives with Answers Pdf free download has been useful to an extent. If you have any other queries of CBSE Class 12 Mathematics Applications of Derivatives Case Study and Passage Based Questions with Answers, feel free to comment below so that we can revert back to us at the earliest possible. By Team Study Rate
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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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Case Study Questions for Class 12 Maths Chapter 6 Application of Derivatives
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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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Chapter 4: Applications of Derivatives
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A rocket launch involves two related quantities that change over time. Being able to solve this type of problem is just one application of derivatives introduced in this chapter. We also look at how derivatives are used to find maximum and minimum values of functions. As a result, we will be able to solve applied optimization problems, such as maximizing revenue and minimizing surface area. In addition, we examine how derivatives are used to evaluate complicated limits, to approximate roots of functions, and to provide accurate graphs of functions.
- 4.0: Prelude to Applications of Derivatives
- 4.0E: Exercises
- 4.1: Related Rates
- 4.1E: Related Rates Exercises
- 4.2: Maxima and Minima
- 4.2E: Maxima and Minima Exercises
- 4.3: Derivatives and the Shape of a Graph
- 4.3E: Shape of the Graph Exercises
- 4.4: Graphing
- 4.4 E: Sketch the GRAPH Exercises
- 4.5: Optimization Problems
- 4.5 E: Optimization Exercises
- 4.6: Linear Approximations and Differentials In this section, we examine another application of derivatives: the ability to approximate functions locally by linear functions. Linear functions are the easiest functions with which to work, so they provide a useful tool for approximating function values. In addition, the ideas presented in this section are generalized later in the text when we study how to approximate functions by higher-degree polynomials Introduction to Power Series and Functions.
- 4.7: The Mean Value Theorem The Mean Value Theorem is one of the most important theorems in calculus. We look at some of its implications at the end of this section. First, let’s start with a special case of the Mean Value Theorem, called Rolle’s theorem.
- 4.8: Antiderivatives
- 4.8E: AntiDerivative & Indefinite Integral Exercises
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Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org .
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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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Class 12 Maths Case Study Questions
Table of Contents
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Download the app to get CBSE Sample Papers 2023-24, NCERT Solutions (Revised), Most Important Questions, Previous Year Question Bank, Mock Tests, and Detailed Notes.
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
More Case Study Questions
These are only some samples. If you wish to get more case study questions for CBSE class 12 maths, install the myCBSEguide App. It has class 12 Maths chapter-wise case studies with solutions.
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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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.
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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.
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.
CBSE 12th Standard Maths Subject Application of Derivatives Case Study Questions 2021 By QB365 on 21 May, 2021 QB365 Provides the updated CASE Study Questions for Class 12 Maths, and also provide the detail solution for each and every case study questions .
Here, we have provided case-based/passage-based questions for Class 12 Mathematics Chapter 6 Applications of Derivatives. Case Study/Passage-Based Questions. Shobhit's father wants to construct a rectangular garden using a brick wall on one side of the garden and wire fencing for the other three sides as shown in the figure. He has 200 ft of ...
Introduction to one-dimensional motion with calculus. Interpreting direction of motion from position-time graph. Interpreting direction of motion from velocity-time graph. Interpreting change in speed from velocity-time graph. Worked example: Motion problems with derivatives. Analyzing straight-line motion graphically.
4.0: Prelude to Applications of Derivatives. A rocket launch involves two related quantities that change over time. Being able to solve this type of problem is just one application of derivatives introduced in this chapter. We also look at how derivatives are used to find maximum and minimum values of functions.
This page titled 4: Applications of Derivatives is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by OpenStax via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request. In this chapter we look at how derivatives are used to find ...
#casestudyquestions #casestudyquestionsforclass12mathIn this video you will learn how to solve case study questions of class 12 math's board exams as per lat...
CBSE 12th Standard Maths Subject Application of Derivatives Case Study Questions With Solution 2021 By QB365 on 21 May, 2021 QB365 Provides the updated CASE Study Questions for Class 12 Maths, and also provide the detail solution for each and every case study questions .
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.
APPLICATION OF DERIVATIVES 195 Thus, the rate of change of y with respect to x can be calculated using the rate of change of y and that of x both with respect to t. Let us consider some examples. Example 1 Find the rate of change of the area of a circle per second with respect to its radius r when r = 5 cm. Solution 2The area A of a circle with radius r is given by A = πr.
In this post, you will get CASE Study Questions of Chapter 6 (Application Of Derivatives) of Class 12th. These Case study Questions are based on the Latest Syllabus for 2021- 22 of the CBSE Board.
The concept of derivatives has been used in small scale and large scale. The concept of derivatives used in many ways such as change of temperature or rate of change of shapes and sizes of an object depending on the conditions etc., Rate of Change of a Quantity. This is the general and most important application of derivative.
4.1: Prelude to Applications of Derivatives. A rocket launch involves two related quantities that change over time. Being able to solve this type of problem is just one application of derivatives introduced in this chapter. We also look at how derivatives are used to find maximum and minimum values of functions.
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 ...
This page titled Chapter 4: Applications of Derivatives is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by OpenStax. In this chapter we look at how derivatives are used to find maximum and minimum values of functions. As a result, we will be able to solve applied optimization problems, such as maximizing ...
Application of Derivatives Class 12 Important Questions with Solutions Previous Year Questions. Rate Measure, Increasing-Decreasing Functions and Approximation. Question 1. The total cost C (x) associated with the production of x units of an item is given by C (x) = 0.005x 3 - 0.02x 2 + 30x + 5000.
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.
Case Study Questions Application of Derivatives. A telephone company in a town has 500 subscribers on its list and collects fixed charges of 300 per subscriber per year. The company proposes to increase the annual subscription and it is believed that for every increase of 1 one subscriber will discontinue the service.
In applications of derivatives class 12 chapter 6, we will study different applications of derivatives in various fields like Science, Engineering, and many other fields. In chapter 6, we are going to learn how to determine the rate of change of quantity, finding the equations of tangents, finding turning points on the graphs for various ...
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.
Download Class PDF. Aug 15, 2021 • 1h 2m • 140 views. In this Session , Vishal Mahajan discuss the Application of Derivatives .This Session will be beneficial Of Class 12 & all aspirants preparing for Competitive Exams.This session will be Conducted in English & Hindi and notes will be provided in English.