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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 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

• Last modified on: 1 year ago
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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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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 . Case study questions are latest updated question pattern from NCERT, QB365 will helps to get  more marks in Exams

QB365 - Question Bank Software

Cbse 12th standard maths application of derivatives case study questions 2021.

12th Standard CBSE

Final Semester - June 2015

Case Study Questions

(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

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Cbse 12th standard maths application of derivatives case study questions 2021 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.

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Home » Resources » Career Guide » AOD (Chapter 6) Revision for Class 12 Maths Board Exams Preparation

AOD (Chapter 6) Revision for Class 12 Maths Board Exams Preparation

Class xii maths tips for board examination preparation – application of derivatives (chapter 6).

Hope you have managed to complete the syllabus once from NCERT and started revision with Chapter 5. We also hope that you have made the study diary and are able to achieve the daily targets. Also, you must have made the formulae registers for Mathematics and other similar subjects. In case you missed my last article, check out  Maths Preparation for Class 12 Board Exams .

Let us take up the next topic Application of derivatives (Chapter 6). As we have already mentioned, the total Calculus has 44 marks, out of which 20 marks are allotted to chapters 5 and 6. Normally 8-12 marks are allotted to chapter 5. So, you can again expect 8-12 marks for this chapter 6. You should definitely expect one 6-mark question in this topic. In most cases, you’ll find one 4-mark question also. 1-mark questions (one or two nos) may or may not be there depending on the marks of chapter 5.

Your target time for first revision should be 6 hours appx for the five exercises (6.1 to 6.5) and the Misc exercise and also little bit of previous years’ questions from reference book (if time permits).

Some important questions from NCERT are as follows:

• Study plan should be pre defined . There are two ways of utilizing the daily 5 hours study till the schools are closed. Either you study two subjects daily for 2.5 hrs each or you give 1 hour each to five subjects. In first case, you must have a routine for the entire week for rotation of the subjects. Follow whatever you are comfortable with.
• Take a break when needed! Take a small break after about every 1.5 hrs. Listening to soft music or having a stroll is a good option.
• Avoid distractions . Use mobile phone as a necessity only, not for chatting and what’s apping. Social networking – not now – become little unsocial. However, keep track of the different competitive examination forms.

Study Resources for CBSE Class 12 Mathematics Chapter 6 – Application of Derivatives (AOD)

NCERT Class 12 – Mathematics, Chapter 6   Theory and Examples  |   Solutions  .

Also, practice CBSE Sample Papers for Class 12 .

This article is a part of our How to prepare for CBSE Class XII Board Exams guide. For more article, check out the links below.

If you have any queries, please ask in comments below.

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Important Tips For Calculus and Integration for Class 12 Math Board Exam

Maths preparation for class 12 board exams 2023 revision plan, related posts.

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Application Of Derivatives Class 12 Previous Year Question Paper PDF

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Application Of Derivatives Class 12 Previous Year Question Paper

If you wish to get good grades in your class 12 unit tests and board exams, then you should solve the Application Of Derivatives Class 12 previous year question paper on SelfStudys. We have provided this practice material in a systematic manner so that you can learn the pattern of the exam with ease. 

Additionally, you will be able to track your preparation with the help of the Application Of Derivatives Class 12 PYQs practice. You will become familiar with the marking scheme of the questions, their difficulty level, and time allocations properly. 

Solutions Of Application Of Derivatives Class 12 PYQs

On our platform, you can explore the PYQs on Class 12 Application Of Derivatives along with solutions. Each answer also includes a step-by-step explanation so that you can understand the reasoning behind the question. You must go through these solutions after completing the questions in this chapter. All the answers are available at the end of the PDF file. 

Why Should All The Students Solve Application Of Derivatives Class 12 PYQs? 

Let’s have a look at why should all the students solve the Class 12 previous year question paper for Application Of Derivatives: 

• Sneak peek into the process that is used to set the Application Of Derivatives Class 12 previous year question paper. 
• Understanding of the structure of the Application Of Derivatives Class 12 PYQs.
• Marking scheme for each question of Class 12 previous year question paper for Application Of Derivatives. 
• Work as a Best Revision Tool: The PYQs on Class 12 Application Of Derivatives are very beneficial for all the students who want to do a thorough revision after doing all the preparation. This is also the best way for students to know their strengths and weaknesses in the chapter and work on them. 
• Boosts the confidence of the students: Students will get the idea of the exam structure and type of questions by regularly practicing the Application Of Derivatives Class 12 previous year question paper which in turn will boost their confidence as they will believe to attempt all the questions in the examination effectively because they have solved them earlier. Also, the concentration levels of students will improve. 
• Ensure effective time management skills: By practicing Application Of Derivatives Class 12 PYQs, students will also get familiar with the time duration of the examination. They can use various strategies and manage their time well in the exam.
• Assist in Self Assessment: Students will reflect on their own learning process and will be more responsible by solving the Class 12 previous year question paper for Application Of Derivatives.

How To Download Class 12 Previous Year Question Paper For Application Of Derivatives? 

Let’s have a look at how you can download PYQs on Class 12 Application Of Derivatives PDF: 

• Reach to the selfstudys.com website 

• Click on the three dots at the left, scroll down, and click on the CBSE option which is clearly visible.

• Tap once again on the CBSE option, and click on the Previous Year Question Paper option.

• You will be redirected to the CBSE Previous Year Papers webpage. Click on the option stating 12th PYP Chapter Wise.

•  A list of subjects will appear. Click on the “Maths” subject to download the Application Of Derivatives Class 12 Previous Year Question Paper .

• Now on the new page, you have the option to download the Application Of Derivatives Class 12 PYQs in normal or HD PDF quality. 

Significance Of PYQs On Class 12 Application Of Derivatives 

Let’s have a look at the significance of the Class 12 previous year question paper for Application Of Derivatives.

• Practicing PYQs on Class 12 Application Of Derivatives gives all the students an idea about the type of questions asked, the duration of the exam, the marking scheme, and various other details. 
• Students will get to know the weightage of each topic of the chapter which will help them to cover the syllabus accordingly. 
• The Application Of Derivatives Class 12 previous year question paper consists of plenty of questions that help students to have a thorough knowledge of every topic. 

Tips To Follow While Solving Application Of Derivatives Class 12 Previous Year Question Paper

Solving Application Of Derivatives Class 12 PYQs is always a good idea for all the students to prepare for their exams. Regular practice of these questions will enable students to solve any question irrespective of its difficulty.

Below are some tips that every student must follow to get the most out of the Class 12 previous year question paper for Application Of Derivatives.  

• Start by reading the complete question paper. Have a look at each and every question and try to remember the answer to that question. 
• Choose any section from the PYQs on Class 12 Application Of Derivatives and start attempting questions from that section first. Do not attempt questions in a badly organized manner. 
• In case, you don’t know the answer to any question of the Application Of Derivatives Class 12 previous year question paper, leave some space and skip them for the time being. Attempt the questions for which you are completely sure of the answer. 
• Divide the time for every section. Have awareness of the clock while attempting Application Of Derivatives Class 12 PYQs. 
• It is advisable for all the students to thoroughly review the Class 12 previous year question paper for Application Of Derivatives after completing it and look for errors to correct them. 

How To Make The Most Of Application Of Derivatives Class 12 PYQs?

Through this section, students can learn how to make the most of PYQs on Class 12 Application Of Derivatives.

• Make a study schedule: While preparing for the Application Of Derivatives Class 12 previous year question paper, all students must make a study schedule and then follow the same. Making a study schedule helps a student to be consistent in their exam preparation. 
• Take breaks while studying: Practising Application Of Derivatives Class 12 PYQs constantly in a room can exhaust a student's mind and make it difficult to focus. It is suggested that students should take 10-15 minutes of break from their study to relax which will help them to concentrate better. 
• Avoid using slang: There is a lot of difference between spoken english and written english. The use of chat slang will be considered as a spelling mistake so, avoid using such words while solving Class 12 previous year question paper for Application Of Derivatives. 
• Take proper rest: Exam stress and fear can affect the student’s performance in board exams. It is advisable for all the students to take proper rest. 

Solving PYQs on Class 12 Application Of Derivatives helps student to boost their confidence. They will believe that they can attempt all the answers in the examination effectively because they have solved these questions. 

A Complete Study Plan For Solving Class 12 Previous Year Question Paper For Application Of Derivatives

If a student really wants to score good marks in the Class 12 Examination, they should follow a strict schedule for solving the Application Of Derivatives Class 12 previous year question paper. The schedule involves: 

• Go through the syllabus: The first thing a student must do is to go through the complete syllabus and learn about the chapter properly. It is recommended that students should avoid last-minute preparations and study for the syllabus every day. 
• Go through study material: After completing the syllabus, students should go through study materials to track their preparation which includes Application Of Derivatives Class 12 PYQs and other important documents. By going through the study material in this chapter, they can know where they are lacking and can work on that area. 
• Conduct Mock Tests: Once a student has completed the syllabus, they should involve themselves in mock tests based on the Class 12 previous year question paper for Application Of Derivatives. Mock tests will help students to analyze themselves better. They can also help them build a strategy for which questions to attempt first, how to master time management and improve their concentration.

Common Mistakes Which Students Make While Practising PYQs On Class 12 Application Of Derivatives 

Let’s look at the common mistakes that students make while practicing PYQs on Class 12 Application Of Derivatives. 

• Studying for long: One of the most common mistakes that students make while preparing for the CBSE Class 12 Board Exams is to study for long and not take breaks in between. It is advisable for all the students to plan their schedule while attempting the Application Of Derivatives Class 12 previous year question paper and include power breaks in it.
• Not doing written practice: All the students preparing for this chapter must do writing practice of formulas as it helps to memorize the concepts and leads to great learning. It is also beneficial for cognitive skills as it requires re-focusing of attention. Also, writing after learning makes information stick in the memory. 
• Not reading the entire question paper: On average, students have the habit of not reading the Application Of Derivatives Class 12 PYQs which results in making various mistakes. They do not organize their thoughts before writing the answer. It is advisable for all the students to organize their thoughts before writing as it helps to improve their productivity. 
• Mugging up at the last minute: Leaving topics and formulas related to the Class 12 previous year question paper for Application Of Derivatives for the last few days is the worst mistake students make. They keep thinking that they will study them later but instead of studying, they do last-minute mugging for that topics. This can cause stress for students while taking exams as they can forget all the concepts.

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Question 6 - Case Based Questions (MCQ) - Chapter 6 Class 12 Application of Derivatives

Last updated at April 16, 2024 by Teachoo

There is a bridge whose length of three sides of a trapezium other than base are equal to 10 cm.

Based on the above information answer the following:.

This question is inspired from Example 37 - Chapter 6 Class 12 (AOD) - Maths

What is the value of DP?

(a) √( 100 - x 2 )  , (b) √( x 2 - 100 ), (c) 100 - x 2   , (d) x 2 − 100.

What is the area of trapezium A(x)?

(a) (x - 10 )√( 100 - x 2 )  , (b) ( x + 10) √( 100 - x 2 ), (c) ( x - 10 ) (100 - x 2 ), (d) ( x + 10)(100 - x 2 ).

If A'(x) = 0, then what are the values of x?

(a) 5,-10 , (b) - 5, 10, (c) - 5,-10 .

What is the value of maximum Area?

(a) 75 √2  cm 2  , (b) 75 √3  cm 2, (c) 75 √5  cm 2   , (d) 75 √7  cm 2  .

Question There is a bridge whose length of three sides of a trapezium other than base are equal to 10 cm. Based on the above information answer the following: Question 1 What is the value of DP? (A) √(100−𝑥^2 ) (B) √(𝑥^2−100) (C) 100−𝑥^2 (D)〖 𝑥〗^2 − 100 In Δ ADP By Pythagoras theorem DP2 + AP2 = AD2 DP2 + x2 = 102 DP2 + x2 = 100 DP2 = 100 – 𝑥2 DP = √(𝟏𝟎𝟎 −𝒙𝟐) So, the correct answer is (A) Question 2 What is the area of trapezium A(x)? (A) (𝑥 −10)√(100−𝑥^2 ) (B) (𝑥+10)√(100−𝑥^2 ) (C) (𝑥−10)(100−𝑥^2) (D) (𝑥+10)(100−𝑥^2 ") " Let A be the area of trapezium ABCD A = 1/2 (Sum of parallel sides) × (Height) A = 𝟏/𝟐 (DC + AB) × DP A = 1/2 (10+2𝑥+10) (√(100−𝑥2)) A = 1/2 (2𝑥+20) (√(100−𝑥2)) A = (𝒙+𝟏𝟎) (√(𝟏𝟎𝟎−𝒙𝟐)) So, the correct answer is (B) Question 3 If A'(x) = 0, then what are the values of x? (A) 5,−10 (B) −5, 10 (C) −5,−10 (D) 5,10 A = (𝒙+𝟏𝟎) (√(𝟏𝟎𝟎−𝒙𝟐)) Since A has a square root It will be difficult to differentiate Let Z = A2 = (𝑥+10)^2 (100−𝑥2) Where A'(x) = 0, there Z’(x) = 0 So, the correct answer is (A) Differentiating Z Z =(𝑥+10)^2 " " (100−𝑥2) Differentiating w.r.t. x Z’ = 𝑑((𝑥 + 10)^2 " " (100 − 𝑥2))/𝑑𝑘 Using product rule As (𝑢𝑣)′ = u’v + v’u Z’ = [(𝑥 + 10)^2 ]^′ (100 − 𝑥^2 )+(𝑥 + 10)^2 " " (100 − 𝑥^2 )^′ Z’ = 2(𝑥 + 10)(100 − 𝑥^2 )−2𝑥(𝑥 + 10)^2 Z’ = 2(𝑥 + 10)[100 − 𝑥^2−𝑥(𝑥+10)] Z’ = 2(𝑥 + 10)[100 − 𝑥^2−𝑥^2−10𝑥] Z’ = 2(𝑥 + 10)[−2𝑥^2−10𝑥+100] Z’ = −𝟒(𝒙 + 𝟏𝟎)[𝒙^𝟐+𝟓𝒙+𝟓𝟎] Putting 𝒅𝒁/𝒅𝒙=𝟎 −4(𝑥 + 10)[𝑥^2+5𝑥+50] =0 (𝑥 + 10)[𝑥^2+5𝑥+50] =0 (𝑥 + 10) [𝑥2+10𝑥−5𝑥−50]=0 (𝑥 + 10) [𝑥(𝑥+10)−5(𝑥+10)]=0 (𝒙 + 𝟏𝟎)(𝒙−𝟓)(𝒙+𝟏𝟎)=𝟎 So, 𝑥=𝟓 & 𝒙=−𝟏𝟎 So, the correct answer is (A) Question 4 What is the value of maximum Area? (A) 75 √2 cm^2 (B) 75 √3 cm^2 (C) 75 √5 cm^2 (D) 75 √7 cm^2 We know that Z’(x) = 0 at x = 5, −10 Since x is length, it cannot be negative ∴ x = 5 Finding sign of Z’’ for x = 5 Now, Z’ = −4(𝑥 + 10)[𝑥^2+5𝑥+50] Z’ = −4[𝑥(𝑥^2+5𝑥+50)+10(𝑥^2+5𝑥+50)] Z’ = −4[𝑥^3+5𝑥^2+50𝑥+10𝑥^2+50𝑥+500] Z’ = −𝟒[𝒙^𝟑+𝟏𝟓𝒙^𝟐+𝟏𝟎𝟎𝒙+𝟓𝟎𝟎] Differentiating w.r.t x Z’’ = 𝑑(−4[𝒙^𝟑 + 𝟏𝟓𝒙^𝟐 + 𝟏𝟎𝟎𝒙 + 𝟓𝟎𝟎])/𝑑𝑘 Z’’ =−4[3𝑥^2+15 × 2𝑥+100] Z’’ =−4[3𝑥^2+30𝑥+100] Putting x = 5 Z’’ (5) = −4[3(5^2) +30(5) +100] = −4 × 375 = −1500 < 0 Hence, 𝑥 = 5 is point of Maxima ∴ Z is Maximum at 𝑥 = 5 That means, Area A is maximum when x = 5 Finding maximum area of trapezium A = (𝑥+10) √(100−𝑥2) = (5+10) √(100−(5)2) = (15) √(100−25) = 15 √75 = 75√𝟑 cm2 So, the correct answer is (C)

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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• NCERT Exemplar
• Maths Exemplar Class 12
• Application Of Derivatives

NCERT Exemplar Solutions for Class 12 Maths Chapter 6 Application of Derivatives

Subject-wise textbooks of the NCERT Exemplar for Class 12 are important for students to prepare for the board exam. Solving questions from the Exemplar will strengthen conceptual understanding among students. To help students understand the subject better, NCERT Exemplar Solutions , prepared by subject experts at BYJU’S, is the right resource. Also, these solutions are in accordance with the latest CBSE guidelines.

The sixth chapter of NCERT Exemplar Solutions for Class 12 Maths is the Application of Derivatives. In this chapter, students will learn how to solve problems of derivatives based on the rate of change of quantitative, increasing and decreasing functions, tangents and normals, approximation, maxima and minima, and maximum and minimum values of a function in a closed interval. Students who aspire to attain a strong grip over the concepts of this chapter can access the solutions PDF of NCERT Exemplar Solutions for Class 12 Maths Chapter 6 Application of Derivatives from the link below.

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Exercise 6.3 Page No: 135

Short Answer (S.A.)

1. A spherical ball of salt is dissolving in water in such a manner that the rate of decrease of the volume at any instant is proportional to the surface. Prove that the radius is decreasing at a constant rate.

Given, a spherical ball of salt

Then, the volume of ball V = 4/3 πr 3 where r = radius of the ball

Now, according to the question we have

dV/dt ∝ S, where S = surface area of the ball

Therefore, the radius of the ball is decreasing at constant rate.

2. If the area of a circle increases at a uniform rate, then prove that perimeter varies inversely as the radius.

We know that the area of circle, A = πr 2 , where r = radius of the circle

And, perimeter = 2πr

According to the question, we have

Therefore, it’s seen that the perimeter of the circle varies inversely as the radius of the circle.

3. A kite is moving horizontally at a height of 151.5 meters. If the speed of kite is

10 m/s, how fast is the string being let out; when the kite is 250 m away from

the boy who is flying the kite? The height of boy is 1.5 m.

Speed of the kite(V) = 10 m/s

Let FD be the height of the kite and AB be the height of the kite and AB be the height of the boy.

Now, let AF = x m

So, BG = AF = x

And, dx/dt = 10 m/s

From the figure, it’s seen that

GD = DF – GF = DF – AB

Now, in ∆ BDG

BG 2 + GD 2 = BD 2 (By Pythagoras Theorem)

x 2 + (150) 2 = (250) 2

x 2 + 22500 = 62500

x 2 = 62500 – 22500 = 40000

Let initially the length of the string be y m

So, in ∆ BDG

BG 2 + GD 2 = BD 2

x 2 + (150) 2 = y 2

Differentiating both sides w.r.t., t, we have

Therefore, the rate of change of the length of the string is 8 m/s.

4. Two men A and B start with velocities v at the same time from the junction of

two roads inclined at 45° to each other. If they travel by different roads, find

the rate at which they are being separated.

Let’s consider P to be any point at which the two roads are inclined at an angle of 45 o .

Now, two men A and B are moving along the roads PA and PB respectively with same speed ‘V’.

∠APB = 45 o and they move with the same speed.

So, ∆APB is an isosceles triangle.

Now, draw PQ ⊥ AB.

We have, AB = y

So, AQ = y/2 and PA = PA = x (assumption)

And, ∠APQ = ∠BPQ = 45 o /2 = 22.5 o

Now, in right ∆APQ

sin 22.5 o = AQ/AP

5. Find an angle q, 0 < q < /2, which increases twice as fast as its sine.

Therefore, the required angle is π/3.

6. Find the approximate value of (1.999) 5 .

(1.999) 5 = (2 – 0.001) 5

Let x = 2 and ∆x = -0.001

Also, let y = x 5

Differentiating both sides w.r.t, x, we get

dy/dx = 5x 4 = 5(2) 4 = 80

Now, ∆y = (dy/dx). ∆x = 80. (-0.001) = -0.080

And, (1.999) 5 = y + ∆y

= x 5 – 0.080 = (2) 5 – 0.080 = 32 – 0.080 = 31.92

Therefore, approximate value of (1.999) 5 is 31.92

7. Find the approximate volume of metal in a hollow spherical shell whose internal and external radii are 3 cm and 3.0005 cm, respectively.

The internal radius r = 3 cm

And, external radius R = r + ∆r =3.0005 cm

∆r = 3.0005 – 3 = 0.0005 cm

Let y = r 3 ⇒ y + ∆y = (r + ∆r) 3 = R 3 = (3.0005) 3

Differentiating both sides w.r.t., r, we get

Therefore, the approximate volume of the metal in the shell is 0.018π cm 3

8. A man, 2m tall, walks at the rate of m/s towards a street light which is m above

the ground. At what rate is the tip of his shadow moving? At what rate is the length of the

shadow changing when he is m from the base of the light?

Let AB is the height of street light post and CD is the height of the man such that

AB = 5(1/3) = 16/3 m and CD = 2 m

Let BC = x length (the distance of the man from the lamp post)

And CE = y is the length of the shadow of the man at any instant.

It’s seen from the figure that,

Now, taking ratio of their corresponding sides, we have

Differentiating both sides w.r.t, t, we have

So, the length of shadow is decreasing at the rate of 1 m/s.

Now, let u = x + y

(where, u = distance of the tip of shadow from the light post)

On differentiating both sides w.r.t. t, we get

Therefore, the tip of the shadow is moving at the rate of m/s towards the light post and the length of shadow decreasing at the rate of 1 m/s.

9. A swimming pool is to be drained for cleaning. If L represents the number of litres of water in the pool t seconds after the pool has been plugged off to drain and L = 200 (10 – t ) 2 . How fast is the water running out at the end of 5 seconds? What is the average rate at which the water flows out during the first 5 seconds?

Given, L = 200(10 – t) 2 where L represents the number of liters of water in the pool.

On differentiating both the sides w.r.t, t, we get

dL/dt = 200 x 2(10 – t) (-1) = -400(10 – t)

But, the rate at which the water is running out

= -dL/dt = 400(10 – t)

Now, rate at which the water is running after 5 seconds will be

= 400 x (10 – 5) = 2000 L/s (final rate)

T = 0 for initial rate

= 400 (10 – 0) = 4000 L/s

So, the average rate at which the water is running out is given by

= (Initial rate + Final rate)/ 2 = (4000 + 2000)/ 2 = 6000/2 = 3000 L/s

Therefore, the required rate = 3000 L/s

10. The volume of a cube increases at a constant rate. Prove that the increase in its surface area varies inversely as the length of the side.

Let’s assume x to be the length of the cube.

So, the volume of the cube V = x 3 …. (1)

Given that, dV/dt = K

Now, on differentiating the equation (1) w.r.t. t, we get

dV/dt = 3x 2 . dx/dt = K (constant)

So, dx/dt = K/3x 2

Surface area of the cube, S = 6x 2

Differentiating both sides w.r.t. t, we get

Therefore, the surface area of the cube varies inversely as the length of the side.

11. x and y are the sides of two squares such that y = x – x 2 . Find the rate of change of the area of second square with respect to the area of first square.

Let’s consider the area of the first square A 1 = x 2

And, area of the second square be A 2 = y 2

Now, A 1 = x 2 and A 2 = y 2 = (x – x 2 ) 2

Differentiating both A 1 and A 2 w.r.t. t, we get

Therefore, the rate of change of area of the second square with respect to first is 2x 2 – 3x + 1.

12. Find the condition that the curves 2 x = y 2 and 2 xy = k intersect orthogonally.

It’s seen that the given curves are equation of two circles.

2 x = y 2 ….. (1) and

2 xy = k ….. (2)

We know that, two circles intersect orthogonally if the angle between the tangents drawn to the two circles at the point of their intersection is 90 o .

Now, differentiating equations (1) and (2) w.r.t. t, we get

Now, solving equations (1) and (2), we have

Putting the value of y in equation (1),

2x = (k/2x) 2 ⇒ 2x = k 2 /4x 2

8(1) 3 = k 2

Therefore, the required condition is k 2 = 8.

13. Prove that the curves xy = 4 and x 2 + y 2 = 8 touch each other.

Given curves are equations of two circles,

xy = 4 ….. (i) and

x 2 + y 2 = 8 …. (ii)

Different equation (i) w.r.t., x

Therefore, the point of contact of the two circles are (2, 2) and (-2, 2).

14. Find the co-ordinates of the point on the curve √ x + √ y = 4 at which tangent is equally inclined to the axes.

Equation of the curve is given by, √ x + √ y = 4

Now, let (x 1 , y 1 ) be he required point on the curve

So, √ x 1 + √ y 1 = 4

On differentiating on both the sides w.r.t. x 1 , we get

On squaring on both the sides, we get

y 1 /x 1 = 1 ⇒ y 1 = x 1

Now, putting the value of y1 in the given equation of the curve.

√ x 1 + √ y 1 = 4

√ x 1 + √ x 1 = 4 ⇒ 2√ x 1 = 4 ⇒ √ x 1 = 2 ⇒ x 1 = 4

As y 1 = x 1

Therefore, the required point is (4, 4).

15. Find the angle of intersection of the curves y = 4 – x 2 and y = x 2 .

The given curves are y = 4 – x 2 …. (i) and y = x 2 …… (ii)

And, we know that the angle of intersection of two curves is equal to the angle between the tangents drawn to the curves at their point of intersection.

Now, differentiating equations (i) and (ii) w.r.t x, we have

dy/dx = -2x ⇒ m 1 = -2x

m 1 is the slope of the tangent to the curve (i).

and dy/dx = 2x ⇒ m 2 = 2x

m 2 is the slope of the tangent to the curve (ii).

So, m 1 = -2x and m 2 = 2x

On solving equation (i) and (ii), we get

4 – x 2 = x 2 ⇒ 2x 2 = 4 ⇒ x 2 = 2 ⇒ x = ±√2

So, m 1 = -2x = -2√2 and m 2 = 2x = 2√2

Let θ be the angle of intersection of two curves

16. Prove that the curves y 2 = 4 x and x 2 + y 2 – 6 x + 1 = 0 touch each other at the point (1, 2).

Given curve equations are: y 2 = 4 x …. (1) and x 2 + y 2 – 6 x + 1 = 0 ….. (2)

Now, differentiating (i) w.r.t. x, we get

2y.(dy/dx) = 4 ⇒ dy/dx = 2/y

Slope of tangent at (1, 2), m 1 = 2/2 = 1

Differentiating (ii) w.r.t. x, we get

2x + 2y.(dy/dx) – 6 = 0

2y. dy/dx = 6 – 2x ⇒ dy/dx = (6 – 2x)/ 2y

Hence, the slope of the tangent at the same point (1, 2)

⇒ m 2 = (6 – 2 x 1)/ (2 x 2) = 4/4 = 1

It’s seen that m 1 = m 2 = 1 at the point (1, 2).

Therefore, the given circles touch each other at the same point (1, 2).

17. Find the equation of the normal lines to the curve 3 x 2 – y 2 = 8 which are parallel to the line x + 3 y = 4.

Given curve, 3 x 2 – y 2 = 8

Differentiating both sides w.r.t. x, we get

6x – 2y. dy/dx = 0 ⇒ -2y(dy/dx) = -6x ⇒ dy/dx = 3x/y

So, slope of the tangent to the given curve = 3x/y

Thus, the normal to the curve = -1/(3x/y) = -y/3x

Now, differentiating both sides of the given line x + 3y = 4, we have

1 + 3.(dy/dx) = 0

dy/dx = -1/3

As the normal to the curve is parallel to the given line x + 3y = 4

We have, -y/3x = -1/3 ⇒ y = x

On putting the value of y in 3x 2 – y 2 = 8, we get

3x 2 – x 2 = 8

2x 2 = 8 ⇒ x 2 = 4 ⇒ x = ±2

Thus, the points on the curve are (2, 2) and (-2, -2).

Now, the equation of the normal to the curve at (2, 2) is given by

Therefore, the required equations are x + 3y = 8 and x + 3y = -8.

18. At what points on the curve x 2 + y 2 – 2 x – 4 y + 1 = 0, the tangents are parallel to the y -axis?

Given, the equation of the curve is x 2 + y 2 – 2 x – 4 y + 1 = 0 ….. (i)

Differentiating both the sides w.r.t. x, we get

Now putting the value of y in equation (i), we get

x 2 + (2) 2 – 2x – 8 + 1 = 0

x 2 – 2x + 4 – 8 + 1 = 0

x 2 – 2x – 3 = 0 ⇒ x 2 – 3x + x – 3 = 0

x(x – 3) + 1(x – 3) = 0 ⇒ (x – 3)(x + 1) = 0

x = -1 or 3

Therefore, the required points are (-1, 2) and (3, 2).

19. Show that the line x/a + y/b = 1, touches the curve y = b . e -x/a at the point where

the curve intersects the axis of y.

Given curve equation, y = b . e -x/a and line equation x/a + y/b = 1

Now, let the coordinates of the point where the curve intersects the y-axis be (0, y 1 )

Now differentiating y = b . e -x/a both sides w.r.t. x, we get

Therefore, the equation of the curve intersects at (0, b) which is on the y-axis.

20. Show that f ( x ) = 2 x + cot -1 x + log [√(1 + x 2 ) – x] is increasing in R.

On squaring both the sides, we get

4x 4 + 1 + 4x 2 ≥ 1 + x 2

4x 4 + 4x 2 – x 2 ≥ 0

4x 4 + 3x 2 ≥ 0

x 2 (4x 2 + 3) ≥ 0

The above is true for any value of x ∈ R.

Therefore, the given function is an increasing function over R.

21. Show that for a ³ 1, f ( x ) = √3 sin x – cos x – 2 ax + b is decreasing in R.

f ( x ) = √3 sin x – cos x – 2 ax + b, a ³ 1

On differentiating both sides w.r.t. x, we get

f’ ( x ) = √3 cos x + sin x – 2 a

For increasing function, f’ ( x ) < 0

Therefore, the given function is decreasing in R.

22. Show that f ( x ) = tan –1 (sin x + cos x ) is an increasing function in (0, π/4).

Given, f ( x ) = tan –1 (sin x + cos x ) in (0, π/4).

Differentiating both sides w.r.t. x, we got

Therefore, the given function f(x) is an increasing function in (0, π/4).

23. At what point, the slope of the curve y = – x 3 + 3 x 2 + 9 x – 27 is maximum? Also find the maximum slope.

Given, curve y = – x 3 + 3 x 2 + 9 x – 27

dy/dx = -3x 2 + 6x + 9

Let slope of the curve dy/dx = z

So, z = -3x 2 + 6x + 9

dz/dx = -6x + 6

For local maxima and local minima,

-6x + 6 = 0 ⇒ x = 1

d 2 z/dx 2 = -6 < 0 Maxima

Putting x = 1 in equation of the curve y = (-1) 3 + 3(1) 2 + 9(1) – 27

= -1 + 3 + 9 – 27 = -16

Maximum slope = -3(1) 2 + 6(1) + 9 = 12

Therefore, (1, -16) is the point at which the slope of the given curve is maximum and maximum slope = 12.

24. Prove that f ( x ) = sin x + √3 cos x has maximum value at x = π/6.

Therefore, the given function has maximum value at x = π/6 and the maximum value is 2.

Long Answer (L.A.)

25. If the sum of the lengths of the hypotenuse and a side of a right angled triangle is given, show that the area of the triangle is maximum when the angle between them is π/3.

Let AC = x, BC = y

So, AB = √(x 2 + y 2 )

Let z = x + y (given)

Now, the area of ∆ABC = ½ x AB x BC

Therefore, the area of the given triangle is maximum when the angle between its hypotenuse and a side is π/3.

26. Find the points of local maxima, local minima and the points of inflection of the function f ( x ) = x 5 – 5 x 4 + 5 x 3 – 1. Also find the corresponding local maximum and local minimum values.

Given, f ( x ) = x 5 – 5 x 4 + 5 x 3 – 1

Differentiating the function,

f’ ( x ) = 5 x 4 – 20 x 3 + 15 x 2

For local maxima and local minima, f’ ( x ) = 0

The maximum value of the function at x = 1

f(x) = (1) 5 – 5(1) 4 + 5(1) 3 – 1

= 1 – 5 + 5 – 1 = 0

The minimum value at x = 3 is

f(x) = (3) 5 – 5(3) 4 + 5(3) 3 – 1

= 243 – 405 + 135 – 1

= 378 – 406 = -28

Therefore, the function has its maxima at x = 1 and the maximum value = 0 and its has minimum value at x = 3 and its minimum value is -28.

27. A telephone company in a town has 500 subscribers on its list and collects fixed charges of Rs 300/- per subscriber per year. The company proposes to increase the annual subscription and it is believed that for every increase of Rs 1/- one subscriber will discontinue the service. Find what increase will bring maximum profit?

Let’s consider that the company increases the annual subscription by Rs x.

So, x is the number of subscribers who discontinue the services.

Total revenue, R(x) = (500 – x) (300 + x)

= 150000 + 500x – 300x – x 2

= -x 2 + 200x + 150000

Differentiating both sides w.r.t. x, we get R’(x) = -2x + 200

For local maxima and local minima, R’(x) = 0

-2x + 200 = 0 ⇒ x = 100

R’’(x) = -2 < 0 Maxima

So, R(x) is maximum at x = 100

Therefore, in order to get maximum profit, the company should increase its annual subscription by Rs 100.

28. If the straight-line x cos α + y sin α = p touches the curve x 2 /a 2 + y 2 /b 2 = 1, then prove that a 2 cos 2 α + b 2 sin 2 α = p 2 .

The given curve is x 2 /a 2 + y 2 /b 2 = 1 and the straight-line x cos α + y sin α = p

Differentiating equation (i) w.r.t. x, we get

Therefore, a 2 cos 2 α + b 2 sin 2 α = p 2 .

29. An open box with square base is to be made of a given quantity of card board of area c 2 . Show that the maximum volume of the box is c 3 / 6√3 cubic units.

Let x be the length of the side of the square base of the cubical open box and y be its height.

So, the surface area of the open box

30. Find the dimensions of the rectangle of perimeter 36 cm which will sweep out a volume as large as possible, when revolved about one of its sides. Also find the maximum volume.

Let’s consider x and y to be the length and breadth of given rectangle ABCD.

According to the question, the rectangle will be resolved about side AD which making a cylinder with radius x and height y.

So, the volume of the cylinder V = πr 2 h = πx 2 y …. (1)

36 = 2(x + y)

y = 18 – x ….. (ii)

Putting the value of y in the equation (i), we get

V = πx 2 (18 – x) = π(18x 2 – x 3 )

dV/dx = π(36x – 3x 2 ) …. (iii)

For local maxima and local minima dV/dx = 0

π(36x – 3x 2 ) = 0

36x – 3x 2 = 0

3x(12 – x) = 0

x ≠ 0 and 12 – x = 0 ⇒ x = 12

From equation (ii), we have

y = 18 – 12 = 6

Differentiating equation (iii) w.r.t. x, we get

d 2 V/dx 2 = π(36 – 6x)

d 2 V/dx 2 = π(36 – 6 x 12) = π(36 – 72)

= -36 π < 0 maxima

Now, volume of the cylinder so formed = πx 2 y

= π x (12) 2 x (6) = π(144) 2 x 6 = 864π cm 3

Therefore, the required dimension are 12 cm and 6 cm and the maximum volume is 864π cm 3 .

31. If the sum of the surface areas of cube and a sphere is constant, what is the ratio of an edge of the cube to the diameter of the sphere, when the sum of their volumes is minimum?

Surface area of cube = 6x 2

And, surface area of the sphere = 4πr 2

Now, their sum is

So, it is the minima.

Therefore, the required ratio is 1: 1 when the combined volume is minimum.

32. AB is a diameter of a circle and C is any point on the circle. Show that the area of Δ ABC is maximum, when it is isosceles.

Let consider AB be the diameter and C is any point on the circle with radius r.

Squaring on both the sides, we get

Therefore, the area of Δ ABC is minimum when it is an isosceles triangle.

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Class 12 Maths Application of Derivatives Extra Questions

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Class 12 Chapter 6 Maths Extra Questions

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Application of Derivatives Chapter 6 Important Questions

Chapter 6 Application of Derivatives

• Neither decreasing nor increasing
• None of these

The slope of the tangent to the curve x = a sint, y = a {tex}\left\{ {\cos t + \log (\tan \frac{t}{2})} \right\}{/tex} at the point ‘t’ is

• {tex}\tan \frac{t}{2}{/tex}
• none of these

The function f (x) = x 2 – 2x is strict decreasing in the interval

• {tex} [1,\infty ) {/tex}
• {tex} ({\text{ }}-\infty ,{\text{ }}1){/tex}

The equation of the tangent to the curve y 2 = 4ax at the point (at 2 , 2at) is

• ty = x + at 2
• tx + y =at 3
• ty = x – at 2
• The maximum value of  {tex}{\left( {\frac{1}{x}} \right)^x}{/tex} is ________.
• The minimum value of f if f(x) = sin x in [ {tex}\frac {-\pi}2,\frac {\pi}2{/tex} ] is ________.
• The equation of normal to the curve y = tan x at (0, 0) is ________.

Find the approximate value of f(3.02) where f(x) = 3x 2 + 5x + 3.

If the line ax+by+c=0 is a normal to the curve xy=1,then show that either a>0,b<0 or a<0,b>0

Find the interval in which the function f(x) = x 2 e -x is increasing.

The volume of a sphere is increasing at the rate of 3 cubic centimeter per second. Find the rate of increase of its surface area, when the radius is 2 cm.

Find the approximate value of {tex}{\left( {1.999} \right)^5}{/tex} .

Show that the function f(x) = 4x 3 – 18x 2 + 27x – 7 is always increasing on R.

Show that the height of the cylinder of maximum volume that can be inscribed in a sphere of radius a is {tex}\frac{{2a}}{{\sqrt 3 }}{/tex} .

A particle moves along the curve 6y = x 3 + 2. Find the points on the curve at which the y-coordinate is changing 8 times as fast as the x – coordinate.

Find the equation of tangent to the curve {tex} y = \frac { x – 7 } { x ^ { 2 } – 5 x + 6 }{/tex} at the point, where it cuts the X-axis.

• Show that semi – vertical angle of right circular cone of given surface area and maximum volume is {tex}{\sin ^{ – 1}}\left( {\frac{1}{3}} \right){/tex} .
• (d) 0,  Explanation: {tex}f'(t) = t{e^{ – t}}( – 1) + {e^{ – t}} \Rightarrow f'(1) = – {e^{ – 1}} + {e^{ – 1}} = 0 {/tex}
• (a) Neither decreasing nor increasing,  Explanation: f(x) = x 2 {tex}\Rightarrow {/tex} f'(x) = 2x for all x in R. Since f ‘(x) = 2x > 0 for x >0, and f ‘ (x) = 2x< 0 for x < 0, therefore on R, f is neither increasing nor decreasing. Infact , f is strict increasing on [ 0, {tex}\infty {/tex} ) and strict decreasing on (- {tex}\infty ,0\;{/tex} ].
• (d) cot t,  Explanation: Given, {tex} x=asint,y=a\left\{ { \cos t+\log (\tan \frac { t }{ 2 } ) } \right\}{/tex} {tex} { { dx } }{ { dt } } =a\cos t,\frac { { dy } }{ { dt } } {/tex} {tex}=a\left[ -\sin t+\frac { 1 }{ \tan \frac { t }{ 2 } } .{ sec }^{ 2 }\frac { t }{ 2 } .\frac { 1 }{ 2 } \right] {/tex} {tex}=a\left[ -\sin t+\frac { 1 }{ 2sin\frac { t }{ 2 } .cos\frac { t }{ 2 } } \right] {/tex} {tex}=a\left[ -\sin t+\frac { 1 }{ sint } \right] =a\frac { { cos }^{ 2 }t }{ sint } {/tex} Slope of the tangent {tex}=\frac { { dy } }{ { dx } } =\frac { { \frac { { dy } }{ { dt } } } }{ { \frac { { dx } }{ { dt } } } } =\frac { { a\frac { { cos }^{ 2 }t }{ sint } } }{ { a\cos t } } =\cot t{/tex}
• (d) {tex} ({\text{ }}-\infty ,{\text{ }}1){/tex} ,  Explanation: f ‘ (x ) = 2x – 2 = 2 ( x – 1) <0 if x < 1 i.e. x {tex}x \in \left( { – \infty ,1} \right) {/tex} . Hence f is strict decreasing in {tex}\left( { – \infty ,1} \right){/tex}
• (a) ty = x +at 2 ,  Explanation: {tex}{ y }^{ 2 }=4ax{/tex} {tex}\\ \Rightarrow 2y\frac { dy }{ dx } =4a{/tex} {tex}\\ \Rightarrow \frac { { dy } }{ { dx } } =\frac { 2a }{ y }{/tex} {tex} \Rightarrow \frac { { dy } }{ { dx } } {/tex} at {tex}(a{ t^{ 2 } },2at){/tex} is {tex}\frac { { 2a } }{ { 2at } } =\frac { 1 }{ t } {/tex} {tex}\Rightarrow {/tex}  Slope of tangent {tex}=m=\frac { 1 }{ t } {/tex} Hence, equation of tangent is {tex} y-{ y }_{ 1 }=m\left( x-{ x }_{ 1 } \right) {/tex} {tex}\\ \Rightarrow y-2at=\frac { 1 }{ t } (x-a{ t^{ 2 } }){/tex} {tex}\Rightarrow yt-2a{ t^{ 2 } }=x-a{ t^{ 2 } }{/tex} {tex}\Rightarrow yt=x+a{ t^{ 2 } }{/tex}
• {tex}{e^{\frac{1}{e}}}{/tex}
• {tex}x = 3,\Delta x = 0.02{/tex} {tex}f(x + \Delta x) = f(x) + f'(x)\Delta x{/tex} {tex}f(x + \Delta x) = (3{x^2} + 5x + 3) + (6x + 5) \times 0.02{/tex} Put {tex}x = 3,\Delta x = 0.02{/tex} f(3.02)={3(9)+5(3)+3}+{6(3)+5}×0.02 =45+0.46 f(3.02) = 45.46
• we have, xy =1 {tex} \Rightarrow y = \frac{1}{x}{/tex} {tex}\therefore\ \frac{dy}{dx}=-\frac{1}{x^2}{/tex} The slope of the normal = x 2 If ax+by+c=0 is normal to the curve xy=1,then {tex}x^2=-\frac{a}{b} \ [\because slope\ of\ normal\ =-\frac{coeff.\ of\ x}{coeff. of\ y}]{/tex} {tex}\therefore \; – \frac{a}{b} > 0{/tex} {tex}\Rightarrow\ a>0,b<0 \ or\ a<0,b>0{/tex}
• f(x) = x 2 e -x Differentiating w.r.t x, we get, f'(x) = {tex}-x^2e^{-x}+2xe^{-x}=xe^{-x}(2-x){/tex} For increasing function, f'(x) {tex}\geq0{/tex} {tex}xe^{-x}(2-x)\geq0{/tex} {tex}x(2-x)\geq0{/tex} [ {tex}\because\ e^{-x}{/tex} is always positive] {tex}x(x-2)\leq0{/tex} [ since – ( x – 2) will change the inequality ) Here x < 0 & (x – 2) > 0 {tex}\Rightarrow{/tex} x < 0 & x > 2 {tex}\Rightarrow{/tex} 0 < x < 2 But when x > 0 & (x – 2) < 0 {tex}\Rightarrow{/tex} x > 0 & x < 2 {tex}0\le x \leq\ 2{/tex}
• Let r be the radius of sphere and V be its volume. Then V = {tex}\frac { 4 } { 3 } \pi r ^ { 3 }{/tex} ……..(i) Given, {tex}\frac { d V } { d t }{/tex} = 3 cm 3 /s Differentiating (i) both sides w.r.t x,we get, {tex}\frac { d V } { d t } = \frac { 4 } { 3 } \pi \left( 3 r ^ { 2 } \right) \frac { d r } { d t }{/tex} {tex}\Rightarrow \quad 3 = \frac { 4 } { 3 } ( 3 \pi r ^2 ) \frac { d r } { d t }{/tex} {tex}\Rightarrow \quad \frac { d r } { d t } = \frac { 3 } { 4 \pi r ^ { 2 } }{/tex} …….(ii) Now, let S be the surface area of sphere, then S = {tex}4 \pi r ^ { 2 }{/tex} {tex}\Rightarrow \quad \frac { d S } { d t } = 4 \pi ( 2 r ) \frac { d r } { d t }{/tex} {tex}\Rightarrow \quad \frac { d S } { d t } = 8 \pi r \left( \frac { 3 } { 4 \pi r ^ { 2 } } \right){/tex} [using Eq.(ii)] {tex}\Rightarrow \quad \left( \frac { d S } { d t } \right) = \frac { 6 } { r }{/tex} when r = 2, then  {tex}\frac { d S } { d t } = \frac { 6 } { 2 }{/tex} = 3 cm 2 /s Therefore,the rate of inrcrease of the surface area of sphere is 3 cm {tex}^2{/tex} /s when it’s radius is 2 cm
• Let x = 2 and {tex}\Delta x = – 0.001\,\left[ {\because 2 – 0.001 = 1.999} \right]{/tex} let y = x 5 On differentiating both sides w.r.t. x, we get {tex}\frac{{dy}}{{dx}} = 5{x^4}{/tex} Now, {tex}\Delta y = \frac{{dy}}{{dx}}.\Delta x = 5{x^4} \times \Delta x{/tex} {tex} = 5 \times {2^4} \times \left[ { – 0.001} \right]{/tex} {tex} = – 80 \times 0.001 = – 0.080{/tex} {tex}\therefore {\left( {1.999} \right)^5} = y + \Delta y{/tex} {tex}= {2^5} + \left( { – 0.080} \right){/tex} = 32 – 0.080 = 31.920
• Here, f(x) =4x 3 – 18x 2 + 27x – 7 On differentiating both sides w.r.t. x, we get f'(x) = 12x 2 – 36x + 27 {tex}\Rightarrow{/tex} f'(x) = 3(4x 2 -12 + 9) {tex}\Rightarrow{/tex} f'(x) = 3(x – 3) 2 {tex}\Rightarrow{/tex} f'(x) {tex}\geq{/tex} 0 Since, a perfect square number cannot be negative] {tex}\therefore{/tex} Given function f(x) is an increasing function on R.
• Given curve is 6y = x 3 + 2 …(i) so, {tex}6\frac{{dy}}{{dt}} = 3{x^2}\frac{{dx}}{{dt}}{/tex} {tex}\Rightarrow 6 \times 8\frac{{dx}}{{dt}} = 3{x^2}\frac{{dx}}{{dt}}\,\,\left[ {\because \frac{{dy}}{{dt}} = 8\frac{{dx}}{{dt}}} \right]{/tex} {tex}\Rightarrow 16 = {x^2}{/tex} {tex}\Rightarrow x = \pm 4{/tex} Put the value of x in equation (1) When x = 4 6y = ( 4 ) 3 + 2 {tex}\Rightarrow{/tex} 6y = 64 + 2 {tex}\Rightarrow{/tex} 6y = 66 {tex}\therefore{/tex} {tex}y = \frac{{66}}{6} = 11{/tex} So, point is (4, 11) Now, When x = – 4 6y = ( – 4} 3 + 2 = – 64 + 2 {tex}\therefore{/tex} {tex}y = \frac{{ – 62}}{6}= \frac{-31}{3}{/tex} So the point is {tex}\left( { – 4,\frac{{ – 31}}{3}} \right){/tex}
• Given equation of curve is {tex} y = \frac { x – 7 } { x ^ { 2 } – 5 x + 6 }{/tex} …….(i) On differentiating both sides w.r.t. x, we get {tex} \frac { d y } { d x } = \frac { \left( x ^ { 2 } – 5 x + 6 \right) \cdot 1 – ( x – 7 ) ( 2 x – 5 ) } { \left( x ^ { 2 } – 5 x + 6 \right) ^ { 2 } }{/tex} {tex}\left[ \because \frac { d } { d x } \left( \frac { u } { v } \right) = \frac { v \frac { d u } { d x } – u \frac { d v } { d x } } { v ^ { 2 } } \right]{/tex} {tex}\Rightarrow \frac { d y } { d x } = \frac { \left[ \left( x ^ { 2 } – 5 x + 6 \right) – y \left( x ^ { 2 } – 5 x + 6 \right) (2x +5)\right] } { \left( x ^ { 2 } – 5 x + 6 \right) ^ { 2 } }{/tex} {tex}\Rightarrow \frac { d y } { d x } = \frac { 1 – ( 2 x – 5 ) y } { x ^ { 2 } – 5 x + 6 }{/tex} [dividing numerator and denominator by x 2 – 5x + 6] Also, given that curve cuts X-axis, so its y-coordinate is zero. Put y = 0 in Eq. (i), we get {tex}\frac { x – 7 } { x ^ { 2 } – 5 x + 6 } = 0{/tex} {tex}\Rightarrow{/tex} x= 7 So, curve passes through the point (7, 0). Now, slope of tangent at (7,0) is {tex}m = \left( \frac { d y } { d x } \right) _ { ( 2,0 ) } = \frac { 1 – 0 } { 49 – 35 + 6 } = \frac { 1 } { 20 }{/tex} Hence, the required equation of tangent passing through the point (7, 0) having slope 1/20 is y – 0 = {tex}\frac{1}{20}{/tex} (x – 7) {tex} \Rightarrow{/tex} 20y = x – 7 {tex} \therefore{/tex} x – 20y = 7

Chapter Wise Important Questions Class 12 Maths Part I and Part II

• Relations and Functions
• Inverse Trigonometric Functions
• Determinants
• Continuity and Differentiability
• Application of Derivatives
• Application of Integrals
• Differential Equations
• Vector Algebra
• Three Dimensional Geometry
• Linear Programming
• Probability

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NCERT Solutions for Class 6, 7, 8, 9, 10, 11 and 12

NCERT Exemplar Class 12 Maths Chapter 6 Application of Derivatives

June 6, 2022 by Bhagya

NCERT Exemplar Class 12 Maths Chapter 6 Application of Derivatives are part of NCERT Exemplar Class 12 Maths . Here we have given Exemplar Problems for Class 12 Maths Chapter 6 Application of Derivatives PDF.

NCERT Exemplar Class 12 Maths Chapter 6 Application of Derivatives Solutions is given below.

• NCERT Exemplar Class 12 Maths Solutions
• Chapter 1 Relations and Functions
• Chapter 2 Inverse Trigonometric Functions
• Chapter 3 Matrices
• Chapter 4 Determinants
• Chapter 5 Continuity and Differentiability
• Chapter 6 Application of Derivatives
• Chapter 7 Integrals
• Chapter 8 Applications of Integrals
• Chapter 9 Differential Equations
• Chapter 10 Vector Algebra
• Chapter 11 Three Dimensional Geometry
• Chapter 12 Linear Programming
• Chapter 13 Probability

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