case study class 12 maths application of derivatives

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

• Post author: studyrate
• Post published:
• Post category: 12 board / Class 12
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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.

Join our Telegram Channel, there you will get various e-books for CBSE 2024 Boards exams for Class 9th, 10th, 11th, and 12th.

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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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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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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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Class 12 Maths Case Study Questions

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

Why Case Studies in CBSE Syllabus?

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

Case Study Questions in Maths

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

Focus on concepts

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

Easy Questions with a Practical Approach

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

Practice Questions Regularly

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

12 Maths Case-Based Questions

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

Case Study Question – 1

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

Case Study Question – 2

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

Case Study Questions Matrices -1

Answer Key:

Case Study Questions Matrices – 2

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

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

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

• (a) 3  ×  3

Case Study Questions Determinants – 01

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

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

• (b) Shyam Lal
• (a) Ram Lal

Case Study Questions Determinants – 02

Case study questions application of derivatives.

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

Case Study Questions Vector Algebra

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

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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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)

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NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives

Ncert solutions for class 12 maths chapter 6 – free pdf download.

The NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives provides answers for all the questions listed under this chapter which is included in the Class 12 Maths CBSE Syllabus for 2023-24. These NCERT Solutions include a complete set of questions and answers organised with an advanced level of difficulty, which provides students with sufficient opportunity to apply their skills. Students can download these NCERT Solutions for Class 12 Maths  and strengthen their skills.

Further, students can clarify their doubts while solving problems at any time by simply downloading the PDF version of the NCERT Solutions for Class 12 Maths . Clarify doubts instantly by downloading the NCERT Solutions for Class 12 Maths Chapter 6 from the link provided below.

Download PDF of NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives

Access answers to ncert class 12 maths chapter 6 application of derivatives.

This chapter of NCERT Solutions for Class 12 Maths mainly has a set of topics like The rate of change of quantities, Increasing and decreasing functions, Tangents and normals, Approximations, Maxima and minima and many more. Some of the important concepts are:

1. A function f is said to be

(a) increasing on an interval (a, b) if x 1 < x 2 in (a, b) ⇒ f(x 1 ) < f(x 2 ) for all x 1 , x 2 ∈ (a, b)

(b) decreasing on (a,b) if x 1 < x 2 in (a, b) ⇒ f(x 1 ) > f(x 2 ) for all x 1 , x 2 ∈ (a, b)

(c) constant in (a, b), if f (x) = c for all x ∈ (a, b), where c is a constant

2. First Derivative Test

3. Second Derivative Test

Students can utilise the NCERT Solutions for Class 12 Maths Chapter 6 for quick references to comprehend key and complex topics.

NCERT Solutions for Class 12 Maths Chapter 6 Exercises

Get detailed NCERT Solutions for all the questions listed under the below exercises:

Exercise 6.1 Solutions : 18 Questions (10 Long, 6 Short, 2 MCQs)

Exercise 6.2 Solutions : 19 Questions (10 Long, 7 Short, 2 MCQs)

Exercise 6.3 Solutions : 27 Questions (14 Long, 11 Short, 2 MCQs)

Exercise 6.4 Solutions : 9 Questions (7 Short, 2 MCQs)

Exercise 6.5 Solutions : 29 Questions (15 Long, 11 Short, 3 MCQs)

Miscellaneous Exercise Solutions : 24 Questions (14 Long, 4 Short, 6 MCQs)

The main topics covered in  NCERT Solutions for Class 12 Chapter 6 Application of Derivatives are:

Key Features of NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives

• NCERT Solutions will help in boosting students’ confidence levels.
• The content is well-structured so that it becomes easier for students to learn and understand.
• NCERT Solutions are based on the latest CBSE syllabus for 2023-24.
• This chapter will help the students to strengthen their foundation on the application of derivatives.
• NCERT Solutions are a valuable aid for students in their assignments and competitive exams.

Disclaimer – 

Dropped Topics –  6.4 Tangents and Normals, 6.5 Approximations, Examples 45, 46, Ques. 1, 4–5 and 20–24 (Miscellaneous Exercise), Page 245 Points 4–10 in the Summary

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Application of Derivatives Class 12 Notes Maths Chapter 6

January 9, 2024 by Sastry CBSE

CBSE Class 12 Maths Notes Chapter 6 Application of Derivatives

Rate of Change of Quantities: Let y = f(x) be a function of x. Then, \(\frac { dy }{ dx }\) represents the rate of change of y with respect to x. Also, [latex s=1]\frac { dy }{ dx }[/latex]x = x0 represents the rate of change of y with respect to x at x = x 0 .

Let I be an open interval contained in the domain of a real valued function f. Then, f is said to be

• increasing on I, if x 1 < x 2 in I ⇒ f(x 1 ) ≤ f(x 2 ), ∀ x 1 , x 2 ∈ I.
• strictly increasing on I, if x 1 < x 2 in I ⇒ f(x 1 ) < f(x 2 ), ∀ x 1 , x 2 ∈ I.
• decreasing on I, if x 1 < x 2 in I ⇒ f(x 1 ) ≥ f(x 2 ), ∀ x 1 , x 2 ∈ I.
• strictly decreasing on I, if x 1 < x 2 in f(x 1 ) > f(x 2 ), ∀ x 1 , x 2 ∈ I.

Let x 0 be a point in the domain of definition of a real-valued function f, then f is said to be increasing, strictly increasing, decreasing or strictly decreasing at x 0 , if there exists an open interval I containing x0 such that f is increasing, strictly increasing, decreasing or strictly decreasing, respectively in I. Note: If for a given interval I ⊆ R, function f increase for some values in I and decrease for other values in I, then we say function is neither increasing nor decreasing.

Let f be continuous on [a, b] and differentiable on the open interval (a, b). Then,

• f is increasing in [a, b] if f'(x) > 0 for each x ∈ (a, b).
• f is decreasing in [a, b] if f'(x) < 0 for each x ∈ (a, b).
• f is a constant function in [a, b], if f'(x) = 0 for each x ∈ (a, b).

Note: (i) f is strictly increasing in (a, b), if f'(x) > 0 for each x ∈ (a, b). (ii) f is strictly decreasing in (a, b), if f'(x) < 0 for each x ∈ (a, b).

Monotonic Function: A function which is either increasing or decreasing in a given interval I, is called monotonic function.

Approximation: Let y = f(x) be any function of x. Let Δx be the small change in x and Δy be the corresponding change in y. i.e. Δy = f(x + Δx) – f(x).Then, dy = f'(x) dx or dy = \(\frac { dy }{ dx }\) Δx is a good approximation of Δy, when dx = Δx is relatively small and we denote it by dy ~ Δy. Note: (i) The differential of the dependent variable is not equal to the increment of the variable whereas the differential of the independent variable is equal to the increment of the variable. (ii) Absolute Error The change Δx in x is called absolute error in x.

Note: If a tangent line to the curve y = f(x) makes an angle θ with X-axis in the positive direction, then \(\frac { dy }{ dx }\) = Slope of the tangent = tan θ. dx

Equations of Tangent and Normal The equation of tangent to the curve y = f(x) at the point P(x 1 , y 1 ) is given by y – y 1 = m (x – x 1 ), where m = \(\frac { dy }{ dx }\) at point (x 1 , y 1 ).

The equation of normal to the curve y = f(x) at the point Q(x 1 , y 1 ) is given by y – y 1 = \(\frac { -1 }{ m }\) (x – x 1 ), where m = \(\frac { dy }{ dx }\) at point (x 1 , y 1 ).

If slope of the tangent line is zero, then tanθ = θ, so θ = 0, which means that tangent line is parallel to the X-axis and then equation of tangent at the point (x 1 , y 1 ) is y = y 1 .

If θ → \(\frac { \pi }{ 2 }\), then tanθ → ∞ which means that tangent line is perpendicular to the X-axis, i.e. parallel to the Y-axis and then equation of the tangent at the point (x 1 , y 1 ) is x = x 0 .

Maximum and Minimum Value: Let f be a function defined on an interval I. Then, (i) f is said to have a maximum value in I, if there exists a point c in I such that f(c) > f(x), ∀ x ∈ I. The number f(c) is called the maximum value of f in I and the point c is called a point of a maximum value of f in I. (ii) f is said to have a minimum value in I, if there exists a point c in I such that f(c) < f(x), ∀ x ∈ I. The number f(c) is called the minimum value of f in I and the point c is called a point of minimum value of f in I. (iii) f is said to have an extreme value in I, if there exists a point c in I such that f(c) is either a maximum value or a minimum value of f in I. The number f(c) is called an extreme value off in I and the point c is called an extreme point.

Local Maxima and Local Minima (i) A function f(x) is said to have a local maximum value at point x = a, if there exists a neighbourhood (a – δ, a + δ) of a such that f(x) < f(a), ∀ x ∈ (a – δ, a + δ), x ≠ a. Here, f(a) is called the local maximum value of f(x) at the point x = a. (ii) A function f(x) is said to have a local minimum value at point x = a, if there exists a neighbourhood (a – δ, a + δ) of a such that f(x) > f(a), ∀ x ∈ (a – δ, a + δ), x ≠ a. Here, f(a) is called the local minimum value of f(x) at x = a.

The points at which a function changes its nature from decreasing to increasing or vice-versa are called turning points. Note: (i) Through the graphs, we can even find the maximum/minimum value of a function at a point at which it is not even differentiable. (ii) Every monotonic function assumes its maximum/minimum value at the endpoints of the domain of definition of the function.

Every continuous function on a closed interval has a maximum and a minimum value.

Let f be a function defined on an open interval I. Suppose cel is any point. If f has local maxima or local minima at x = c, then either f'(c) = 0 or f is not differentiable at c.

Critical Point: A point c in the domain of a function f at which either f'(c) = 0 or f is not differentiable, is called a critical point of f.

First Derivative Test: Let f be a function defined on an open interval I and f be continuous of a critical point c in I. Then,

• if f'(x) changes sign from positive to negative as x increases through c, then c is a point of local maxima.
• if f'(x) changes sign from negative to positive as x increases through c, then c is a point of local minima.
• if f'(x) does not change sign as x increases through c, then c is neither a point of local maxima nor a point of local minima. Such a point is called a point of inflection.

Second Derivative Test: Let f(x) be a function defined on an interval I and c ∈ I. Let f be twice differentiable at c. Then, (i) x = c is a point of local maxima, if f'(c) = 0 and f”(c) < 0. (ii) x = c is a point of local minima, if f'(c) = 0 and f”(c) > 0. (iii) the test fails, if f'(c) = 0 and f”(c) = 0.

Note (i) If the test fails, then we go back to the first derivative test and find whether a is a point of local maxima, local minima or a point of inflexion. (ii) If we say that f is twice differentiable at o, then it means second order derivative exists at a.

Absolute Maximum Value: Let f(x) be a function defined in its domain say Z ⊂ R. Then, f(x) is said to have the maximum value at a point a ∈ Z, if f(x) ≤ f(a), ∀ x ∈ Z.

Absolute Minimum Value: Let f(x) be a function defined in its domain say Z ⊂ R. Then, f(x) is said to have the minimum value at a point a ∈ Z, if f(x) ≥ f(a), ∀ x ∈ Z.

Note: Every continuous function defined in a closed interval has a maximum or a minimum value which lies either at the end points or at the solution of f'(x) = 0 or at the point, where the function is not differentiable.

Let f be a continuous function on an interval I = [a, b]. Then, f has the absolute maximum value and/attains it at least once in I. Also, f has the absolute minimum value and attains it at least once in I.

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12th Class Mathematics Applications of Derivatives Question Bank

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

A) 1 done clear

B) -1 done clear

C) 0 done clear

D) 2 done clear

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

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

A) x = 1 done clear

B) x = 2 done clear

C) x = 3 done clear

D) x = 4 done clear

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

B) 2 done clear

C) 3 done clear

D) -1 done clear

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

A) 2 done clear

B) 1 done clear

C) -2 done clear

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

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

C) 1 done clear

D) 0 done clear

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

A) - 1 done clear

C) 2 done clear

D) 4 done clear

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

A) 0 done clear

C) -1 done clear

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

C) - 1 done clear

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

A) 1/9 done clear

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

C) 1/2 done clear

D) not defined done clear

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

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

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

D) none of these done clear

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

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

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

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

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

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

B) continuous and differentiable at x = 3 done clear

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

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

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

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

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

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

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

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

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

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

D) None of these. done clear

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

A) 12 done clear

B) 32 done clear

C) 36 done clear

D) 10 done clear

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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Application of Derivatives Class 12 Notes PDF (Short & Handwritten)

Revision notes make you aware of those topics that you might have missed during your regular classes. Thus, Application of Derivatives is an extremely important chapter of class 12 Maths and so, all students who have opted for Maths in their intermediate should refer to the Application of Derivatives Class 12 Notes. The revision notes work as a reference that help students like you to revise the concepts and formulas which you have studied earlier from your Mathematics textbook.

Therefore, to help you better learn the Application of Derivatives concepts and techniques to solve questions, we have mentioned here the PDF of revision notes. The Application of Derivatives notes in the PDF file can be downloaded for free of cost from this page.

Application of Derivatives Notes PDF

The PDF file is easy to access and often free, it is popular among students as they can carry them anywhere they want into their device like smartphone, Laptop, etc. Keeping in mind the convenience of students, our subject matter experts have prepared the Application of Derivatives Notes PDF which is very illustrative and can help you recall all of your previous learning of class 12 Maths Application of Derivatives.

The Application of Derivatives PDF notes can be a fun and time saving study resource that can help students to cover all the important and highlighted key points of the chapter.

Topic wise Class 12 Application of Derivatives Notes

Our Maths experts have organised the Class 12 Application of Derivatives Notes in a topic wise manner. Because of that students are able to find an organised study resource that enables them to focus on study and avoid clutter in learning.

The topic wise Application of Derivatives notes PDF is an ideal study resource that improves memory power and increases the attention. The notes are considered a great tool to develop a deeper understanding in the concepts explained. Keeping in mind this, our subject experts have prepared our own revision notes for Application of Derivatives.

How to Download Application of Derivatives Class 12 Notes in PDF for Free?

Follow the below-given steps to download Application of Derivatives Class 12 Notes in PDF for Free.

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Application of Derivatives Class 12 Notes That Can Help You Score Better Marks

We all agree that students need to do lots of practice in order to master Maths; however, it isn’t an easy task because without a proper understanding of the concepts it wouldn’t be easy to solve even a single question from the chapter Application of Derivatives. But the class 12 Application of Derivatives notes that we provide here are a very ideal study resource to develop a higher level of understanding in the topic.

A good grip on the concepts of class 12 Application of Derivatives can assist candidates to easily score better marks due to two reasons:

• Revision notes are short yet precise that help students to cover more important concepts in a lesser time.
• Research suggests that repeatedly memorising something boosts depth of mental processing as a result; thus, it develops deeper levels of analysis that produce more elaborate, longer-lasting, and stronger memory.

In conclusion, we can say that Notes of Application of Derivatives along with a regular practice of questions can help students to score better marks as they can remember the topics for a longer period of time.

Other than Notes of Application of Derivatives You Can Use

In case, if you make your mind to not only focus on Application of Derivatives class 12 notes then you can use -

• MCQ Test of Application of Derivatives: Similar to notes, solving the Mock test of Application of Derivatives can help you recall your previous learning from the chapter. If you want to level up your preparation level and want to challenge your conceptual understanding of the Application of Derivatives you can use the CBSE MCQ test.
• Chapter End Questions of Application of Derivatives: The Application of Derivatives notes PDF that we provide here are prepared referring to the NCERT Class 12 Maths Book so, those who want to use other study resources than revision notes of Application of Derivatives can use the NCERT Books. The book contains exercises that enable the learners to solve various questions to cross-check how well they understood the topic.
• Highlighted Points of Topic: Every single chapter contains the key points that grab attention to readers, even such highlighted points are considered important. Thus, if you want to take a pause for a while from using the Application of Derivatives PDF notes then you can go back to your preferred textbook to read those highlighted points of the topic.

The Right Time To Use Application of Derivatives Class 12 Notes

There is no right or wrong time to use Application of Derivatives Class 12 Maths Notes as long as you are sincere about your study; however, there are 3 most important times that we think all students should revise whatever they have studied in Application of Derivatives. 

• On a weekly basis after completing the chapter Application of Derivatives: There’s a few students who dedicate themselves in learning Maths during regular classes; however, experts suggests that revision just after completing the chapter Application of Derivatives can help students to deepen their understanding in the concepts that were explained in the lesson. Thus, the right time to use the Application of Derivatives Class 12 Notes can be on a weekly basis after finishing the chapter’s basics subtopics.
• During Board Exam Preparation: The importance of revision Notes of Application of Derivatives can be understood during the board exam preparation because it saves time and allows students to cover the complete chapter in a lesser time. Also, the board exam preparation time often becomes a mess but the notes are neat, clean and organised in a manner that helps students to be more productive.
• While preparing for competitive exams like JEE: Class 12 Application of Derivatives is not limited to board exams only, rather it plays a crucial role in national level competitive exam preparation. Therefore, the use of Application of Derivatives Class 12 Notes PDF while preparing for competitive exams like JEE can be a great time.

What You Definitely Need to Revise Class 12 Application of Derivatives?

The Application of Derivatives is an important part of higher level Maths, because in standard like 12, students have to study various topics at the same time, it becomes a challenge to handle so many topics and subjects at the same time. Thus, here we have listed what you definitely need to revise Class 12 Application of Derivatives for the ease of your study and exam preparation.

• A Short Notes of Application of Derivatives: You wouldn’t like to use the revision Notes of Application of Derivatives that look like bulky textbooks, cluttered with so much information in an unorganised manner, right? So, you need a short yet precise Short Notes of Application of Derivatives that can help you recall your learnings from the chapter within a few minutes. You can find such Notes of Application of Derivatives here on Selfstudys website.
• Easy To Understand Definitions: Unclear and lengthy definitions never work, and it becomes very challenging to retain such definitions for a longer time. Therefore, you need revision notes that can help you easily understand the definitions of Application of Derivatives. Easy to understand definitions help you retain them for a longer time and give you an ability to recall them wherever you need them to solve a particular question related to Application of Derivatives.
• Some Questions of Application of Derivatives to Practise: Without solving the questions of Application of Derivatives a revision is incomplete. If you really want to revise the topics, you must solve the questions based on Application of Derivatives. Such questions can be found from Notes of Application of Derivatives that we provide here or from the NCERT Class 12 Maths textbooks. Our Maths experts have mentioned the questions of Application of Derivatives especially in Class 12 Maths notes so that a learner can use them to practise.

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