case study questions from application of derivatives

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

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

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

Applications of Derivatives Case Study Questions With answers

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

Case Study/Passage-Based Questions

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

Answer: (b) area

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

Answer: (c) y+2x=200

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

Answer: (c) 200x – 2×2

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

Answer: (a) 50 ft

(v) Maximum area of garden will be

Answer: (c) 5000sq.ft

Hope the information shed above regarding Case Study and Passage Based Questions for Class 12 Maths Chapter 6 Applications of Derivatives with Answers Pdf free download has been useful to an extent. If you have any other queries of CBSE Class 12 Mathematics Applications of Derivatives Case Study and Passage Based Questions with Answers, feel free to comment below so that we can revert back to us at the earliest possible. By Team Study Rate

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

Class 12th Maths - Application of Derivatives Case Study Questions and Answers 2022 - 2023 Study Materials Sep-08 , 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.

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

Application of Derivatives Class 12 Maths Important Questions Chapter 6

July 14, 2022 by Sastry CBSE

Get access to Class 12 Maths Important Questions Chapter 6 Application of Derivatives, Application of Derivatives Class 12 Important Questions with Solutions Previous Year Questions will help the students to score good marks in the board examination.

Application of Derivatives Class 12 Important Questions with Solutions Previous Year Questions

Rate Measure, Increasing-Decreasing Functions and Approximation

Question 1. The total cost C(x) associated with the production of x units of an item is given by C(x) = 0.005x 3 – 0.02x 2 + 30x + 5000. Find the marginal cost when 3 units are produced, where by marginal cost we mean the instantaneous rate of change of total cost at any level of output. (CBSE 2018) Answer: We have, C(x) = 0.005x 3 – 0.02x 2 + 30x + 5000 Clearly, the marginal cost, MC (x) = \(\frac{d}{d x}\)C(x) = \(\frac{d}{d x}\)(0.005x 3 – 0.02x 2 + 30x + 500) = 0.005 × 3x 2 – 0.02 × 2x + 30 + 0 = 0.015x 2 – 0.04x + 30

Now, marginal cost when 3 units arei produced MC(3)= 0.015(9) – 0.04(3) + 30 = 0135 – 0.12 + 30= 30.015

Question 2. The total revenue received from the sale of x units of a product is given by R(x) = 3x 2 + 36x + 5 in rupees. Find the marginal revenue when x = 5, where by marginal revenue we mean the rate of change of total revenue with respect to the number of items sold at an instant. (CBSE 2018 C) Answer: Marginal Rcvcnuc (MR) = \(\frac{d R}{d x}=\frac{d}{d x}\)(3x 2 + 36x + 5) = 6x + 36 ∴ When x = 5 Marginai Revenue (MR) = 6 × 5 + 36 = 66

Question 4. Show that the function f(x) = x 3 – 3x 2 + 6x – 100 is increasing on R. (All India 2017) Answer: Given, f(x) = x 3 – 3x 2 + 6x – 100 On differentiating both sides w.r.t. x, we get f'(x) = 3x 2 – 6x + 6 = 3x 2 – 6x + 3 + 3 = 3(x 2 – 2x + 1) + 3 = 3(x – 1) 2 + 3 > 0 ∴ f'(x) > 0 This show that function f(x) is increasing on R. Hence proved.

Question 6. Show that the function f(x) = 4x 3 – 18x 2 + 27x – 7 is always increasing on R. (Delhi 2017) Answer: We have, f(x) = 4x 3 – 18x 2 + 27x – 7 On differentiating both sides w.r.t. x, we get f(x) = 12x 2 – 36x + 27 ⇒ f'(x) = 3(4x 2 -12x + 9) ⇒ f'(x) = 3(2x – 3) 2 ⇒ f(x) > 0 ⇒ For any x ∈ R, (2x – 3) 2 > 0 Since, a perfect square number cannot be negative. ∴ Given function f(x) is an increasing function on R.

Question 9. Find the intervals in which the function f(x) = -2x 3 – 9x 2 – 12x + 1 is (i) strictly increasing (ii) strictly decreasing, (CBSE 201B C) Answer: Given, f(x) = -2x 3 – 9x 2 – 12x + 1 On differentiating both sides w.r.t. x, we get f'(x) = – 6x 2 – 18x – 12 ⇒ f'(x) = – 6(x 2 + 3x + 2) ⇒ f'(x) = – 6(x 2 + 2x + x + 2) ⇒ – 6 [x(x + 2) + 1 (x + 2)] ⇒ – 6 (x + 2) (x + 1)

Hence, f(x) is strictly increasing in the interval (- 2, -1) and f(x) is strictly decreasing in the interval (- ∞, – 2) ∪ (-1, ∞).

Question 10. The length x of a rectangle is decreasing at the rate of 5 cm/min and the width y is increasing at the rate of 4 cm/min. When x = 8 cm and y = 6 cm, find the rate of change of (i) the perimeter. (ii) area of rectangle. (All India 2017) Answer: Using the relation, perimeter of rectangle, P = 2(x + y) and area of rectangle, A = xy, differentiate both sides with respect to t and put them in rate of change value and get the result. Given that length x of a rectangle is decreasing at the rate of 5 cm/min. ∴ \(\frac{d y}{d x}\) = – 5cm/min ………..(i) Also, the breadth y of rectangle is increasing at the rate of 4 cm/min. ∴ \(\frac{d y}{d x}\) = 4 cm/min ……(ii)

(i) Here, we have to find rate of change of perimeter, i.e. dP/dt and we know that, perimeter P = 2 (x + y)

On differentiating both sides w.r.t. t, we get \(\frac{d P}{d t}\) = 2\(\left(\frac{d x}{d t}+\frac{d y}{d t}\right)\) ⇒ \(\frac{d P}{d t}\) = 2(-5 + 4) = 2(-1) = -2cm/min [from Eqs. (i) and (ii)] Hence, perimeter of rectangle is decreasing at the rate 2 cm/min.

(ii) Here, we have to find rate of change of area \(\frac{d A}{d t}\) We know that, area of rectangle A = xy On differentiating both sides w.r.t. t, we get \(\frac{d A}{d t}=x \cdot \frac{d y}{d t}+y \cdot \frac{d x}{d t}\) [by using product rule of derivative]

Question 11. The side of an equilateral triangle is increasing at the rate of 2 cm/s. At what rate is its area increasing, when the side 22. of the triangle is 20 cm? (Delhi 2015) Answer: Let a be the side of an equilateral triangle and A be the area of an equilateral triangle. Then, \(\frac{d a}{d x}\) = 2 cm/s

We know that, area of an equilateral triangle, A = \(\frac{\sqrt{3}}{4}\)a 2

Question 12. Find the intervals in which the function f(x) = 3x 4 – 4x 3 – 12x 2 + 5 is (i) strictly increasing. (ii) strictly decreasing. (Delhi 2014) Answer: First, find the first derivative and put equal to zero, we get different values of x and then divide the real line into disjoint intervals. Further check sign of f(x) in a given interval, if f'(x) > 0, then it is strictly increasing and if f'(x) < 0, then it is strictly decreasing. f(x) = 3x 4 – 4x 3 – 12x 2 + 5 On differentiating both sides w.r.t. x, we get f'(x) = 12x 3 -12x 2 – 24x For strictly increasing or strictly decreasing, put f'(x) = 0, we get 12x 3 – 12x 2 – 24x = 0 ⇒ 12x [x 2 – 2x + x – 2] = 0 12x (x + 1) (x – 2) = 0 ∴ x = 0, -1 or 2 Now, we find intervals in which f(x) is strictly increasing or strictly decreasing.

We know that, a function f(x) is said to be strictly increasing , if f'(x) > 0 and it is said to be strictly decreasing , if f'(x) < 0. So, the given function f(x) is (i) strictly increasing on the intervals (-1, 0) and (2, ∞). (ii) strictly decreasing on the intervals (-∞, -1) and (0, 2).

Question 13. Find the intervals in which the function given by ; f(x) = \(\frac{3}{10}\) x 4 – \(\frac{4}{5}\)x 3 – 3x 2 + \(\frac{36x}{5}\) + 11 is (i) strictly increasing. (ii) strictly decreasing. (All India 2014C) Answer: (i) Strictly increasing in (-2, 1) and (3, ∞). (ii) Strictly decreasing in (-∞,- 2) and (1, 3).

Question 14. The sides of an equilateral triangle are increasing at the rate of 2 cm/s. Find the rate at which the area increases, when the side is 10 cm? (All India 2014C) Answer: 10√ 3 cm 2 /s

Question 15. Find the value(s) of x for which y = [x(x – 2)] 2 is an increasing function. (All India 2014; Delhi 2010) Answer: Given function is y = [x(x – 2)] 2 = [x 2 – 2x] 2 . On differentiating both sides w.r.t. x, we get \(\frac{d y}{d x}\) = 2(x 2 – 2x) – (x 2 – 2x) = 2(x 2 – 2x) (2x – 2) = 4x(x – 2) (x -1)

On putting \(\frac{d y}{d x}\) = 0, we get 4x(x – 2)(x – 1) = 0 ⇒ x = 0, 1 and 2

Now, we find interval in which f(x) is strictly increasing or strictly decreasing.

Hence, y is strictly increasing in (0, 1) and (2, ∞). Also, y is a polynomial function, so it continuous at x = 0, 1 and 2. Hence, y is increasing in [0, 1] ∪ [2, ∞).

Question 16. Using differentials, find the approximate value of (3.968) 3/2 . (Delhi 2014C) Answer: Let y = f(x) = (x) 3/2 On differentiating both sides w.r.t. x, we get Let x = 4 and x + Δx = 3.968 Then, Δx = -0.032 Now, f(x+ Δx) 3/2 ≈ f(x) + f'(x)Δx (x+ Δx) 3/2 ≈ (x) 3/2 + \(\frac{3}{2}\).(x) 1/2 .(-0.032) ⇒ (4 – 0.032) 3/2 ≈ (4) 3/2 + \(\frac{3}{2}\)(4) 1/2 (-0.032) [put x = 4] ⇒ (3.968) 3/2 ≈ 8 + \(\frac{3}{2}\) .2.(-0.032) ⇒ (39368) 3/2 ≈ 8 – 0.096 ⇒ (3.968) 3/2 ≈ 7.904

Question 17. Find the intervals in which the function f(x) = \(\frac{3}{2}\) x 4 – 4x 3 – 45x 2 + 51 is (i) strictly increasing. (ii) strictly decreasing. (Foreign 2014C) Answer: (i) Strictly increasing in (-3, 0) and (5, ∞). (ii) Strictly decreasing in (-∞,- 3) and (0,5).

Question 18. Find the approximate value of f(3.02), upto 2 places of decimal, where f(x) = 3x 2 + 15x + 3. (Foreign 2014) Answer: First, split 3.02 into two parts x and Ax, so that x + Δx = 3.02 and f(x + Δx) = f(3.02) Now, write f(x + Δx) = f(x) + Δx . f'(x) and use this result to find the required value. Given function is f(x) = 3x 2 + 15x + 3 On differentiating both sides w.r.t. x, we get f'(x) = 6x + 15 Let x = 3 and Δx = 0.02 So that f(x + Δx) – f(302) By using f(x + Δx) ~ f(x)+ Δx f'(x), we get f(x + Δx) = 3x 2 + 15x + 3 + (6x + 15) Δx f(3 + 0.02) = 3(3) 2 + 15(3) + 3+ [6(3) + 15] (0.02) = 27 + 45 + 3 + 33(0.02) = 75 + 0.66 = 75.66 Hence, f (3.02) ≈ 75.66

Hence, the function is (i) increasing in [0, \(\frac{\pi}{4}\)] and [\(\frac{5\pi}{4}\), 2π] (ii) decreasing in \(\left[\frac{\pi}{4}, \frac{5 \pi}{4}\right]\)

Question 23. Find the intervals in which the function given by f(x) = x 4 – 8x 3 + 22x 2 – 24x + 21 is (i) increasing, (ii) decreasing. (All Delhi 2012C) Answer: Given function is f(x) = x 4 – 8x 3 + 22x 2 – 24x + 21 On differentiating both sides w.r.t. x, we get f(x) = 4x 3 – 24x 2 + 44x – 24 = 4(x 3 – 6x 2 + 11 x – 6) = 4(x – 1) (x 2 – 5x + 6) = 4(x – 1) (x – 2) (x – 3) Put f'(x) = 0 ⇒ 4(x – 1)(x – 2)(x – 3) = 0 ⇒ x = 1, 2, 3 So, the possible intervals are (-∞, 1), (1, 2), (2, 3) and (3, ∞). For interval (- ∞, 1), f'(x) < 0 For interval (1, 2), f'(x) > 0 For interval (2, 3), f'(x) < 0 For interval (3, ∞), f'(x) > 0.

Also, as f(x) is a polynomial function, so it is continuous at x = 1, 2, 3…………. Hence, (i) function increases in [1, 2] and [3, ∞). (ii) function decreases in (-∞, 1 ] and [2, 3]

Question 24. Sand is pouring from the pipe at the rate of 12 cm3/s. The falling sand forms a cone on a ground in such a way that the height of cone is always one-sixth of radius of the base. How fast is the height of sand 5 cone increasing when the height is 4 cm? (Delhi 2011) Answer: Let V be the volume of cone, h be the height and r be the radius of base of the cone. Given, \(\frac{d V}{d t}\) = 12 cm 3 /s ……(i)

Also, height of cone = \(\frac{1}{6}\) × (radius of base of cone) ∴ h = \(\frac{1}{6}\)r or r = 6h ………..(ii)

We know that, volume of cone is given by V = \(\frac{1}{3}\)πr 2 h …(iii)

Question 25. If the radius of sphere is measured as t 9 cm with an error of 0.03 cm, then find the approximate error in calculating its surface area. (All India 2011) Answer: Let S be the surface area, r be the radius of the sphere. Given, r = 9 cm Then, dr = Approximate error in radius r = 0.03 cm and dS = Approximate error in surface area Now, we know that surface area of sphere is given by S = 4πr 2 On differentiating both sides w.r.t. r, we get \(\frac{d S}{d r}\) = 4π × 2r = 8πr dS = 8πr × dr ⇒ dS = 8π × 9 × 0.03 [∵ r = 9 cm and dr = 0.03 cm ] ⇒ dS = 72 × 0.03π ∴ dS = 2.16π cm 2 /cm Hence, approximate error in surface area is 2.16π cm 2 /cm.

Question 26. Find the intervals in which the function f(x) = sin x + cos x, 0 ≤ x ≤ 2π is strictly l increasing and strictly decreasing. (Foreign 2011) Answer: Strictly increasing in the intervals [0, \(\frac{\pi}{4}\)) and (\(\frac{5 \pi}{4}\), 2] strictly decreasing in the intervals \(\left(\frac{\pi}{4}, \frac{5 \pi}{4}\right)\)

Question 27. Show that the function f(x) = x 3 – 3x 2 + 3x, x ∈ R is increasing on R. (All India 2011C) Answer: We know that, a continuous function y = f(x) is said to be increasing on R, if \(\frac{d y}{d x}\) ≥ 0, ∀ x ∈ R. Given, y = x 3 – 3x 2 + 3x

Question 28. Find the intervals in which the function f(x) = (x – 1 ) 3 (x – 2) 2 is (i) increasing, (ii) decreasing. (All India 2011C) Answer: Given, f(x) = (x – 1 ) 3 (x – 2) 2

On differentiating both sides w.r.t. x, we get f'(x) = (x – 1) 3 .\(\frac{d}{d x}\)(x – 2) 2 + (x – 2) 2 .\(\frac{d}{d x}\)(x – 1) 3 ⇒ f'(x) = (x – 1) 3 – 2(x – 2)+ (x – 2) 2 3(x – 1) 2 = (x – 1) 2 (x – 2)[2(x – 1) + 3(x – 2)] = (x – 1) 2 (x – 2) (2x – 2 + 3x – 6) ⇒ f’(x) = (x – 1) 2 (x – 2){3x – 8)

Now, put f'(x) = 0 ⇒ (x – 1) 2 (x – 2) (5x – 8) = 0 Either (x – 1) 2 = 0 or x – 2 = 0 or 5x – 8 = 0 ∴ x = 1, \(\frac{8}{5}\), 2 Now, we find intervals and check in which interval f(x) is strictly increasing and strictly decreasing.

We know that, a function f(x) is said to be an strictly increasing function, if f'(x) > 0 and strictly decreasing, if f'(x) < 0. So, the given function f(x) is increasing on the intervals (-∞, 1) (1, \(\frac{8}{5}\)) and (2, ∞) and decreasing on continuous at x = 1, -, 2 Hence, f(x) is (i) increasing on intervals I (ii) decreasing on interval Note: Every strictly increasing (strictly decreasing) function is incresing (decreasing) but converse need not be true.

Question 29. Find the intervals in which the function f(x) = 2x 3 + 9x 2 + 12x + 20 is (i) increasing (ii) decreasing. (Delhi 2011c) Answer: Given function is f(x) = 2x 3 + 9x 2 + 12x + 20 On differentiating both sides w.r.t. x, we get f(x) = 6x 2 + 18x + 12 Put f'(x) = 0, we get 6x 2 + 18x + 12 = 0 ⇒ 6(x 2 + 3x + 2) = 0 ⇒ 6 (x + 1) (x + 2) = 0 ⇒ (x + 1) (x + 2) = 0 ⇒ x + 1= 0 or x + 2 = 0 x = – 2, -1 Now, we find intervals and check in which interval f(x) is strictly increasing and strictly decreasing.

We know that, a function f(x) is said to be an strictly increasing function, if f'(x) > 0 and strictly decreasing, if f'(x) < 0. So, given function is increasing on intervals (-∞,- 2) and (-1, ∞) and decreasing on interval (-2, -1). Since, f(x) is a polynomial function, so it is continuous at x = -1, – 2. Hence, given function is (i) increasing on intervals (-∞, – 2] and [-1, ∞). (ii) decreasing on interval [-2,-1 ].

Question 30. Find the intervals in which the function f(x) = 2x 3 – 9x 2 + 12x -15 is (i) increasing. (ii) decreasing. (Delhi 2011c) Answer: (i) The function increasing on intervals (- ∞, 1]and [2, ∞). (ii) The function decreasing on interval [1, 2]

Question 31. Find the intervals in which the function f(x) = 2x 3 – 15x 2 + 36x + 17 is increasing or decreasing. (All India 2010c) Answer: The function increasing on (- ∞, 2] and [3, ∞) and decreasing on [2, 3]

Question 32. Find the intervals in which the function f(x) = 2x 3 – 9x 2 + 12x + 15 is (i) increasing. (ii) decreasing. (All India 2010C) Answer: (i) The function increasing on (- ∞,1] and [2, ∞). (ii) The function decreasing on [1, 2]

Question 34. Find the intervals in which the function f(x) = sin 3x – cos 3x, 0 < x < π, is strictly increasing or strictly decreasing. (Delhi 2016) Answer: Given, f(x) = sin 3x – cos 3x, 0 < x < π On differentiating both sides w.r.t. x, we get f'(x) = 3cos 3x + 3sin 3x

While f'(x) < 0 in \(\frac{\pi}{4}\) < x < \(\frac{7\pi}{12}\) and \(\frac{11 \pi}{12}\) < x < π So, f(x) is strictly decreasing in the intervals \(\left(\frac{\pi}{4}, \frac{7 \pi}{12}\right)\) and (\(\frac{11 \pi}{12}\), π)

Question 35. Prove that the function f defined by f(x) = x 2 – x + 1 is neither increasing nor decreasing in (-1, 1). Hence, find the intervals in which f(x) is (i) strictly increasing. (ii) strictly decreasing. (Delhi 2014C) Answer: Given function is f(x) = x 2 – x + 1. On differentiating both sides w.r.t. x, we get f'(x) = 2x – 1 On putting f'(x) = 0 ⇒ 2x – 1 = 0 ⇒ x = \(\frac{1}{2}\)

Now, we find intervals in which f(x) is strictly increasing or strictly decreasing.

Here, f(x) is strictly increasing on (\(\frac{1}{2}\), ∞) and f(x) is strictly decreasing on (- ∞, \(\frac{1}{2}\)) ⇒ f(x) is strictly increasing on (\(\frac{1}{2}\), 1) and f(x) is strictly decreasing on (-1, \(\frac{1}{2}\)) ∴ f'(x) does not have same sign throughout the interval (-1, 1). Thus, f(x) is neither increasing nor decreasing in (-1, 1).

Question 36. Find the intervals in which the function f(x) = 20 – 9x + 6x 2 – x 3 is (i) strictly increasing. (ii) strictly decreasing. (All India 2010) Answer: Given function is f(x) = 20 – 9x + 6x 2 – x 3 . On differentiating both sides w.r.t. x, we get f'(x) = – 9 + 12x – 3x 2 On putting f'(x) = 0, we get -9 + 12x – 3x 2 = 0 ⇒ -3(x 2 – 4x + 3) = 0 ⇒ -3(x – 1)(x – 3) = 0 ⇒ (x – 1)(x – 3) = 0 ⇒ x – 1 = 0 or x – 3 = 0 ⇒ x = 1 or 3 Now, we find intervals in which f(x) is strictly increasing or strictly decreasing.

We know that, a function f(x) is said to be strictly increasing when f'(x) > 0 and it is said to be strictly decreasing, if f'(x) < 0. So, the given function f(x) is (i) strictly increasing on the interval (1, 3) and (ii) strictly decreasing on the intervals (-∞, 1) and (3, ∞).

Tangents and Normals

Since, point (x 1 , y 1 ) is on the curv, therefore y 1 = \(\sqrt{3 x_{1}-2}\) ⇒ y 1 = \(\sqrt{3 \times \frac{41}{48}-2}=\frac{3}{4}\) Hence, the point is (x 1 , y 1 )

Now, equation of the tangent is given by y – y 1 = m(x – x 1 ) ⇒ y – \(\frac{3}{4}\) = 2(x – \(\frac{41}{48}\)) ⇒ \(\frac{4 y-3}{4}=\frac{48 x-41}{24}\) ⇒ \(\frac{24}{4}\)(4y – 3) = 48x – 41 ⇒ 6(4y – 3) = 48x – 41 ⇒ 48x – 41 – 24y + 18 = 0 ⇒ 48x – 24y = 23, which is the required equation of the tangent.

On differentiating w.r.t. x, we get 2x = 4\(\frac{d y}{d x}\) ⇒ \(\frac{d y}{d x}=\frac{2 x}{4}=\frac{x}{2}\)

Slope of normal = \(-\frac{1}{\frac{d y}{d x}}=-\frac{1}{\frac{x}{2}}=-\frac{2}{x}\) Let (h, k) be the point where normal and curve intersect.

Question 3. Find the equations of the tangent and the normal to the curve 16x 2 + 9y 2 = 145 at the point (x 1 , y 1 ), where x 1 = 2 and y 1 > 0. (CBSE 2018) Answer: Given equation of curve is 16x 2 + 9y 2 = 145 Clearly, when x = 2, then 16(2) 2 + 9y 2 =145 9y 2 = 145 – 64 ⇒ 9y 2 = 81 ⇒ y 2 = 9 ⇒ y = ±3 [taking square root on both sides] But it is given that y 1 > 0 y = 3

Question 4. Find the angle of intersection of the curves x 2 + y 2 = 4 and (x – 2) 2 + y 2 = 4, at the point in the first quadrant. (CBSE 2018C) Answer: Given curves are x 2 + y 2 = 4 and (x – 2) 2 + y 2 = 4 From Eqs. (i) and (ii), we get x 2 + 4 – 4x + y 2 = 4 ⇒ 4 – 4x = 0 ⇒ x = 1

Question 6. Find the equation of tangents to the curve y = x 3 + 2x – 4 which are perpendicular to the line x + 14y – 3 = 0. (All India 2016) Answer: Given equation of curve is y = x 3 + 2x – 4

On differentiating both sides w.r.t. x, we get \(\frac{d y}{d x}\) = 3x 2 + 2

∴ The slope of required tangent is m 1 = \(\frac{d y}{d x}\) = 3x 2 + 2

Now, slope of line x + 14y – 3 = 0 or y = \(-\frac{x}{14}+\frac{3}{14}\) is m 2 = \(-\frac{1}{14}\)

Since, the required tangent is perpendicular to the line x + 14y -3=0. ∴ m 1 m 2 = -1 ⇒ (3x 2 + 2) × \(\left(-\frac{1}{14}\right)\) = -1 ⇒ 3x 2 + 2 = 14 ⇒ 3x 2 = 12 ⇒ x 2 = 4 ⇒ x =± 2

When x = 2, then y = 2 3 + 2 × 2- 4= 8 + 4 – 4 = 8

When x = – 2, then y = (-2) 3 + 2 × (-2) – 4 = -8 – 4 – 4 = -16 ∴ Points of contact are (2, 8) and (- 2, -16).

Now, equation of tangent at point (2, 8) is ⇒ y – 8 = \(\left(\frac{d y}{d x}\right)_{(2,8)}\) (x – 2) ⇒ y – 8 = ( 3 × 2 2 + 2)(x – 2) ⇒ y – 8 = 14(x – 2) ⇒ y = 14x – 20

and equation of tangent at point (- 2, -16) is ⇒ y + 16 = \(\frac{d y}{d x}_{(-2,-16)}\) (x + 2) ⇒ y + 16 = [3 (- 2) 2 + 2](x + 2) ⇒ y + 16 = 14 (x + 2) ⇒ y = 14x + 12 Hence, the required equation of tangents are y = 14x – 20 and y = 14x + 12

Question 8. Find the point on the curve 9y 2 = x 3 , where the normal to the curve makes equal intercepts on the axes. (Foreign 2015) Answer: Given curve is 9y 2 = x 3 …(i) On differentiating both sides w.r.t. x, we get 9 × 2y\(\frac{d y}{d x}\) = 3x 2 ⇒ \(\frac{d y}{d x}=\frac{3 x^{2}}{18 y}=\frac{x^{2}}{6 y}\) Let (x 1 , y 1 ) be the required point on the curve (i).

Question 9. Find the equations of the tangent and normal to the curves x = a sin 3 θ and y = a cos 3 θ at θ = \(\frac{\pi}{4}\). (Delhi 2014) Answer: Given, x = a sin 3 θ On differentiating both sides w.r.t. θ, we get \(\frac{d x}{d \theta}\) = a(3sin 2 θ cos θ) ⇒ \(\frac{d x}{d \theta}\) = 3a sin 2 θ cos θ and y = a cos 3 θ

Question 10. Find the equations of the tangent and normal to the curves \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1 at the point (√2a, b). (Delhi 2014) Answer: Given equation of curves is \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1

On differentiating both sides w.r.t. x, we get \(\frac{2 x}{a^{2}}-\frac{2 y}{b^{2}} \cdot \frac{d y}{d x}\) = 0 ⇒ \(\frac{d y}{d x}=\frac{b^{2} x}{a^{2} y}\)

Slope of the tangent at point (√2a, b) is \(\left(\frac{d y}{d x}\right)_{(\sqrt{2} a, b)}=\frac{\sqrt{2} a b^{2}}{b a^{2}}=\frac{\sqrt{2} b}{a}\)

Hence, the equation of the tangent at point (√2a, b) is y – b = \(\frac{\sqrt{2} b}{a}\)(x – √2a) ⇒ a (y- b) = √2 b (x – √2a) ⇒ ay – ab = √2 bx – 2ab ⇒ √2bx – ay – ab = 0

Now, the slope of the normal at point (√2a, b) = \(\frac{-1}{\text { Slope of tangent }}=\frac{-1}{\sqrt{2} b / a}\)

Hence, the equation of the normal at point(√2a, b) is (y – b) = \(-\frac{a}{\sqrt{2} b}\)(x – √2a) ⇒ √2b(y – b) = -a (x – √2a) ⇒ √2 by – √2 b 2 = -ax + √2 a 2 ∴ ax + √2by – √2 (a 2 + b 2 ) = 0

Question 11. Find the points on curve y = x 3 – 11x + 5 at which equation of tangent is y = x – 11. (Delhi 2012C) Answer: First, find the slope of tangent to the given curve and of given equation of tangent, then equate them to get value of x. Put value of x in given curve to find required points.

Given, equation of curve is y = x 3 – 11x + 5 ………(i)

Slope of the tangent at any point (x, y) is given by \(\frac{d y}{d x}\) \(\frac{d y}{d x}\) = 3x – 11 ………….(ii)

Also, slope of the tangent y = x – 11 is 1. ∴ \(\frac{d y}{d x}\) = 1 ⇒ 3x 2 – 11 = 1 [frpm Eq. (i)] ⇒ 3x 2 = 12 ⇒ x 2 = 4 ⇒ x = ±2

When x = 2, then y = (2) 3 -11(2) + 5 = 8 – 22 + 5 = – 9 When x = – 2, then y = (-2) 3 -11 (-2) + 5 = -8 + 22 + 5 = 19 Since, the points (-2, 19) does not lies on the line y = x – 1 Hence, the required points on the curve are (2, -9).

Question 12. Find the points on the curve x 2 + y 2 – 2x – 3 = 0 at which tangent is parallel to X-axis. (Delhi 2011) Answer: As tangent is parallel to X-axis, put \(\frac{d y}{d x}\) = 0 and find value of x from it. Then, put this value of x in the equation of the given curve and find value of y.

Given equation of curve is x 2 + y 2 – 2x – 3 = 0

On putting \(\frac{d y}{d x}\) = 0, we get 1 – x = 0 ⇒ x = 1 Now, on putting x = 1 in Eq. (i), we get 1 + y 2 – 2 – 3 = 0 y 2 = 4 ⇒ y = ±2 Hence, the required points are (1, 2) and (1, -2).

Question 13. Find the points on the curve y = x 3 at which the slope of the tangent is equal to y-coordinate of the point. (Foreign 2011) Answer: First, determine the derivative and put \(\frac{d y}{d x}\) = y and then find the value of x from it. Further, put this value of x in the equation of the given curve and find the value of y.

Given equation of curve is y = x 3 . …….(i) On differentiating both sides w.r.t. x, we get \(\frac{d y}{d x}\) = 3x 2

∴ Slope of tangent at any point (x, y) is \(\frac{d y}{d x}\) = 3x 2

Now, given that Slope of tangent = y-coordinate of the point ⇒ \(\frac{d y}{d x}\) = y ⇒ 3x 2 = y [∵\(\frac{d y}{d x}\) = 3x 2 ] ⇒ 3x 2 = x 3 [∵ y = x 3 ] ⇒ 3x 2 – x 3 = 0 ⇒ x 3 (3 – x) = 0 ⇒ Either x 2 = 0 or 3 – x = 0 x = 0, 3

Now, on putting x = 0 and 3 in Eq. (i), we get y = (0) 3 = 0 [at x = 0] and y = (3) 3 = 27 Hence, the required points are (0, 0) and (3, 27).

Question 17. Find the equation of tangent to the curve y = x 4 – 6x 3 + 13x 2 – 10x + 5 at point x = 1, y = 0. (Delhi 2011C) Answer: Given equation of curve is y = x 4 – 6x 3 + 13x 2 – 10x + 5

On differentiating both sides w.r.t. x, we get \(\frac{d y}{d x}\) = 4x 3 – 18x 2 + 26x – 10

Slope of a tangent at point (1, 0) is m = \(\left[\frac{d y}{d x}\right]_{x=1}\) = 4 – 18 + 26 – 10 = 2

∴ Equation of tangent at point (1,0) having slope 2 is y – 0 = 2(x – 1) ⇒ y = 2x – 2 Hence, required equation of tangent is 2x – y = 2

Question 18. Find the points on the curve y = [x(x – 2)] 2 , where the tangent is parallel to X-axis. (Delhi 2010) Answer: We have to find the points on the given curve where the tangent is parallel to X-axis. We know that, when a tangent is parallel to X-axis, then \(\frac{d y}{d x}\) = 0 ⇒ \(\frac{d}{d x}\)(x 2 – 2x) 2 = 0 ⇒ 2(x 2 – 2x)(2x – 2) = 0 ⇒ x = 0, 1, 2 When x = 0, then y = [0(-2)] 2 = 0 When x = 1, then y = [1 – 2(1)] 2 = 1 When x = 2, then y = [2 2 – 2 × 2] 2 = 0 Hence, the tangent is parallel to X-axis at the points (0, 0), (1, 1) and (2, 0).

Question 19. Find the equation of tangent to the curve y = \(\frac{x-7}{x^{2}-5 x+6}\) at the point, where it cuts the X-axis. (All India 2010C, 2010) Answer: Given equation of curves is y = \(\frac{x-7}{x^{2}-5 x+6}\) ……….(i)

Hence, the required equation of tangent passing through the point (7, 0) having slope 1/20 is y – o = \(\frac{1}{20}\)(x – 7) ⇒ 20y = x – 7 ∴ x – 20y = 7

Question 20. Find the equations of the normal to the curve y = x 3 + 2x + 6, which are parallel to line x + 14y + 4 = 0. (Delhi 2010) Answer: First, find the slope of normal to curve, i.e. \(\frac{-1}{d y / d x}\) and put them equal to the slope of line, simplify them and find the values of x and y. Further, use the equation y – y 1 = Slope of normal (x – x 1 ).

Given equation of curve is y = x 2 + 2x + 6 ……..(i) and the given equation of line is x + 14y + 4= 0

On differentiating both sides of Eq. (i) w.r.t. x, we get \(\frac{d y}{d x}\) = 3x 2 + 2 Slope of normal = \(\frac{-1}{\left(\frac{d y}{d x}\right)}=\frac{-1}{3 x^{2}+2}\)

Also, slope of the line x + 14y + 4 = 0 is – \(\frac{1}{14}\) [∵ slope of the line Ax + By + C = 0 is – \(\frac{A}{B}\)] We know that, if two lines are parallel, then their slopes are equal. ∴ \(-\frac{1}{3 x^{2}+2}=-\frac{1}{14}\) ⇒ 3x 2 + 2 = 14 ⇒ 3x 2 = 12 ⇒ x 2 = 4 ⇒ x = ± 2

When x = 2, then from Eq. (i), y = (2) 3 + 2(2) + 6 = 8 + 4 + 6 = 18 and when x = – 2, then from Eq. (i), y = (-2) 3 + 2(-2) + 6 = -8 – 4 + 6 = – 6

∴ Normal passes through (2, 18) and (-2, -6). Also, slope of normal = – \(\frac{1}{14}\) Hence, equation of normal at point (2, 18) is y – 18 = – \(\frac{1}{14}\)(x – 2) ⇒ 14y – 252 = – x + 2 ⇒ x + 14 y = 254

and equation of normal at point (-2, -6) is y + 6 = – \(\frac{1}{14}\) (x + 2) ⇒ 14y + 84 = -x – 2 ⇒ x + 14y = -86 Hence, the two equations of normal are x + 14y = 254 and x + 14y = – 86.

Question 21. Find the angle of intersection of the curves y 2 = 4ax and x 2 = 4by. (Foreign 2016) Answer: Given equations of curves are y 2 = 4ax …(i) and x 2 = 4 by … (ii) Clearly, the angle of intersection of curves (i) and (ii) is the angle between the tangents to the curves at the point of intersection. So, let us first find the intersection point of given curves.

On substituting the value of y from Eq. (ii) in Eq. (i), we get \(\left(\frac{x^{2}}{4 b}\right)^{2}\) = 4ax ⇒ \(\frac{x^{4}}{16 b^{2}}\) = 4ax ⇒ x 4 = 64ab 2 x ⇒ x 4 – 64ab 2 x = 0 ⇒ x(x 2 – 64ab 2 ) = 0 x = 0 or x = 4a 1/3 b 2/3 Clearly, when x = 0, then from Eq. (i), y = 0 and when x = 4a 1/3 b 2/3 , then from Eq. (i), y 2 =16a 4/3 b 2/3 ⇒ y = 4a 2/3 b 1/3

Thus, the points of intersection are (0, 0) and (4a 1/3 b 2/3 , 4a 2/3 b 1/3 ). Now, let us find the angle of intersection at (0, 0) and (4a 1/3 b 2/3 , 4a 2/3 b 1/3 ). Let be the slope of tangent to the curve (i) and m 2 be the slope of tangent to the curve (ii).

On squaring and adding Eqs. (iii) and (iv), we get cos 2 (x 1 + y 1 ) + sin 2 (x 1 + y 1 ) = 1 + y 1 2 1 + y 2 1 = 1 ⇒ y 2 1 = o ⇒ y 1 = o

On putting y 1 = 0 in Eq. (iii), we get cosx 1 = 0 ⇒ x 1 = \(\frac{\pi}{2}, \frac{3 \pi}{2}, \frac{-\pi}{2}, \frac{-3 \pi}{2}\) [∵ -2π ≤ x ≤ 2π] But only x 1 = \(\frac{\pi}{2}\) and \(\frac{-3 \pi}{2}\) satisfy Eq. (iv).

Hence, the points of contact are (\(\frac{\pi}{2}\), o) and (\(\frac{-3 \pi}{2}\), 0) ∴ Equations of tangents are y – 0 = –\(\frac{1}{2}\)(x – \(\frac{\pi}{2}\)) and y -0 = –\(\frac{1}{2}\)(x + \(\frac{-3 \pi}{2}\))

Question 23. Find the value of p for which the curves x 2 = 9p (9 – y) and x 2 = p (y + 1) cut each other at right angles. (All India 2015) Answer: Given equations of curves are x 2 = 9p (9 – y) …….(i) and x 2 = p (y + 1) ………….(ii)

As, these curves cut each other at right angle, therefore their tangent at point of intersection are perpendicular to each other. So, let us first find the point of intersection and slope of tangents to the curves. From Eqs. (i) and (ii), we get 9p(9 – y) =p(y + 1) ∴ 9(9 – y) = y + 1 [∵ p ≠ 0, as if p = 0, then curves becomes straight, which will be parallel] ⇒ 81 – 9y = y + 1 ⇒ 80 = 10y ⇒ y = 8

On substituting the value of yin Eq. (i),we get x 2 = 9p ⇒ x = + 3 Thus, the point of intersection are (3√p, 8) and (-3√p, 8)

Now, consider Eq.(i),we get \(\frac{x^{2}}{9 p}\) = 9 – y ⇒ y = 9 – \(\frac{x^{2}}{9 p}\)

On differentiating both sides w.r.t. x, we get \(\frac{d y}{d x}=\frac{-2 x}{9 p}\) ……(iii)

From Eq. (ii), we get \(\frac{x^{2}}{p}\) = y + 1 ⇒ y = \(\frac{x^{2}}{p}\) – 1

On differentiating both sides w.r.t. x, we get \(\frac{d y}{d x}=\frac{2 x}{p}\) ………(iv)

Now, for intersection point (3√p, 8), we have slope of tangent to the first curve = \(\frac{-2(3 \sqrt{p})}{9 p}=\frac{-6 \sqrt{p}}{9 p}\) [using Eq. (iii)]

and slope of tangent to the second curve = \(\frac{2(3 \sqrt{p})}{p}=\frac{6 \sqrt{p}}{p}\) [using Eq. (iv)]

∵ Tangents are perpendicular to each other. Then, Slope of first curve x Slope of second curve = -1 ∴ \(\frac{-6 \sqrt{p}}{9 p} \times \frac{6 \sqrt{p}}{p}\) = -1 ⇒ \(\frac{4}{p}\) = 1 ⇒ p = 4

And for intersection point (3√p, 8), we have slope of tangent to the first curve = \(\frac{-2(-3 \sqrt{p})}{9 p}=\frac{6 \sqrt{p}}{9 p}\) [using Eq. (iii)]

and slope of tangent to the second curve = \(\frac{2(-3 \sqrt{p})}{p}=\frac{-6 \sqrt{p}}{p}\) [using Eq. (iv)]

Tangents are perpendicular to each other. Then, \(\frac{6 \sqrt{p}}{9 p} \times \frac{-6 \sqrt{p}}{p}\) = -1 [∵ m 1 m 2 = -1] ⇒ \(\frac{4}{p}\) = 1 ⇒ p = 4 Hence, the value of p is 4.

Question 24. Find the equations of the tangent to the curve y = x 2 – 2x + 7 which is (i) parallel to the line 2x – y + 9 = 0. (ii) perpendicular to the line 5y – 15x = 13. (Delhi 2014C) Answer: Given equation of curve is y = x 2 – 2x + 7

On differentiating both sides w.r.t. x, we get \(\frac{d y}{d x}\) = 2x – 2

(i) Given, equation of the line is 2x – y + 9 = 0 ⇒ y = 2x + 9 which is of the form y = mx + c.

∴ Slope of the line is m = 2 If a tangent is parallel to the line, then slope of tangent is equal to the slope of the line. Therefore, \(\frac{d y}{d x}\) = m ⇒ 2x – 2 = 2 ⇒ x = 2 When x = 2, then from Eq. (i), we get y = 2 2 – 2 × 2 + 7 ⇒ y = 7 The point on the given curve at which tangent is parallel to given line is (2, 7) and the equation of the tangent is y – 7 = 2(x – 2) [∵ y – y 1 =m(x- x 1 )] ⇒ 2x – y + 3 = 0 Hence, the equation of the tangent line to the given curve which is parallel to line 2x – y + 9 = 0 is y – 2x – 3 = 0.

(ii) The equation of the given line is 5y – 15x = 13 ⇒ y = \(\frac{15 x+13}{5}\) = 3x + \(\frac{13}{5}\) which is of the form y = mx + c

Question 25. Find the equation of the normal at a point on the curve x 2 = 4y, which passes through the point (1, 2). Also, find the equation of the corresponding tangent. (Delhi 2013) Answer: Given curve is x 2 = 4y. On differentiating both sides w.r.t. x,we get 2x = 4\(\frac{d y}{d x}\) ⇒ \(\frac{d y}{d x}\) = \(\frac{d y}{d x}\) Let (h, k) be the coordinates of the point of contact of the normal to the curve x2 = 4y. Then, slope of the tangent at (h, k) is given by \(\left(\frac{d y}{d x}\right)_{(h, k)}=\frac{h}{2}\) and slope of the normal ay (h, k) = \(\frac{-1}{d y / d x}=\frac{-2}{h}\)

Therefore, the equation of normal at (h, k) is y – k = \(\frac{-2}{h}\)(x – h) ………….(i) [∵ Equation of normal in slope form is y – y 1 = \(\frac{-1}{m}\)(x – x 1 )] Since, it passes through the point (1, 2), so on putting x =1 and y = 2, we get 2 – k = \(\frac{-2}{h}\)(1 – h) ⇒ k = 2 + Since, (h, k) also lies on the curve x 2 = 4y, therefore h 2 = 4k ………(iii)

On solving Eqs.(ii) and (iii), we get h = 2 and k = 1

Substituting the values of h and k in Eq.(i), the required equation of normal is y – 1 = \(\frac{-2}{2}\)(x – 2) ⇒ x + y = 3

Now, equation of tangent at (h, k) is y – k = \(\frac{h}{2}\)(x – h)

On putting h = 2 and k = 1, we get y – 1 = \(\frac{2}{2}\)(x – 2) ⇒ y – 1 = x – 2 ∴ y = x – 1

Question 26. Find the equations of tangents to the curve 3x 2 – y 2 = 8, which passes through the point [\(\frac{4}{3}\), 0 ]. (All India 2013) Answer: First, differentiate the given curve with respect to x and determine \(\frac{d y}{d x}\). Then, find the equation of tangent at (x, y,}. Now, as this equation is passes through given point (x0, y0), so this point will satisfy the tangent and curve also. Further, simplify it and get the required equations of tangent.

Given equation of curve is 3x 2 – y 2 = 8

Now, on solving Eqs. (iii) and (iv), we get 4h = 8 ⇒ h = 2

On putting h = 2 in Eq. (iv), we get 3 (2) 2 – k 2 = 8 ⇒ k 2 = 4 ⇒ k = ±2

Now, putting the values of h and k in Eq. (h), we get ⇒ y = (±2) = \(\frac{3(2)}{\pm 2}\)(x – 2) ⇒ y + 2 = ±3(x – 2) ⇒ y = ± 3(x – 2) ± 2 ⇒ y = ± {3(x – 2) + 2} ⇒ y = ± (3x – 6 + 2) ⇒ y = ± (3x – 4) Hence, y = – 3x + 4 and y = 3x – 4 are two required equations of tangent.

Question 27. Find all the points on the curve y = 4x 3 – 2x 5 at which the tangent passes through the origin. (Delhi 2013C) Answer: Given curve is y = 4x 3 – 2x 5 . ………(i)

Let any point on the curve be (x 1 ,y 1 ). So, it satisfies Eq. (i). ∴ y 1 = 4x 1 3 – 2x 1 5 ………(ii)

On differentiating both sides of Eq. (i), we get \(\frac{d y}{d x}\) = 12x 2 – 10x 4

Equation of tangent at point (x 1 , y 1 ) is y – y 1 = \(\left(\frac{d y}{d x}\right)_{\left(x_{1}, y_{1}\right)}\) (x – x 1 ) ⇒ y – y 1 = [12(x 1 ) 2 – 10(x 1 ) 4 ](x – x 1 )

Since, it passes through the origin. ∴ 0 – y 1 = (12x 1 2 – 10x 1 4 )(0 – x 1 ) ⇒ y 1 = (12x 1 2 – 10x 1 4 )x 1 ………(iii)

From Eqs. (ii) and (iii), we get (12x 2 1 – 10x 4 1 )x 1 = 4x 1 3 – 2x 1 5 ⇒ 2x 1 3 (6 – 5x 1 2 ) = 2x 1 3 (2 – x 1 2 ) ⇒ 2x 1 3 (4 – 4x 1 2 ) = 0 ⇒ x 1 = 0 or 4 – 4x 1 2 = 0 ∴ x 1 = 0 or x 1 = 1

On putting the values of x, = 0,1 and -1 respectively in Eq. (ii), we get At x 1 = 0, y 1 = 0 At x 1 = 1, y 1 = 4(1) 3 -2(1) 5 = 4 – 2 = 2 and at x 1 = (-1), y 1 = 4(-1) 3 – 2(-1) 5 = 4 (-1) – 2 (-1) = -4+ 2 = -2 Hence, all points on the curve at which the tangent passes through origin, are (0, 0), (1, 2) and (-1,-2).

Question 28. Prove that the curves x = y 2 and xy = k cut at the right angles, if 8k 2 = 1. (Delhi 2013C) Answer: Given equations of curves are x = y 2 …(i) and xy = k …(ii)

On cubing both sides, we get 8k 2 = 1 Hence Proved.

Question 29. Prove that all normals to the curves x = a cos t + at sin t and y = a sin t – at cos t are at a constant distance ‘a’ from the origin. (All India 2013C) Answer: Given equations of curves are x = a cost + at sin t and y = asin t – at cost

On differentiating x and y separately w.r.t. t, we get \(\frac{d x}{d t}\) = -asin t + a(t cost + sin t) = -a sin t + a tcos t+ a sin t = at cost and \(\frac{d y}{d t}\) = acos t – a[t(-sin t) + cos t)] = a cost + at sin t – acost = at sin t

Now \(\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{a t \sin t}{a t \cos t}\) = tan t = Slope of normal at any point t = \(\frac{-1}{\tan t}\) = -cot t

Now, equation of normal at any point t is given by y – (asin t – at cos t) = – cot t[x – (a cos t + atsint)] ⇒ y – asin t + at cos t = \(\frac{-\cos t}{\sin t}\) (x – acost – at sin t) ⇒ y sint – a sin 2 t + at cost sint = -x cost + acos 2 t + at sint cost ⇒ x cost + y sint = a(cos 2 t + sin 2 t) ⇒ x cost + ysint = a [∵ sin 2 θ + cos 2 θ = 1] ∴ xcos t + ysin t – a = 0 Now, the distance of normal from the origin is given by \(\frac{|0 \cdot \cos t+0 \cdot \sin t-a|}{\sqrt{\cos ^{2} t+\sin ^{2} t}}=\frac{|-a|}{\sqrt{1}}\) = a

Question 30. Find the equations of tangent and normal to the curve x = 1 – cos θ, y = θ – sin θ at θ = \(\frac{\pi}{4}\) Answer: First, differentiate the given curve with respect to and then determine \(\frac{d y}{d x}\) at θ = \(\frac{\pi}{4}\). Further, use the formula, equation of tangent at (x 1 , y 1 ) is y – y 1 = m(x – x 1 ) and equation of normal at (x 1 , y 1 ) is y – y 1 = –\(\frac{1}{m}\) (x – x 1 ).

Maxima and Minima

Question 2. The sum of the perimeters of a circle and square is k, where k is some constant. Prove that the sum of their areas is least, when the side of the square is double the radius of the circle. (Delhi 2014C) Answer: Let r be the radius of circle and x be the side of a square. Then, given that Perimeter of square + Perimeter of circle = k (constant) i.e. 4x + 2πr = k ⇒ x = \(\frac{k-2 \pi r}{4}\) ……(i)

Let A denotes the sum of their areas. ∴ Area, A = area of a square + area of circle ∴ A = x 2 + πr 2 …(ii)

Question 3. A tank with rectangular base and rectangular sides, open at the top is to be constructed, so that its depth is 2 m and volume is 8 m 3 . If building of tank cost ₹ 70 per sq m for the base and ₹ 45 per sq m for sides. What is the cost of least expensive tank? (Delhi 2019) Answer: Let x m be the length, y m be the breadth and h=2m be the depth of the tank. Let ₹ H be the total cost for building the tank. Now, given that h = 2 m and volume of tank = 8 m 3 . Clearly, area of the rectangular base of the tank = length × breadth = xy m 2

and the area of the four rectangular sides = 2 (length + breadth) × height = 2 (x + y) × 2= 4 (x + y) m 2

∴ Total cost, H = 70 × xy + 45 × 4 (x + y) ⇒ H = 70xy + 180 (x + y) …………(i)

Also, volume of tank = 8 m 3 ⇒ l × b × h = 8 ⇒ x × y × 2 = 8 ⇒ y = \(\frac{4}{x}\) …..(ii)

Question 6. Show that the altitude of the right circular cone of maximum volume that can he inscribed in a sphere of radius r is \(\frac{4r}{3}\) Also, find the maximum volume in terms of volume of the sphere. (Delhi 2019, 2016) Or Show that the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius r is \(\frac{4r}{3}\). Also, show that the maximum volume of the cone is \(\frac{8}{27}\) of the volume of the sphere. (All India 2014) Answer: Let R be the radius and h be the height of the cone, which inscribed in a sphere of radius r. ∴ OA = h – r

In ΔOAB, by Pythagoras theorem, we have r 2 = R 2 + (h – r) 2 ⇒ r = R 2 + h 2 + r 2 – 2rh ⇒ R 2 = 2rh – h 2 ……….(i)

Note that, to admit maximum light, area of window should be maximum. Here, area of window A = area of rectangle + area of semicircular region = 2x × y + \(\frac{1}{2}\)πx 2 ⇒ A = 2x\(\left(\frac{10-x(\pi+2)}{2}\right)\) + \(\frac{1}{2}\) πx 2 [from Eq. (i)] ⇒ A = 10x – x 2 (π + 2) + \(\frac{1}{2}\) πx 2

On differentiating both sides w.r.t. x, we get dA \(\frac{d A}{d x}\) = 10 – 2x (π + 2) + πx = 10- 2x π – 4x + πx = 10 – πx – 4x ……(iii)

For maximum, Put \(\frac{d A}{d x}\) = 0 ⇒ 10 = πx + 4x ⇒ x = \(\frac{10}{\pi+4}\)

Question 8. Show that the surface area of a closed cuboid with square base and given volume is minimum, when it is a cube. (All India 2017) Answer: Let V be the fixed volume of a closed cuboid with length x, breadth x and height y. Then, V = x × x × y ⇒ y = \(\frac{V}{x^{2}}\) …………(i)

Let C denotes the cost of the box. C = 2x 2 × 5 + 4xy × 2.50 = 10x 2 + 10xy = 10x (x + y) = 10x(x + \(\frac{1024}{x^{2}}\)) = 10x 2 + \(\frac{10240}{x}\) ……..(i)

Now, when θ = tan -1 √2, then tan 2 θ = 2 ⇒ sin 2 θ = 2cos 2 θ

Now, perimeter (P) of ΔABC = AB + BC + AC = AE + BE + BD + DC + AF + FC = (AE +AF) + (BE + BD + DC + FC) = 2AE + 4BD …(i)

Consider ΔOEA, in this we have AE = \(\frac{O E}{\tan \theta}=\frac{r}{\tan \theta}\) and OA = \(\frac{r}{\sin \theta}\) and in ΔADB we have BD = AD tan θ = (AO + OD) tanθ = [\(\frac{r}{\sin \theta}\) + r)tan θ Now, P = 2.\(\frac{r}{\sin \theta}\) + 4(\(\frac{r}{\sin \theta}\) + r)tan θ [From Eq. (i)] ⇒ P(θ) = r(2 cot θ + 4 sec θ + 4 tan θ) …….(ii)

Question 15. Find the local maxima and local minima of the function f(x) = sin x – cos x, 0 ≤ x ≤ 2π. Also, find the local maximum and local minimum values. (Delhi 2015) Answer: We have, f(x) = sin x – cos x, 0< x < 2π On differentiating both sides w.r.t. x, we get f'(x) = cosx + sinx ……….(i) For local maximum and local minimum, Put f'(x) = 0, i.e. cosx + sinx = 0 ⇒ cosx = – sinx ⇒ tanx = -1 ⇒ x = π – \(\frac{\pi}{4}\) or 2π – \(\frac{\pi}{4}\) ⇒ x = \(\frac{3 \pi}{4}\) or \(\frac{7 \pi}{4}\)

Question 18. A point on the hypotenuse of a right triangle is at distances a and b from the sides of the triangle. Show that the minimum length of the hypotenuse is (a 2/3 + b 2/3 ) 3/2 . (Delhi 2015C) Answer: Let P be a point on the hypotenuse AC of right angled ΔABC. Such that PL ⊥ AB and PL = a and PM ⊥ BC and PM = b. Let ∠APL = ∠ACB = 6 [say] Then, AP = a sec θ, PC = b cosec θ

Let l be the length of the hypotenuse, then l = AP + PC ⇒ l = a sec θ + b cosec θ, 0 < θ< \(\frac{\pi}{2}\)

On putting V 2 = 2π 2 r 6 /9 in Eq. (i), we get 2π 2 r 6 = π 2 r 4 h 2 ⇒ 2r 2 = h 2 ⇒ h = √2r ⇒ \(\frac{h}{r}\) = √2 ⇒ cot θ = √2 [from the figure, cot θ = \(\frac{h}{r}\).] ∴ θ = cot -1 √2 Hence, the semi-vertical angle of the right circular cone of given volume and least curved surface area is cot -1 √2. Hence proved.

Note: If square of any area is maximum (or minimum), then area is also maximum (or minimum).

Question 29. Show that the right circular cone of least curved surface and given volume has an altitude equal to √2 times the radius of the base. (Delhi 2011) Answer: Let C denotes the curved surface area, r be the radius of base, h be the height and V be the volume of right circular cone. We know that, volume of cone is given by V = \(\frac{1}{3}\)πr 2 h ⇒ h = \(\frac{3 V}{\pi r^{2}}\) …(i) Also, the curved surface area of cone is given by C = ml, where l = \(\sqrt{r^{2}+h^{2}}\) is the slant height of cone. ∴ C = πr\(\sqrt{r^{2}+h^{2}}\)

Let A be the area of rectangle. ∴ A = (2x) (2y) [∵ area of rectangle = length x breadth] ⇒ A = 4 xy ⇒ A = 4x\(\sqrt{b^{2}-x^{2}}\) [∵ y = \(\sqrt{b^{2}-x^{2}}\)]

Question 32. Show that of all the rectangles with a given perimeter, the square has the largest area. (Delhi 2011) Answer: Let x and y be the lengths of two sides of a rectangle. Again, let P denotes its perimeter and A be the area of rectangle. Then, P = 2 (x + y) [∵ perimeter of rectangle = 2(l + b)] ⇒ P = 2x + 2y ⇒ y = \(\frac{P-2 x}{2}\) …..(i)

We know that, area of rectangle is given by A = xy ⇒ A = x\(\left(\frac{P-2 x}{2}\right)\) [by using Eq. (i)] ⇒ A = \(\left(\frac{P-2 x}{2}\right)\)

On differentiating both sides w.r.t. x, we get \(\frac{d A}{d x}=\frac{P-4 x}{2}\)

For maxima or minima, put \(\frac{d A}{d x}\) = 0 ⇒ \(\frac{P-4 x}{2}\) = 0 ⇒ P = 4x ⇒ 2x + 2y = 4x [∵ P = 2x + 2y] ⇒ x = y

So, the rectangle is a square. Also, \(\frac{d^{2} A}{d x^{2}}=\frac{d}{d x}\left(\frac{P-4 x}{2}\right)\) = \(-\frac{4}{2}\) = -2 < 0 ⇒ A is maximum. Hence, area is maximum, when rectangle is a square. Hence proved.

Question 35. Find the point on the curve y 2 = 2x which is at a minimum distance from the point (1, 4). (All India 2011) Answer: First, consider any point on the curve, use the, formula of distance between two points. Then, square both sides and eliminate one variable with i the help of given equation. Further, apply concept : of maxima and minima to find the required point.

On differentiating both sides w.r.t. y, we get \(\frac{d Z}{d y}=\frac{4 y^{3}}{4}\) – 8 = y 3 – 8

For maxima or minima, put \(\frac{d Z}{d y}\) = 0 ⇒ y 3 – 8 = 0 ⇒ y 3 = 8 ⇒ y = 2 Also, \(\frac{d^{2} Z}{d y^{2}}=\frac{d}{d y}\)(y 3 – 8) = 3y 2

On putting y = 2, we get \(\left(\frac{d^{2} Z}{d y^{2}}\right)_{y=2}\) = 3(2) 2 = 12 >0 \(\frac{d^{2} Z}{d y^{2}}\) > 0

∴ Z is minimum and therefore PQ is also minimum as Z = PQ 2 . On putting y = 2 in the given equation, i.e. y 2 = 2x, we get (2) 2 = 2x ⇒ 4 = 2x ⇒ x = 2 Hence, the point which is at a minimum distance from point (1, 4) is P (2, 2).

Question 36. A wire of length 28 m is to be cut into two pieces. One of the two pieces is to be made into a square and the other into a circle. What should be the lengths of two pieces, so that the combined area of circle and square is minimum? (All India 2010) Answer: First, find length of circular part (or its circumference) and calculate the length of square part (or its perimeter). Add these two terms and equate it to 28 m and apply second derivative test to check minimum.

Let x m be the side of the square and r be the radius of circular part. Then, Length of square part = perimeter of square = 4 × Side = 4 × and length of circular part = circumference of circle = 2πr Given, length of wire = 28 ⇒ 4x + 2πr = 28 ⇒ 2x + πr = 14 ∴ x = \(\frac{14-\pi r}{2}\) ……(i)

Question 37. Show that the volume of the largest cone that can be inscribed in a sphere of radius R is \(\frac{8}{27}\) of the volume of the sphere. (Delhi 2010C) Answer: Let R be the radius and h be the height of the cone, which inscribed in a sphere of radius r. ∴ OA = h – r

Question 38. Find the maximum area of an isosceles triangle inscribed in the ellipse \(\frac{x^{2}}{25}+\frac{y^{2}}{16}\) = 1, with its vertex at one end of the major axis. (Delhi 2010c) Answer: Given equation of ellipse is \(\frac{x^{2}}{25}+\frac{y^{2}}{16}\) = 1

Note: If A 2 is maximum/minimum, then A is also maximum/minimum.

Question 39. Show that the right circular cylinder, open at the top and of given surface area and maximum volume is such that its height is equal to the radius of the base. (Delhi 2010) Answer: Let V be the volume, S be the total surface area of a right circular cylinder which is open at the top. Again, let r be the radius of base and h be the height. Now, S = 2πrh + πr 2 [∵ cylinder is open at top] ⇒ h = \(\frac{S-\pi r^{2}}{2 \pi r}\) ……….(i)

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Class 12 Maths Case Study Questions

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Download the app to get CBSE Sample Papers 2023-24, NCERT Solutions (Revised), Most Important Questions, Previous Year Question Bank, Mock Tests, and Detailed Notes.

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

Why Case Studies in CBSE Syllabus?

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

Case Study Questions in Maths

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

Focus on concepts

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

Easy Questions with a Practical Approach

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

Practice Questions Regularly

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

12 Maths Case-Based Questions

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

Case Study Question – 1

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

Case Study Question – 2

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

Case Study Questions Matrices -1

Answer Key:

Case Study Questions Matrices – 2

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

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

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

• (a) 3  ×  3

Case Study Questions Determinants – 01

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

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

• (b) Shyam Lal
• (a) Ram Lal

Case Study Questions Determinants – 02

Case study questions application of derivatives.

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

Case Study Questions Vector Algebra

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

More Case Study Questions

These are only some samples. If you wish to get more case study questions for CBSE class 12 maths, install the myCBSEguide App. It has class 12 Maths chapter-wise case studies with solutions.

12 Maths Exam pattern

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

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

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

12 Maths Prescribed Books

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

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12th Class Mathematics Applications of Derivatives Question Bank

Done case based (mcqs) - applications of derivatives total questions - 15.

A) \[4x-\frac{1}{2}{{x}^{2}}\] done clear

B) \[4\text{ }-\text{ }x\] done clear

C) \[\text{x }-\text{ 4}\] done clear

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

question_answer 2) What is the number of days it will take for the plant to grow to the maximum height?

A) 4 done clear

B) 6 done clear

C) 7 done clear

D) 10 done clear

question_answer 3) What is the maximum height of the plant?

A) 12 cm done clear

B) 10 cm done clear

C) 8 cm done clear

D) 6 cm done clear

question_answer 4) What will be the height of the plant after 2 days?

A) 4 cm done clear

B) 6 cm done clear

D) 10 cm done clear

question_answer 5) If the height of the plant is 7/2 cm, the number of days it has been exposed to the sunlight is

A) 2 done clear

B) 3 done clear

C) 4 done clear

D) 1 done clear

A) 37500 done clear

B) 12.5 done clear

C) -12.5 done clear

D) -37500 done clear

question_answer 7) What will be the maximum profit?

A) Rs 38,28,125 done clear

B) Rs 38281.25 done clear

C) Rs 39,000 done clear

D) None. done clear

question_answer 8) Check in which interval the profit is strictly increasing.

A) \[\left( 12.5,\infty  \right)\] done clear

B) for all real numbers done clear

C) for all positive real numbers done clear

D) (0, 12.5) done clear

question_answer 9) When the production is 2 units what will be the profit of the company?

B) 37,730 done clear

C) 37,770 done clear

D) None done clear

question_answer 10) What will be production of the company when the profit is Rs  38250?

A) 15 done clear

B) 30 done clear

C) 2 done clear

D) data is not sufficient to find done clear

A) \[\pm \frac{1}{4}\] done clear

B) \[\pm \frac{1}{2}\] done clear

C) ±1 done clear

question_answer 12) Find the slope of the normal based on the position of the stick

A) 360 done clear

B) -360 done clear

C) \[\frac{1}{360}\] done clear

D) \[-\frac{1}{360}\] done clear

question_answer 13) What will be the equation of the tangent at the critical point if it passes through (2, 3)?

A) \[x+360y=1082\] done clear

B) \[y=360x-717\] done clear

C) \[x=717y+360\] done clear

D) none of these done clear

question_answer 14) Find the second order derivative of the function at x =5.

A) 598 done clear

B) 1176 done clear

C) 3588 done clear

D) 3312 done clear

question_answer 15) At which of the following intervals will \[f\text{ }\left( x \right)\]be strictly increasing?

A) \[\left( -\infty ,\text{ }-1/2 \right)\cup \text{  }\left( 1/2,\text{ }\infty  \right)\] done clear

B) \[\left( -1/2,\,\,0 \right)\cup \left( 1/2,\text{ }\infty  \right)\] done clear

C) \[\left( 0,\,\,1/2 \right)\cup \left( 1/2,\text{ }\infty  \right)\] done clear

D) \[\left( -\infty ,\,-\,1/2 \right)\cup \left( 0,\text{ 1/2} \right)\] done clear

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Case Based (MCQs) - Applications of Derivatives

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