HSC MATHSOMTEX CLASSES 8TH YEAR OMTEX CLASSES 8 Years of Success MATHEMATICS & STATISTICS NAME :- ______________________________ STANDARD: -S.Y.J.C (Second year junior college) CLASSES :- OMTEX CLASSES FOR PRIVATE CIRCULATION ONLY “You don’t know what you can do until you try” “IF YOU ARE SATISFIED WITH OUR TEACHING TELL TO OTHERS IF NOT TELL TO US” 1 HSC MATHS OMTEX CLASSES 8TH YEAR Preface It gives us great pleasure to present this thoroughly revised edition of OMTEX MATHEMATICS & STATISTICS for Standard XII, prepared according to the pattern prescribed by the board. A thorough study and practice of this edition with the help of Omtex guidance (teaching + coaching) will enable the students to pass the HSC Examination with flying colours. Meticulous care has been taken to make this edition of OMTEX MATHEMATICS & STATISTICS perfect and useful in every respect. However, suggestions, if any, for its improvement are most welcome. Omtex classes [REG. NO: - 760076951] Note: - No part of this book may be copied, adapted, abridged or translated, stored in any retrieval system, computer system, photographic or other system or transmitted in any form or by any means without a prior written permission of the Omtex classes. 2 HSC MATHS OMTEX CLASSES 8TH YEAR MATHS – II CH. NO. 1. THEORY OF ATTRIBUTES EX. NO. 1 NINE SQUARE TABLE OF TOTAL FREQUENCIES 1. Find the missing frequencies in the following data of two attributes A and B. N = 800, AB = 120, B = 500, A = 300. 2. For a data for 2 attributes, it is given that N = 500, A = 150, B = 100, AB = 60, find the other class frequencies. 3. In a population of 10,000 adults, 1290 are literate, 1390 are unemployed and 820 are literate unemployed. Find the number of (i) literate employed. (ii) literates, (iii) employed. 4. In a co – educational school of 200 students contained 150 boys. An examination was conducted in which 120 passed. If 10 girls failed, find the number of (i) boys who failed, (ii) girls who passed. 3 (AB). A = 100. (AC) = 72. EX. 3 IMPORTANT POINTS TO REMEMBER If If If 𝐀𝐁 = 𝐀𝐁 > 𝐀𝐁 < 𝐀 (𝐁) 𝐍 𝐀 (𝐁) . If Q = 0 then A and B have no association OR A and B are independent attributes. NO. αβ = 25. A and B are independent attributes OR A and B have no association. Aβγ = 360. EX. find (A).HSC MATHS OMTEX CLASSES 8TH YEAR 5. N = 100. . 𝐍 𝐀 (𝐁) 𝐍 YULE’S COEFFICIENT OF ASSOCIATION. AB = 60. AB = 140. αβC = 160. Aβ = 200. B and C. N = 500. 3. 1. (AB) = 125. find AβC and ABγ . (A)=224. then. N = 500. (AB) = 30. B = 150. αβγ = 32. . A = 100. B = 300. Discuss the association of A and B if i. ii. 𝐐 = 𝐀𝐁 𝛂𝛃 − 𝐀𝛃 (𝛂𝐁) 𝐀𝐁 𝛂𝛃 + 𝐀𝛃 (𝛂𝐁) PROPERTIES OF Q Q always lies between – 1 to 1. N = 1000. N = 1000. Aβ = 30. 4 . B = 150. Discuss the association between attributes A and B if i. If for 3 attributes A. (C) = 150. ABγ = 180 αBγ = 240. (B) = 301. it is given that (ABC) = 210. β = 400. A = 300. 1. 2. 2. (B). If Q = -1 the A and B have perfect negative association. A and B have positive association. If N = 800. NO. A and B have negative association. (C). then. If Q > 0 then A and B have positive association. AβC = 250. (AB) = 20. If Q < 0 then A and B have negative association. AB = 200. αBC = 280. A = 300. N = 100. If Q = 1 then A and B have perfect positive association. (A) = 40. then. (A) = 50. 4. 6. (B) = 40. αB = 20. 2 Check the consistency of the following data. (AB) = 25. (NC) = 60 and (ABC) = 32. (B) = 60. (AC) and (BC). ii. 7. Of unemployed 140 No. Find Yule’s coefficient of association for the following data.300 literates in the same town. Examine the consistency of the following data and if so. β = 350.HSC MATHS 3. Intelligent husbands with intelligent wives 40 Intelligent husbands with dull wives 100 Dull husbands with intelligent wives 160 Dull husbands with dull wives 190 88 persons are classified according to their smoking and tea drinking habits. OMTEX CLASSES 8TH YEAR 4. Find Yule’s coefficient and draw your conclusion. Find the number of unsuccessful students who did not attend the coaching class. Of literate 130 No. Out of 700 literates in town. Smokers Non – smokers Tea Drinkers 40 33 Non Tea Drinkers 3 12 Show that there is no association between sex and success in examination from the following data. (AB) = 24. 8. Total Adults 10000 Unemployed 1390 Literates 1290 Literate unemployed 820 Comment on the result. N = 500. α = 300. Total No. 6. 150 were criminals. 10. Boys Girls Passed examination 120 40 Failed examination 30 10 300 students appeared for an examination and of these. 9. αβ = 70. Find the association between literacy and unemployment in the following data. Of adults 1000 No. Find Q. 5. find Q. Out of 9. 5 were criminals. N = 200. Also find Q. 200 passed. Of literate unemployed 80 Find the association between literacy and employment from the following data. AB = 60. 130 had attended a coaching class and 75 of these passed. 5 . α = 160. The population of a town for 4 year was as given below. (5.HSC MATHS OMTEX CLASSES CH.4. Find (2.5 0. 𝟐! 𝟑! 1. Find f(x).5) using Forward Difference interpolation formula. NUMERICAL METHOD 8TH YEAR EX.5). (196 students) Marks 0-19 20-39 40-59 60-79 80-99 No. Given the following data. f(5) = 86.5. NEWTON’S FORWARD INTERPOLATION FORMULA.6 8. (16. Estimated values of logarithms upto 1 decimal are given below find log(25). 2. f(7) = 201.3 1. NO. (x2+3x+8) 10. NO. f(3) = 31. find f(5) from the following table.708125) x 00 300 600 900 sinx 0 0.4 1. f(1) = 0.0625) 7.875) 5. find f(x).4375) Year 1980 1982 1984 1986 Population (in Thousand) 52 54 58 63 4.35625) x 10 20 30 40 logx 1 1.5)2 using Newton’s Forward Interpolation Formula.5) and f(2. (6. 39. f(x) is a polynomial in x. Find f(1985) (60.75) x 2 4 6 8 f(x) 4 7 11 18 2.1). Given the following table find f(24)using an appropriate interpolation formula. Of Candidates 41 62 65 50 17 6 .83) x 1 2 3 4 f(x) 7 18 35 58 11.078125) 6. (1. Using Newton’s Interpolation formula. using Newton’s Interpolation formula. Estimated values of sinx up to 1 decimal are given below find sin(450). 𝒇 𝒙 = 𝒇 𝒙𝟎 + 𝒖𝚫𝒇 𝒙𝟎 + 𝒖 𝒖 − 𝟏 𝚫𝟐 𝒇 𝒙𝟎 𝒖 𝒖 − 𝟏 (𝒖 − 𝟐)𝚫𝟑 𝒇 𝒙𝟎 + + … … … …. For a function f(x). Construct a table of values of the function f x = x 2 for x = 0.25. In an examination the number of candidates who scored marks between certain limits were as follows.1. f(2) = 18. (0.2. f(0) = 1. Estimate the number of candidates getting marks less than 70. if f(0) = 8. (8. Find f(2. Estimate f(1. 7. (3x2+2x+2. f(1) = 3. (485.3. f(3) = 25. For a function f(x). 1.87 1 9. Also find f(1. f(1) = 12.84) x 20 30 40 50 f(x) 512 439 346 243 3. f(2) = 11. The function y = f(x) is given by the points (7.1 Under 50 years 94.3 Under 49 years 84. (10. determine the percentage number of criminals under 35 years. (5.405%) Age % number of criminals Under 25 years 52 Under 30 years 67. Given log10 10 = 1. A company started selling a new product x in the market. log10 15 = 1. The profit of the company per year due to this product is as follows. log10 12 = 1.HSC MATHS OMTEX CLASSES 8TH YEAR EX.139) 7 .2 and log10 20 = 1.1). (9) x -1 0 3 f(x) 3 1 19 2.3). BY LAGRANGE’S INTERPOLATION FORMULA 𝒇 𝒙 = 𝒙 − 𝒙𝟏 𝒙 − 𝒙𝟐 𝒙 − 𝒙𝟑 𝒇 𝒙𝟎 + 𝒙𝟎 − 𝒙𝟏 𝒙𝟎 − 𝒙𝟐 𝒙𝟎 − 𝒙𝟑 𝒙 − 𝒙𝟎 𝒙 − 𝒙𝟐 𝒙 − 𝒙𝟑 𝒇 𝒙𝟏 + 𝒙𝟏 − 𝒙𝟎 𝒙𝟏 − 𝒙𝟐 𝒙𝟏 − 𝒙𝟑 𝒙 − 𝒙𝟎 𝒙 − 𝒙𝟏 𝒙 − 𝒙𝟑 𝒇 𝒙𝟐 + 𝒙𝟐 − 𝒙𝟎 𝒙𝟐 − 𝒙𝟏 𝒙𝟐 − 𝒙𝟑 𝒙 − 𝒙𝟎 𝒙 − 𝒙𝟏 𝒙 − 𝒙𝟐 𝒇 𝒙𝟑 𝒙𝟑 − 𝒙𝟎 𝒙𝟑 − 𝒙𝟏 𝒙𝟑 − 𝒙𝟐 1. In lakh) 4 5 5 5 th year by using Lagrange’s Interpolation formula.1). (77.3. 9). (8. By using suitable interpolation formula estimate f(2) from the following table. Using the Lagrange’s Interpolation formula.1904) Year 1st 2nd 7th 8th Profit (Rs. NO.5 using Lagrange’s formula. LAGRANGE’S INTERPOLATION FORMULA. By suing Lagrange’s Interpolation formula. (3. Find the value of y at x = 9.1. (35) x 0 1 2 5 f(x) 2 3 10 147 3.625) 6. estimate f(x) when x = 3 from the following table. (1. 2. find log10 13 = ? [Values are approximate and rounded off to 1 decimal place]. (9. Find the profit of the company in the 6 4.4 5. Also what can you say about f(x). f(x) = 4x – 3 b. ii. 2. assuming the interval of difference to be 1. 5. 7. u5 = 3. Given u4 = 0. x 0 1 2 3 4 f(x) 3 2 7 24 59 6. 4 ∆𝒇 𝒙 = 𝒇 𝒙 + 𝒉 − 𝒇 𝒙 𝛁𝒇 𝒙 = 𝒇 𝒙 − 𝒇(𝒙 − 𝒉) SHIFT OPERATOR IS REPRESENTED BY ‘E’ ∆= 𝑬 − 𝟏 𝑬𝒏 𝒇 𝒙 = 𝒇 𝒙 + 𝒏𝒉 1. u7 = 45. u6 = 9 and the second difference are constant. 22. taking the interval of differentiating equal to 1. u4 = 12. 11. iii. 2 . By constructing a difference table. Given f x = x 2 + 3x + 5 taking the interval of differentiating equal to 1. f x = x 2 + x. 58. Find ∆f x and ∆2 f x . 4. if u3 = 5. iv. u5 = 21. find 6th and 7th term of the sequence 6. 2. 3. Form the difference table for f(x) = x2 +5 taking values for x = 0. 3. FORWARD DIFFERENCE TABLE 8TH YEAR 1. Find ∆f x and ∆2 f x . f x = x x − 1 x − 2 x − 3 . f x = x 2 − 2x + 4. find 7th and 8th term of the sequence 8. Estimate f(5) from the following table. Find ∆f x in each of the following case. From the table? x 0 1 2 3 4 5 f(x) 0 3 8 15 24 35 4. 3. i. Given f x = x 2 − 8x + 2. 8. NO. 27.HSC MATHS OMTEX CLASSES EX. Obtain the difference table for the data. By constructing a difference table. obtain the 6th term of the series 7. Find u9. 1 . EX. Find u2. By constructing a difference table. u6 = 32. 14. 38. Write down the forward difference table of the following polynomials f(x) for x = 0(1)5 a. 28. f(x) = x2 – 4x – 4. 9. 3. 32. NO. f x = 2x + 3. Find ∆2 f 1 if f x = x x + 1 x + 2 . 18. 44. 8 . 11. 41. 18. u5 = 76. ∆n f x are in geometric progression. 11. ∆2 f x . iii. iv. Estimate the missing term by using " ∈ "and "∆" from the following table. ∆2 𝑥 2 + 5 ii. Assuming that the difference interval h = 1. Evaluate i. Evaluate i. Given: u2 = 13. iv. i. x 0 1 2 3 4 y 1 3 9 . v. … … … … … . u4 = 49. Compute ∆3 u2 + ∆2 u3 . ∆𝑒 8. 𝑓 4 = 𝑓 3 + ∆𝑓 2 + ∆2 𝑓 1 + ∆3 𝑓 1 . ∆4 (𝑎𝑏 𝑥 ) iii. ii. prove the following. ∆ 𝑠𝑖𝑛 𝑎𝑥 + 𝑏 iii. 13. u3 = 28. ∆4 𝑓 𝑥 = 𝑓 𝑥 + 4 − 4𝑓 𝑥 + 3 + 6𝑓 𝑥 + 2 − 4𝑓 𝑥 + + 𝑓 𝑥 . u5 = 8. 𝑓 5 = 𝑓 4 + ∆𝑓 3 + ∆2 𝑓 2 + ∆3 𝑓 1 + ∆4 𝑓 1 . Prove the following: i. find∆5 u0 . u1 = 12. x 1 2 3 4 5 6 7 y 2 4 8 . 14. 15. ∆f(x). ∆3 𝑓 𝑥 = 𝑓 𝑥 + 3 − 3𝑓 𝑥 + 2 + 3𝑓 𝑥 + − 𝑓 𝑥 ii. Given: u2 = 13.HSC MATHS OMTEX CLASSES 8TH YEAR 5. Show that ∆𝑙𝑜𝑔𝑓 𝑥 = 𝑙𝑜𝑔 1 + 10. x 1 2 3 4 5 f(x) 2 5 7 . 9 . 𝑓 𝑎 + 3 = 𝑓 𝑎 + 3∆𝑓 𝑎 + 3∆2 𝑓 𝑎 + ∆3 𝑓 𝑎 . 𝑓 2 = 𝑓 1 + ∆𝑓 0 + ∆2 𝑓 −1 + ∆3 𝑓 −1 . 𝑓 𝑎 + 5 = 𝑓 𝑎 + 5∆𝑓 𝑎 + 10∆2 𝑓 𝑎 + 10∆3 𝑓 𝑎 + 5∆4 𝑓 𝑎 + ∆5 𝑓 𝑎 . u4 = 49. u2 = 81. u4 =100. 12. ∆3 5𝑒 𝑥 ii. u3 = 28. If f x = ex . find ∆2 u2 . a. Given: u0 = 3. Show that 𝑒 𝑥 . ∆𝑐𝑜𝑠 + 𝑏) (𝑎𝑥 7. ∈𝑒 𝑥 ∆2 𝑒 𝑥 = 𝑒 𝑥 ∆𝑓 𝑥 𝑓 𝑥 9.81 b. iii. Show that f x .32 64 128 c. ∆2 3 𝑥 ∈ ∆2 𝑥 3 ∈𝑥 3 1 ∆2 𝑥 𝑥 ∆2 ∈ vi.32 6. 5 [ 𝑦 + 𝑦𝑛 + 2 𝑦1 + 𝑦2 +. [1.HSC MATHS BY TRAPEZOIDAL RULE. = BY SIMPSON’S 𝑏−𝑎 𝑎 𝑛 𝟏 𝒓𝒅 𝟑 𝑏 RULE. [0. 3 𝑦0 + 𝑦𝑛 + 3 𝑦1 + 𝑦2 + 𝑦4 + 𝑦5 . [0.10] in 10 3. +𝑦𝑛−1 3 0 + 2 𝑦2 + 𝑦4 + 𝑦6 +. . = 𝑛 NOTE: This rule is applicable only when ‘n’ is even. Solve 6 𝑑𝑥 0 1+𝑥 number of division is 6. +𝑦𝑛−1 ) 2 0 8TH YEAR 𝑦 𝑑𝑥 = WHERE :. [115] 4. … … … . Find the approximate value of equal parts. Using trapezoidal rule calculate the approximate value of ordinates. +𝑦𝑛−1 8 + 2 𝑦3 + 𝑦6 +. [2. … … … + 𝑦𝑛−2 𝑦 𝑑𝑥 = 𝑎 𝑏−𝑎 WHERE :. … … … .335] 1 2 𝑥 𝑑𝑥 0 by using trapezoidal rule by dividing the interval [0. Evaluate 6 𝑑𝑥 0 1+𝑥 2 10 . = 𝑛 NOTE: This rule is applicable only when ‘n’ is a multiple of 3. NO. Find the approximate value of 3 2 𝑥 𝑑𝑥 0 3 4 𝑥 𝑑𝑥 −3 by taking 7 equidistant using trapezoidal rule where 𝑛 = 6.02] by using trapezoidal rule by taking 7 equidistant ordinates. Find the approximate value of equal parts. 1.34] 1 2 𝑥 𝑑𝑥 0 by using trapezoidal rule by dividing the interval [0.1] in to 5 2. .9125] 5. … … … . 𝑦 + 𝑦𝑛 + 4 𝑦1 + 𝑦3 + 𝑦5 +. .4095] 6. BY SIMPSON’S 𝟑 𝒕𝒉 𝟖 𝑏 RULE. [0. … … … + 𝑦𝑛 −3 𝑦 𝑑𝑥 = 𝑎 𝑏−𝑎 WHERE :. 𝑏 OMTEX CLASSES EX. 7 2.6 2.9833] x 1 1.4 2. [261] x 8TH YEAR 8. 11 . [1. A curve is drawn to pass through the points given by the following tables. 13.HSC MATHS 7.5 3 3. [705 sq.7 2. Evaluate 6 𝑑𝑥 0 1+𝑥 2 1 3 𝑟𝑑 rule calculate the approximate value of 3 𝑥 4 𝑑𝑥 −3 by taking 7 equidistant by using Simpson’s 1 3 𝑟𝑑 rule.5 2 2. 10.8 3. Calculate 6 (3𝑥 2 0 OMTEX CLASSES + 2𝑥 + 1) 𝑑𝑥 with 7 ordinates use trapezoidal rule.8 3.7 2.8 3.6 2.5 4 𝑡 rule estimate the area bounded by the y 2 2. Where y=d = depth of the river in meter from one bank at a 9. Using Simpson’s ordinates.4 2. A curve is drawn to pass through the points given by the following tables. Using Simpson’s ordinates.365] 12. [8. Using trapezoidal rule estimate approximate area of cross section of river 80 m wide and the depth (d) in meters at a distance (x) of the river from one bank to another is given .6 2. The cross section of the river 80 m wide is given by distance x.6 2.1 Using Simpson’s and 𝑥 = 4. [7.175] Using Simpson’s 3 8 3 8 1 3 𝑟𝑑 rule estimate the area bounded by the curve. [99] 14. on the x – axis and the lines 𝑥 = 1 and 𝑥 = 4. [98] 11.5 2 2.6 2. on the x – axis and the lines 𝑥 = 1 𝑡 rule calculate the approximate value of 3 𝑥 4 𝑑𝑥 −3 by taking 7 equidistant x 1 1. A curve is drawn to pass through the points given by the following tables.5 3 3.] x 1 1. on the x – axis and the lines 𝑥 = 1 and 𝑥 = 4. m.5 3 3. Take 6 equidistant.4 2.5 2 2.5 4 y 2 2.1 curve.5 4 y 2 2. [8.6 2.] 80 𝑦 0 0 10 20 30 40 50 60 70 80 7 9 12 15 14 8 3 d 0 4 dx.1 Using trapezoidal rule estimate the area bounded by the curve.075 sq unit. It is observed that on an average.1. Only 4 games. (3124/3125) 11. BINOMIAL AND POISSON DISTRIBUTION EX. An unbiased coin is tossed 6 times. If he fires 5 shots. Find the probability of getting 3 heads. find the probability that a. n = 100. (80/243) 8. find the probability distribution if success denotes finding a white ball. 2. Atleast 4 are swimmers/ (2304/3125) 6. p = 1/3 b. Also find that atleast 1 person out of 3 is smoker. Find the probability of getting a. (1/100000) (99954/100000) 9. An ordinary coin is tossed 4 times. p and q. The probability that a student is not a swimmer is 1/5. If four balls are drawn in this manner. what is the probability that A would win. n = 10. 1 person out of 5 is a smoker. (61/125) 10. Find Mean and Variance of Binomial Distribution. (11/32) 3. An urn contains 2 white and 3 black balls. A biased coin in which P(H) = 3 and P(T) = 3 is tossed 4 times. programs. (432/3125) 5. BINOMIAL DISTRIBUTION NOTE: . (5/16) 2. 3. (496/729) b. Exactly twice (31/625) b. 1 1. 1 2 1. If a ball is drawn 5 times in this way. 4 or more games. At least one is popular. a. A bag contains 7 white and 3 black balls. n = 12.1 12 . (1323/100000) EX. what is the probability of hitting the target a. If getting a head is success then find the probability distribution. Find the probability of A’s winning atleast 4 games out of 5. (64/125) b. In a new series of 6 games. A ball drawn is always replaced in the bag. find the probability of we get 2 white and 3 black balls. find the probability that all the lines are busy. If the chances that any of the 5 telephone lines are busy at any instant are 0. Out of five students considered. A has won 20 out of 30 games of chess with B. (729/15625) b. If a. 2. Atleast once. Exactly four problems out of 6. Exactly 1 head(1/4) b. If 3 new programs are introduced. NO. find the probability that a.HSC MATHS OMTEX CLASSES 8TH YEAR CH. Find the probability of getting atleast 4 heads. 4 are swimmers. (112/243) 7. A marks man’s chance of hitting a target is 4/5. Exactly 3 tails(1/4) 4. No problems out of 6. the probability of A’s winning is 2/3. None is popular. In a certain tournament. Also find the probability that not more than three lines are busy.For a binomial variate parameter means n. 12. It is noted that out of 5 T. 3. only one is popular. (256/625) b. A ball is drawn. What is the probability of ‘A’ solving a. (64/125) (61/125). its colour noted and is replaced in the urn.V. p = 0. Find the probability that no person out of 3 is a smoker. On an average ‘A’ can solve 40% of the problems. in 6 trials of a coin. p = 2/5 c. n = 150 p = 0. b.05 b. Exactly 3 b. 3. In a Poisson distribution the probability of 0 successes is 10%. More than 2 individuals get a reaction. 12. The number of complaints received in a super market per day is a random variable. S.For a Poisson distribution Mean = Variance = 𝛌. 3. 𝐏 𝐱 = For a Poisson variate parameter is known as 𝛌 and𝛌 = 𝐧𝐩. σ = 3. The average number of incoming telephone calls at a switch board per minute is 2. find p(2) & p(x ≤ 2). VAriance = 10. x = 10. The probability that a car passing through a particular junction will make an accident is 0.00005.D. Among 10000 can that pass the junction on a given day.HSC MATHS OMTEX CLASSES 8TH YEAR 4.3. Find the probability of exactly 2 complaints received on a given day. Find its mean. Find the probability that during a given period 2 or more telephone calls are received. 7. At most 2 customers appear. No customer appears. In the following situations of a Binomial variate x. c. POISSON DISTRIBUTION Note: For a random variable x with a Poisson distribution with the parameter𝛌. 𝛌𝐱 𝐞−𝛌 𝐱! 13 . find the probability that two car meet with an accident. find p(2) & p x ≤ 3 . In a manufacturing process 0. 6. 5. 4. Find the probability that at most 2 items are defective. find its mean and variance and also find p(3). 1. find mean. 11. If a random variable x follows Poisson distribution such that p(1) = p(2). variance = 5 d.2 EX.0001. the probability of success is given by. In a Poisson distribution. who appear at the counter of a bank in 1 minute is 2. 8. having a Poisson distribution with λ= 3. For a Poisson distribution with λ = 0. If 20000 individuals are given the injection find the probability that more than 2 having reaction. The probability that an individual will have a reaction after a particular drug is injected is 0. Mean = 12. 10. find the probability that a. Mean = 6. The probability that a person will react to a drug is 0. = 2. if p(2) = p(3). What is the probability that a consignment of 400 bolts will have exactly 5 defective bolts? 9. 13. b. if a. For a Poisson distribution with λ = 3.5% of the goods produced are defective. In a sample of 400 goods. A machine producing bolts is known to produce 2% defective bolts. The average customers. Find n and p for a binomial distribution. If 𝐧 ≥ 𝟏𝟎𝟎 & 𝛌 ≤ 𝟏𝟎.7.25 Note: .001 out of 2000 individuals checked. x = 6. can they be approximated to a Poisson Variate? a. 14. n = 400 p = 0. 2. Find the probability that in a given minute a. B. 1 1. [22 units] A B C D I 12 20 11 5 II 1 16 2 14 III 28 9 8 5 IV 10 17 15 1 14 . A Departmental Store has 4 workers to pack their items. he cannot be assigned the corrections of that division. How should the manager of the store assign the job to the workers. The timing in minutes required for each workers to complete the packing per item sold is given below. so as to minimize the total time of packing?[9 minutes] Books Toys Crockery Cattery A 2 10 9 7 B 12 2 12 2 C 3 4 6 1 D 4 15 4 9 3. For an examination. C & D. the answer papers of the divisions I. III and IV are to be distributed amongst 4 teachers A. No. NO. Solve the following minimal assignment problem.HSC MATHS OMTEX CLASSES 8TH YEAR CH. [13 days] A B C D I .5 2 6 II 4 5 3 8 III 6 6 2 5 IV 1 6 3 4 5. [21 units] A B C D 1 16 1 6 11 2 25 10 0 10 3 10 25 2 14 4 15 7 14 10 2. Also. for every teacher to asses the papers of the various divisions is listed below find the allocation of the work so as to minimize the time required to complete the assessment. 4. If the time required in days. Solve the following minimal assignment problem. Solve the following minimal assignment problem. It is a policy decision of the department that every teacher corrects the papers of exactly one division. [14 units] A B C D I 12 1 11 5 II 3 11 10 8 III 3 4 6 1 IV 2 13 11 7 6. [16 units] A B C D 1 3 4 6 5 2 5 6 10 9 3 1 2 3 2 4 4 10 6 4 4. A’s son is in Division I. ASSIGNMENT PROBLEMS AND SEQUENCING Ex. A Departmental head has four subordinates and four task to be performed. The time each man would take to perform each task is given below. II. since Mr. [11 days] CASES EMPLOYEES E1 E2 E3 E4 E5 I 5 3 6 4 3 II 2 4 3 2 6 III 4 4 2 4 IV 2 5 1 3 7 V 6 7 2 5 3 11. 376] JOBS MACHINISTS M1 M2 M3 M4 A 62 71 87 48 B 78 84 92 61 C 50 61 111 87 D 101 73 71 77 E 82 59 81 80 10. How should the horses be assigned to the riders so as to minimise the expected loss? Also find the minimum expected loss. The number of penalty points to be expected when each rider rides each horse is shown below.) of sending material to ‘five’ terminals by ‘four’ trucks. The expected profits for each job done by each machinist are given below. Minimise the following assignment problem. Find the assignment of jobs to the machinists that will results in maximum profit. Determine he optimal assignment of cases to the employees so that the total number of days required completing these ‘five’ cases will be minimum. [16 units] I II III IV 8. The cost (in hundreds of Rs. [‘One’ truck is assigned to only ‘one’ terminal] Which terminal will ‘not’ receive material from the truck company? What is the minimum cost?[Rs. A Chartered Accountants’ firm has accepted ‘five’ new cases. Also find the maximum profit. [14 penalty points] HORSES RIDERS R1 R2 R3 R4 H1 12 1 11 5 H2 3 11 10 8 H3 3 4 6 1 H4 2 13 11 7 9. A team of 4 horses and 4 riders has entered the jumping show contest. where ‘-‘means that the particular employee cannot be assigned the particular case. Find the assignment of trucks to terminals which will minimize the cost. The estimated number of days required by each of their ‘five’ employees for each case are given below. [One machinist can be assigned only ‘one’ job][Rs. Also find the minimum number of days. The owner of a small machine shop has ‘four’ machinists available to assign jobs for the day. ‘Five’ jobs are offered to be done on the day.HSC MATHS OMTEX CLASSES A 2 9 10 7 B 13 12 2 6 C 3 6 4 1 D 4 13 15 9 8TH YEAR 7. 800] TRUCKS TERMINALS T1 T2 T3 T4 T5 A 3 7 3 5 5 B 6 1 8 2 7 C 2 4 5 6 6 D 6 4 8 3 2 15 . incurred by a company is given below. Solve the following problem for minimum elapsed time.HSC MATHS EX. Ticked column. Job M1 M2 A 7 4 B 2 6 A 5 2 C 3 5 B 1 6 D 2 4 C 9 7 E 7 3 D 3 8 F 4 1 E 10 4 G 5 4 2. Solve the following for minimum elapsed time and idling time for each machine. Job M1 M2 1 2 6 2 5 8 3 4 7 4 9 4 5 6 3 6 8 9 7 7 3 8 5 8 9 4 11 4. Job M1 M2 3. Also state the idling time for each machine. NO. Solve the following problem for minimum elapsed time. Job Machine A Machine B Machine C 1 8 5 4 2 10 6 9 3 6 2 8 4 7 3 6 5 11 4 5 5. Job Machine A Machine B Machine C A 2 3 5 B 7 2 6 C 6 1 4 D 3 4 10 E 8 0 4 F 7 3 5 G 9 2 11 IMPORTANT POINTS TO REMEMBER U R T C Un ticked row. 2 OMTEX CLASSES 8TH YEAR 1. Solve the following problem for minimum elapsed time. Also state the idling time for each machine. 16 . Find the sequence that minimises the total elapsed time. required to complete the following jobs on two machineries. Also state the idling time for the machine. Also state the idling time for each machine. Solve the following problems for minimum elapsed time. Job Machine A Machine B Machine C 1 8 3 8 2 3 4 7 3 7 5 6 4 2 2 9 5 5 1 10 6 1 6 9 6. Age group 0-30 30-60 60 & above Population A 4000 8000 3000 Deaths in A 180 120 200 Population B 7000 9000 4000 Deaths in B 250 320 230 3.R.R.D. NO.25 per thousand.) 𝐂. Age group 0-25 25-50 50-75 Above 75 Population 5000 7000 6000 2000 No. For the following data. Age group 0-25 25-50 50-75 Above 75 Population A in thousands 60 70 40 30 Deaths in A 250 120 180 200 Population B in thousands 20 40 30 10 Deaths in B 120 100 160 170 4. = 𝐍𝐮𝐦𝐛𝐞𝐫 𝐨𝐟 𝐝𝐞𝐚𝐭𝐡𝐬 × 𝟏𝟎𝟎𝟎 𝐒𝐢𝐳𝐞 𝐨𝐟 𝐭𝐡𝐞 𝐩𝐨𝐩𝐮𝐥𝐚𝐭𝐢𝐨𝐧 1.HSC MATHS OMTEX CLASSES 8TH YEAR CH. 5. 𝐃. VITAL STATISTICS. For the following data. Find x if the C. Find x if the C.R. Compare the crude death rate of the two given population.75 Age group 0-20 20-40 40-60 Above 60 Population in thousands 58 71 41 30 Deaths 195 130 x 245 17 . 𝐑.D. find the crude death rate. = 3. of deaths 800 600 500 100 2. Compare the crude death rate of the two given population. For the following data.D. = 31. MORTALITY RATES AND LIFE TABLE CRUDE DEATH RATE (C. Age group Population Deaths 0-35 4000 80 35-70 3000 120 Above 70 1000 x 5. HSC MATHS OMTEX CLASSES SPECIFIC DEATHS RATES (S.) for population A and B of the following. Age – group 0-30 30-60 60-80 Above 80 Population A in thousands 30 60 50 20 Deaths in A 150 120 200 400 Population B in thousands 50 100 90 70 Deaths in B 200 140 270 350 STANDARD DEATHS RATES (S. Find the Standard Deaths Rates for the following data: Age – group 0-30 30-60 Above 60 Population A in thousands 60 90 50 Deaths in A 240 270 250 Standard Population in thousands 20 30 20 2.) for population A and B for the following. = 𝐍𝐮𝐦𝐛𝐞𝐫 𝐨𝐟 𝐝𝐞𝐚𝐭𝐡𝐬 𝐢𝐧 𝐭𝐡𝐞 𝐬𝐩𝐞𝐜𝐢𝐟𝐢𝐜 𝐚𝐠𝐞 𝐠𝐫𝐨𝐮𝐩 × 𝟏𝟎𝟎𝟎 𝐏𝐨𝐩𝐮𝐥𝐚𝐭𝐢𝐨𝐧 𝐢𝐧 𝐭𝐡𝐚𝐭 𝐚𝐠𝐞 𝐠𝐫𝐨𝐮𝐩 1.R. 𝐒𝐏 𝚺𝐒𝐏 𝐍𝐨𝐭𝐞: 𝐦 = 𝐬𝐩𝐞𝐜𝐢𝐟𝐢𝐜 𝐝𝐞𝐚𝐭𝐡 𝐫𝐚𝐭𝐞 = SP = Standard Population 𝐃 × 𝟏𝟎𝟎𝟎 𝐏 1.) for the following data.D.) 8TH YEAR 𝐒. 𝐃. Age – group 0-25 25-50 50-75 Over 75 Population A in thousands 66 54 55 25 Deaths in A 132 108 88 100 Population B in thousands 34 58 52 16 Deaths in B 102 116 78 80 Standard Population in thousands 40 60 80 20 18 . 𝐑. Find the Age specific deaths rates (S. Age – group 0-30 30-60 60 and above Population A in thousands 50 90 30 Deaths in A 150 180 200 Population B in thousands 60 100 20 Deaths in B 120 160 250 3.D. 𝐓. 𝐑. Find the Age Specific deaths rates (S.R. Age group Population No.D.T. of deaths 0-15 6000 150 15-40 20000 180 40-60 1000 120 Above 60 4000 160 2. 𝐃.D.R.D.R. Find the Standard Deaths Rates for the following data.) 𝐒. = 𝚺𝐦.R. Find the age Specific deaths rates (S. ? ? 4. Taking A. Compare the standardized death rates for the population A and B for the given data.5 4.? ? 19 . Age lx dx qx px Lx Tx e0x 56 400 ? ? ? ? 3200 ? 57 250 ? ? ? ? ? ? 58 120 . Fill in the blanks in the following table marked by ‘?’ sign. Taking A. x 0 1 2 3 4 5 6 lx 50 36 21 12 6 2 0 3. Age – group 0-20 20-40 40-75 Above 75 Population A in thousands 7 15 10 8 Deaths in A 140 150 110 240 Population B in thousands 9 13 12 6 Deaths in B 270 260 300 150 𝒅𝒙 𝒍𝒙 LIFE TABLES 𝒑𝒙 = 𝟏 − 𝒒𝒙 𝒍𝒙 + 𝒍𝒙+𝟏 𝟐 𝑻𝒙 𝒍𝒙 Age 𝒙 𝒍𝒙 𝒅𝒙 = 𝒍𝒙 − 𝒍 𝒙+𝟏 𝒒𝒙 = 𝑳𝒙 = 𝑻𝒙 = 𝚺𝑳𝒙 𝒆𝟎 = 𝒙 1. Age lx dx qx px Lx Tx e0x 50 60 ? ? ? ? 240 ? 51 50 .HSC MATHS OMTEX CLASSES 8TH YEAR 3. as the standard population. Construct the life tables for the following data. Age – group 0-30 30-60 Above 60 Population A in thousands 5 7 3 Deaths in A 150 210 120 Population B in thousands 6 8 2. Construct the life tables for the rabbits from the following data.5 Deaths in B 240 160 7. Compare the standardized death rates for the population A and B for the given data. as the standard population. Fill in the blanks in the following tabled marked by ‘?’ sign. x 0 1 2 3 4 5 6 lx 10 9 7 5 2 1 0 2. [Ans. Find Index number.435] Security at 2000 2003 2006 Stock market P0 P1 P1 A 160 180 210 B 2400 35 8 C 800 550 850 D 3500 2000 4000 E 150 600 220 20 . 107.1. [Ans.4 130 90. 109. SIMPLE AGGREGATIVE METHOD FORMULA INDEX NUMBER 𝐏𝟎𝟏 = 𝐏𝟏 𝐏𝟎 × 𝟏𝟎𝟎. 1. 1.2 60. Find the index number for the year 2003 and 2006 by taking the base year 2000.7 88. [Ans. INDEX NUMBER 8TH YEAR EX. FIND THE INDEX NUMBER.7 Prices in Prices in 1990 (P0) 2002 (P1) A 12 38 B 28 42 C 10 24 D 16 30 E 24 46 3.HSC MATHS OMTEX CLASSES CH.1 85. 75.73] Commodities I II III IV V 2. Find Index number. 180] Commodities Prices in Prices in 2002 (P0) 2003 (P1) 21. Find Index number. 137. 6.9 100.3 55. I.6 30. 48.375] Commodities Prices Prices Prices in in in 2000 2003 2006 Trucks 800 830 850 Cars 176 200 215 Three wheelers 100 127 115 Two wheelers 44 43 43 4.5 70. [Ans. NO. NO. 315] Food Units 2004 2005 Items P0 P1 Potato Kg 10 12 Onion Kg 12 25 Tomato Kg 12 25 Eggs Dozen 24 2 Banana Dozen 18 20 II. 110. Compute the Index Number. Index Number = 112. Index Number = 120. 𝐱 = 15] Commodity Base Year Current Year P0 P1 I 3 5 II 16 25 III 40 35 IV 7 10 V 14 x 3. Index Number = 180. 𝐱 = 100] Commodity Base Year Current Year P0 P1 I 40 60 II 80 90 III 50 70 IV x 110 V 30 30 21 . FIND THE VALUE OF X IN EACH CASE.5.15] Real Estate 1990 1998 2006 Area wise A 100 65 B 35 22 C 5 7 D 12 11 6. [Ans. 126. [Ans. 250 75 12 25 2006 P1 14 16 16 26 24 THE INDEX NUMBER BY THE METHOD OF AGGREGATES IS GIVEN IN EACH OF THE FOLLOWING EXAMPLE. 1. [Ans. 238.HSC MATHS OMTEX CLASSES 8TH YEAR 5. X = 10] Commodity Base year Current Year P0 P1 A 12 38 B 28 41 C x 25 D 26 36 E 24 40 2. Calculate Index Number. 69. [Ans. [Ans.526.078. 2. WEIGHTED AGGREGATIVE INDEX NUMBERS.HSC MATHS OMTEX CLASSES 8TH YEAR EX. WALSCH’S PRICE INDEX NUMBER 𝑷𝟎𝟏 𝑾 = 𝒑𝟏 𝒒𝟎 𝒒𝟏 𝒑𝟎 𝒒𝟎 𝒒𝟏 × 𝟏𝟎𝟎 𝒑𝟏 𝒒𝟎 × 𝒑𝟎 𝒒𝟎 𝒑𝟏 𝒒𝟏 × 𝟏𝟎𝟎 𝒑𝟎 𝒒𝟏 𝒑𝟏 𝒒𝟎 + 𝒑𝟎 𝒒𝟎 𝟐 𝒑𝟏 𝒒𝟎 + 𝒑𝟎 𝒒𝟎 + 𝒑𝟏 𝒒𝟏 𝒑𝟎 𝒒𝟏 × 𝟏𝟎𝟎 𝒑𝟏 𝒒𝟏 × 𝟏𝟎𝟎 𝒑𝟎 𝒒𝟏 𝒑𝟏 𝒒𝟎 × 𝟏𝟎𝟎 𝒑𝟎 𝒒𝟎 𝒑𝟏 𝒒𝟏 × 𝟏𝟎𝟎 𝒑𝟎 𝒒𝟏 22 . LASPEYRE’S PRICE INDEX NUMBER 𝑷𝟎𝟏 𝑳 = 2. NO. DORBISH . FORMULA 1. PAASCHE’S PRICE INDEX NUMBER 𝑷𝟎𝟏 𝑷 = 3.BOWLEY’S PRICE INDEX NUMBER 𝑷𝟎𝟏 𝑫 − 𝑩 = 4. FISHER’S PRICE INDEX NUMBER 𝑷𝟎𝟏 𝑭 = 6. MARSHALL – EDGEWORTH PRICE INDEX NUMBER 𝑷𝟎𝟏 𝑴 − 𝑬 = 5. 2. Ans. 132. 134. [Ans. Paasche’s. Calculate Price Index Number by using Walsch’s Method.5 60 2 4. Find Walsch’s Price Index Number. 132. find k. Ans. 132. Find x.83 Commodities Base Year Current Year Price Quantity Price Quantity A 5 4 7 1 B 2 6 3 6 C 10 9 12 4 3.1 {using log table} Commodities Base Year Current Year Price Quantity Price Quantity A 20 3 25 4 B 30 5 45 2 C 50 2 60 1 D 70 1 90 3 23 .HSC MATHS OMTEX CLASSES 8TH YEAR 1.05] Commodities Base Year Current Year Price Quantity Price Quantity A 20 3 25 4 B 30 5 45 2 C 50 2 60 1 D 70 1 90 3 2. Ans. Commodity P0 Q0 P1 Q1 A 4 6 6 5 B 4 k 4 4 6. Dorbish – Bowley’s and Marshall – Edgeworth Index Numbers. Find Fisher’s Price Index Number. 130. Ans. x = 4 Commodities 1960 1965 Price Quantity Price Quantity A 1 10 2 5 B 1 5 X 2 5. For the following data find Laspeyre’s. For the following the Laspeyre’s and Paasche’s index number are equal. 126. 116. The ratio of Laspeyre’s and Paasche’s Index number is 28:27.1.21 Commodities Base Year Current Year Price Quantity Price Quantity I 10 4 20 9 II 40 5 3 5 III 30 1 50 4 IV 50 0. for the current year and weights (W). [Ans. NO. 150] Group Food Clothes Fuel & Lighting House Rent Miscellaneous I 200 150 140 100 120 W 6 4 3 3 4 5. [Ans.3] Group Food Clothes Fuel & Lighting House Rent Miscellaneous I W 320 140 270 160 210 20 15 18 22 25 4. [Ans.HSC MATHS OMTEX CLASSES 8TH YEAR EX. COST OF LIVING INDEX NUMBER = 2. Find the cost of living index number. 221. 137. 1. The price relatives I. Find the cost of living Index number. FAMILY BUDGET METHOD. [x = 3] Commodity Food Clothes Fuel & Lighting House Rent Miscellaneous I W 200 6 150 4 140 x 100 3 120 4 24 . Find x if the cost of living index number is 150. for the base year are given below find the cost of living Index number. COST OF LIVING INDEX NUMBER THERE ARE TWO METHODS TO CONSTRUCT COST OF LIVING INDEX NUMBER. Taking the base year as 1995. AGGREGATIVE EXPENDITURE METHOD. 3. COST OF LIVING INDEX NUMBER = NOTE: W = 𝒑𝟎 𝒒𝟎 I= 𝑷𝟏 𝒑𝟏 𝒒𝟎 𝒑𝟎 𝒒𝟎 𝑰𝑾 𝑾 × 𝟏𝟎𝟎 𝑷𝟎 × 𝟏𝟎𝟎 1.5] Group 1995 2000 Price Quantity Price Food 23 4 25 Clothes 15 5 20 Fuel and Lighting 5 9 8 House Rent 12 5 18 Miscellaneous 8 6 13 2. construct the cost of living index number for the year 2000 from the following data. 208] Group 1995 2000 Price Quantity Price Food 90 5 200 Clothes 25 4 80 Fuel and Lighting 40 3 50 House Rent 30 1 70 Miscellaneous 50 6 90 3. [Ans. 7. Calculate the two regression lines. 25 . 8. Production 120 115 120 124 126 121 Price Rs/unit 13 15 14 13 12 14 6.D.HSC MATHS OMTEX CLASSES 8TH YEAR CH. Find the line of Regression of y on x. Exp x Sales y Rs. 3. Adv. The following are production and Price Statistics of a certain commodity. The regression co – efficient of x on y is 0. x 1 2 3 y 2 1 6 Hence find the most likely value of y when x = 4. For a bivariate data the mean of x series is 35 and the mean of y series is 29. What should be advertisement expenditure if the company wants to obtain sales target of 120 lakhs. x independent variable 2 4 5 6 8 11 18 12 10 8 7 5 y dependent variable 7. Estimate x when y = 15. You are given below the following information about advertising and sales.8 a. For a bivariate data the means of y series is 40 and mean of x series is 35.6. Lakhs) 90 Mean 10 12 3 S.D. find the regression line of y on x. Estimate the value of x when y = 25. 2. Lakhs (Rs. Compute the appropriate regression equation for the data. Find the regression equation of x on y. Correlation co − eficient = 0. The Regression co – efficient of y on x is 1. For a certain biariate data the following incormation are available x y Mean 13 17 3 2 S. For the following data. estimate y when x = 10. find the regression equation of y on x and estimate y when x = 10.56.2. Estimate the value of y when x = 28. REGRESSION ANALYSIS 1. From the following data. 4. b. Correlation between x and y is 0. NO. x 1 2 3 4 5 6 y 2 4 7 6 5 6 5. Find the estimated price when the production is 125 units. y 2 = 340. x ii. y iii. Find x. x = 40. 60 crores? 10. x = 500. x 2 = 6500. σy = 20. n = 8. = 9800. 10 crores? b. x − x 2 = 32. y = 32. Hence Find x. y−y x − x y − y = 10000. In crores) 40 Mean 6 10 1. y = 9600. find x and r. r 12. y = 40. Identify the Regression of x on y. in crores (Rs. r = 0. y and r. Estimate the likely sales for a proposed adv. σx = 60. 20. xy = 214. byx iv. x 2 = 220. The Mean ofMarks in Accountancy is 44 and 9 variance of Marks in Statistics is 16 th of the variance of marks in Accountancy. y 2 = 10000. For a group of 30 couples the regression line of the age of the wife y on the age of the husband x is given by 3y − 4x + 60 = 0. Estimate y when x = 12. What should be the adv. If the two regression line for a bivariate data are 2x = y + 15 x on y and 4y = 3x + 25. 18. Exp x Sales y Rs. coefficient correlation r = 0. y gave the following results x = 8500. y − y 2 = 16 and x − x y − y = 6. of Rs. 19. find i.5 S. x and expenditrue on food and entertainment Rs. y = 1000.9 a. obtain the regression lines. exp. variance of x = 9. 25 15. bxy v. 14. σx 𝜎𝑦 2 = 9 . y and r. 11.D. If the regression line of x on y is 2x − y − 4 = 0 and the regression line of y on x is x − y = 1. An inquiry of 50 families to study the relationship between expenditure on accomodation Rs. Find both the regression co – efficient and the measure of the acute angle between the regression lines n=10. . x−x 2 2 = 20000. y = 300. From the following data available from n = 5. Also find σx if σy = 2. N = 50. 16. y and S. From the data given below find the regression equation of y on x and x on y. The following data about the sales and adv. expenditure of a firm is given below. D. find x. xy = 7900 26 . Find the Mean marks of statistics and correlation of co − efficient. 13. Adv. of y. x = 250. pairs of values of x and y. x = 30.HSC MATHS OMTEX CLASSES 8TH YEAR 9. Estimate the value of x when y = 5. Regression Equations are 8x − 10y + 66 = 0 & 40x − 18y = 214.6 17. For 50 students of a class the regression equation of marks in statistics x on the marks in accountancy y is 3y − 5x + 18 = 0. exp. if the firm proposes a sales target of Rs. If two regression lines for a bivariate data are 10x + 3y − 62 = 0 and 6x + 57 − 50 = 0. If y = 40. P. in order to produce number of radios of each type. Form the LPP to minimize the cost. Pineapple and juice. If one unit of A contains 150 units of vitamins. NO. LINEAR PROGRAMMING 1. 4 on radio I and Rs.P. Formulate the above as LPP to minimize the costs. A pineapple farm produced two products. 2. To produce radio I require 2 hours in plant A and 3 hours in plant B. Formulate the above problem as L. If each unit of type A brings a profit of Rs. 4. 5. Two foods A & B are available at cost of Rs. 4 & Rs. To produce radio II requires 3 hours in plant A & 1 hour in plant B. Food A contains 6 units of vitamins & 7 units of minerals per gram & its cost 24 paisa per gram. A diet for a sick person must contain at least 3000 units of vitamins. If manufacturer makes a profit of Rs.HSC MATHS OMTEX CLASSES 8TH YEAR CH. 50. The amounts of materials. A company produce two types of presentation goods A & B that require gold & silver. A chair requires 2 units of wood & 1 units of board. 8 per unit respectively.P. The company can procure (purchase) 12 gms of silver & 8 gms of gold. In order to earn maximum profit. 12 on radio II. 6. 200. A manufacturer makes two types of Radio I and II. formulate above problem as L. 8. he gets a profit of Rs. From this material. 35 & that of type B Rs. Each unit of type A requires 3 gms of silver & 1 gm of gold while that of B requires 2 gms of gold & 2 gms of silver.P. 75 units of minerals & 1500 calories. labour and equipment required to produce each product & weekly availability of each of these resources are given below: Product Labour Materials Equipments Juice 12 2 50 Pineapple 4 2 25 Availability 1800 500 10000 27 . 2 units of minerals & 35 calories. If he sells one chair. 1 unit of minerals and 30 calories & one unit of B contains 100 units of vitamins. Food B contains 8 units of vitamins & 12 units of minerals per gram & its cost 40 paisa per gram. 3. A table requires 3 units of wood & 4 units of board. A carpenter has 60 units of wood & 50 units of board. 100 & if he sells one table he gets a profit of Rs. formulate this problem as LPP to get the maximum profit. he can prepare chairs and tables. to maximize the profit. Plant A can be operated at the most 15 hours a day & plant B can be operated for at the most 12 hours a day. The daily requirement of vitamins & minerals are 96 units & 128 units respectively. each of which must be processed through two machines A and B. 100 per ton and that for the product Q is Rs. The maximum availability of the machines A and B per day are 12 & 10 hours respectively. form the LPP and find the solution set graphically. Formulate this problem as a linear programming problem for maximum profit. He makes this compound by mixing two compounds.HSC MATHS OMTEX CLASSES 8TH YEAR 7. A Business firm produces two types of products P and Q. 9. The plant consist of three production departments A. C. The number of man hours of labour required in production of each unit of these products & available time is given below. A small factory produces two types of product requiring two process in melting and in machine shop. so that it has atleast 10 litres of A. 400 & Rs. B. The equipment of each department can be used for 8 hours a day. 70 per ton. Product P requires 2 hours in department A and 1 hour in department C per ton. 28 . where as a scooter requires 3 hours in machine A and 6 hours in machine B. 3 litres of B & no C. Each unit of compound I has 4 litres of A. Each unit of Compound II has 1 litres of A. Formulate the problem to minimize the cost. The unit cost of the compounds I & II are Rs. Product Q requires 1 hour in department B and 1 hour in department C per ton. A toy manufacturers produces bicycles and scooters. 600. The average profit for the product P is Rs. Product Melting Machine shop A 10 8 B 6 4 Available time 1000 hrs 600 hrs 8. 10. B and C. A. For manufacturing a bicycle require 4 hours in machine A and 2 hours in machine B. 2 litres of B & 4 litres of C. 12 litres of B & 20 litres of C. A chemist has a compound to be made using three basic elements. 1. the statement ``2 + 3 = 5'' has truth value T. then 𝑝⋁𝑞 is the statement ``The wall is red or the lamp is on (or both)'' whereas𝑝 → 𝑞 is the statement ``If the lamp is on then the wall is red''. we often use letters. such as p. ``not''. ``It is not the case that 2 + 3 = 5'' is the negation of the statement above. If p is the statement ``The wall is red'' and q is the statement ``The lamp is on''.HSC MATHS OMTEX CLASSES 8TH YEAR MATHS – I CH.. Other examples of compound statements are: If you finish your homework then you can watch T. Of course. truth values and truth tables A statement is an assertion that can be determined to be true or false. or the sun is shining or it is raining and the sun is shining.V. In symbolic logic. The statement ~𝑝⋀ 𝑞 translates to ``The wall isn't red and the lamp is on''. ``or''. For example. For example. Thus. Connective Not And Or If …… then …. 29 . if and only if …… Symbol Formal name ~ ⋀ ⋁ Negation Conjunction Disjunction Conditional Bi – conditional → ↔ Note that the connective ``or'' in logic is used in the inclusive sense (not the exclusive sense as in English). I have read this and I understand the concept. q and r to represent statements and the following symbols to represent the connectives. it is stated more simply as ``2 + 3 ≠ 5''. the logical statement ``It is raining or the sun is shining '' means it is raining. LOGIC Statements. Statements that involve one or more of the connectives ``and''. This is a question if and only if this is an answer. The truth value of a statement is T if it is true and F if it is false. NO. ``if then'' and `` if and only if '' are compound statements (otherwise they are simple statements). And assuming that ‘not sad’ is happy. Given p ≡ x is an irrational number. Write the following statements in symbolically. ii. Ram is tall or Shyam is intelligent. translate the following statements into symbolic form. 7. ~p ⋁q where q: Sachin is healthy. 1 SENTENCE CONNECTIVES Negation (not) (~) Disjunction (or) (v) Conjunction (and)(but)(as well as) (𝚲) If then (→) If and only if (↔) 1.S. If a man is happy. 9. 5. Let p : Riyaz passes B. i. ~p → q 3. then he is rich. No. a + b 2 = a2 + b2 if and only if ab = 0. Hari is either intelligent or hard working. ii. i. ii. “Kiran failed or Kiran passed as well as he is happy” 4. q : Riyaz gets a job. 6. Akhila likes mathematics but not chemistry. Express the following statements in verbal form: i. p → q ⋀ r ii.M. If a man is not rich. “A person is successful only if he is a politician or he has good connections”. i. Write the verbal statement for the following. 2. i. 𝐔𝐬𝐢𝐧𝐠 𝐭𝐡𝐞 𝐬𝐭𝐚𝐭𝐞𝐦𝐞𝐧𝐭𝐬: P: Kiran passed the examination. p ⋀ q ⋀ ~r 8. i. Bangalore is a garden city and Mumbai is a metropolitan city. IF the question paper is not easy then we shall not pass. ii. q ≡ x is the square of an integer. ii. Write the following statements in symbolic form.HSC MATHS OMTEX CLASSES 8TH YEAR Ex. p⋀q p: Sachin is smart. r : Riyaz is happy. Write a verbal sentence to describe the following. represent the following statement in symbolic form. Write the following statements in symbolic form. Using appropriate symbols. p ⋀ ~q ii. 30 . Express the following in the symbolic form i. then he is not happy. S: Kiran is sad. [ p ⋁ (~ q ⋀ ~p)] → p viii. [ p ⋁ (~ q ⋀ p)] → p iii. Construct the truth table and determine whether the statement is tautology. ~p ↔ ~q p ∶ Rama is young. ~( ~p ⋀ ~q ) ⋀ ∼ 𝑟 q 𝒑⋁𝒒 T T F T T T F F 31 . Truth table for Disjunction ∨ 𝒑 T T F F 3. 2 Truth Table 1. Ex. Write the following statements in verbal form using p & q. vi. ~( p ⋀ q) iv. p → ~q b. Let p: Rohit is tall. Truth table for Conjunction (∧) 𝒑 q 𝒑 ∧ 𝒒 T T T T F F F T F F F F 1. No. p → (q → p) v. contradiction or neither. i. 𝐆𝐢𝐯𝐞 𝐚 𝐯𝐞𝐫𝐛𝐚𝐥 𝐬𝐭𝐚𝐭𝐞𝐦𝐞𝐧𝐭 𝐟𝐨𝐫 a. p ⋁ (~q ⋀ p). ~p ⋀(~q) b. q ∶ Rama is intellignet. ~( ~p ⋀ ~q ) x. ~ ( p ↔ q). a.HSC MATHS OMTEX CLASSES 8TH YEAR 10. Truth Table for negation (~) 𝑷 ~𝑷 T F F T 2. ( p → ~q) → (q ⋀ ~q) ix. vii. p ⋁(~p ⋀ q) 11. q: Rohit is handsome. ( p → q) ⋀ (q ⋀ ~q) ii. HSC MATHS OMTEX CLASSES 8TH YEAR 2. xviii. Some students are obedient. b. iv. No quadrilateral is a triangle. All doctors are honest. xx. D = Set of dogs. Some mathematicians are wealthy. Can you conclude that some poets are wealthy? vi. Some right-angled triangles are isosceles. Some poets are mathematicians. If a quadrilateral is a rhombus. Some students are lazy. Some dogs are wild. If U = set of all animals. W = Set of all wild animals. Using truth table prove that. Sunday implies a holiday. xvii. Do as directed. Prove that the statement pattern ( p ⋀ q) ⋀ (~p⋁~q) is a contradiction. iii. iv. Write the disjunction of the statements: India is a democratic country. Prove that p ⋀ (~p ⋁ q) ≡ p ⋀ q. All students are intelligent. All triangles are polygons. xiii. All students are lazy. Some students are intelligent. Write the truth table for “Disjunction”. xiv. Show that the statements p → q and ~( p ⋀ ~q) are equivalent. Observe the diagram and state whether the following statements are true or false a. No artist is cruel. All wild animals are dogs. xi. 3. xv. ix. Prove that the following statements are logically equivalent: p → q ≡ ~q → ~p ii. Some students are hard working. xix. iii. Some poets are intelligent. Using the truth table. Using truth table show that. No poet is intelligent. vi. Show that p ↔ q ≡ ( p → q ) ⋀ ( q → p ). p → q ≡ (~p ⋁ q) vii. ii. viii. then it is a parallelogram. v. Some parallelograms are rectangles. p → q ≡ (~q) → (~p) viii. Some doctors are honest. ix. vii. Show that the following pairs of statements are equivalent: p ⋀ q and ~ (p → ~q). i. France is in India. No politician is honest. Represent the following statements by Venn Diagrams: i. xvi. x. 32 . v. xii. No.HSC MATHS OMTEX CLASSES 8TH YEAR Ex. Inverse & contra positive Suppose (𝒑 → 𝒒) is a statement then Converse ≡ 𝒒 → 𝒑 Inverse ≡ ~𝒑 → ~𝒒 Contra positive ≡ ~𝒒 → ~𝒑 1. 3. I like Mathematics or English. All politicians are corrupt. 13. Formula ~ 𝒑⋁𝒒 = ~𝒑 ∧ ~𝒒 ~ 𝒑 ∧ 𝒒 = ~𝒑⋁~𝒒 ~ 𝒑 → 𝒒 = 𝒑 ∧ ~𝒒 ~ 𝒑 ↔ 𝒒 = 𝒑 ∧ ~𝒒 ⋁ 𝒒 ∧ ~𝒑 1. 4. 2. 16. No men is animal. Write the negation of the following. 12. 15. 9. 6. Rajesh is intelligent or 𝟐 is not an irrational number. No. 3. Everest is not in Nepal and Tokyo is in India. If two numbers are equal then their squares are unequal. 4. Ex. 10. 14. All students are hardworking. 11. If weather is humid then it will rain. 33 . 2. 5. 8. 𝒂 + 𝒃 𝟐 = 𝒂𝟐 + 𝒃𝟐 if and only if 𝒂𝒃 = 𝟎. If a quadrilateral is a rectangle then it is a parallelogram. Converse. If 2+5=10 then 4+7 = 21 All students are sincere. 3. 7. If x is zero then we cannot divide by x. 𝒑 ∧ 𝒒 ⋁𝒒 𝒑⋁(𝒒 ∧ ~𝒑) ~𝒑 ∧ 𝒒 ⋁ 𝒑 ∧ ~𝒒 𝒑 ∧ 𝒒 → 𝒑 𝒑⋁𝒒 ↔ 𝒑 ∧ 𝒒 4 is an irrational. The sky is blue. All rational number are natural numbers. Dual : In dual statements (⋀) will be converted into (⋁) and (⋁) will be converted into (⋀) and negation should be kept as it is. 6. 5. Ex. 1. 8. 7. 3 + 3 = 6. 2. London is in Mumbai. 4. 5. 3. Dual of the statements. I don’t like fried rice or cold – drinks. Sit down. 𝑝⋁ 𝑞⋀𝑟 ≡ (𝑝⋁𝑞)⋀(𝑝⋁𝑟) 2. 3 + 9 = 18. X + 5 = 7. 9. You are intelligent. 1. ∼ (𝑝⋀𝑞) ≡∼ 𝑝⋁ ∼ 𝑞 3. 4.HSC MATHS OMTEX CLASSES 8TH YEAR Ex. I like ice – cream and cream – biscuits. 6. 34 . State whether the sentences are statements in logic or not also state their truth value. 23 is a perfect square. 𝑙𝑖𝑚𝑥→3 𝑥−1 8. 𝑙𝑖𝑚→0 10. 1. 𝑙𝑖𝑚𝑥→0 9. 𝑙𝑖𝑚→0 4 2 3 𝑥−2 2 3 10−𝑥 −2 𝑥−2 𝑎+ 6 −𝑎 6 4. 𝑙𝑖𝑚𝑦→𝑏 3. 𝑙𝑖𝑚𝑥→𝑎 𝑥 −8 −𝑎 −8 ( Ex No 4 [Rationalizing] 1. 𝑙𝑖𝑚𝑥→2 8. 𝑙𝑖𝑚𝑥→2 𝑥 2 −5𝑥+6 (−32) 11. 𝑙𝑖𝑚𝑥→−3 7. 𝑙𝑖𝑚𝑥→1 𝑥 3 −8𝑥 2 +16𝑥 𝑥 3 −𝑥−16 𝑥 2 +2𝑥 𝑥−3 2𝑥 2 − 𝑥 3 −𝑥−24 𝑥 7 +𝑥 4 −2 5. 𝑙𝑖𝑚𝑥→3 𝑥+6−3 𝑥 2 −9 2𝑥+1−3 243 𝑥 −6 −𝑎 −6 3𝑎 2 4 1 𝑎+ 8 −𝑎 8 8𝑎 7 𝑥+1 ( 𝑥 3 −4−2 𝑥 2 +4𝑥+5− 𝑥 2 +1) 1 2 1 36 1 21 6. 𝑙𝑖𝑚𝑥→2 5. LIMIT Ex. [Algebraic Limits] − 𝑥 2 −2𝑥 9𝑥 2 6. NO.HSC MATHS OMTEX CLASSES 8TH YEAR CH. 𝑙𝑖𝑚𝑥→ −1 7. 𝑙𝑖𝑚𝑥→2 4. 𝑙𝑖𝑚𝑥→ 6. 𝑙𝑖𝑚𝑥→3 𝑥 3 −6𝑥 2 +9𝑥 [𝑛𝑜𝑡 𝑑𝑒𝑓𝑖𝑛𝑒𝑑] 13. 𝑙𝑖𝑚𝑥→1 𝑥 3 −2𝑥+1 (11) 12. 𝑙𝑖𝑚𝑦→3 2𝑦 2 −3𝑦−9 1. 𝑙𝑖𝑚𝑥→1 4𝑥 3 −𝑥 (3) 2 33 𝑥 3 −2𝑥 2 −4𝑥+8 10. 𝑙𝑖𝑚𝑥→3 𝑥 3 −5𝑥−12 2𝑥 2 −9−3 𝑥 4 −64𝑥 (11) 240 20−𝑥 2 −4 𝑎+𝑥− 𝑎−𝑥 4𝑥 𝑥+2𝑎− 3𝑎 𝑥 2 −𝑎 2 𝑥+− 𝑥 −6 1 4 𝑎 1 1 4 3𝑎 3 𝑥 2 +9−5 2+𝑥− 𝑥 2 +𝑥−7 9−𝑥 2 10. 𝑙𝑖𝑚𝑥→1 15. 𝑙𝑖𝑚𝑥→4 Ex No 3 1 𝑥−2 1 𝑥−3 1 Ex No 2. 𝑙𝑖𝑚𝑥→2 𝑥 8. 𝑙𝑖𝑚𝑥→2 ) 9. No. 𝑙𝑖𝑚𝑥→2 2. 𝑙𝑖𝑚𝑥→4 5. 𝑙𝑖𝑚𝑥→2 𝑥 3 −𝑥 2 −𝑥−2 2. 𝑙𝑖𝑚𝑥→4 𝑥 2 −𝑥−12 3. 𝑙𝑖𝑚𝑥→2 1. 𝑙𝑖𝑚𝑥→3 3. 𝑙𝑖𝑚𝑥→3 𝑥 12 −312 2 − 12 6𝑎5 1 5. 𝑙𝑖𝑚𝑥→2 𝑥 3 −5𝑥 2 +8𝑥−4 4 4. 𝑙𝑖𝑚𝑥→3 4. 𝑙𝑖𝑚𝑥→2 1 𝑥 2 +4𝑥+3 𝑥 2 −5𝑥+6 1 𝑥 2 −9 + 𝑥 2 +8𝑥+15 𝑥 3 −27 1 1 − 𝑥 3 −27 (0) 1 𝑥 2 −3𝑥+2 1 1 − 𝑥 2 +𝑥−12 169 42 −2 1 − 𝑥 𝑥−2 𝑥 2 −5𝑥+6 1 𝑥 2 3𝑥−4 − 𝑥 2 −3𝑥+2 (1) 𝑥−2 + 2𝑥 2 −7𝑥+6 −3 + 𝑥 2 −13𝑥+36 (− 25 ) 𝒍𝒊𝒎 𝒙𝒏 − 𝒂𝒏 = 𝒏𝒂𝒏−𝟏 𝒙→𝒂 𝒙 − 𝒂 6. 2 𝑥 2 −3 2𝑥+4 (−3) 𝑥 3 +6𝑥 2 +9𝑥 3 𝑙𝑖𝑚𝑥→−3 𝑥 3 +5𝑥 2 +3𝑥−9 4 𝑥 5 −243 𝑥 2 −9 135 4 2 9 2 1 2 3 2 2 2 𝑦 2 −4𝑦+3 𝑥 2 + 2𝑥−4 0 𝑥−1 𝑥−1 7 2 (5) 7. [Algebraic Limits] 1. 𝑙𝑖𝑚𝑥→𝑎 𝑥 10 −𝑎 10 (10𝑎 7 ) 2. 𝑙𝑖𝑚𝑥→3 𝑥 3 +𝑥 2 −36 8𝑥 3 −1 𝑥 3 −𝑥−24 𝑥 3 +𝑥 2 −12 16 7 26 9. 𝑙𝑖𝑚𝑥→4 14. 𝑙𝑖𝑚𝑥→𝑎 1 2 5 2. 𝑙𝑖𝑚𝑥→3 8. 2. 𝑙𝑖𝑚𝑥→ 𝑥 2 +3 3𝑥−12 3 𝑥 4 −16 𝑥 4 −9 5 12 3. 𝑙𝑖𝑚𝑥→2 𝑦 5 −𝑏 5 𝑦 9 −𝑏 9 9𝑏 4 𝑥 7 −128 7 𝑥 6 −64 𝑥 8 −38 3 𝑥 3 −𝑎 3 3 𝑥 2 −22 1 1 𝑥 3 −23 𝑥 2 −4 1 1 3 26 5 ( 5 ) 7. 𝑙𝑖𝑚→0 2 𝑥 35 . 𝑙𝑖𝑚𝑥→0 6. 𝑙𝑖𝑚𝑥→0 11. 𝑙𝑖𝑚𝑥→0 13. 𝑙𝑖𝑚𝑥→0 𝒂𝒙 − 𝟏 = 𝒍𝒐𝒈𝒂 𝒙→𝟎 𝒙 𝑥 𝑥 𝑎 −𝑏 8. 𝑙𝑖𝑚𝑥→1 20. 𝑙𝑖𝑚𝑥→0 𝑥𝑡𝑎𝑛𝑥 Ex. 𝑙𝑖𝑚𝑥→0 𝑎 𝑥 −1 1−𝑐𝑜𝑠𝑥 𝑡𝑎𝑛𝑥 𝑥 3 10 𝑥 −2𝑥 −5𝑥 +1 16. 𝑙𝑖𝑚𝑥→0 𝑥 2 +𝑥 𝑠𝑖𝑛 𝑥−1 𝑥 2 +𝑥−2 𝑥 2 3 (8) 1 24 5 1−𝑐𝑜𝑠𝑥 𝑠𝑖𝑛𝑥 1−𝑐𝑜𝑠𝑥 𝑥 3 1−𝑐𝑜𝑠 3𝑥 9 ( 2) 1 2 4𝑥 4 𝑡𝑎𝑛 3𝑥 3 2𝑥 2 𝑥 2 4 𝑠𝑖𝑛 4𝑥𝑠𝑖𝑛 6𝑥 5𝑥 2 𝑥𝑐𝑜𝑠𝑥 +𝑠𝑖𝑛𝑥 𝑥 2 +𝑡𝑎𝑛𝑥 3𝑠𝑖𝑛 2𝑥+2𝑥 2𝜃+3𝑠𝑖𝑛𝜃 5. 𝑙𝑖𝑚𝑥→0 20. 𝑙𝑖𝑚𝑥→3 Ex. 𝑙𝑖𝑚𝑥→2 15. 𝑙𝑖𝑚𝑥→0 3. 𝑙𝑖𝑚𝑥→0 𝑥 0 7. 6. 𝑙𝑖𝑚𝑥→0 3𝑠𝑖𝑛𝑥 +𝑠𝑖𝑛 4𝑥 𝑎𝑏 𝑥 −𝑎 𝑥 𝑏 𝑥+1 4 𝑥 −3𝑥 𝑥−1 𝑒 8𝑥 − 𝑒 5𝑥 − 𝑒 3𝑥 +1 𝑐𝑜𝑠 4𝑥−𝑐𝑜𝑠 10𝑥 7. 𝑙𝑖𝑚𝑥→0 7. 𝑙𝑖𝑚𝑥→0 1 + 2𝑥 2. 𝑙𝑖𝑚𝑥→0 2𝑥 −1 2 𝑥 2 𝑒 𝑥 +3𝑥 +4𝑥 −3 𝑥 𝑎 𝑥 +𝑏 𝑥 −2𝑥 +1 𝑥 4 𝑥 −1 1−𝑐𝑜𝑠𝑥 𝑥 3 18. 𝑙𝑖𝑚𝑥→0 = 𝑒 7 = 𝑒 8 =− 1 𝑒 4 13 1 = 𝑒 3 = 1 5 2𝑥 1 10 𝑒 7 𝑙𝑜𝑔 1+𝑝𝑥 = 𝑝 36 . 𝑙𝑖𝑚𝑥→0 10. 𝑙𝑖𝑚𝑥→0 4. 𝑙𝑖𝑚𝑥→0 𝑠𝑖𝑛𝑥𝑙𝑜𝑔 19. 𝑙𝑖𝑚𝑥→0 𝑥 2 (25) Ex. 𝑙𝑖𝑚𝑥→0 6.HSC MATHS 𝑥+ 3 − 𝑥 3 6+𝑥− 10−𝑥 𝑥 2 −4 𝑎+− 𝑎 𝑎+ 1 2𝑎 3 𝑥 2 1 8 2 OMTEX CLASSES 𝑥 3 + 𝑥+2−10 49 𝑥 2 −4 16 𝑥 2 + 𝑥+6− 12 37 𝑥 2 −9 36 8TH YEAR 11. 𝑙𝑖𝑚𝑥→0 𝑒 3𝑥 −𝑒 𝑥 17. 𝑙𝑖𝑚𝑥→0 𝑡𝑎𝑛 3𝑥 𝑠𝑖𝑛 2 𝑠𝑖𝑛 2𝑥 2 3 2 8 9 5 8 𝑥 2 2 1−𝑐𝑜𝑠𝑚𝑥 𝑚 2 𝑥 2 𝑥𝑡𝑎𝑛𝑥 2 14. 𝑙𝑖𝑚𝑥→0 4. 𝑙𝑖𝑚𝑥→0 19. 𝑙𝑖𝑚𝑥→0 8. 𝑙𝑖𝑚𝑥→0 18. 𝑙𝑖𝑚𝑥→0 3𝑥+2𝑡𝑎𝑛 3𝑥 15. 𝑙𝑖𝑚𝑥→0 (5) 𝑥 10. 𝑙𝑖𝑚𝑥→2 13. 𝑙𝑖𝑚𝑥→0 5. 𝑙𝑖𝑚𝑥→0 12. 𝑙𝑖𝑚→0 14. 𝑙𝑖𝑚𝜃→0 3𝜃+5𝑡𝑎𝑛𝜃 𝒍𝒊𝒎 21. 𝑙𝑖𝑚𝑥→0 3. 𝑙𝑖𝑚𝑥→0 2. 𝑙𝑖𝑚𝑥→0 𝑥𝑠𝑖𝑛𝑥 6𝑥 −3𝑥 −2𝑥 +1 𝑥 2 5𝑥 +5−𝑥 −2 15. 𝑙𝑖𝑚𝑥→0 13. 𝑙𝑖𝑚𝑥→0 17. 7. [Logarithmic Limits] 𝑒 𝑎 +𝑥 −𝑒 𝑎 𝑥 𝑎 𝑥 −𝑏 𝑥 𝑥 32𝑥 −23𝑥 𝑠𝑖𝑛𝑥 5𝑥 −3𝑥 4 𝑥 −1 3𝑥 −2𝑥 2 1−𝑐𝑜𝑠 2𝑥 7𝑥 +8𝑥 +9𝑥 −3𝑥 +1 𝑎 3𝑥 −𝑎 2𝑥 −𝑎 𝑥 +1 𝑥 5𝑥 1. 𝑙𝑖𝑚𝑥→0 14. 𝑙𝑖𝑚𝑥→0 𝑡𝑎𝑛𝑥 9. 𝑙𝑖𝑚𝑥→0 𝑥𝑐𝑜𝑡𝑥 1 𝑠𝑖𝑛𝑥 6. 𝑙𝑖𝑚𝑥→0 𝑆𝑖𝑛 3 2𝑥 𝑥 3 𝑥 𝑠𝑖𝑛 2 2 16. 5 Trigonometric Limits 𝑠𝑖𝑛 25𝑥 𝑥 𝑠𝑖𝑛𝜋𝑥 𝑥 𝑠𝑖𝑛 5𝑥 𝒍𝒊𝒎 1. No. 𝑙𝑖𝑚𝑥→0 1 − 5𝑥 3 4𝑥 7 5 𝑥 3 𝑥 4 𝑥 𝒙→𝟎 𝟏 𝒙 = 𝒆 1+3𝑥 𝑥 1−4𝑥 4𝑥+1 𝑥 1−4𝑥 4−8𝑥 𝑥 4+5𝑥 𝑥 1 = 𝑒 10 = 𝑒 12 20 5. 𝑙𝑖𝑚𝑥→0 11. 𝑙𝑖𝑚𝑥→0 1 + 4𝑥 3. 𝑙𝑖𝑚𝑥→0 12. 𝑙𝑖𝑚𝑥→0 1 + 4. 𝑙𝑖𝑚𝑥→0 2. Exponential Limits 𝒍𝒊𝒎 𝟏 + 𝒙 1. 𝑙𝑖𝑚→0 12. 𝑙𝑖𝑚𝑥→0 25 𝜋 5 𝒔𝒊𝒏𝒙 𝒕𝒂𝒏𝒙 = 𝟏 & 𝒍𝒊𝒎 = 𝟏 𝒙→𝟎 𝒙 𝒙→𝟎 𝒙 2 +5𝑥 𝑠𝑖𝑛 𝑥 9. 𝑙𝑖𝑚𝑥→0 1−𝑐𝑜𝑠𝑥 (2) 8. 𝑙𝑖𝑚𝑥→0 12. 𝑙𝑖𝑚𝑥→0 𝑐𝑜𝑠 12𝑥−𝑐𝑜𝑠 4𝑥 = 𝑠𝑒𝑐𝑥 −1 𝑥 2 𝑠𝑒𝑐𝑥 +1 𝑠𝑖𝑛 𝑥 + −𝑠𝑖𝑛𝑥 3 𝑥𝑡𝑎𝑛𝑥 𝑐𝑜𝑠 8𝑥−𝑐𝑜𝑠 2𝑥 = (24) 17. 𝑙𝑖𝑚𝑥→𝑎 14. 8. 𝐼𝑓 𝑓 𝑥 = 3𝑥 + 𝑥 2 𝑓𝑖𝑛𝑑 𝑙𝑖𝑚→0 2. 3. 𝑙𝑖𝑚𝑥→𝑎 𝑐𝑜𝑡𝑥 −𝑐𝑜𝑡𝑎 = (− 𝑠𝑖𝑛2 𝑎) Ex. 𝑙𝑖𝑚→0 11. 𝑙𝑖𝑚𝑥→2 𝑥 − 1 18. 𝑙𝑖𝑚𝑥→1 18. 𝑙𝑖𝑚𝑥→0 9. 𝑙𝑖𝑚𝑥→𝜋 3 = 4 3 = (−𝑠𝑖𝑛𝑎) 12. 𝑙𝑖𝑚𝑥→0 𝑥 Ex. Using first principle find 𝐟 ′ 𝐱 or Find 1. 2. 𝑙𝑖𝑚𝑥→0 10. 𝑙𝑖𝑚𝑥→0 8. 𝑙𝑖𝑚𝑥→1 2 1−𝑥 𝑐𝑜𝑠𝑒𝑐𝑥 −1 1+𝑐𝑜𝑠 𝜋𝑥 1−𝑥 1−𝑡𝑎𝑛𝑥 2 2 𝜋 −𝑥 2 =(𝜋) = = 1 2 5.HSC MATHS OMTEX CLASSES 𝑙𝑜𝑔 1+3𝑥 8TH YEAR 𝑙𝑜𝑔𝑥 −𝑙𝑜𝑔 3 𝑥−3 𝑙𝑜𝑔𝑥 −1 𝑥−𝑒 1 9. 𝑓 𝑥 = 2𝑥 + 1 9. 𝑓 13. 𝑓 𝑓 𝑓 𝑓 𝑥 𝑥 𝑥 𝑥 = 𝑥 = 𝑥 2 = 𝑥 3 1 = 𝑥 = 𝐥𝐢𝐦𝐡→𝟎 𝐟 𝐱+𝐡 −𝐟(𝐱) 𝐡 11.1) 𝑥 𝑥 +1 𝑙𝑜𝑔 10+𝑙𝑜𝑔 10 =3 14. 𝑓 𝑥 = 𝑥 Ex. 𝑓 𝑥 = 𝑐𝑜𝑠𝑥 𝑓 2+ −𝑓 2 1. 𝑙𝑖𝑚𝑥→𝜋 16. 𝑓 𝑥 𝑥 𝑥 𝑥 = 𝑐𝑜𝑠5𝑥 = 𝑠𝑖𝑛2 𝑥 = 𝑎2𝑥 = 𝑙𝑜𝑔 3𝑥 + 2 5. 𝑙𝑖𝑚𝑥→𝑒 = 𝑒 1 𝑥 −2 1 𝑥 −4 1 =3 1 16. 𝑙𝑖𝑚𝑥→0 4. 𝑙𝑖𝑚𝑥→3 =5 = 10 2 2 15. 𝑙𝑖𝑚𝑥→0 11. 𝑙𝑖𝑚→0 10. 𝟕 = 2𝑥 2 − 3𝑥 + 5 𝟗 𝑓 3+ −𝑓 3 𝐼𝑓 𝑙𝑖𝑚→0 𝑖𝑠 𝑡𝑜 𝑏𝑒 𝑓𝑖𝑛𝑑 𝑜𝑢𝑡. 3. 𝑙𝑖𝑚𝑥→1 𝑥 𝑥 −1 = 𝑒 17. 𝑙𝑖𝑚𝑥→𝑎 2𝑠𝑖𝑛𝑥 −𝑠𝑖𝑛 2𝑥 𝑡𝑎𝑛𝑥 −𝑠𝑖𝑛𝑥 𝑥 3 𝑥 𝑙𝑜𝑔 7+𝑥 −𝑙𝑜𝑔 7−𝑥 =1 =7 = 𝑒 = 𝑒 = (1) 1 2 13. 𝑙𝑖𝑚𝑥→𝜋 19. 𝑓 14. 𝑙𝑖𝑚𝑥→0 𝑥 𝑙𝑜𝑔 5+𝑥 −𝑙𝑜𝑔 5−𝑥 𝑥 𝑙𝑜𝑔 10+𝑙𝑜𝑔 (𝑥+0. 𝑙𝑖𝑚𝑥→𝜋 20. 9. 4. 𝑙𝑖𝑚𝑥→𝜋 2 4 4 2 𝜋 2 2 = 𝜋−4𝑥 𝑐𝑜𝑠𝑥 −𝑠𝑖𝑛𝑥 = 16 𝑡𝑎𝑛 𝑥+ −𝑡𝑎𝑛𝑥 𝑐𝑜𝑠𝑥 −𝑐𝑜𝑠𝑎 𝑥−𝑎 𝑥−𝑎 = (𝑐𝑜𝑠𝑥) = (𝑠𝑒𝑐 𝑥) 𝜋−4𝑥 3𝑐𝑜𝑠𝑥 +𝑐𝑜𝑠 3𝑥 𝜋−2𝑥 3 5+𝑐𝑜𝑠𝑥 −2 𝜋−𝑥 2 3−𝑡𝑎𝑛𝑥 𝜋−3𝑥 = 1 = 2 2 1 2 1 8 21. 𝑓 𝑥 = 𝑐 1 7. 𝑤𝑒𝑟𝑒 𝑓 𝑥 𝑓 𝑥 −𝑓 1 𝟏 𝐹𝑖𝑛𝑑 𝑙𝑖𝑚𝑥→1 𝑥 2 −1 𝑤𝑒𝑟𝑒 𝑓 𝑥 = 𝑥 2 + 3 𝟒 𝑓 −3+ −𝑓 −3 1 𝟏 𝐹𝑖𝑛𝑑 𝑙𝑖𝑚→0 𝑤𝑒𝑟𝑒 𝑓 𝑥 = − 𝑥−5 𝟔𝟒 37 . Trigonometric Limits 1. 𝑙𝑖𝑚𝑥→0 𝑐𝑜𝑠 4𝑥−𝑐𝑜𝑠 10𝑥 = 𝑐𝑜𝑠 4𝑥−𝑐𝑜𝑠 8𝑥 7. 10. 𝑙𝑖𝑚𝑥→0 2. 4. 𝑙𝑖𝑚𝑥→𝜋 2 𝑡𝑎𝑛𝑥 − 𝑡𝑎𝑛𝑎 𝑥−𝑎 𝑠𝑒𝑐𝑥 −𝑡𝑎𝑛𝑥 𝑠𝑖𝑛 𝜋𝑥 𝜋 −𝑥 2 =2 1 2 𝑠𝑒𝑐 2 𝑎 𝑡𝑎𝑛 𝑎 𝑥 3 𝑠𝑖𝑛 3𝑥−𝑠𝑖𝑛 5𝑥 𝑥 𝑐𝑜𝑠 3𝑥−𝑐𝑜𝑠𝑥 𝑥 2 𝑥 2 = = 1 2 = −2 = (−4) 1 42 15 1 32 15. 𝑓 𝑥 = 𝑥 8. 𝑓 12. 𝑙𝑖𝑚𝑥→4 𝑥 − 3 13. 6. 𝑓 𝑥 = 𝑠𝑖𝑛𝑥 10. 𝑙𝑖𝑚𝑥→0 3. 𝑙𝑖𝑚𝑥→0 6. 𝑙𝑖𝑚𝑥→𝜋 22. 𝑓𝑜𝑟 𝑥 ≠ 3. 𝑓 𝑥 = 𝑠𝑖𝑛𝑥𝑙𝑜𝑔 (1+𝑥) = 2𝑙𝑜𝑔3 3𝑥 −1 2 4. 𝑎𝑡 𝑥 = 0. 𝑓𝑜𝑟 𝑥 ≠ 0. 𝑓𝑜𝑟 0 ≤ 𝑥 < 2. 𝑎𝑡 𝑥 = 0. 𝑓𝑜𝑟 𝑥 = 0. 𝑓 𝑥 = 5. 𝑎𝑡 𝑥 = 1. 𝑓 𝑥 = 5𝑐𝑜𝑠𝑥 −1 𝜋 −𝑥 2 = 2𝑙𝑜𝑔5 9. 𝑎𝑡 𝑥 = 0. 𝑓𝑜𝑟 𝑥 = 1. 𝑓𝑜𝑟 𝑥 = 0. 3 3 𝜋 𝑓𝑜𝑟 𝑥 ≠ 2 . 𝑓 𝑥 = 2𝑥 + 3 =4 13. 𝑎𝑡 𝑥 = 4. 𝑎𝑡 𝑥 = . 𝑓𝑜𝑟 𝑥 ≠ 4. 1. 𝑓 𝑥 = 7. 𝑓 𝑥 = = 𝑥𝑐𝑜𝑠𝑥 +3𝑡𝑎𝑛𝑥 𝑥 2 +𝑠𝑖𝑛𝑥 𝑥+6−3 𝑥 2 −9 =4 1 2 6. 𝑓𝑜𝑟 𝑥 = 0. 𝑓𝑜𝑟 𝑥 = 0. 𝑓 𝑥 = 𝑥 2 −16 𝑥−4 =9 11. 𝑎𝑡 𝑥 = . 𝑓𝑜𝑟0 ≤ 𝑥 < 2. No. 𝑎𝑡 𝑥 = 2. 𝜋 𝜋 𝑓𝑜𝑟 𝑥 = . 𝑓 𝑥 = = 𝑒 5𝑥 −𝑒 2𝑥 𝑠𝑖𝑛 3𝑥 3−𝑡𝑎𝑛𝑥 𝜋−3𝑥 =1 4 3 8. 𝑓𝑜𝑟 𝑥 ≠ 0. 𝑎𝑡 𝑥 = 0. 𝑎𝑡 𝑥 = 3.HSC MATHS 5. 𝑓𝑜𝑟 𝑥 = 4. 2 2 𝑓𝑜𝑟 𝑥 ≠ 0. 𝑓 𝑥 = 𝑥 2 − 𝑥 − 1 = 4𝑥 + 1 12. 𝑓 𝑥 = 𝑥 + 3 − 2 =2 3. 𝑓𝑜𝑟 0 ≤ 𝑥 < 3. 𝑒 3𝑥 −1 𝑠𝑖𝑛𝑥 𝑥 2 1. 𝑓𝑜𝑟 𝑥 ≠ 3 . 𝑓𝑜𝑟 2 ≤ 𝑥 ≤ 4 . 𝑓𝑜𝑟 𝑥 = 0. 𝑓𝑜𝑟 𝑥 ≠ 1. 𝑓𝑜𝑟 3 ≤ 𝑥 ≤ 6. 𝐹𝑖𝑛𝑑 OMTEX CLASSES 𝑓 2+ −𝑓(2) 𝑤𝑒𝑟𝑒 𝑓 1+ −𝑓(1) 𝑙𝑖𝑚→0 𝑤𝑒𝑟𝑒 𝑓 𝑥 8TH YEAR − 𝟗 𝟏 𝑓 𝑥 = 𝑥 2 +2 = 𝑥+1 (−𝟏) 𝑥+5 1 Continuity Ex. Discuss the continuity for the following functions and if the function discontinues. 𝐸𝑣𝑎𝑙𝑢𝑎𝑡𝑒 𝑙𝑖𝑚→0 6. 𝑓 𝑥 = 𝑓𝑜𝑟 𝑥 ≠ 0. 𝑓𝑜𝑟 𝑥 = 𝜋 𝜋 . 𝑓𝑜𝑟 2 ≤ 𝑥 ≤ 5. 𝑎𝑡 𝑥 = 3. determine whether the discontinuity is removable. 𝑎𝑡 𝑥 = 2. 𝑎𝑡 𝑥 = 0. 38 𝜋 =4 2. 𝑓𝑜𝑟 𝑥 = 3. 𝑓 𝑥 = 5 𝑙𝑜𝑔 3 = 𝑙𝑜𝑔2 5𝑥 −3𝑥 2𝑥 −1 10. 𝑓𝑜𝑟 𝑥 ≠ 0. 𝑓 𝑥 = 𝑥 2 − 𝑥 + 5 = 2𝑥 + 5 . 𝑓𝑜𝑟 𝑥 ≠ 0. 𝑓𝑜𝑟 𝑥 ≠ 0. 𝑓𝑖𝑛𝑑 𝑓 0 . 𝑖𝑠 𝑐𝑜𝑛𝑡𝑖𝑛𝑢𝑜𝑢𝑠 𝑎𝑡 𝑥 = 0. 1.HSC MATHS OMTEX CLASSES 8TH YEAR Ex. 𝑖𝑠 𝑐𝑜𝑛𝑡𝑖𝑛𝑢𝑜𝑢𝑠 𝑎𝑡 𝑥 = 0. 𝑓𝑜𝑟 𝑥 = 0. 39 . 𝐼𝑓 𝑓 𝑥 = 𝑐𝑜𝑠 3𝑥−𝑐𝑜𝑠𝑥 𝑥 2 1−𝑐𝑜𝑠𝑘𝑥 𝑓𝑜𝑟 𝑥𝑡𝑎𝑛𝑥 15 𝑥 −3𝑥 −5𝑥 +1 . = 2 𝑥 2 + 1 + 𝛽 𝑓𝑜𝑟 𝑥 > 0. 𝐼𝑓 𝑓 𝑥 = 3𝑠𝑖𝑛𝑥 −1 2 𝑥𝑙𝑜𝑔 (1+𝑥) . 2. 𝑥 ≠ 0. 𝐼𝑓 𝑓 𝑥 = 5. 3. 𝑓𝑜𝑟 𝑥 < 0 𝑓𝑖𝑛𝑑 𝛼&𝛽 𝑖𝑓 𝑓 0 = 2. 𝑓𝑜𝑟 𝑥 < 0 𝑓𝑖𝑛𝑑 𝛼&𝛽 𝑖𝑓 𝑓 2 = 4. 𝑓𝑜𝑟 𝑥 < 0. 9. 𝐼𝑓 𝑓 𝑥 𝑖𝑠 𝑐𝑜𝑛𝑡𝑖𝑛𝑢𝑜𝑢𝑠 𝑎𝑡 𝑥 = 0 𝑤𝑒𝑟𝑒 𝑓(𝑥) = = 𝑘 𝑎 𝑥 − 𝑎−𝑥 . 𝑓𝑖𝑛𝑑 𝑓 0 . 𝑖𝑠 𝑐𝑜𝑛𝑡𝑖𝑛𝑢𝑜𝑢𝑠 𝑎𝑡 𝑥 = 0. 𝑥𝑡𝑎𝑛𝑥 3𝑥 −1 𝑒 . = 2 𝑥 2 + 1 + 𝛽 𝑓𝑜𝑟 𝑥 ≥ 0. 𝑓𝑜𝑟 𝑥 ≠ 0. 𝐼𝑓 𝑓 𝑥 𝑖𝑠 𝑐𝑜𝑛𝑡𝑖𝑛𝑢𝑜𝑢𝑠 𝑎𝑡 𝑥 = 0 𝑤𝑒𝑟𝑒 𝑓 𝑥 = 𝑥 2 + 𝛼. = 𝑙𝑜𝑔 + 𝑏𝑥) (1 4𝑥 8. 𝐼𝑓 𝑓 𝑥 = = 1. 𝑥 𝑓𝑜𝑟 𝑥 ≠ 0. 𝑓𝑜𝑟 𝑥 ≠ 0. 𝑖𝑓 𝑓 0 = 3. 𝑓𝑖𝑛𝑑 𝑎 & 𝑏. 𝑓𝑜𝑟 𝑥 = 0 𝑓𝑖𝑛𝑑 𝑘. 𝑎𝑥 7𝑠𝑖𝑛𝑥 −1 2 6. 𝑓𝑜𝑟 𝑥 > 0 𝑖𝑠 𝑐𝑜𝑛𝑡𝑖𝑛𝑢𝑜𝑢𝑠 𝑎𝑡 𝑥 = 0. 𝐼𝑓 𝑓 𝑥 𝑖𝑠 𝑐𝑜𝑛𝑡𝑖𝑛𝑢𝑜𝑢𝑠 𝑎𝑡 𝑥 = 0 𝑤𝑒𝑟𝑒 𝑓 𝑥 = 𝑥 2 + 𝛼. 2. 𝑖𝑠 𝑐𝑜𝑛𝑡𝑖𝑛𝑢𝑜𝑢𝑠 𝑎𝑡 𝑥 = 0. 𝑓𝑖𝑛𝑑 𝑘. 𝑓𝑖𝑛𝑑 𝑓 0 . 𝐼𝑓 𝑓 𝑥 = 𝑥𝑙𝑜𝑔 (1+5𝑥) 4. 𝑓𝑖𝑛𝑑 𝑓 0 . 𝐼𝑓 𝑓 𝑥 = 7. No. 𝑖𝑠 𝑐𝑜𝑛𝑡𝑖𝑛𝑢𝑜𝑢𝑠 𝑎𝑡 𝑥 = 0. 5𝑎 𝑥 + 𝑥 3 + 𝑙𝑜𝑔𝑥 2 EX. 32. 39. 3. NO. 4𝑥 3 − 5𝑥 2 + 8𝑥 − 1 3. 24. 19. 9. 20. 𝑠𝑖𝑛𝑥 1−𝑐𝑜𝑠𝑥 1+𝑠𝑖𝑛𝑥 1−𝑠𝑖𝑛𝑥 1+𝑡𝑎𝑛𝑥 1−𝑡𝑎𝑛𝑥 𝑒 𝑥 −1 𝑒 𝑥 +1 3+𝑠𝑖𝑛𝑥 1+3𝑠𝑖𝑛𝑥 1+𝑐𝑜𝑠𝑥 𝑥𝑠𝑖𝑛𝑥 𝑐𝑜𝑠𝑥 +𝑠𝑖𝑛𝑥 𝑐𝑜𝑠𝑥 −𝑠𝑖𝑛𝑥 1−𝑠𝑖𝑛 2𝑥 1+𝑠𝑖𝑛 2𝑥 1−𝑐𝑜𝑠 2𝑥 1+𝑐𝑜𝑠 2𝑥 𝑥𝑠𝑖𝑛𝑥 −𝑐𝑜𝑠𝑥 𝑥𝑐𝑜𝑠𝑥 +𝑠𝑖𝑛𝑥 2 40. 35. 10. If 𝑦 = 𝑡𝑎𝑛𝑥. 4. 12. 𝑡𝑎𝑛−1 𝑥 + 𝑠𝑒𝑐 −1 𝑥 𝑥 + 1 2 𝑥 𝑥 2 + 𝑠𝑖𝑛−1 𝑥 + 𝑐𝑜𝑠 −1 𝑥 7.HSC MATHS OMTEX CLASSES 8TH YEAR CH. 29. 5𝑥 𝑐𝑜𝑠𝑥 𝑥 𝑥+1 1+𝑥 1−𝑥 1+ 𝑥 1− 𝑥 18. 𝑃𝑟𝑜𝑣𝑒 𝑡𝑎𝑡 𝑑𝑥 = 𝑠𝑒𝑐 2 𝑥 𝑏𝑦 𝑢𝑠𝑖𝑛𝑔 𝑡𝑒 𝑟𝑢𝑙𝑒 𝑜𝑓 𝑑𝑥 EX. 11. NO. 14. 25. 6. 𝑐𝑜𝑠𝑥 𝑥. 1. 16. 15. 33. 4. 15. 5𝑥 𝑒 𝑥 𝑒 𝑥 𝑥 3 4𝑥 𝑥 4 𝑥 −3 3𝑥 𝑒 𝑥 𝑠𝑖𝑛𝑥 𝑥𝑠𝑖𝑛−1 𝑥 𝑒 𝑥 𝑠𝑒𝑐𝑥 𝑠𝑒𝑐𝑥. 𝑥 𝑠𝑖𝑛𝑥 + 3𝑥 𝑡𝑎𝑛𝑥 41. 21. 𝑑𝑦 𝑥 2 +1 𝑥 2 −1 3𝑥−5 2𝑥+3 𝑥 3 −5𝑥+2 2𝑥+1 𝑥 2 +𝑥−1 𝑥 2 +𝑥−3 𝑥 𝑥−1 𝑥−2 𝑠𝑖𝑛𝑥 𝑙𝑜𝑔𝑥 𝑐𝑜𝑠𝑥 1+𝑥 𝑒 𝑥 𝑙𝑜𝑔𝑥 𝑥 2 +𝑥−1 1+𝑠𝑖𝑛𝑥 𝑥+ 𝑥 𝑥+1 𝑥 1+𝑙𝑜𝑔𝑥 𝑠𝑖𝑛𝑥 1+𝑐𝑜𝑠𝑥 𝑑 𝑢 𝑣 30. 10. 14. 𝑥 2 + 1 5 3 2 8. 38. 37. 𝑥 𝑥 2 + 1 9. 1 3 2𝑥 3 − 5𝑥 + 1 𝑎2 − 𝑥 2 1 𝑥 2 +1 1 2𝑥+1 2𝑥 2 −5𝑥+1 2𝑥 + 1 + 1−𝑥 1+𝑥 1 𝑥 2 +3 𝑥 + 𝑥 𝑠𝑖𝑛 3𝑥 𝑐𝑜𝑠 8𝑥 + 5 𝑠𝑖𝑛2 𝑥 𝑠𝑖𝑛3 𝑥 𝑠𝑖𝑛𝑥 𝑠𝑖𝑛−1 𝑐𝑜𝑠𝑥 40 . 𝑡𝑎𝑛𝑥 𝑥𝑡𝑎𝑛𝑥 2 𝑥 𝑙𝑜𝑔𝑥. 1. 1. 𝑒 𝑥 𝑠𝑖𝑛𝑥. 𝑥 − 1 (𝑥 − 2) 2𝑥 + 1 𝑐𝑜𝑠𝑥. 3. 8. 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 2. 27. 12. 1 3 + 1 𝑥 − 𝑥 𝑛 4 9. NO. 5 𝑥 1 8. 5. 17. 5. 36. 7. 7. 𝑎𝑥 𝑏 + 𝑥 3 − 𝑙𝑜𝑔𝑥 6. 1. 13. 34. 22. 𝑥 𝑎 + 𝑎 𝑥 + 𝑒 𝑥 + 𝑎𝑎 4. 11. 3. 16. DIFFERENTIATION (DERIVATIVES) EX. 2. 23. 3. NO. 6. 7𝑥 3 + 𝑥 + 𝑠𝑖𝑛𝑥 − 2𝑐𝑜𝑠𝑥 4 5. 28. 2. 26. 31. 13. 10. 2. 𝐃𝐢𝐟𝐟𝐞𝐫𝐞𝐧𝐭𝐢𝐚𝐭𝐞 𝐭𝐡𝐞 𝐟𝐨𝐥𝐥𝐨𝐰𝐢𝐧𝐠 𝐰𝐢𝐭𝐡 𝐫𝐞𝐬𝐩𝐞𝐜𝐭 𝐭𝐨 𝐱. 𝑆. 𝑥 𝑎 2 OMTEX CLASSES 38. 36. 𝑑𝑥 = 1 − 𝑦 2 . 4. 𝑑𝑦 53. 6. 37. 𝑡𝑎𝑛−1 𝑥 2 19. 𝑙𝑜𝑔 𝑒 𝑥 48. 𝑠𝑖𝑛𝑥 1 8TH YEAR 𝑙𝑜𝑔 𝑙𝑜𝑔𝑥 𝑙𝑜𝑔𝑥 2 𝑠𝑖𝑛2 𝑒 3𝑥 𝑠𝑖𝑛(𝑚 𝑠𝑖𝑛−1 𝑥) 𝑒 𝑚 2𝑠𝑖𝑛𝑥 𝑠𝑖𝑛2 𝑥 + 𝑐𝑜𝑠 2 𝑥 32𝑥 + 𝑙𝑜𝑔 𝑥 − 3 𝑙𝑜𝑔 𝑠𝑖𝑛𝑥 + 𝑐𝑜𝑠𝑥 𝑠𝑒𝑐𝑥 + 𝑡𝑎𝑛𝑥 1 𝑎 𝑐𝑜𝑠 𝑥+𝑏 𝑠𝑖𝑛 𝑥 1 + 𝑥 2 + 𝑥 1 𝑥+ 1+𝑥 2 1 𝑠𝑖𝑛 −1 𝑥 𝑦 = 7𝑥+𝑥 𝑒 𝑠𝑖𝑛𝑥 +𝑐𝑜𝑠𝑥 𝑙𝑜𝑔 −1 𝑥) (𝑡𝑎𝑛 𝑙𝑜𝑔 𝑠𝑒𝑐𝑥 + 𝑡𝑎𝑛𝑥 (𝑙𝑜𝑔 𝑥)3 44.HSC MATHS 17. 1 − 𝑥 2 𝑝𝑢𝑡 𝑥 = sin 𝜃 𝑜𝑟 cos 𝜃 41 . 𝑎 39. 𝑙𝑜𝑔 𝑒 3𝑥 49. 32. 26. 31. 𝑒 𝑐𝑜𝑠𝑥 𝑥 3 𝑐𝑜𝑠 𝑥 2 𝑒 𝑎𝑥 𝑠𝑖𝑛 𝑏𝑥 + 𝑐 𝑥 2 + 𝑎 2 + 𝑥 2 + 𝑏 2 𝑥 2 +𝑒 3𝑥 𝑆. NO. 5. 34. 29. 21. 𝑠𝑖𝑛−1 (𝑠𝑖𝑛 𝜃) = 𝜃 𝑐𝑜𝑠 −1 𝑐𝑜𝑠 𝜃 = 𝜃 𝑡𝑎𝑛−1 𝑡𝑎𝑛 𝜃 = 𝜃 𝑐𝑜𝑠𝑒𝑐 −1 𝑐𝑜𝑠𝑒𝑐 𝜃 = 𝜃 𝑠𝑒𝑐 −1 𝑠𝑒𝑐 𝜃 = 𝜃 𝑐𝑜𝑡 −1 𝑐𝑜𝑡 𝜃 = 𝜃 𝑎 𝑏 −1 𝑎 𝑠𝑖𝑛 𝑏 −1 𝑎 𝑐𝑜𝑠 𝑏 𝑏 𝑎 −1 𝑏 = 𝑐𝑜𝑠𝑒𝑐 𝑎 −1 𝑏 = 𝑠𝑒𝑐 𝑎 7. 𝑙𝑜𝑔 𝑙𝑜𝑔 𝑙𝑜𝑔𝑥 46. 𝑥 𝑎 2 −𝑥 2 2 + 𝑎 2 𝑥 𝑠𝑖𝑛−1 𝑎 2 3 47. If 𝑦 = 𝑐𝑜𝑠𝑒𝑐 𝑥 − 𝑐𝑜𝑡 𝑥 . 22. 𝑡𝑎𝑛−1 𝑥 18. 𝑠𝑖𝑛−1 20. 𝑙𝑜𝑔 51. 28. 43. 27. If 𝑦 𝑥−2 5 𝑥+3 𝑥+1 𝑥−5 1−𝑐𝑜𝑠2𝑥 1+𝑐𝑜𝑠2𝑥 𝑐𝑜𝑠 𝑥 1−𝑠𝑖𝑛 𝑥 𝑠𝑖𝑛𝑥 1+𝑐𝑜𝑠𝑥 𝑒 𝑥 −𝑒 −𝑥 = 𝑒 𝑥 +𝑒 −𝑥 𝑎 𝑥 2 . 33. 42. INVERSE [FORMULAE] Formulae 1. 8. 30. 35. 𝑙𝑜𝑔 50. 𝑑𝑥 = 2𝑠𝑖𝑛𝑥 𝑑𝑦 𝑦 EX. 1. 24. 𝑡𝑎𝑛−1 = 𝑐𝑜𝑡 −1 Standard substitutions. 9. 𝑙𝑜𝑔 52. 4. 23. 𝑇. 2. 𝑙𝑜𝑔 𝑥 + 𝑥 2 − 𝑎2 45. 41. 3. 25. 40. 𝑇. 40. 6. 46. 𝑐𝑜𝑠𝑒𝑐 −1 11. 𝑡𝑎𝑛−1 26. 𝑠𝑖𝑛 1 1+𝑥 2 2𝑥 𝑠𝑖𝑛−1 1+𝑥 2 −1 −1 4𝑥 1+4𝑥 2 18. 𝑡𝑎𝑛−1 29. 3. 𝑡𝑎𝑛−1 30. 2𝑥 1 − 𝑥 2 𝑝𝑢𝑡 𝑥 = sin 𝜃 EX. 𝑐𝑜𝑠 −1 1 − 𝑥 2 3. 12. 2𝑥 1−𝑥 2 2 −1 1−𝑥 𝑠𝑖𝑛 1+𝑥 2 1+𝑥 2 𝑠𝑒𝑐 −1 1−𝑥 2 −1 𝑎 2 +𝑥 2 𝑎 1+𝑥 2 1−𝑥 2 𝑥 𝑥 1−𝑥 2 1−𝑐𝑜𝑠 2𝑥 1+𝑐𝑜𝑠 2𝑥 42. 𝑐𝑜𝑡 −1 33. 48. 𝑡𝑎𝑛 −1 31. 49. 𝑡𝑎𝑛−1 32. 𝑠𝑖𝑛−1 1 − 𝑥 2 2. 4 𝑥 1−4𝑥 3 1+𝑥 2 𝑐𝑜𝑡 −1 𝑥−𝑥 2 1+𝑥𝑠𝑖𝑛𝑥 𝑥−𝑠𝑖𝑛𝑥 𝑥 −1 𝑡𝑎𝑛 1+20𝑥 𝑥+ 𝑥 𝑡𝑎𝑛−1 1− 𝑥 3 8𝑥 𝑡𝑎𝑛−1 1−16𝑥 2 𝑐𝑜𝑠 5𝑥+𝑠𝑖𝑛 5𝑥 𝑐𝑜𝑡 −1 𝑐𝑜𝑠 5𝑥−𝑠𝑖𝑛 5𝑥 −1 1−3𝑥 3 3𝑥−𝑥 3 1+𝑥 𝑠𝑒𝑐 −1 2 𝑥 2𝑥 2 𝑠𝑖𝑛−1 1+𝑥 4 8𝑥 𝑠𝑖𝑛−1 1+16𝑥 2 1+25𝑥 2 𝑐𝑜𝑠𝑒𝑐 −1 10𝑥 1−𝑥 𝑐𝑜𝑠 −1 1+𝑥 1+𝑥+ 1−𝑥 𝑠𝑖𝑛−1 2 −1 −1 1 + 𝑥 2 + 𝑥 36. 𝑥 + 3 4 2𝑥 + 1 5 2 3𝑥 − 1 1 3 2. 45. 𝑡𝑎𝑛−1 16. 𝑥 1−2𝑥 3 2−3𝑥 4 3 2 42 .HSC MATHS 2. 4. 𝑡𝑎𝑛−1 50. 𝑠𝑖𝑛−1 5. INVERSE [SUMS] 1. 3𝑥 − 4𝑥 3 8. 𝑐𝑜𝑠 −1 1 − 2𝑥 2 9. 47. 𝑠𝑒𝑐 14. 𝑐𝑜𝑡 𝑐𝑜𝑠𝑒𝑐𝑥 + 𝑐𝑜𝑡𝑥 43. 6. 39. 𝑡𝑎𝑛−1 25. 𝑡𝑎𝑛−1 22. 𝑡𝑎𝑛 1+𝑥 2 −1 𝑥 𝑠𝑖𝑛 𝑥 𝑡𝑎𝑛−1 1+𝑐𝑜𝑠𝑥 −1 𝑥− 𝑠𝑒𝑐𝑥 + 𝑡𝑎𝑛𝑥 𝑎 2 −𝑥 2 𝑥+ 𝑎 2 −𝑥 2 𝑥 2 𝑎 2 −𝑥 2 2𝑐𝑜𝑠𝑥 +3𝑠𝑖𝑛𝑥 13 𝑐𝑜𝑠𝑥 + 3𝑠𝑖𝑛𝑥 2 𝑏𝑥 +𝑎 𝑏−𝑎𝑥 𝑎+𝑏𝑐𝑜𝑠𝑥 𝑏−𝑎𝑐𝑜𝑠𝑥 5𝑥 1−6𝑥 2 𝑎𝑠𝑖𝑛𝑥 +𝑏𝑐𝑜𝑠𝑥 𝑎𝑐𝑜𝑠𝑥 −𝑏𝑠𝑖𝑛𝑥 3−2𝑡𝑎𝑛𝑥 2+3𝑡𝑎𝑛𝑥 𝑥 1+56𝑥 2 24. 𝑡𝑎𝑛−1 𝑠𝑒𝑐𝑥 + 𝑡𝑎𝑛𝑥 37. 𝑡𝑎𝑛−1 17. 𝑠𝑖𝑛 7. 𝑠𝑖𝑛 𝑥 + 𝑐𝑜𝑠 1 − 𝑥 EX. 𝑐𝑜𝑠 −1 28. 5. NO. 1 + 𝑥 2 𝑝𝑢𝑡 𝑥 = tan 𝜃 𝑜𝑟 cot 𝜃 𝑥 2 − 1 𝑝𝑢𝑡 𝑥 = sec 𝜃 𝑜𝑟 𝑐𝑜𝑠𝑒𝑐 𝜃 2𝑥 1+𝑥 2 1−𝑥 2 1+𝑥 2 3𝑥−𝑥 3 1−3𝑥 2 OMTEX CLASSES 8TH YEAR 𝑝𝑢𝑡 𝑥 = tan 𝜃 𝑝𝑢𝑡 𝑥 = tan 𝜃 𝑝𝑢𝑡 𝑥 = tan 𝜃 7. 𝑡𝑎𝑛−1 20. 27. 𝑐𝑜𝑠 −1 15. 𝑐𝑜𝑡 −1 44. 𝑐𝑜𝑠 −1 10. LOGARITHMIC FUNCTIONS 1. 1−𝑠𝑖𝑛𝑥 1+𝑠𝑖𝑛𝑥 1−𝑥 1+𝑥 34. NO. 4. 𝑡𝑎𝑛−1 21. 𝑠𝑖𝑛−1 2𝑥 1 − 𝑥 2 4. 𝑡𝑎𝑛−1 35. 𝑠𝑒𝑐 −1 1 1−2𝑥 2 1 23. 𝑡𝑎𝑛 −1 13. 𝑡𝑎𝑛−1 38. 41. 𝑡𝑎𝑛−1 19. 5. 4. 16. 𝑥 10. 𝑠𝑜𝑤 𝑡𝑎𝑡 𝑑𝑥 = 𝑥 𝑥 2 −𝑦 2 𝑑𝑦 𝑦 𝑠𝑖𝑛−1 𝑥 2 +𝑦 2 = 𝑙𝑜𝑔 𝑎. 𝑥 2 + 𝑦 2 + 2𝑔𝑥 + 2𝑓𝑦 + 𝑐 = 0 𝑥 3 + 𝑦 3 = 3𝑎𝑥𝑦 𝑥 3 + 𝑦 3 = 3𝑎𝑥 2 𝑦 𝑎𝑥 2 + 2𝑥𝑦 + 𝑏𝑦 2 + 2𝑔𝑥 + 2𝑓𝑦 + 𝑐 = 0. Show that 21. If 23. 1 + 𝑥 3 𝑥 𝑥 29. 𝑥 𝑒 𝑎𝑥 𝑠𝑒𝑐𝑥 . show that 𝑑𝑥 = 𝑥 𝑥 2 −𝑦 2 𝑑𝑦 𝑐𝑜𝑠 −1 𝑥 2 +𝑦 2 = 2𝑘. If 26. NO. 20. 18. 6. 9. If 24. 𝑥 𝑥 8.HSC MATHS 3 1 OMTEX CLASSES 17. 𝑥 𝑥 + 𝑥 28. If 𝑒 𝑥 + 𝑒 𝑦 = 𝑒 𝑥+𝑦 show that = − 𝑒 𝑦 𝑒 𝑥 19. 3. 𝑙𝑜𝑔𝑥 −1 8TH YEAR 3. 6.𝑙𝑜𝑔𝑥 1−2𝑥 𝑥 𝑙𝑜𝑔𝑥 𝑥 + 𝑥 𝑠𝑖𝑛 𝑥 𝑥 𝑥 + 𝑠𝑖𝑛𝑥 𝑥 (𝑠𝑖𝑛 𝑥)𝑙𝑜𝑔𝑥 𝑙𝑜𝑔𝑥 𝑥 𝑐𝑜𝑠 𝑥 𝑠𝑖𝑛𝑥 𝑐𝑜𝑠𝑥 𝑥 1+𝑥 2 𝑡𝑎𝑛𝑥 𝑥 1+𝑥 2 𝑠𝑖𝑛 −1 𝑥 1−𝑥 2 𝑥 𝑠𝑖𝑛𝑥 1+𝑥 2 𝑥 7. 16. If 𝑑𝑦 𝑥 𝑥 2 −1+1 = −𝑦 2 𝑑𝑥 𝑥 𝑥 2 −1 𝑥+𝑦 𝑑𝑦 𝑦 𝑠𝑒𝑐 𝑥−𝑦 = 𝑎. (𝑡𝑎𝑛−1 𝑥) 𝑥 11. 𝑥 𝑙𝑜𝑔 18. 𝑥 𝑐𝑜𝑠 −1 𝑥 𝑥 𝑐𝑜𝑠𝑥 𝑐𝑜𝑠𝑥 𝑥 𝑥 𝑥 𝑥 𝑠𝑖𝑛 −1 𝑥 𝑥 𝑙𝑜𝑔𝑥 26. If 𝑦 = 𝑥 2 𝑥+1 𝑥 then show that 𝑑𝑥 = 𝑑𝑦 𝑥 2 𝑥+1 𝑥 𝑥+2 𝑥+1 + 𝑙𝑜𝑔 𝑥 2 𝑥+1 EX. 2. 14. 8. If 𝑥𝑦 + 𝑦𝑠𝑒𝑐 −1 𝑥 = 1. 𝑥 + 𝑎 𝑥 + 𝑥 𝑎 27. 15. show that 𝑑𝑥 = 𝑥 43 . 5. show that 𝑦 𝑑𝑥 = 𝑥 𝑡𝑎𝑛2 𝑘 𝑥 2 −𝑦 2 𝑑𝑦 1−𝑡𝑎𝑛𝑎 𝑡𝑎𝑛−1 𝑥 2 +𝑦 2 = 𝑎. 22. 14. 17. show that 𝑦 𝑑𝑥 = 𝑥 1+𝑡𝑎𝑛𝑎 𝑑𝑦 𝑦 𝑥 4 𝑦 5 = 𝑥 + 𝑦 9 . 23. 5. 13. 2𝑥−1 2 𝑥+1 3 3𝑥−2 3 𝑎+𝑥 𝑎−𝑥 5 𝑥 2 + 5 6. 4. 12. 𝑠𝑖𝑛−1 𝑥 𝑐𝑜𝑠 𝑥 12. 25. show that 𝑑𝑥 = 𝑥 𝑥 3 −𝑦 3 𝑑𝑦 𝑦 𝑙𝑜𝑔 𝑥 3 +𝑦 3 = 𝑎. 15. 𝑦 = 𝑐𝑜𝑠 𝑥 + 𝑦 𝑥 = 𝑦 + 𝑡𝑎𝑛−1 𝑦 𝑥 + 𝑦 = 𝑠𝑖𝑛 𝑥𝑦 𝑥 + 𝑦 = 𝑥 𝑦 = 𝑥 2 𝑒 𝑥𝑦 𝑥𝑦 = 𝑙𝑜𝑔 𝑥𝑦 𝑥 𝑦 = 𝑠𝑖𝑛𝑥 𝑦 𝑦 = 𝑥. IMPLICIT FUNCTION 1. 𝑠𝑖𝑛𝑦 2 2 2 10. 𝑦 𝑥 2 + 𝑦 2 = 𝑥 + 𝑦 𝑥 2 𝑦 2 = 𝑥 2 − 𝑦 2 𝑥 + 𝑒 𝑥 = 𝑦 + 𝑒 𝑦 𝑒 𝑥 + 𝑒 𝑦 = 𝑥 + 𝑦 𝑥𝑦 = 𝑒 𝑥−𝑦 𝑑𝑦 𝑑𝑥 11. If 25. 7. 21. 19. 𝑙𝑜𝑔𝑥 𝑥 𝑥 9. If 22. 13. 24. 𝑥 3 + 𝑦 3 = 𝑎3 20. 7. NO. 𝑦 = 𝑎 𝑐𝑜𝑠 3 𝜃 . If 𝑑𝑦 𝑦 = 𝑥 1−𝑥𝑐𝑜𝑠𝑦 𝑑𝑥 𝑑𝑦 𝑙𝑜𝑔𝑥 𝑥 𝑦 = 𝑒 𝑥−𝑦 . If 32. 𝑦 = 𝑒 𝑠𝑖𝑛 2𝑡 . 𝑦 = 𝑒 𝑠𝑖𝑛 2𝑡 . If 𝑒 𝑥 = 𝑥 𝑦 . show that 𝑑𝑥 = 1+𝑙𝑜𝑔𝑥 2 𝑑𝑦 𝑦 1+𝑥𝑦 𝑦 = 𝑥. If 𝑥 = 𝑐𝑜𝑠𝑥𝑦. show that 𝑑𝑥 = 𝑥 𝑑𝑦 𝑑𝑦 𝑦 𝑠𝑖𝑛 2 (𝑎+𝑦) 𝑠𝑖𝑛𝑎 8TH YEAR 28. If 𝑦 = 𝑥 𝑙𝑜𝑔𝑦 . If 𝑦 = 𝑥. 𝑦 = 1+𝑚 3 . 𝑦 = 𝑎 𝑡𝑎𝑛 𝜃. If 𝑥 = 1+𝑚 3 . 𝑃𝑟𝑜𝑣𝑒 𝑡𝑎𝑡 𝑎 𝑎𝑚 𝑑𝑦 𝑑𝑦 𝑑𝑥 + 𝑐𝑜𝑡 𝜃 = 0 𝑑𝑦 𝜃 8. If 𝑦 = 1 + 𝑥𝑒 𝑦 . show that 𝑑𝑥 = 1−𝑦𝑙𝑜𝑔𝑘 𝑑𝑦 𝑑𝑥 33.HSC MATHS 27. 𝑃𝑟𝑜𝑣𝑒 𝑡𝑎𝑡 𝑑𝑥 + 𝑏 𝑐𝑜𝑡 𝜃 = 0 𝑑𝑦 5.r. If 𝑥 = 𝑠𝑖𝑛2 𝜃 and 𝑦 = 𝑡𝑎𝑛 𝜃 . show that 30.to 𝑠𝑖𝑛 𝑥 𝑑𝑦 1 𝑡 2. Prove that 38. 44 . If 𝑥 3 𝑦 𝑘 = 𝑥 + 𝑦 3+𝑘 OMTEX CLASSES . If 𝑥 = 𝑎𝑐𝑜𝑠 𝜃 . If 𝑥 = 12. If 𝑙𝑜𝑔 𝑥 2 + 𝑦 2 = 𝑡𝑎𝑛−1 40. 𝑦 = 𝑎 1 − 𝑐𝑜𝑠𝜃 . Differentiate 𝑡𝑎𝑛𝑥 w. 𝑠𝑖𝑛 𝑎 + 𝑦 . 𝑒 𝑥𝑦 . If 𝑥 = 𝑎 𝜃 + 𝑠𝑖𝑛𝜃 . 𝑠𝑜𝑤 𝑡𝑎𝑡 𝑑𝑥 + 𝑡𝑎𝑛 2𝑡 𝑑𝑦 𝑦𝑙𝑜𝑔𝑥 𝑒 𝑐𝑜𝑠 2𝑡 . 𝑓𝑖𝑛𝑑 𝑑𝑦 𝑑𝑥 𝑎𝑡 𝜃 = 𝑑𝑦 𝜋 4 6.r. If 𝑥 𝑦 = 5𝑥−𝑦 . 𝑥 + 𝑦 = 𝑎 𝑦 𝑥 𝑚 = 𝑑𝑦 − 1+𝑦 1−𝑥 2 𝑥 1−𝑥 2 𝑒 𝑦 35. If 𝑥 = 𝑎 𝑠𝑒𝑐 𝜃 . 𝑠𝑜𝑤 𝑡𝑎𝑡 𝑑𝑥 = 3. 𝛉. 𝑠𝑜𝑤 𝑡𝑎𝑡 𝑑𝑥 = 10. If 𝑥 = 13. 𝑦 = 𝑏 𝑠𝑖𝑛 𝜃 . show that 37.to 𝑥 3 4. 𝐦 are parameters 1. PARAMETER FUNCTIONS In the following problems 𝐭. If 31. show that 𝑑𝑥 = 29. 𝑆𝑜𝑤 𝑡𝑎𝑡 = − 𝑑𝑥 𝑥𝑙𝑜𝑔𝑦 2𝑏𝑡 1−𝑡 2 𝑑𝑢 𝑏 2 𝑣 2 . 𝑠𝑜𝑤 𝑡𝑎𝑡 𝑑𝑥 = = 𝑡𝑎𝑛 2 9. show that 34. 𝑦 𝑑𝑦 show that 𝑑𝑥 𝑥 𝑑𝑦 𝑥𝑙𝑜𝑔 5−𝑦 show that 𝑑𝑥 = 𝑥𝑙𝑜𝑔 5𝑥 = 𝑥+𝑦 𝑥−𝑦 EX. If 𝑠𝑖𝑛 𝑦 = 𝑥. show that 𝑑𝑥 = 2−𝑦 36. Differentiate 𝑥 2 + 1 w. show that 𝑑𝑥 = 𝑥 1−𝑥𝑦 𝑑𝑦 𝑦𝑙𝑜𝑔𝑘 𝑦 = 𝑘 𝑥+𝑦 . show that 𝑑𝑥 = 𝑐𝑜𝑠𝑒𝑐 𝜃 7. 𝑣 = 𝑎 1+𝑡 2 . 𝑦 = 2𝑎𝑡. If 𝑢 = 2𝑚 3 −1 3𝑚 2 2 𝑎 1−𝑡 2𝑎𝑡 𝑑𝑦 𝑡 2 −1 2 . Prove that 𝑑𝑥 𝑑𝑦 𝑑𝑥 𝑦 2 𝑥 𝑦−𝑥 𝑙𝑜𝑔𝑥 −1 𝑙𝑜𝑔 𝑥 2 𝑙𝑜𝑔𝑦 2 𝑙𝑜𝑔𝑦 −1 = = = 39. 𝑠𝑜𝑤 𝑡𝑎𝑡 𝑑𝑥 = 2𝑡 1+𝑡 𝑑𝑦 𝑒 𝑠𝑖𝑛 2𝑡 −𝑐𝑜𝑠 2𝑡 𝑒 𝑐𝑜𝑠 2𝑡 . 𝑦 = 1+𝑡 2 . 𝑠𝑜𝑤 𝑡𝑎𝑡 𝑑𝑣 = − 𝑎 2 𝑢 1+𝑡 = 0. 𝑠𝑖𝑛𝑦. If 𝑒 𝑦 = 𝑑𝑦 𝑑𝑥 𝑑𝑦 𝑦 𝑥 . If 𝑥 = 𝑎𝑡 2 . If 𝑥 = 𝑎 𝑠𝑖𝑛3 𝜃 . If 𝑥 = 11. 𝑦 = 2𝑠𝑖𝑛𝜃 − 𝑠𝑖𝑛2𝜃.HSC MATHS 14. 3𝜃 2 20. Differentiate 𝑐𝑜𝑠 −1 2𝑥 − 1 with respect to 1 − 𝑥 2 17. 𝑠𝑜𝑤 𝑡𝑎𝑡 𝑑𝑥 = 𝑐𝑜𝑡𝜃 18. with respect to 𝑠𝑒𝑐 −1 𝑥 𝑑𝑦 16. Differentiate 𝑥. 𝑠𝑖𝑛𝑥 with respect to 𝑡𝑎𝑛𝑥 𝑑𝑦 𝑐 2 𝑐𝑜𝑠𝑥 1+𝑠𝑖𝑛𝑥 2 𝑡𝑎𝑛𝜃 + 𝑐𝑜𝑡𝜃 . Differentiate 𝑙𝑜𝑔𝑥 15. If 𝑥 = 𝑎 𝑠𝑖𝑛𝜃 − 𝜃 𝑐𝑜𝑠 𝜃 . If 𝑥 = 2𝑐𝑜𝑠𝜃 − 𝑐𝑜𝑠2𝜃 . 𝑦 = 19. 𝑠𝑜𝑤 𝑡𝑎𝑡 𝑑𝑥 = 𝑡𝑎𝑛 45 . If 𝑥 = 𝑐𝑙𝑜𝑔 𝑡𝑎𝑛𝜃 . 𝑦 = 𝑎 𝑐𝑜𝑠 𝜃 + 𝜃𝑠𝑖𝑛𝜃 . Differentiate 𝑡𝑎𝑛−1 𝑥 OMTEX CLASSES 8TH YEAR with respect to 𝑙𝑜𝑔𝑥. 𝑡𝑒𝑛 𝑠𝑜𝑤 𝑡𝑎𝑡 𝑑𝑦 𝑑𝑥 = − 𝑐𝑜𝑡 2𝜃. 001 14. Find approximately. Find approximately. the value of cos(89030’).HSC MATHS OMTEX CLASSES 8TH YEAR Ex.389 8. 1 Approx.7 4 13. log3 = 1. 3 3 4.1 3 2.0986. the value of log10 1016 given log10 e = 2. 9. given e = 2.01 given log3 = 1. the value of tan(450 30′ ) given 10 = 0.0866.0986 and 31. the value of 26. the value of e2.07 4 6.5 16. 3 3. Find approximately the value of log e 101 given log e 10 = 203026 11. Find approximately. No. the value of 100. Find approximately. 10. given 10 = 0. Find approximately. Find approximately. Find approximately.01 .0175c . Find approximately.0175c 17. Find approximately.96 to four decimal places. 4. 𝐑𝐚𝐝𝐢𝐚𝐧 =𝐜 CH. the value of 5 32. the value of log 9. 12.002 .0175c 19. 2 Error 𝐕𝐨𝐥𝐮𝐦𝐞 𝐨𝐟 𝐭𝐡𝐞 𝐜𝐮𝐛𝐞 = 𝐬𝐢𝐝𝐞 𝐕𝐨𝐥𝐮𝐦𝐞 𝐨𝐟 𝐭𝐡𝐞 𝐬𝐩𝐡𝐞𝐫𝐞 = 𝟑 𝐕𝐨𝐥𝐮𝐦𝐞 𝐨𝐟 𝐭𝐡𝐞 𝐡𝐞𝐦𝐢𝐬𝐩𝐡𝐞𝐫𝐞 = 𝟒 𝟑 𝛑𝐫 𝟑 𝟐 𝟑 𝛑𝐫 𝟑 𝐒𝐮𝐫𝐟𝐚𝐜𝐞 𝐚𝐫𝐞𝐚 𝐨𝐟 𝐭𝐡𝐞 𝐬𝐩𝐡𝐞𝐫𝐞 = 𝟒𝛑𝐫 𝟐 𝐒𝐮𝐫𝐟𝐚𝐜𝐞 𝐚𝐫𝐞𝐚 𝐨𝐟 𝐭𝐡𝐞 𝐡𝐞𝐦𝐢𝐬𝐩𝐡𝐞𝐫𝐞 = 𝟑𝛑𝐫 𝟐 𝐕𝐨𝐥𝐮𝐦𝐞 𝐨𝐟 𝐭𝐡𝐞 𝐜𝐲𝐥𝐢𝐧𝐝𝐞𝐫 = 𝛑𝐫 𝟐 𝐡 𝐕𝐨𝐥𝐮𝐦𝐞 𝐨𝐟 𝐭𝐡𝐞 𝐜𝐨𝐧𝐞 = 𝟏 𝟐 𝛑𝐫 𝐡 𝟑 𝐒𝐮𝐫𝐟𝐚𝐜𝐞 𝐚𝐫𝐞𝐚 𝐨𝐟 𝐭𝐡𝐞 𝐜𝐮𝐛𝐞 = 𝟔 𝐬𝐢𝐝𝐞 𝐀𝐫𝐞𝐚 𝐨𝐟 𝐭𝐡𝐞 𝐬𝐪𝐮𝐚𝐫𝐞 = 𝐬𝐢𝐝𝐞 𝐀𝐫𝐞𝐚 𝐨𝐟 𝐭𝐡𝐞 𝐫𝐞𝐜𝐭𝐚𝐧𝐠𝐥𝐞 = 𝐥𝐛 𝟑 𝐬𝐢𝐝𝐞 𝟒 𝟐 𝟐 𝐕𝐨𝐥𝐮𝐦𝐞 𝐨𝐟 𝐭𝐡𝐞 𝐜𝐮𝐛𝐨𝐢𝐝 = 𝐥𝐛𝐡 𝐀𝐫𝐞𝐚 𝐨𝐟 𝐚𝐧 𝐞𝐪𝐮𝐢𝐥𝐚𝐭𝐞𝐫𝐚𝐥 𝐭𝐫𝐢𝐚𝐧𝐠𝐥𝐞 = 𝟐 46 .0866. 80.083 15.1.999 7. the value of e1. tan−1 0. sin300 = 0. APPLICATION OF DERIVATIVES 𝑫𝒆𝒈𝒓𝒆𝒆 =𝟎 𝑴𝒊𝒏𝒖𝒕𝒆𝒔 =′ 𝑺𝒆𝒄𝒐𝒏𝒅𝒔 = " 𝟏𝟎 = 𝟔𝟎′ 1.0175c cos300 = 0.3026. given 10 = 0. Find approximately. the value of 4.1 given e2 = 7. the value of sin 310 . Find approximately. the value of f x = 2x 3 + 7x + 1 at x = 2. No. Find approximately. Find approximately. Find approximately.001 . given 10 = 0.71828. Find approximately.5 Ex. 63 5. the value of 511 3 . 64.5 18. Find approximately. the value of 28 to three decimal place. the value of 5x 2 + 80 x 1 1 at x = 5. Find approximately. the value of cos(30030’). sin300 = 0.1 4 & 3. the value of 997. cos300 = 0. Find approximately. the value of tan−1 1. 3% in the measurement of the radius of spherical balloon.06 cm. Find approximately a. Approximate error b. Find the corresponding error in calculating the volume of the cube. Relative error c. If the radius of a spherical balloon increases 0. Find the approximate % increase in its volume. 5. The side of a square is 5 meter is incorrectly measured as 5. where g is a constant. 3.01cm. It is 8. where P = Pressure. If an edge of a cube is measured as 2m with an possible error of 0.01cm. Find approximately the error in calculating the volume. Find approximate error in the calculated value of its area. if an error of 2% is made in measuring its radius. 4. Relative error c. 10. Under ideal conditions a perfect gas satisfies the equation PV = K. Percentage error in calculating the volume. 47 . Relative error c. 9.05 per unit.06cm. Percentage error in calculating the volume. Radius of a sphere is measures as 25 cm with an error of 0. find the %error in its volume. Consequent error b. 6. The diameter of a spherical ball is found to be 2cm with a possible error of 0. Side of an equilateral triangle is measured as 6cm with a possible error of 0. 13. 12.2%. Percentage error in the surface area of the sphere. Radius of a sphere is found to be 24cm with the possible error of 0. If an error of 2cm is made in measuring the edge. Radius of the sphere is measured as 12 cm with an error of 0. if the error in the measurement of length (L) is 1. Find the approx error in the surface area of the cube having an edge of 3m. Also find the percentage error. 7. Find the % error in the measure of period. V = Volume and K = Constant. Find approximately the possible error in the calculated value of the volume of the ball.082mm.5 cm.1%. Find a. The volume of a cone is found by measuring its height and diameter of base as 7 cm and 5 cm respectively. Find up to one decimal place the resulting error in the calculation of the area of sphere. Time (T) for completing certain length (L) is given by the equation T = 2π l g 1. If K = 60 and Pressure is found by measurement to be 1. If an error of 0. Find a.5 unit with error of 0. Approximate error b. Find the approximate % error in calculating the volume of a sphere.4mm. Find the consequent error in the volume. 2. 11. 14.11 meters.HSC MATHS OMTEX CLASSES 𝒅𝒚 8TH YEAR 𝑨𝒑𝒑𝒓𝒐𝒙𝒊𝒎𝒂𝒕𝒆 𝒆𝒓𝒓𝒐𝒓 = 𝜹𝒚 = 𝒅𝒙 𝜹𝒙 𝑹𝒆𝒍𝒂𝒕𝒊𝒗𝒆 𝒆𝒓𝒓𝒐𝒓 𝒊𝒏 𝒙 = 𝑹𝒆𝒍𝒂𝒕𝒊𝒗𝒆 𝒆𝒓𝒓𝒐𝒓 𝒊𝒏 𝒚 = 𝜹𝒙 𝒙 𝜹𝒚 𝒚 𝜹𝒙 𝒙 𝜹𝒚 𝒚 𝑷𝒆𝒓𝒄𝒆𝒏𝒕𝒂𝒈𝒆 𝒆𝒓𝒓𝒐𝒓 𝒊𝒏 𝒙 = 𝑷𝒆𝒓𝒄𝒆𝒏𝒕𝒂𝒈𝒆 𝒆𝒓𝒓𝒐𝒓 𝒊𝒏 𝒚 = × 𝟏𝟎𝟎 × 𝟏𝟎𝟎 found that the diameter is measured incorrectly to the extent of 0. Divide 100 in two part. such that sum of twice of one part and square of the other is minimum. 8x 3 − 75x 2 + 150x 2. 6. No. Divide 70 in two part. Product of two natural numbers is 36. The sum of their square is minimum. 4. 11. i. Find two Natural numbers x and y such that i. 12. Their product is maximum ii. The perimeter of a rectangle is 100 cm. 48 . 8. 16. Output ′Q′. 13. Find its dimensions when its areas is maximum. Find them when their sum is minimum. x + y = 60 and xy 3 is maximum. 18 and the sum of the square is minimum. such that i. Product of two Natural Number is 144. ii. such that the sum of their squares is minimum. Find its dimensions so as to have the largest volume. 15. Find them when their sum is minimum. Where 3 x is the input. Find the length of its sides when its area is maximum. Find two Natural Number whose sum is i. 14. A metal wire 36cm long is bent to form a rectangle. Find the position of the point P on seg AB of length 12cm. Perimeter of a rectangle is 48cm. is given by Q = 10 + 60x + 7x 2 2 − x3 . 16 and the sum of the cube is minimum. ii. 17. x + y = 6 and x 2 y is maximum. 5.HSC MATHS OMTEX CLASSES 8TH YEAR Ex. 7. Divide 10 in two part. 3. Find its dimensions so as to use the minimum area of sheet metal. so that the product of their square of one part and fourth power of the other is maximum. 10. Find Input for which output ‘Q’ is maximum. 9. An open tank with a square base is to be constructed so as to hold 4000 cu.mt. A box with a square base and open top is to be made from a material of area 192 sq. iii. Find the maximum volume of a right circular cylinder if the sum of its radius and height is 6 mts. Divide 12 in two parts. 2x 3 − 15x 2 + 36x + 10 iii. 30 and product is maximum. x 3 − 9x 2 + 24x ii. cm. Examine each of the function for Maximum and Minimum. Find the length of sides when its area is maximum. MAXIMA AND MINIMA 1. so that AP2 + BP2 is minimum. of water. 3. 19. 𝑠𝑖𝑛2 𝑥 𝑐𝑜𝑠 2 𝑥 11. 28. 𝑥 2 + 2𝑥 + 3 20. 1−𝑐𝑜𝑠 2𝑥 28. 1 1−𝑥 2 2 𝑥 2𝑥 3 1 𝑥 4 𝑥 𝑥 3 −2𝑥 2 +5𝑥−7+ 𝑥𝑎 𝑥 𝑥 1 𝑥 𝑥+1 (𝑥+2) 2 1 𝑥 34. 9. Integrate the following functions 1. 𝑒 5𝑥 + 3−5𝑥 15. 1−𝑐𝑜𝑠𝑥 16. 5. 3𝑥 2 +5 41. 4𝑥 3𝑥 16. 43. 1−𝑐𝑜𝑠𝑥 18. 𝑐𝑜𝑠 3 𝑥 9. 𝑠𝑖𝑛3 𝑥𝑐𝑜𝑠 3 𝑥 12. 𝑠𝑖𝑛2 𝑥 2. 31. 1−𝑐𝑜𝑠 2𝑥 17. 5. 𝑠𝑒𝑐 2 3𝑥 − 1 31. 1+𝑐𝑜𝑠𝑥 32. 𝑠𝑖𝑛𝑥 − 𝑐𝑜𝑠𝑥 30. 39. 1 5𝑥 2 +4 1 4𝑥 2 +25 1 1 44. 𝑥+2 𝑥+3 𝑥 32. 𝑐𝑜𝑠𝑥 10. 6. 𝑐𝑜𝑡 2 𝑥 7. 53𝑥 13. 1 + 𝑥 1 2 1 1 4−9𝑥 2 1 5−3𝑥 2 1 1 1 2𝑥 + 1 + 1 3𝑥−2 3 𝑥 25. 25−9𝑥 2 45. 3. 38. 3 23. 10. INDEFINITE INTEGRATION Ex. 23. 𝑥 − 1 2 𝑥 19. 𝑠𝑖𝑛 2 𝑥𝑐𝑜 𝑠 2 𝑥 1 2 1+𝑐𝑜𝑠𝑥 2 1−2𝑐𝑜𝑠𝑥 𝑠𝑖𝑛 2 𝑥 3𝑐𝑜𝑠𝑥 −4 24. 𝑠𝑖𝑛𝑥𝑠𝑒𝑐 2 𝑥 𝑐𝑜𝑠𝑥 33. 𝑡𝑎𝑛2 𝑥 6. 4 3 𝑥 2 1 18. 20. 𝑐𝑜𝑠 3𝑥 5. 𝑠𝑒𝑐𝑥 +𝑡𝑎𝑛𝑥 27. 7. 21. 1 + 𝑐𝑜𝑠2𝑥 1 − 𝑐𝑜𝑠2𝑥 1 + 𝑐𝑜𝑠𝑥 1 + 𝑠𝑖𝑛2𝑥 1 + 𝑠𝑖𝑛𝑥 1 − 𝑠𝑖𝑛𝑥 1 1+𝑐𝑜𝑠𝑥 1 1 1 25. 1+𝑐𝑜𝑠𝑥 15. 1 37.HSC MATHS OMTEX CLASSES 8TH YEAR CH. 2. 22. NO. 33. 4𝑥 3 2. 𝑐𝑜𝑠 2 𝑥𝑠𝑖 𝑛 2 𝑥 29. 4𝑥 52𝑥 𝑥− 𝑥−1 1 𝑎+𝑥− 𝑎 1 3𝑥+10− 3𝑥−7 𝑥+1 𝑥−2 𝑥+2 𝑥+3 𝑥+ 1−𝑥 2 𝑥 1−𝑥 2 40. 3𝑥 2 +4 42. 𝑒 𝑥 2𝑥 17. 𝑠𝑖𝑛 2 2 𝑥 2 14. 4. 7 12. 21. 𝑐𝑜𝑡 𝑥 − 𝑠𝑖𝑛 5𝑥 + 3 + 1 . 𝑠𝑖𝑛𝑥. 𝑠𝑖𝑛𝑥𝑐𝑜𝑠𝑥𝑠𝑖𝑛2𝑥 13. 1−𝑐𝑜𝑠 2𝑥 49 . 24. 𝑠𝑖𝑛 2 𝑥 𝑠𝑖𝑛 2 𝑥 𝑐𝑜𝑠 2𝑥 𝑡𝑎𝑛𝑥 4. 8. 4−9𝑥 2 Ex. 𝑥 + 𝑥 26. No. 1 Integrate the following functions 1. 𝑐𝑜𝑠 2 𝑥 3. 1+𝑠𝑖𝑛𝑥 𝑐𝑜𝑠𝑥 𝑠𝑖𝑛𝑥 26. 3𝑥 2𝑥 14. 1 3𝑥+5 𝑥 −3 + 𝑥 + 2 1 36. 1−𝑠𝑖𝑛𝑥 34. No. 30. 9+𝑥 2 11. 𝑠𝑖𝑛 2 𝑥 35. 29. 𝑡𝑎𝑛2 3𝑥 − 𝑠𝑖𝑛 4𝑥 + 3 − 𝑐𝑜𝑠𝑒𝑐 2 𝑥 2 𝑥 3 2𝑥 + 5 1 3𝑥−2 1 3𝑥−2 3 22. 𝑠𝑖𝑛3 𝑥 8. 27. No. Integrate the following functions Note: . 1. 5𝑥 2 −6𝑥+3 2𝑥+1 𝑥 2 −1 𝑥 10. 𝑡𝑎𝑛 −1 𝑠𝑖𝑛𝑥 1−𝑐𝑜𝑠𝑥 𝑠𝑖𝑛 2𝑥 1+𝑐𝑜𝑠2𝑥 Ex.𝑙𝑜𝑔𝑥 10. 𝑡𝑎𝑛−1 𝑐𝑜𝑠𝑥 1+𝑠𝑖𝑛𝑥 37. 3𝑥 2 𝑡𝑎𝑛2 𝑥 3 7. 𝑥 2 +6𝑥+10 46. 29. 𝑥 𝑐𝑜𝑠𝑥 𝑛 5. 𝑠𝑖𝑛 2 39. 4. 𝑐𝑜𝑠𝑥 1+𝑠𝑖𝑛𝑥 𝑥 𝑒 𝑥 1+𝑥 𝑥𝑒 𝑥 1 𝑙𝑜𝑔𝑥 3 2 33. [ 𝑡𝑎𝑛 −1 𝑥) 1+𝑥 2 2 30. 𝑥𝑐𝑜 𝑠 2 23. 𝑥 2 +5𝑥−1 44. Integration by Substitution 1. 6. 𝑥𝑐𝑜 𝑠 2 34. 𝑙𝑜𝑔 1 1+ 𝑥 𝑥 2 25. 12. − 𝑥 2 𝑥 𝑥+1 𝑙𝑜𝑔 (𝑡𝑎𝑛 ) 𝑠𝑖𝑛𝑥 𝑐𝑜𝑡𝑥 𝑠𝑖𝑛𝑥 𝑐𝑜𝑠 3 𝑥 𝑠𝑖𝑛𝑥 2 27. 𝑡𝑎𝑛−1 38. 𝑥+3𝑠𝑖𝑛𝑥 48. 3. 2𝑥 2 +3 5𝑥 4𝑥+10 15. 𝑠𝑖𝑛 𝑥 𝑥 𝑛 −1 2 18. 25−𝑐𝑜𝑠 2 𝑥 11. 19. 3. 28. 8. 𝑥 𝑥 1 𝑙𝑜𝑔𝑥 2+𝑙𝑜𝑔𝑥 2 1+𝑥 2 1 2+3𝑙𝑜𝑔𝑥 2 2 1 4. 𝑐𝑜𝑠𝑥 𝑠𝑒𝑐𝑥 . 𝑠𝑖𝑛𝑥𝑠𝑒𝑐 𝑥 37. 𝑥. 1+𝑠𝑖𝑛 2 𝑥 𝑠𝑒𝑐𝑥 1−𝑡𝑎𝑛𝑥 2 1 1 𝑐𝑜𝑠𝑥 38. 45. 2. 𝑒 𝑥 𝑒 2𝑥 −4 𝑒 𝑥 𝑒 2𝑥 +1 𝑒 𝑥 4−𝑒 2𝑥 𝑠𝑖𝑛𝑥 20.HSC MATHS OMTEX CLASSES 8TH YEAR 35. 2𝑥𝑒 𝑥 3. 𝑥𝑙𝑜𝑔𝑥 . 7. 𝑐𝑜𝑠 2 21. 𝑥 2 𝑠𝑒𝑐 2 𝑥 3 6. 𝑙𝑜𝑔𝑥 3 𝑥 𝑙𝑜𝑔𝑥 41. 𝑥𝑠𝑖𝑛𝑥 2 2. 36. 𝑥𝑡𝑎 𝑛 −1 𝑥 2 1+𝑥 4 𝑡𝑎𝑛𝑥 𝑠𝑒𝑐𝑥 +𝑐𝑜𝑠𝑥 1 1 𝑛 1 𝑎 𝑥 2𝑥+1 24. 9. 𝑥 𝑥+1 𝑥+3 𝑥−3 2𝑥+3 𝑥+1 2𝑥+1 3𝑥−2 5. No. 40. 2+𝑥 2−𝑥 𝑥 2 +1 𝑥−1 2𝑥 2 +𝑥 𝑥−1 𝑥 3 +5𝑥 2 +2𝑥+3 2𝑥−1 9. 6+4𝑥−𝑥 2 47. 𝑠𝑖𝑛 4 𝑥 31. 4. 𝑥+1 2 𝑥 2 −2𝑥+3 𝑥 −1 2 Ex. 8. 𝑒 𝑡𝑎𝑛𝑥 𝑠𝑒𝑐 2 𝑥 𝑥+1 𝑥+𝑙𝑜𝑔𝑥 2 2𝑥 (𝑐𝑜𝑠 −1 𝑥) 2 1−𝑥 2 𝑐𝑜𝑠𝑥 𝑥) 3 42. 3 35.𝑐𝑜𝑠𝑒𝑐𝑥 𝑙𝑜𝑔 𝑡𝑎𝑛𝑥 2−𝑥 2𝑥+6 (𝑠𝑖𝑛 −1 1−𝑥 2 𝑠𝑖𝑛 (𝑡𝑎𝑛 −1 𝑥) 1+𝑥 2 50 . 𝑠𝑖𝑛−1 𝑐𝑜𝑠𝑥 36. 14. 𝑥 2 +𝑥+5 43. 𝑥 2 +1 11. 𝑠𝑖𝑛3 𝑥 𝑐𝑜𝑠𝑥 17. 13. 12.Whenever the degree (Highest Power of a polynomial equation) of the numerator is greater than or equal to the degree of the denominator then divide the numerator by denominator. 𝑥𝑠𝑖 𝑛 2 2 𝑙𝑜𝑔𝑥 𝑒 𝑥 𝑥+1 𝑥𝑒 𝑥 22.𝑙𝑜𝑔 26. 2𝑥𝑠𝑖𝑛𝑥 𝑐𝑜𝑠𝑥 16. 32. 𝑠𝑖𝑛 19. 6.𝑐𝑜𝑠 2 𝑥 29. 1−𝑡𝑎𝑛𝑥 66. 𝑙𝑜𝑔 55. 𝑒 𝑥 +1 63. 1−𝑡𝑎𝑛 2 𝑥 28. 74. 32. 𝑠𝑖𝑛 24. 50. 77. 𝑠𝑖𝑛𝑥 . 𝑒 𝑥 −1 62. 𝑒 𝑥 16−𝑒 2𝑥 𝑥 2 1+𝑥 6 𝑥 𝑥 4 +25 𝑠𝑒𝑐 2 𝑥 3 𝑡𝑎𝑛 2 𝑥+2 𝑠𝑖𝑛𝑥 𝑠𝑒𝑐 2 𝑥 1 1 1+𝑡𝑎𝑛 2 𝑥 14. Integration of the type 𝐟 𝐱 1. 𝑐𝑜𝑠 2𝑥 𝑠𝑖𝑛𝑥 +𝑐𝑜𝑠𝑥 2 𝑥 𝑥 2 −𝑎 2 𝑥 1−𝑥 2 2𝑥+1 𝑥62+𝑥−5 2𝑥+3 𝑥 2 +3𝑥−1 𝑒 𝑥 𝑒 𝑥 +1 1 𝑥 𝑎+𝑏𝑙𝑜𝑔𝑥 𝑠𝑖𝑛𝑥𝑐𝑜𝑠𝑥 2−3 𝑠𝑖𝑛 2 𝑥 𝑒 𝑥 −𝑠𝑖𝑛𝑥 𝑒 𝑥 +𝑐𝑜𝑠𝑥 𝑠𝑖𝑛 2𝑥 𝑎 2 𝑠𝑖𝑛 2 𝑥+𝑏 2 𝑐𝑜𝑠 2 𝑥 𝑒 2𝑥 −1 𝑒 𝑥 +1 1 1+𝑒 𝑥 1−𝑒 −𝑥 𝑥 𝑒−1 +𝑒 𝑥 −1 𝑥 𝑒 +𝑒 𝑥 1+𝑠𝑖𝑛 2𝑥 𝑐𝑜𝑡𝑥 52. 𝑐𝑜𝑠 51 .HSC MATHS OMTEX CLASSES 1 8TH YEAR 49. 𝑠𝑖𝑛 23. 𝑥 34. 30. 𝑎𝑐𝑜𝑠 2 𝑥+𝑏𝑠𝑖 𝑛 2 𝑥 Ex. 31. 25−𝑐𝑜𝑠 2 𝑥 35. 54. 𝑒 2 −1 1 𝑥 1+𝑡𝑎𝑛𝑥 1−𝑡𝑎𝑛𝑥 1 𝑥 1 69. 𝑐𝑜𝑠 20. 4 𝑠𝑖𝑛 2 𝑥−3 37. 𝑐𝑜𝑠 25. 𝑐𝑜𝑠 21. 72. 𝑐𝑜𝑠 17. 1+𝑒 −𝑥 60. 𝑥 2 −𝑥+1 𝑥+1 2 𝐚𝐱 + 𝐛𝐝𝐱 or 1 1−𝑠𝑖𝑛𝑥 𝑠𝑖𝑛𝑥 𝑥+𝑎 𝑐𝑜𝑠𝑥 𝑥−𝑎 𝑠𝑖𝑛𝑥 𝐟 𝐱 𝐚𝐱+𝐛 𝐝𝐱 27. 3𝑡𝑎𝑛𝑥 +1 1 𝑐𝑜𝑠 2 𝑥 59. 51. 9. 39. 5. 78. 𝑥+𝑠𝑖𝑛 2 𝑥 53. 𝑠𝑖𝑛 4. 1+𝑡𝑎𝑛𝑥 67. 76. No. 7. 12. 1+𝑡𝑎𝑛𝑥 68. 4 𝑡𝑎𝑛 2 𝑥−9 36. 15. 71. 5. 𝑠𝑖𝑛𝑥 4 𝑠𝑖𝑛 2 𝑥+5 𝑎+𝑥 𝑎−𝑥 𝑐𝑜𝑠𝑥 𝑙𝑜𝑔𝑥 2 +9𝑥 𝑐𝑜𝑠𝑥 22. 56. 11. 𝑥 2𝑥 − 1 16. 10. 73. 𝑐𝑜𝑠 18. 𝑥 2 𝑥 + 1 3. 13. 𝑠𝑖𝑛𝑥 +𝑐𝑜𝑠𝑥 𝑠𝑖𝑛𝑥 58. 𝑠𝑖𝑛 26. 57. 2𝑥+𝑥𝑙𝑜𝑔𝑥 1 𝑒 𝑥 +𝑒 −𝑥 2 𝑥 1+𝑥 1−𝑥 𝑥 3 3+𝑥 2 3−𝑥 2 𝑠𝑖𝑛 2𝑥 𝑠𝑖𝑛𝑥 1 𝑒 𝑥 64. 70. 𝑥+ 65. 𝑐𝑜𝑠 (2 𝑠𝑖𝑛 2 𝑥+𝑠𝑖𝑛𝑥 −3)𝑐𝑜𝑠𝑥 𝑠𝑖𝑛𝑥 −1 (2𝑒 2𝑥 +9𝑒 𝑥 +5)𝑒 𝑥 𝑒 𝑥 +1 𝑡𝑎𝑛 1+ 𝑥 𝑥 𝑐𝑜𝑡 1+ 𝑥 𝑥 𝑡𝑎𝑛 2+3𝑙𝑜𝑔𝑥 𝑥 𝑠𝑒𝑐 𝑙𝑜𝑔𝑥 𝑥 𝑠𝑒𝑐 𝑥 𝑥 1 1−𝑐𝑜𝑠𝑥 1 1+𝑠𝑖𝑛𝑥 𝑥−𝑎 𝑠𝑖𝑛 𝑥−𝑎 𝑥+𝑎 𝑐𝑜𝑠𝑥 𝑥−𝑎 𝑐𝑜𝑠 𝑥+𝑎 𝑥−𝑎 𝑠𝑖𝑛 𝑥−𝑎 𝑥−𝑏 𝑐𝑜𝑠 𝑥−𝑎 𝑥−𝑏 1 𝑥−𝑎 𝑠𝑖𝑛 𝑥−𝑏 1 𝑥−𝑎 𝑐𝑜𝑠 𝑥−𝑏 1 𝑥−𝑎 𝑐𝑜𝑠 𝑥−𝑏 1 𝑥−𝑎 𝑠𝑖𝑛 𝑥−𝑏 33. 𝑒 2𝑥 +1 61. 75. 4+𝑠𝑖𝑛 2 𝑥 38. 8. 2𝑥 + 1 𝑥 + 1 2. 5+3cos 2x 13+3cosx +sinx 1 3+2cos 2x 52 . 2x−1 x 2 −x+3 Ex. 7. 14. 8. 6. 2. 6. 4. 10. 4. 13. 7. 8. 3. 4−2𝑥−𝑥 2 1 𝑥 2 +4𝑥+3 1 3𝑥 2 −4𝑥+2 1 3+4𝑥−4𝑥 2 Ex. 2. 9. 3. No. 8. Integration of the type 𝐦𝐱+𝐧 𝐚𝐱 𝟐 +𝐛𝐱+𝐜 𝐝𝐱 OR 2x+1 9−4x 2 3x+5 𝐦𝐱+𝐧 𝐚𝐱 𝟐 +𝐛𝐱+𝐜 𝐝𝐱 12. 6. 𝟏 𝒂+𝒃𝒄𝒐𝒔𝒙 1 sinx +cosx 1 cosx −sinx 1 𝒅𝒙 OR 𝟏 𝒂𝒔𝒊𝒏𝒙+𝒃𝒄𝒐𝒔𝒙+𝒄 1 4+5sin 2x 1 𝒅𝒙 1. x 2 +4x+5 x 3 +5x 2 +12x+10 x 2 +4x+5 x−1 10. 4.HSC MATHS OMTEX CLASSES 8TH YEAR Ex. 5. 3+4x−3x 2 11. 8. 2. 9. Integration of the type 𝟏 𝐚𝐱 𝟐 +𝐛𝐱+𝐜 𝐝𝐱 OR 5 𝟏 𝐚𝐱 𝟐 +𝐛𝐱+𝐜 𝐝𝐱 9. No. 7. 10. 2x−3 x 2 −3x+4 2x+1 x 2 +3x−4 x+4 3−x 1. 6. No. 1 𝑥 2 +6𝑥+10 1 𝑥 2 +𝑥+1 5 4𝑥 2 +4𝑥−15 1 15+4𝑥−4𝑥 2 5. [Important] Integration of the type 𝟏 𝒂+𝒃𝒔𝒊𝒏𝒙 1 3−2sinx 1 5+4cosx 1 5−3cosx 1 3+2sinx 𝒅𝒙 OR 5. 3. 2x+3 x 2 +3x+1 x−3 x 2 −6x+4 1−x 3+2x−x 2 2x+1 x 2 +3x+5 3x+7 2x 2 +3x−2 x+3 x 2 +4 7. 1 9+8𝑥−𝑥 2 𝑒 𝑥 𝑒 2𝑥 +4𝑒 𝑥 +13 1. 9. 3. 𝑐𝑜𝑠𝑒𝑐 6 𝑥 24. 𝑙𝑜𝑔 𝑥 + 4 9. 𝑥𝑒 𝑥 3. 𝑐𝑜𝑠 5 𝑥 𝑠𝑖𝑛𝑥 14. 1+𝑐𝑜𝑠 2𝑥 8. 𝐮𝐯𝐝𝐱 = 𝐮 1. 𝑠𝑖𝑛5 𝑥𝑐𝑜𝑠 3 𝑥 18. 𝑥𝑠𝑒𝑐 2 𝑥 𝑥 15. 5. 11. 1. [Important] Integration of the type 𝟏 𝐚+𝐛𝐬𝐢𝐧𝟐 𝐱 1 3−2 sin 2 x 1 2+3 sin 2 x 𝐝𝐱 OR 3. 𝑠𝑖𝑛6 𝑥 4. 8. [Important] Integration of the type 1. 𝑠𝑖𝑛3 𝑥𝑐𝑜𝑠 4 𝑥 7. 𝑐𝑜𝑠𝑒𝑐 4 𝑥 22.HSC MATHS OMTEX CLASSES 8TH YEAR Ex. 𝑠𝑖𝑛𝑥. 6. 𝑠𝑒𝑐 4 𝑥 20. 𝑡𝑎𝑛3 𝑥 15. 4. 𝑒 𝑥 𝑐𝑜𝑠 𝑥 𝑒 𝑥 𝑐𝑜𝑠𝑥 𝑒 𝑥 𝑠𝑖𝑛2𝑥 𝑒 4𝑥 𝑠𝑖𝑛3𝑥 𝐯𝐝𝐱 − 𝐯𝐝𝐱 . 10. 𝑙𝑜𝑔 𝑙𝑜𝑔𝑥 𝑥 10. 𝑡𝑎𝑛5 𝑥 23. 7. 𝑥+1 𝑥 𝑥 2 −4 1 1+𝑥+𝑥 2 +𝑥 3 𝑥−1 𝑥+1 2 𝑥+1 𝑥−1 2 9. 𝑠𝑖𝑛5𝑥𝑠𝑖𝑛7𝑥 Ex. 𝑠𝑖𝑛𝑥𝑐𝑜𝑠𝑥𝑐𝑜𝑠3𝑥 28. 2 sin 2 x+3 cos 2 x 1 a 2 sin 2 x+b 2 cos 2 x Ex. 𝐝 𝐮 𝐝𝐱 𝐝𝐱 14. 𝑐𝑜𝑠𝑒𝑐 8 𝑥 25. 𝟏 𝐚+𝐛𝐜𝐨𝐬 𝟐 𝐱 1 1+7 cos 2 x 1 5−cos 2 x 𝐝𝐱 OR 𝟏 𝐚𝐬𝐢𝐧𝟐 𝐱+𝐛𝐜𝐨𝐬 𝟐 𝐱+𝐜 1 𝐝𝐱 1. 𝑠𝑖𝑛7 𝑥 6. 𝑡𝑎𝑛−1 𝑥 12. 𝑠𝑖𝑛3 𝑥 𝑠𝑖𝑛𝑥 11. 𝑐𝑜𝑠 3 𝑥 17. 1 𝑥+1 𝑥+2 𝑥 𝑥−1 𝑥+2 𝑥 2 +1 𝑥+1 𝑥+2 𝑥−3 3𝑥−2 𝑥 2 −3𝑥+2 5. 𝑠𝑖𝑛5 𝑥𝑐𝑜𝑠 2 𝑥 13. No. 𝑐𝑜𝑠 5 𝑥 3. 𝑐𝑜𝑠 4 𝑥 𝑠𝑖𝑛 3 𝑥 12. 12. 𝑥𝑠𝑖𝑛𝑥 2. 𝑡𝑎𝑛4 𝑥 21. 𝑙𝑜𝑔 𝑥 2 + 4 𝑙𝑜𝑔𝑥 𝑥 3 3 16. No. 𝑐𝑜𝑡𝑥 𝑠𝑖𝑛𝑥 16. 2. 𝑠𝑖𝑛3𝑥𝑐𝑜𝑠4𝑥 31. 𝑐𝑜𝑠3𝑥𝑐𝑜𝑠4𝑥 27. 𝑠𝑖𝑛5 𝑥 9. No. 6. 𝑐𝑜𝑠5𝑥𝑐𝑜𝑠7𝑥 32. 𝑥 𝑙𝑜𝑔𝑥 13. No. 𝑠𝑖𝑛5𝑥𝑐𝑜𝑠3𝑥 26. 4. [Important] Integrate the following . [Important] Integrate the following. 𝑠𝑖𝑛3 𝑥𝑐𝑜𝑠 3 𝑥 10. 4. 10. 𝑙𝑜𝑔𝑥 2 53 . 7. 5. 1 𝑥+1 2 𝑥 2 +1 𝑥 3 +2𝑥 2 +6 𝑥 2 +𝑥−2 Ex. 𝑠𝑖𝑛2 𝑥𝑐𝑜𝑠 3 𝑥 8. 𝑐𝑜𝑠5𝑥𝑐𝑜𝑠3𝑥 30. 6. 11. 2. 𝑐𝑜𝑠 7 𝑥 5. 𝑐𝑜𝑠 4 𝑥 2. 𝑐𝑜𝑡 3 𝑥 19. 4𝑠𝑖𝑛3𝑥𝑐𝑜𝑠2𝑥 29. HSC MATHS OMTEX CLASSES 8TH YEAR Ex. No. 13. [Important] Integrate the following 𝐞𝐱 𝐟 𝐱 + 𝐟 ′ 𝐱 𝐝𝐱 = 𝐞𝐱 𝐟 𝐱 + 𝐜 1. 𝑒 𝑥 𝑠𝑖𝑛𝑥 + 𝑐𝑜𝑠𝑥 2. 𝑒 𝑥 1 𝑥 + 𝑙𝑜𝑔𝑥 3. 𝑒 𝑥 𝑐𝑜𝑡𝑥 + 𝑙𝑜𝑔 𝑠𝑖𝑛𝑥 𝑑𝑥 4. 𝑒 𝑥 𝑐𝑜𝑠𝑒𝑐𝑥 1 − 𝑐𝑜𝑡𝑥 𝑑𝑥 5. 𝑠𝑖𝑛 𝑙𝑜𝑔𝑥 + 𝑐𝑜𝑠 𝑙𝑜𝑔𝑥 6. 𝑡𝑎𝑛 𝑙𝑜𝑔𝑥 + 𝑠𝑒𝑐 2 𝑙𝑜𝑔𝑥 7. 𝑒 𝑥 𝑠𝑒𝑐𝑥 1 + 𝑡𝑎𝑛𝑥 8. 𝑙𝑜𝑔 𝑙𝑜𝑔𝑥 + 𝑙𝑜𝑔𝑥 1 Ex. No. 1. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 1 𝑥 𝑑𝑥 0 3 1 𝑥 3 𝑑𝑥 1 9 1 𝑑𝑥 4 𝑥 3 ∞ −𝑥 𝑒 𝑑𝑥 0 2 𝑑𝑥 𝑑𝑥 1 3𝑥−2 1 𝑑𝑥 𝑑𝑥 −1 1+𝑥 2 1 1−𝑥 2 𝑑𝑥 0 1+𝑥 2 2 𝑑𝑥 𝑑𝑥 0 𝑥− 𝑥−1 0 2 5𝑥 𝑑𝑥 0 𝑥 2 +4 𝜋 2 DEFINITE INTEGRATION 11. 12. 13. 14. 15. 16. 17. 18. 19. 𝜋 4 𝜋 2 0 𝑐𝑜𝑠𝑥 𝑑𝑥 𝑠𝑖𝑛5𝑥𝑐𝑜𝑠3𝑥𝑑𝑥 𝑠𝑖𝑛3 𝑥 𝑑𝑥 1 + 𝑐𝑜𝑠𝑥𝑑𝑥 𝑒𝑡𝑎𝑛𝑥𝑠𝑒𝑐 2 𝑥𝑑𝑥 (𝑠𝑖𝑛 −1 𝑥)3 1−𝑥 2 2 20. 21. 22. 23. 24. 25. 26. 27. 0 𝜋2 0 𝜋2 𝜋 3 𝜋 4 0 0 2 4−𝑥 2 𝑑𝑥 1 𝑥 2 𝜋 𝑠𝑖𝑛𝑥 2 𝑑𝑥 0 1+𝑐𝑜𝑠𝑥 3 𝜋 𝑥 𝑒 𝑠𝑖𝑛2𝑥𝑑𝑥 0 𝜋 𝑑𝑥 2 𝑑𝑥 0 5+3𝑐𝑜𝑠𝑥 1 1−𝑥 2 𝑑𝑥 −1 1+𝑥 2 𝑑𝑥 0 1−2𝑥 2 1−𝑥 2 1−𝑡𝑎𝑛𝑥 𝑑𝑥 0 1+𝑡𝑎𝑛𝑥 1 𝑙𝑜𝑔𝑥𝑑𝑥 0 𝜋 4 1 2 1 2 𝑑𝑥 𝑠𝑖𝑛𝑥𝑐𝑜𝑠𝑥𝑑𝑥 0 𝑎 𝑎 2 −𝑥 2 𝑎 2 𝜋2 𝑠𝑖𝑛𝑥. 𝑐𝑜𝑠𝑥𝑑𝑥 𝑥 2 𝑑𝑥𝑑𝑥 6 36−𝑥 2 3 𝑥 2 54 HSC MATHS OMTEX CLASSES 8TH YEAR Ex. No. 2. [Important] PROPERTIES 1. 2. 3. 4. 5. 6. 𝒃 𝒇 𝒂 𝒃 𝒇 𝒂 𝒃 𝒇 𝒂 𝒂 𝒇 𝟎 𝒃 𝒇 𝒂 𝒙𝒅𝒙 = 𝒃𝒇 𝒂 𝒕𝒅𝒕 𝒙 𝒅𝒙 𝒄 𝒇 𝒃 𝒙𝒅𝒙 = − 𝒙 𝒅𝒙 = 𝒙 𝒅𝒙 = 𝒙 𝒅𝒙 = 𝒙 𝒅𝒙 = 𝒂𝒇 𝒃 𝒄𝒇 𝒂 𝒂 𝒇 𝟎 𝒃 𝒇 𝒂 𝒙𝒅𝒙 + 𝒙𝒅𝒙 𝒂− 𝒙 𝒅𝒙 𝒂 + 𝒃 − 𝒙 𝒅𝒙 𝒙 𝒅𝒙 + 𝒂 𝒇 𝟎 𝟐𝒂𝒇 𝟎 𝒂𝒇 𝟎 𝟐𝒂− 𝒙 𝒅𝒙 7. 𝒂𝒇 –𝒂 𝒙𝒅𝒙 = 𝟐 𝒂𝒇 𝟎 𝒙𝒅𝒙 𝒐𝒏𝒍𝒚 𝒊𝒇 𝒇 𝒙 𝒊𝒔 𝒆𝒗𝒆𝒏 𝒂𝒏𝒅 = 𝟎 𝒊𝒇 𝒇 𝒙 𝒊𝒔 𝒐𝒅𝒅. 1. 2. 3. 4. 5. 6. 2 𝑥 𝑑𝑥 1 𝑥+ 3−𝑥 2 𝑥+2 𝑑𝑥 1 𝑥+2+ 5−𝑥 4 5−𝑥 𝑑𝑥 5 𝑥−4+ 5−𝑥 4 5 𝑥+4 𝑑𝑥 0 4 𝑥+4+ 4 9−𝑥 7. 8. 9. 10. 11. 12. 13. 0 1+𝑡𝑎𝑛𝑥 𝜋 3 𝜋 6 𝜋2 1 1 𝑑𝑥𝑑𝑥 14. 15. 16. 17. 18. 1+ 𝑐𝑜𝑡𝑥 𝑎𝑑𝑥 0 𝑥+ 𝑎 2 −𝑥 2 ∞ 𝑑𝑥 0 1+𝑥 1+𝑥 2 0 1 𝑥 1 0 4 𝑥 0 4−𝑥 𝜋 4 0 𝜋2 𝜋 2 𝑎𝑠𝑖𝑛𝑥+𝑏𝑐𝑜𝑠𝑥 𝑠𝑖𝑛𝑥 +𝑐𝑜𝑠𝑥 𝑡𝑎𝑛𝑥 𝑑𝑥 0 𝑡𝑎𝑛𝑥 +𝑐𝑜𝑡𝑥 𝑑𝑥 𝜋𝑥𝑠𝑖𝑛𝑥 𝑑𝑥 0 1+𝑠𝑖𝑛𝑥 𝜋 𝑥𝑡𝑎𝑛𝑥 𝑑𝑥 0 𝑠𝑒𝑐𝑥 +𝑡𝑎𝑛𝑥 𝜋 𝑥𝑠𝑖𝑛𝑥 𝑑𝑥 0 1+𝑐𝑜𝑠 2 𝑥 4 𝑑𝑥 0 𝑥+ 16−𝑥 2 3 1 𝑑𝑥 0 𝑥+ 9−𝑥 2 𝑙𝑜𝑔1 + 𝑡𝑎𝑛𝑥 𝑑𝑥 − 𝑥𝑑𝑥 𝑑𝑥 55 HSC MATHS OMTEX CLASSES 8TH YEAR CH. NO. 6. DIFFERENTIAL EQUATION EX. NO. 1. A. Form the differential equations by eliminating the arbitrary constant. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 𝑦 = 𝑥 2 𝑎 + 𝑥 𝑦 = 𝐴. 𝑒 𝑥 𝑦 = 𝑎𝑒 −𝑥 𝑦 = 𝑎𝑥 2 𝑐 𝑦 = 𝑐 2 + 𝑥 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 31. 𝑥 𝑦 + 𝑏 = 1 𝑎 𝑥 2 𝑦 2 + 2=1 𝑎 2 𝑏 3𝑥 𝑥− 𝑎 2 + 𝑦 2 = 𝑎2 𝑦 2 = 4𝑎𝑥 𝑦 2 = 4𝑎 𝑥 + 𝑎 𝑥 2 + 𝑦 2 − 2𝑎𝑥 = 0 𝑦 = 4 𝑥 − 𝑐 2 𝑦 = 𝑎𝑥 2 + 1 𝑥𝑦 = 𝑎 𝑦 = 𝑒 𝑎𝑥 𝑦 = 𝑠𝑖𝑛 𝑎𝑥 𝑦 = 𝑐𝑜𝑠 𝑥 + 𝑎 𝑦 = 𝑚𝑥 + 𝑐 𝑦 = 𝑎𝑥 2 + 𝑏 𝑦= 𝐴𝑒 + 𝐵𝑒 −3𝑥 𝑦 = 𝐴𝑒 𝐵𝑥 𝑦 = 𝑎 𝑥 − 𝑎 𝑦 = 𝐴. 𝑒 2𝑥 + 𝐵. 𝑒 5𝑥 𝑦 = 𝑎𝑒 2𝑥 + 𝑏𝑒 −2𝑥 𝑦 = 𝐴𝑐𝑜𝑠 𝑙𝑜𝑔𝑥 + 𝐵𝑠𝑖𝑛 𝑙𝑜𝑔𝑥 𝑦 = 𝐴𝑐𝑜𝑠7𝑥 − 𝐵𝑠𝑖𝑛7𝑥 𝑒 𝑥 + 𝐶𝑒 𝑦 = 1 𝑥 2 + 𝑐𝑦 2 = 4 𝑦 = 𝐶1 𝑥 2 + 𝐶2 𝑥 𝐴𝑥 3 + 𝐵𝑦 2 = 5 (Note: Important sum use the condition for consistency) 𝑥 2 𝑎 2 + 𝑏 2 = 1 𝑦2 EX. NO. 2. 1. Solve x 2 dx = x 2 + xy + y 2 2. Solve the differential equation ydx − xdy = 0 3. Solve the differential equation dx = e−2y cosx 4. Solve tany dx = sin x + y − sin − y) (x 5. Solve dx = sin x + y + cos x + y by putting x + y = u. x = e and y = e2 . dy dy dy dy dy 6. Find the particular solution of the differential equation y 1 + logx − xlogx dx = 0 when 7. Solve the differential equation x 8. Solve dx = 4x + 3y − 1 2 dy dx dy − y sin = x 2 ex by substituting = v. y x y x by using substitution 4x + 3y − 1 = u. 9. Solve x + 2y + 1 dx − 2x + 4y + 3 dy = 0. 10. Find the particular solution of the differential equation 1 − x dy − 1 + y dx = 0, if y = 4 when x = 2. 56 dx = 1 3 1+ dy 2 dx . Solve dx = 1+x 2 12. Verity that y = ax 2 + b is a solution of x dx 2 − dx = 0. Find the particular solution of the differential equation: x + 1 x = 1 & 𝑦 = 0.E. dx 2 + d2y 1 dy 2 dx − y = 0. Determine the order and degree of the differential equation. dy 4x+6y−2 dy − 1 = 2e−y when 19. Solve the D. Solve the differential equation dx = 2x+3y+3.E. x 2 + y dx − 2xydy = 0 14. dy 2 dx 23. Verify that y = Asin3x + Bcos3x is the general solution of the differential equation d2y dx 2 + 9y = 0. Find the order and degree of the D. Solve y − x dx = y 2 + dx . 57 . Determine the order and degree of the differential equation 5 = 10x − 1 dy dx . 20. 17.HSC MATHS 11. 15. dy dx 18. by taking 2x + 3y = t. dy dy 1+y 2 dy dx OMTEX CLASSES 8TH YEAR = 13. Solve the equation ey + 1 cosxdx + ey sinxdy = 0 d2y dy dy x+y+1 2 x+y −1 2 16. d2y dy 2 dx 21. = y. Determine the order and degree of the D. Solve the D.E. Hence find the particular solution if y = 2 when x = 1. dx 2 + 3 1 − 22.E. 1 Eg. 𝐴 = 914 554 447 778 25 101 (3×3) Diagonal matrix: . 𝐴 = 0 5 0 0 0 1 (3×3) Scalar matrix:. 2 0 0 Eg. 𝐴 = 5 9 3×1 Square matrix: .A square matrix in which non – diagonal element is equal to zero and diagonal element may be or may not be equal to each other. It is denoted by ‘I’. 7. 5. a12 = 5. 225 5874 448 Eg. Eg. a22=7 Row matrix:.A square matrix having every diagonal element equal to one and every non – diagonal element equal to zero is called an unit matrix. 𝐴 = 0 1 0 0 0 1 (3×3) 58 1. 1 0 0 Eg. 6. 𝐴 = 9 5 4 7 2 1 (3×3) Representation of elements 4 5 𝐴 = 8 7 (2×2) Let ‘a’ is used to denote the different element of Matrix A then. 𝐴 = 5 7 (1×2) Column matrix: -A matrix having only one column is called as a column matrix. NO. 2 4 8 Eg. 3.A matrix having only one row is called as a row matrix. 4. 2 0 0 Eg.A rectangular arrangement of number of rows and number of column which is enclosed between two brackets is called ‘Matrix’. 𝐴 = 0 2 0 0 0 2 (3×3) Unit matrix (identity matrix): . a11 = 4. a21 =8. MATRICES Definition: .A square matrix in which non – diagonal element is equal to zero and diagonal element are equal to each other is called a scalar matrix.HSC MATHS OMTEX CLASSES 8TH YEAR CH.A matrix whose number of rows is equal to the number of columns is called a square matrix. 7. 2. . Null matrix or zero matrix : . 𝐴 = 0 5 9 0 0 1 (3×3) b. Determinants of matrix: . 2 0 0 Eg. 2 5 6 Eg. 11 12 11 12 𝐴 = . If A is a given matrix then determinant of matrix 𝑨 is represented by 𝑨 2 −1 −3 1 2 𝐴 = 𝑡𝑒𝑛 𝐴 = −6. 𝐴 = 6 5 0 8 80 1 (3×3) 10. It is denoted by ‘0’ 0 0 0 Eg. 2 7 0 2 9 8 Eg. 13 14 13 14 59 . If A is a given matrix then transpose of matrix A is represented by A’.HSC MATHS OMTEX CLASSES 8TH YEAR 8. 𝐴 = 0 0 0 0 0 0 (3×3) 9. 𝐵 = . 4 6 𝐴 = 𝑡𝑒𝑛 𝐴 ≠ 0 2 −3 14.A transpose matrix is obtained by interchanging row and column from a given matrix. 𝐴 = 9 5 40 𝐴′ = 7 5 6 8 6 1 (3×3) 0 40 1 (3×3) 11. Transpose Matrix: . Equality of two matrices: .singular matrix: . Upper triangular matrix: .If the given matrix is a square matrix then only we can find out determinant value of a given matrix. Lower triangular matrix: .Two matrices are said to be equal only if they are of the same order and their corresponding elements are equal. Non . 4 6 𝐴 = 𝑡𝑒𝑛 𝐴 = 0 2 3 13.A matrix having every element as zero is called as null or zero matrix. 𝐵 = 8 2 6 𝑡𝑒𝑛 𝐴 = 144. Triangular matrix: a. Singular matrix: . 𝑡𝑒𝑛 𝐴 = 𝐵. 5 4 12 0 12 12.A square matrix in which all the elements below the diagonal are zero is called upper triangular matrix.A matrix in which the determinant value is equal to zero is called as singular matrix.A matrix in which the determinant value is not equal to zero is called as non – singular matrix.A square matrix in which all the elements above the diagonal are zero is called lower triangular matrix. 14 −1 k 1. find a and b. find a and b. C = .E = 3 −1 a + b b−a π −1 2 sin 2 . If a − 4b 5 11 5 = . −4 a B. 1. state the values of d21. A= 3 3 . 0 7 3. If B = E. 2 Answer the following questions. Find a. If A = 6 3 is a singular matrix. d13. b. Consider the matrices. d. 3 1 1 6 5 2 a 3 5 7 E= . find a and b.HSC MATHS OMTEX CLASSES 8TH YEAR Ex: 1 A. find a. e. d32. i.B = −1 3 4 2 1 F= −1 2 1 −5 . b and c if 60 . 2. Find a. F = [5 6 −7].H = 2 cosπ 2 4 3 4 a . 6 −5 1 4. C.D = . If A = D. F and H. D = −2 4 1 .G = 1 4 3 −1 a−b 2 3 . D. G = 3 b a 6 Answer the following questions. G. Find which of the following matrices are singular and non – singular. 1 2 3 A= . F and G. B = 6 . 6 −a + b 6 −5 a + 2b 2 − b 2 3 = b+c a−c 1 2 3. If e11 = e12. −8 8 B= 5 −20 . Find a. Which of these are row matrixes? c. For D.C = 2 2 2 3 . find k. 2. Consider the Matrices 4 4 −6 3 2 3 A = 2 −1 3 4 . −4 16 1 C= 3 7 2 3 −1 2 . If A = 4 2 −1 is a singular matrix. iii. State the orders of the matrices A. a. ii. F ′ and G are equal. If G is a triangular matrix. Are matrices A and C. a. A + (B + C) = (A + B) + C b. 3(A + B – C) = 3A +3B – 3C c. Find a. c. −3 4 1 2 2 1 0 .D = 3 5 6 −1 5 −1 5 Answer the following. ii.B = .F = 2 2 .B = obtain the matrix A − 3 B. If A = 6.HSC MATHS OMTEX CLASSES 8TH YEAR Ex: 2 1.B = . iv. find the matrix B.C = . If A = 7. If A = 1 2 .C = 4 3 −5 8 −1 3 −2 3 7 Verify the following. i. A = 2 −1 −3 2 1 6 2 .B = −3 −1 0 2 1 1 Find the matrix C such that A + B + C is a zero matrix. 2. A + B ’ = A’ + B’. If A = 4 3 5 2 −2 b . a. Find x. 3 4 1 2 and 2A + 3B = 0. A + B. A + C. If A + F = 0. 3 7 11 4 5 = . If C – E = I. A + A’ b. 2 1 3 −2 10 1 4 5 + x= 0 −5 3 6 3. find b. y & z if x+y y−z 3 −1 = z − 2x y − x 1 1 61 . 6 3 0 −1 .. 1 x 0 3 1 2 4 2 2 + = y 2 4 4 3 −2 6 5 2 10. If A = 4. B’ +D. find matrix B such that A + B = 0. Can you find. −8 9 1 −3 8. Find the values of x and y satisfying the matrix equation. iii.E = a 4 −3 −5 −2 1 3 −1 2 −1 2 4 8 2 4 . Find the matrix ‘X’ such that 3X + 9. Find x if 5. y. If A = 2 −1 −1 6 4 2 5 7 . secθ tanθ x 3 b. −3 2 −1 = 1 x 2 5 −1 −2 2 secθ −tanθ c. 4 x 1 −2 = 8 c. AI = IA = A.B = −1 0 0 3 −1 0 −1 0 . 1 3 −2 0 5 2 4 x 9 y = 8 6 −3 z −4 3 ii. A B + C = AB + AC. x 3 . T.B = . A = 5 3 1. Find the following products: 4 −2 a. (C). A = 1 2 4. Find x in the following cases.C = verify the following. B= −2 0 −1 0 2 −1 . A = 2.HSC MATHS OMTEX CLASSES 8TH YEAR Ex: 3 1. If A = −3 0 −3 −4 show that A2 − 4A is a scalar matrix. 2 1 2 x 3 y = 2 −1 z 62 . A B − C = AB − AC. B = 1 2 1 1 3 2 3 1 2 0 2 3 −1 2 1 −2 −2 0 5 6 .B = 1 3 5 3 1 −2 . A BC = AB C 𝐛. If A = 2 1 2 2 2 1 0 3 5. Find AB and BA whenever they exist in each of the following cases.C = Then verify the following 4 3 8 3 7 8 𝐚. B ′ A B is a Null Matrix. i. z values in each of the following cases. (A) Find the values of a and b from the matrix equation: 6. a. If A = 3 1 5 . 3 4 1 3 2. Find x. 6 d. x sinθ cosθ sinθ = 5 cosθ x Ex: 4 1. 3 3 3 2 1 −3 −2 1 𝐚. Where I identity matrix 3. 1 2 2 4. B = y . B′ = x y z 4 z 0 S. 2 3 −1 −1 b. 𝐛. 2 2 2 2. (B) Find the values of x and y. 4 1 2 2 1 0 1 3 −2 3 2 3 = 2 x 0 = y 1 6. 3 2 4 1 1 8 a 1 5 b 4 5 −3 5 6. A = 1 1 3.B = . 7 5 6 show that AB = BA. If A = . −2 4 1 2 5 . If A = 10 a. B = 2 3 4 show that AB ≠ BA. 5 3 2 8 1. show that A2 − 5A − 2I is a Zero matrix. 1 1 2 x 3 2 3 OMTEX CLASSES y 3 7 0 7 = a b c −1 z 8TH YEAR x 1 2 2 B = −2 3 . 63 . 0 3 4 1 1 9. 3 4 4 show that AB = 0. Find x. X = y Find the values of x. If A = 14. |B| 3 3 −2 1 3 show that AB is a Non singular matrix. a. −6 2 1 . B= show that 5 10 −8 2 = A2 + AB + B 2 b. A + B A − B = A2 − AB − B 2 . If A = b. −1 −2 1 −1 show that A2 = 2A. If A = 12 8 6. If A = B= −1 −2 5 3 3 4 5 3. y. If A = 3 2 2 0 1 −5 . B= verify that AB = A . a. z if 5A − 3B C = X 1 z 3 1 −6 4 Find the Matrix AB and without computing the Matrix BA. If A = Ex: 5 −3 6 Show that A2 = A. If A = 2 4 . −1 0 −2 0 1 12. z. b. If A = and B = 12 8 −12 1 2 4 3 2. If A = . B= 4 3 6 b. −1 1 2 4 Show that A satisfies the Matrix Equation A2 = 3A + 2I.C = show that BA = CA. A+B 4. B= show that A + B 2 = A2 + BA + B 2 8 −9 −27 4 5 −4 . C = .HSC MATHS 7. 4 1 1 2 . B = 3 0 −2 Verify that AB ≠BA. Show that A2 is a null matrix. If A = 1 2 . If A = 5 2 . y. a. c if 1 8.B = 2 1 2 3 1 13. If A = 3 −1 3 5. 10. B= show that A + B A − B = A2 − B 2 3 −5 2 2 6 18 . If A = 11. If A = 0 1 0 1 . 1 1 14. show that A2 − 5A + 7I = 0. 1. −2y + z = −3. If A = 2 1 2 2 C. Hence find A−1 . B= . If A = −1 3 4 and B = 1 −2 −4 1 −2 −3 −1 2 4 show that A + B 2 = A2 + B 2 1 −1 1 a 8. Ex: 7 A. 3x − 2y + 3z = 4. II. 1. z using Matrix method. find the matrix A + I A − I . y. 1 1 2 2. If A = 6. x + 3y + 4z = 13. Find the inverse of each of the following Matrices by using elementary transformations. by using inverse of a Matrix. 3x + y − 5z = 6. B= 2 −2 −1 3 1 2 1 show that AB . −sinθ cosθ then Show that A2 = 1 3 . exists. 1 2 3 5 2. If A + I = −1 1 3 . 2x − y + z = −7 3. 2x − y = 1. 4x + 2y − z = 3. 1. 1. cosθ sinθ −sinθ cosθ 7. 3. If A = secθ tanθ 0 tanθ secθ 0 0 0 1 3 1 . B= find Matrix X such that AX = B. x − y + z = 1.HSC MATHS OMTEX CLASSES 8TH YEAR 2 −2 −4 −1 2 4 7. find AB 0 2 1 64 . 3 1 2 4 3. −1 3 1 . 2. show that A2 − 4A = 5I. x + y + z = 3. Ex: 6 I. If A = and B = such that A + B 2 −1 4 b 2 = A2 + B 2 than find a & 𝑏. x + 2y − 2z = 5. Hence find A−1 . x − 2t + z = −8. If A = 3 1 7 3 . cosecθ −cotθ −cotθ cosecθ 2 0 1 6. −2 −3 1 4. 2. x + y + 2z = 7. −1 2 2 2 . 5x + 5y + z = 11. If A = cosθ sinθ . 3x + 3y − 4z = 2. x + 4y − 2z = 10. x + 4y + 4z = 15 . −1 −1 cos2θ sin2θ −sin2θ cos2θ −1 0 −2 0 0 . 2x + 2y − 3z = 7. x + 3y + 3z = 12. secθ tanθ tanθ secθ 5. 2. 2x − 2y + 3z = 4. 4. 1 3 4 3. Write down the following equation in the Matrix Form and hence find values of x. 4x − 3y + z = 1. If A = 7. Solve the following equation by the methods of reduction. −1 2 0 6 −x + 3y = 0. 1. 0 −1 2 1 0 1 B. 20. 𝒔𝒊𝒏 𝑨 + 𝒔𝒊𝒏 𝑩 = 𝟐 𝒔𝒊𝒏 𝒔𝒊𝒏 𝑨 − 𝒔𝒊𝒏 𝑩 = 𝟐 𝒄𝒐𝒔 𝒄𝒐𝒔 𝑨 + 𝒄𝒐𝒔 𝑩 = 𝟐 𝒄𝒐𝒔 𝑨+𝑩 𝟐 𝑨+𝑩 𝟐 𝑨+𝑩 𝒄𝒐𝒔 𝒔𝒊𝒏 𝒄𝒐𝒔 𝑨−𝑩 𝟐 𝑨−𝑩 𝟐 𝑨−𝑩 𝟐 𝑨−𝑩 𝟐 5. 2. 5. 2. 𝒕𝒂𝒏𝟐𝜽 = 11. 4. 𝒔𝒊𝒏𝟑𝜽 = 𝟑𝒔𝒊𝒏𝜽 − 𝟒 𝒔𝒊𝒏𝟑 𝜽 12.HSC MATHS ALGEBRAIC FORMULAE 1. 15. 7. 8. 8. 3. 7. 𝒄𝒐𝒔𝟑𝜽 = 𝟒 𝒄𝒐𝒔𝟑 𝜽 − 𝟑𝒄𝒐𝒔𝜽 13. 𝒕𝒂𝒏 𝝅 𝟒 − 𝜽 = 𝟏+𝒕𝒂𝒏𝜽 𝒙+𝒚 𝟏−𝒕𝒂𝒏𝜽 5. 4. 6. 3. 2. 4. 𝒂𝟎 = 𝟏 𝒂 + 𝒃 𝟐 = 𝒂𝟐 + 𝟐𝒂𝒃 + 𝒃𝟐 𝒂 − 𝒃 𝟐 = 𝒂𝟐 − 𝟐𝒂𝒃 + 𝒃𝟐 𝒂𝟐 − 𝒃𝟐 = 𝒂 + 𝒃 (𝒂 − 𝒃) OMTEX CLASSES 8TH YEAR 5. 𝟏 + 𝒄𝒐𝒔𝟐𝜽 = 𝟐 𝒄𝒐𝒔 𝜽 𝟏 − 𝒄𝒐𝒔𝟐𝜽 = 𝟐 𝒔𝒊𝒏𝟐 𝜽 𝟏 + 𝒔𝒊𝒏𝟐𝜽 = 𝒄𝒐𝒔𝜽 + 𝒔𝒊𝒏𝜽 𝟐 𝟏 − 𝒔𝒊𝒏𝟐𝜽 = 𝒄𝒐𝒔𝜽 − 𝒔𝒊𝒏𝜽 𝟐 𝒔𝒊𝒏 𝒂 + 𝒃 = 𝒔𝒊𝒏 𝒂 𝒄𝒐𝒔 𝒃 + 𝒄𝒐𝒔 𝒂 𝒔𝒊𝒏 𝒃 𝒔𝒊𝒏 𝒂 − 𝒃 = 𝒔𝒊𝒏 𝒂 𝒄𝒐𝒔 𝒃 − 𝒄𝒐𝒔 𝒂 𝒔𝒊𝒏 𝒃 𝒄𝒐𝒔 𝒂 + 𝒃 = 𝒄𝒐𝒔 𝒂 𝒄𝒐𝒔 𝒃 − 𝒔𝒊𝒏 𝒂 𝒔𝒊𝒏 𝒃 𝒄𝒐𝒔 𝒂 − 𝒃 = 𝒄𝒐𝒔 𝒂 𝒄𝒐𝒔 𝒃 + 𝒔𝒊𝒏 𝒂 𝒔𝒊𝒏 𝒃 1. + 𝜽 = 𝟏−𝒕𝒂𝒏𝜽 𝟒 𝟏+𝒕𝒂𝒏𝜽 𝟏−𝒙𝒚 𝒙−𝒚 𝒕𝒂𝒏−𝟏 𝟏+𝒙𝒚 = 𝒕𝒂𝒏−𝟏 𝒙 + 𝒕𝒂𝒏−𝟏 𝒚 = 𝒕𝒂𝒏−𝟏 𝒙 − 𝒕𝒂𝒏−𝟏 𝒚 65 . 𝒕𝒂𝒏𝟑𝜽 = 𝟑𝒕𝒂𝒏𝜽−𝒕𝒂𝒏𝟑 𝜽 𝟏−𝟑 𝒕𝒂𝒏𝟐 𝜽 𝟐 𝒄𝒐𝒔𝒆𝒄𝜽 = 𝟏 𝟐𝒕𝒂𝒏𝜽 (𝟏+𝒕𝒂𝒏𝟐 𝜽) 𝒔𝒊𝒏𝜽 𝒔𝒊𝒏𝟐𝜽 = 𝟐𝒔𝒊𝒏𝜽𝒄𝒐𝒔𝜽 = 𝒄𝒐𝒔𝟐𝜽 = 𝒄𝒐𝒔𝟐 𝜽 − 𝒔𝒊𝒏𝟐 𝜽 = 𝟐 𝒄𝒐𝒔𝟐 𝜽 − 𝟏 = 𝟏 − 𝟐 𝒔𝒊𝒏𝟐 𝜽 14. 21. 17. 𝒔𝒊𝒏𝑨 𝒄𝒐𝒔𝑩 = 𝒄𝒐𝒔𝑨 𝒔𝒊𝒏𝑩 = 𝒄𝒐𝒔𝑨 𝒄𝒐𝒔𝑩 = 𝟏 𝟐 𝟏 𝟐 𝟏 𝟐 𝒔𝒊𝒏 𝑨 + 𝑩 + 𝒔𝒊𝒏 𝑨 − 𝑩 𝒔𝒊𝒏 𝑨 + 𝑩 − 𝒔𝒊𝒏 𝑨 − 𝑩 𝒄𝒐𝒔 𝑨 + 𝑩 + 𝒄𝒐𝒔 𝑨 − 𝑩 𝟏 𝟐 𝒄𝒐𝒔 𝑨 − 𝒄𝒐𝒔 𝑩 = −𝟐 𝒔𝒊𝒏 𝟐 𝑨+𝑩 𝟐 𝒔𝒊𝒏 𝒔𝒊𝒏𝑨 𝒔𝒊𝒏𝑩 = − 𝒄𝒐𝒔 𝑨 + 𝑩 − 𝒄𝒐𝒔 𝑨 − 𝑩 1. 𝒕𝒂𝒏 𝝅 𝒕𝒂𝒏𝑨−𝒕𝒂𝒏𝑩 𝒕𝒂𝒏𝑨+𝒕𝒂𝒏𝑩 4. 𝒕𝒂𝒏 𝑨 − 𝑩 = 𝟏+𝒕𝒂𝒏𝑨𝒕𝒂𝒏𝑩 3. 7. 𝒔𝒊𝒏𝟐 𝜽 + 𝒄𝒐𝒔𝟐 𝜽 = 𝟏 𝟏 + 𝒕𝒂𝒏𝟐 𝜽 = 𝒔𝒆𝒄𝟐 𝜽 𝟏 + 𝒄𝒐𝒕𝟐 𝜽 = 𝒄𝒐𝒔𝒆𝒄𝟐 𝜽 𝒕𝒂𝒏𝜽 = 𝒄𝒐𝒕𝜽 = 𝒔𝒆𝒄𝜽 = 𝒔𝒊𝒏𝜽 𝒄𝒐𝒔𝜽 𝒄𝒐𝒔𝜽 𝒔𝒊𝒏𝜽 𝟏 𝒄𝒐𝒔𝜽 = 𝟏−𝒕𝒂𝒏𝟐 𝜽 𝟏+𝒕𝒂𝒏𝟐 𝜽 𝟐𝒕𝒂𝒏𝜽 𝟏−𝒕𝒂𝒏𝟐 𝜽 = = 𝟏 𝒄𝒐𝒕𝜽 𝟏 𝒕𝒂𝒏𝜽 10. 𝒕𝒂𝒏 𝑨 + 𝑩 = 𝟏−𝒕𝒂𝒏𝑨𝒕𝒂𝒏𝑩 2. 𝒕𝒂𝒏−𝟏 6. 𝒂 + 𝒃 𝟑 = 𝒂𝟑 + 𝟑𝒂𝟐 𝒃 + 𝟑𝒂𝒃𝟐 + 𝒃𝟑 𝒂 − 𝒃 𝟑 = 𝒂𝟑 − 𝟑𝒂𝟐 𝒃 + 𝟑𝒂𝒃𝟐 − 𝒃𝟑 𝒂𝟑 + 𝒃𝟑 = 𝒂 + 𝒃 𝒂𝟐 − 𝒂𝒃 + 𝒃𝟐 𝒂𝟑 − 𝒃𝟑 = 𝒂 − 𝒃 𝒂𝟐 + 𝒂𝒃 + 𝒃𝟐 TRIGONOMETRIC FORMULAE 1. 18. 8. 9. 6. 16. 3. 19. 6. 5. 𝒍𝒐𝒈 𝒂𝒎 = 𝒎 𝒍𝒐𝒈 𝒂 𝒂 𝒃 OMTEX CLASSES 4. 𝒕𝒂𝒏 𝝅 − 𝜽 𝟐 𝝅 − 𝜽 𝟐 𝝅 − 𝜽 𝟐 7. 𝒍𝒐𝒈𝒂 𝒂 = 𝟏 7. 𝒍𝒐𝒈 𝒂 − 𝒍𝒐𝒈 𝒃 = 𝒍𝒐𝒈 3. 15. 𝒄𝒐𝒔 𝒙 = − 𝒔𝒊𝒏 𝒙 𝒕𝒂𝒏 𝒙 = 𝒔𝒆𝒄𝟐 𝒙 𝒄𝒐𝒕 𝒙 = −𝒄𝒐𝒔𝒆𝒄𝟐 𝒙 𝒔𝒆𝒄 𝒙 = 𝒔𝒆𝒄 𝒙 𝒕𝒂𝒏 𝒙 𝒄𝒐𝒔𝒆𝒄 𝒙 = −𝒄𝒐𝒔𝒆𝒄 𝒙 𝒄𝒐𝒕 𝒙 𝒍𝒐𝒈 𝒙 = 𝒙 𝒅𝒙 𝒅 𝒅𝒙 𝒅 𝒅 𝟏 𝒂𝒙 = 𝒂𝒙 𝒍𝒐𝒈 𝒂 10. 𝒄𝒐𝒔 6. 𝒂𝒍𝒐𝒈𝒂 𝒃 = 𝒃 5. 3. 𝒂 = 𝒂𝟑 11. 𝒄𝒐𝒔 −𝜽 = 𝒄𝒐𝒔 𝜽 3.HSC MATHS 1. 4. 6. 9. 𝒅𝒙 𝒙 = 𝟏 13. 𝒏 𝟑 4. 𝒍𝒐𝒈 𝒂 + 𝒍𝒐𝒈 𝒃 = 𝒍𝒐𝒈 𝒂𝒃 2. 𝒄𝒐𝒕 8. 𝒔𝒊𝒏 5. 𝒂−𝒎 = 𝒂𝒎 7. 2. 9. 13. 𝒔𝒊𝒏 −𝜽 = − 𝒔𝒊𝒏 𝜽 2. 𝒔𝒊𝒏 𝒙 = 𝒄𝒐𝒔 𝒙 9. 𝒂 𝒂 = 𝒂 𝒂𝒎 𝒂𝒏 𝒃 𝒎 𝒏 = 𝒃𝒎 𝒂𝒎 𝒎+𝒏 𝒂 = 𝒂𝒏 = 𝒂𝒎𝒏 𝟏 = 𝒂𝒎−𝒏 𝒂 = 𝒂𝟐 1. 7. 𝒅𝒙 𝒌 = 𝟎 11. 𝒍𝒐𝒈𝒃 𝒂 = 𝒍𝒐𝒈 𝒃 𝟏 𝟏 𝟏 𝒍𝒐𝒈 𝒂 1. 𝒍𝒐𝒈 𝟏 = 𝟎 8TH YEAR 6. 𝒂𝒃 𝒎 = 𝒂𝒎 𝒃𝒎 3. 𝒂𝒎 = 𝒂−𝒎 8. 11. = 𝒄𝒐𝒔 𝜽 = 𝒔𝒊𝒏 𝜽 = 𝒄𝒐𝒕 𝜽 𝝅 − 𝜽 𝟐 𝝅 𝒄𝒐𝒔𝒆𝒄 − 𝟐 𝝅 𝒔𝒆𝒄 𝟐 − 𝜽 𝝅 𝒔𝒊𝒏 + 𝜽 𝟐 𝝅 𝒄𝒐𝒔 𝟐 + 𝜽 = 𝒕𝒂𝒏 𝜽 𝜽 = 𝒔𝒆𝒄 𝜽 = 𝒄𝒐𝒔𝒆𝒄 𝜽 = 𝒄𝒐𝒔 𝜽 = − 𝒔𝒊𝒏 𝜽 12. 𝒂𝟎 = 𝟏 2. 10. 𝒅𝒙 𝒆𝒙 = 𝒆𝒙 12. 𝒕𝒂𝒏 −𝜽 = −𝒕𝒂𝒏 𝜽 4. 5. 𝒍𝒐𝒈 𝒆 = 𝟏 8. 𝒔𝒊𝒏 𝒄𝒐𝒔 𝒔𝒊𝒏 𝒄𝒐𝒔 𝝅 + 𝜽 𝝅 + 𝜽 𝝅 − 𝜽 𝝅 − 𝜽 = − 𝒔𝒊𝒏 𝜽 = − 𝒄𝒐𝒔 𝜽 = 𝒔𝒊𝒏 𝜽 = − 𝒄𝒐𝒔 𝜽 DERIVATIVES FORMULAE 1. 14. 𝒅 𝒅𝒙 𝒅 𝒅𝒙 𝒅 𝒅𝒙 𝒅 𝒅𝒙 𝒅 𝒅𝒙 𝒅 𝒅𝒙 𝒅 𝒅𝒙 𝒙𝒏 = 𝒏𝒙𝒏−𝟏 8. 𝒂 𝒎 6. 𝒂𝒎 𝒏 𝟏 10. 𝒅𝒙 𝒅 𝒅 𝒅 𝒙 = 𝟐 𝟏 𝒙 66 . 𝒅𝒙 𝒕𝒂𝒏−𝟏 𝒙 = 𝟏+𝒙𝟐 𝒄𝒐𝒔𝒆𝒄−𝟏 𝒙 = Integration formulae 1. 24. 11. 18. 8. 16. 19. 17. 𝒅 𝒅𝒙 𝒅 𝒅 17. 23. 4. 𝟎 𝒅𝒙 = 𝑪 𝒘𝒉𝒆𝒓𝒆 𝒄 𝒊𝒔 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕 𝟏 𝒅𝒙 = 𝒙 + 𝒄 𝒙𝒏 𝒅𝒙 = 𝟏 𝒅𝒙 𝒙 𝟏 𝒅𝒙 𝒙 𝒙𝒏+𝟏 𝒏+𝟏 𝟏 𝒅𝒙 𝟏+𝒙𝟐 𝟏 𝒙 𝒙𝟐 −𝟏 15. 14. 13. 𝒅 𝒅𝒙 𝒅 𝟏 𝒙 𝒙𝟐 −𝟏 −𝟏 𝒙 𝒙𝟐 −𝟏 𝒅 −𝟏 𝒄𝒐𝒔−𝟏 𝒙 = 16. 3. 12. 10. 𝒅𝒙 𝒔𝒊𝒏−𝟏 𝒙 = 15. 𝒅𝒙 𝒔𝒆𝒄−𝟏 𝒙 = 19. 2. = 𝒕𝒂𝒏−𝟏 𝒙 + 𝒄 / − 𝒄𝒐𝒕−𝟏 𝒙 + 𝒄 𝒅𝒙 = 𝒔𝒆𝒄−𝟏 𝒙 + 𝒄 /−𝒄𝒐𝒔𝒆𝒄−𝟏 𝒙 + 𝒄 + 𝒄 𝒕𝒂𝒏𝒙 𝒅𝒙 = 𝒍𝒐𝒈 𝒔𝒆𝒄𝒙 + 𝒄 𝒄𝒐𝒕𝒙 𝒅𝒙 = 𝒍𝒐𝒈 𝒔𝒊𝒏 𝒙 + 𝒄 𝒔𝒆𝒄𝒙 𝒅𝒙 = 𝒍𝒐𝒈 𝒔𝒆𝒄𝒙 + 𝒕𝒂𝒏𝒙 + 𝒄 𝒄𝒐𝒔𝒆𝒄𝒙 𝒅𝒙 = 𝒍𝒐𝒈 𝒄𝒐𝒔𝒆𝒄𝒙 − 𝒄𝒐𝒕𝒙 + 𝒄 𝟏 𝒙𝟐 +𝒂𝟐 𝟏 𝒙𝟐 −𝒂𝟐 𝟏 𝒂𝟐 −𝒙𝟐 = 𝟐 𝒙 + 𝒄 = 𝒍𝒐𝒈𝒙 + 𝒄 𝒆𝒙 𝒅𝒙 = 𝒆𝒙 + 𝒄 𝒂𝒙 𝒅𝒙 = 𝒍𝒐𝒈𝒂 + 𝒄 𝒔𝒊𝒏𝒙 𝒅𝒙 = −𝒄𝒐𝒔𝒙 + 𝒄 𝒄𝒐𝒔𝒙 𝒅𝒙 = 𝒔𝒊𝒏𝒙 + 𝒄 𝒔𝒆𝒄𝟐 𝒙𝒅𝒙 = 𝒕𝒂𝒏𝒙 + 𝒄 𝒄𝒐𝒔𝒆𝒄𝟐 𝒙 𝒅𝒙 = −𝒄𝒐𝒕𝒙 + 𝒄 𝒔𝒆𝒄𝒙 𝒕𝒂𝒏𝒙 𝒅𝒙 = 𝒔𝒆𝒄𝒙 + 𝒄 𝒄𝒐𝒔𝒆𝒄𝒙 𝒄𝒐𝒕𝒙 𝒅𝒙 = −𝒄𝒐𝒔𝒆𝒄𝒙 + 𝒄 𝟏 𝟏−𝒙𝟐 𝒂𝒙 𝒅𝒙 = 𝒍𝒐𝒈 𝒙 + 𝒅𝒙 = 𝒍𝒐𝒈 𝒙 + 𝒅𝒙 = 𝒔𝒊𝒏−𝟏 = 𝒂 𝒕𝒂𝒏−𝟏 = 𝟐𝒂 𝒍𝒐𝒈 = 𝟐𝒂 𝒍𝒐𝒈 𝟏 𝟏 𝟏 𝟏 𝒙 𝒂 𝒙 𝒂 𝒙𝟐 + 𝒂𝟐 + 𝒄 𝒙𝟐 − 𝒂𝟐 + 𝒄 + 𝒄 + 𝒄 + 𝒄 + 𝒄 𝒙 𝒂 𝟏 𝒅𝒙 𝒙𝟐 +𝒂𝟐 𝟏 𝒅𝒙 𝒙𝟐 −𝒂𝟐 𝟏 𝒅𝒙 𝒂𝟐 −𝒙𝟐 𝟏 𝒙 𝒙𝟐 −𝒂𝟐 𝒙−𝒂 𝒙+𝒂 𝒂+𝒙 𝒂−𝒙 𝒅𝒙 = 𝒔𝒊𝒏 −𝟏 𝒙 + 𝒄 /− 𝒄𝒐𝒔 −𝟏 𝒅𝒙 = 𝒂 𝒔𝒆𝒄−𝟏 + 𝒄 67 . 5. 9. 25. 20. 7. 6. 26. 22. 𝒅𝒙 𝒄𝒐𝒕−𝟏 𝒙 = 𝟏+𝒙𝟐 18. 𝒙 + 𝒄 27.HSC MATHS OMTEX CLASSES 𝟏 𝟏−𝒙𝟐 −𝟏 𝟏−𝒙𝟐 𝟏 8TH YEAR 14. 21. 𝟐𝟕𝟎𝟎 𝟑𝝅 𝟐 −𝟏 𝟎 𝑵. 𝑫. 𝑫. 𝑫. 𝑫. 𝑫. 𝟑𝟎𝟎 𝝅 𝟔 𝟏 𝟐 𝟑 𝟐 𝟏 𝟑 𝟐 𝟐 𝟑 𝟑 𝟒𝟓𝟎 𝝅 𝟒 𝟏 𝟐 𝟏 𝟐 𝟏 𝟐 𝟐 𝟏 𝟔𝟎𝟎 𝝅 𝟑 𝟑 𝟐 𝟏 𝟐 𝟑 𝟐 𝟑 𝟐 𝟑 𝟏 𝟑 𝟗𝟎𝟎 𝝅 𝟐 𝟏 𝟎 𝑵. 𝟏 𝑵. 𝑫. −𝟏 𝑵. 𝟏 𝑵. 𝑫 𝟏 𝑵. 𝑫. 𝟎 𝟏𝟖𝟎𝟎 𝝅 𝟎 −𝟏 𝟎 𝑵.HSC MATHS OMTEX CLASSES 8TH YEAR 𝜽 𝐬𝐢𝐧 𝜽 𝐜𝐨𝐬 𝜽 𝐭𝐚𝐧 𝜽 𝐜𝐨𝐬𝐞𝐜 𝜽 𝐬𝐞𝐜 𝜽 𝐜𝐨𝐭 𝜽 𝟎𝟎 𝟎𝟎 𝟎 𝟏 𝟎 𝑵. 𝑫 68 . 𝟎 𝟑𝟔𝟎𝟎 𝟐𝝅 𝟎 𝟏 𝟎 𝑵. 𝑫. −𝟏 𝑵.