Winter 0 Eamination Subject & Code: Basic Maths (70) Model Answer Page No: / Important Instructions to the Eaminers: ) The Answers should be eamined by key words and not as word-to-word as given in the model answer scheme. ) The model answer and the answer written by candidate may vary but the eaminer may try to assess the understanding level of the candidate. ) The language errors such as grammatical, spelling errors should not be given more importance. (Not applicable for subject English and Communication Skills.) ) While assessing figures, eaminer may give credit for principal components indicated in the figure. The figures drawn by the candidate and those in the model answer may vary. The eaminer may give credit for any equivalent figure drawn. 5) Credits may be given step wise for numerical problems. In some cases, the assumed constant values may vary and there may be some difference in the candidate s Answers and the model answer. 6) In case of some questions credit may be given by judgment on part of eaminer of relevant answer based on candidate s understanding. 7) For programming language papers, credit may be given to any other program based on equivalent concept.
Subject & Code: Basic Maths (70) Page No: / ) a) b) Attempt any TEN of the following: 5 Find the value of 6 5 6 6 + 5 6 If A, find the matri B such that 0 A + B 6 A 8 6 B A 8 6 B 8 A + B 0 B A 6 8 6 B 8 c) Find the value of a and b, if a b 5 5 6 a b 6 5 + a b a + b 5 b 6 a b 5 5 6 a + b 6 5 b a
Subject & Code: Basic Maths (70) Page No: / ) d) e) 6 Find the adjoint of matri 7 6 Let A 7 7 C ( A) 6 7 6 adj ( A) 6 Let A 7 A 7 A A 6 A 7 C ( A) 6 7 6 adj ( A) Resolve into partial fractions: A B + + + + A + B Put 0 i. e., + A + 0 A A Put + 0 i. e., B B 0 + B + +
Subject & Code: Basic Maths (70) Page No: / ) A B + + + A + + B A B + + Note for partial fraction problems: The problems of partial fractions could also be solved by the method of equating equal power coefficients. This method is also applicable. Give appropriate marks in accordance with the scheme of marking in the later problems as the solution by this method is not discussed. For the sake of convenience, the solution of the above problem with the help of this method is illustrated hereunder. A B + + + + A + B A + A + B B + 0 A + B + A B A + B A B 0 A + B A B 0 A A B A B + +
Subject & Code: Basic Maths (70) Page No: 5/ ) f) Show that π tanθ tan θ + tan θ g) π tan tanθ π tan θ π + tan tanθ tanθ + tanθ Prove that cos A cos A cos A cos A + A cos Acos A sin Asin A cos A sin ( A) cos A cos cos A + cos cos A cos cos sin A A A A A ( A) cos A cos cos A h) If sin A 0., find the value of sin A. sin sin sin A A A 0. 0. 0.9... * Note (*): Due to the use of advance scientific calculator, writing directly the step (*) is allowed. No marks to be deducted. Given that sin A 0.. A sin 0..578º sin A sin.578º 0.9
Subject & Code: Basic Maths (70) Page No: 6/ ) i) Prove that cos θ sin θ + cos θ cosθ sinθ cos θ sin θ sinθ cos θ + cosθ sin θ + cosθ sinθ cosθ sinθ sin ( θ + θ ) cosθ sinθ sin θ cosθ sinθ sin ( θ ) cosθ sinθ sin θ cos θ cosθ sinθ sinθ cosθ cos θ cosθ sinθ cos θ cos sin cos cos sin sin + + cosθ sinθ cosθ sinθ cos θ + sin θ θ θ θ θ θ θ cos θ sin θ ( θ θ ) cos sin cos θ j) Evaluate without using calculator tan 66º + tan 69º tan 66º tan 69º tan 66º + tan 69º tan 66º + 69º tan 66º tan 69º tan5º tan 90º + 5º tan 80º 5º cot 5º tan 5º
Subject & Code: Basic Maths (70) Page No: 7/ ) k) y Find the slope and y-intercept of the line y 0 a b c a slope m b or 0.75 c y int 6 b y y 0 a b c a slope m b or 0.75 c y int 6 b y y 6 slope m or 0.75 y int 6 l) Find the range of the following:,,, 0, 6,, 7, 0, L S Range L S 0
Subject & Code: Basic Maths (70) Page No: 8/ ) Attempt any FOUR of the following: a) Solve the equations for y and z y z y z y z + 5, +, + 5 7 9 6 by using Cramer s rule. y z + 5 y z + 5 y z + 7 9 6 D + + + 5 5 8 5 7 7 9 6 D D y z or 0.009 008 5 5 + + 5 6 5 8 5 7 7 6 977 or.8 50 5 + + + 5 9 7 7 7 9 89 or. D y.8 y 08.5 D 0.009 Dz. z.970 D 0.009 (Please refer note on the net page)
Subject & Code: Basic Maths (70) Page No: 9/ ) Note: As the use of the advance scientific calculator is permissible, calculating directly the values of fractional quantities e.g., + + + 5 8 5 7 is allowed. The same is also applicable in the net alternative method. No marks to be deducted for such direct calculations. y + 6z 60 0 + 5y 6z 0 8 y + z 5 6 D 0 5 6 5 8 + 0 + 08 + 6 0 70 y 8 60 6 8 5 95 D 0 0 6 690 5 60 0 + 08 + 6 50 590 b) z 60 8 5 5586 D 0 5 0 780 + 60 + 50 590 + 60 0 70 5590 Dy 5586 y 08.55 D 95 Dz 5590 z.970 D 95 If A, find A. A A A + + + 6 8 9 8 6 6 + + + 8 8 + + 8 + 9 (Please check note on net page)
Subject & Code: Basic Maths (70) Page No: 0/ ) Note: In the answer matri of A², if to elements are wrong either in sign or value, deduct marks; if to 6 elements are wrong, you may deduct mark; other deduct all marks. c) If A, B, C 0, verify that A B + C AB + AC. B + C + 0 A( B + C ) + 8 + 6 9 + + 7 8 5 AB + + 6 6 9 + + 6 7 7 AC 0 + + 0 6 + 6 + 0 6 7 AB + AC 7 + 7 8 5 A B + C AB + AC
Subject & Code: Basic Maths (70) Page No: / ) d) If A, find A A 9I +, where I is the unit matri A A A 9 6 + + + 6 6 9 6 + + + + 6 6 + + 9 + 5 7 6 6 9 A 6 9 9 6 0 0 9 0 0 9I 9 0 0 0 9 0 0 0 0 0 9 5 6 9 9 0 0 A A + 9I 6 9 0 9 0 + 7 6 9 6 0 0 9 + 9 5 + 6 + 0 9 + 0 6 0 9 9 0 + + + + 7 + 9 + 0 + 0 6 6 + 9 6 5 8 Note: The above problem could also be solved by taking all the terms simultaneously as follows: A A + 9I 0 0 + 9 0 0 0 0
Subject & Code: Basic Maths (70) Page No: / ) 9 6 + + + 6 6 9 9 0 0 6 9 6 6 9 0 9 0 + + + + + 6 6 + + 9 + 9 6 0 0 9 5 6 9 9 0 0 6 9 0 9 0 + 7 6 9 6 0 0 9 + 9 5 + 6 + 0 9 + 0 6 + 0 9 9 0 + + + 7 + 9 + 0 + 0 6 6 + 9 6 5 8 ++ e) Resolve into partial fractions: + + 5 + + + 5 + y + y + 5y + ( Put y) y + A B + y + y + y + y + y + y + A + y + B Put y + 0 or y + + A + 0 A A Put y + 0 or y + 0 + + B B B
Subject & Code: Basic Maths (70) Page No: / ) y + + y + 5y + y + y + + + + 5 + + + f) Resolve into partial fractions: + 9 + 0 + 9 9 0 0 A B + 9 + + 0 + A + B Put 0 i. e., 0 6A + 0 5 A Put + 0 i. e., 0 0 6B 5 B 0 5 5 + 9 + + 5 5 + + 9 + ) Attempt any FOUR of the following: a) Solve the equations + y + z, + y + z, + y + z by using matri inversion method. + y + z + y + z + y + z
Subject & Code: Basic Maths (70) Page No: / ) A X y K z A 8 6 + 9 C ( A) 8 5 5 5 The minor matri of A is M ( A) 8 5 5 5 the mati of cofactors is, 8 5 C ( A) 5 5 The minors of matri A are A 8 A A 5 A A 5 A A 5 A A
Subject & Code: Basic Maths (70) Page No: 5/ ) the mati of cofactors is, 8 5 C ( A) 5 5 8 5 adj ( A) 5 5 8 5 A 5 5 8 5 X A K 5 5 8 8 8 y z b) Resolve into partial fractions: ( + ) + + A B + C + ( )( + ) + A B + C + + + + + + + + A B C
Subject & Code: Basic Maths (70) Page No: 6/ ) Put + + A + 0 78 0A 9 A 5 Put 0 A ( C ) 0 + 0 0 + + 0 0 + 0 A C 9 0 C 5 9 C 5 C 5 Put A ( B C ) + + + + A B C 9 B 5 5 9 B 5 5 68 B 5 B 5 9 + + 5 + 5 5 + ( + ) Note for Partial Fraction Methods: The above Q. (e) & (f) problems of partial fractions could be solved by the method of equating equal power coefficients also. This method, illustrated in the solution of Q. (e), is also applicable. Give appropriate marks in accordance with the scheme of marking. As this method is very tedious and complicated, hardly someone use this method in such cases. So such solution methods for partial fraction problems are not illustrated herein.
Subject & Code: Basic Maths (70) Page No: 7/ ) c) e + Resolve into partial fractions: e + 7e + 5 e + e + 7e + 5 y + y + 7 y + 5 y + ( y + 5)( y + ) y + 5 e + 5 ( Put e y) e + e + 7e + 5 y + y + 7y + 5 ( Put e y) y + A B + y + 5 y + y + 5 y + y + y + A + y + 5 B 5 Put y + 5 0 y 5 5 + + A + 0 A A Put y + 0 y + 0 + + 5 B 0 B 0 B y + 0 + y + y + y + y + 7 5 5 e + e + e + e + 7 5 5
Subject & Code: Basic Maths (70) Page No: 8/ ) d) Prove that sin( A+ B) sin A.cos B + cos A.sin B QN sin( A + B) OQ QR + RN OQ QR + PM OQ QR PM + OQ OQ QR PQ PM OP + PQ OQ OP OQ cos A.sin B + sin A.cos B e) Note: The above is proved by different ways in several books. Consider all these proof but check whether the method is falling within the scope of curriculum and give appropriate marks in accordance with the scheme of marking. In accordance with the Teacher s Manual published by MSBTE, the result is treated as Fundamental Result which is not proved by the help of any another result. If the above result is proved by students using any another result, suppose using cos (A+B), then this result i.e., cos (A+B) must have been proved first. 5 π Prove that cot ( ) + cos ec cot ( ) tan tan tan +
Subject & Code: Basic Maths (70) Page No: 9/ ) cot ( ) tan + tan tan tan + Let A cos ec 5 cos eca 5 5 cot ( ) + cos ec tan + tan tan tan π + 5 cot ( ) + cos ec tan + cot π Note that the result tan cot π + can be used directly
Subject & Code: Basic Maths (70) Page No: 0/ ) f) tan + tan + tan π Prove that π π + π π π + + tan ( ) π π + π π + + + + tan tan tan tan tan + + tan + + + π + + tan tan tan tan tan π + + tan tan π + π tan tan + tan + tan + tan ( ) tan + π + tan π π + + tan tan + tan tan π ) Attempt any FOUR of the following. a) Without using the calculator, find the value of + sin 0º cos ec 0º + cos 70º tan 0º cot 0º tan 0º sin0º sin 90º + 0º cos 0º
Subject & Code: Basic Maths (70) Page No: / ) sin 0º cos ec0º cos 0º ec cot 0º cos 70º 0 cot0º cot 90º + 0º tan 0º cos 70º cos 90º + 0 sin 0 0 But given that + sin 0º cos ec 0º + cos 70º tan 0º cot 0º + + 0 9 or.5 b) Prove that cos A + cos5 A + cos7 A cos A sin Atan A cos A + cosa + cos5a cosa + cos5a + cos7a cosa + cos 7A + cos5a cos A + cosa + cos5a cos A + cos5a + cosa cos5acos A + cos5a cosacos A + cosa ( A) ( A) cos5a cos + cosa cos + cos5a cosa cos( A + A) cosa
Subject & Code: Basic Maths (70) Page No: / ) cos AcosA sin Asin A cosa cos A sin Atan A c) Prove that (in ABC ), tan A + tan B + tan C tan A tan B tan C We have, A + B + C 80º or π A + B 80º C ( A B) ( C) tan + tan 80º tan A + tan B tan C tan Atan B tan A + tan B tan C tan Atan B [ ] tan A + tan B tan C + tan Atan B tan C tan A + tan B + tan C tan Atan B tan C d) Prove that tanθ tan θ tan θ tan θ tan θ tan θ + θ tanθ + tan θ tanθ tan θ tanθ tanθ + tan θ tanθ tan tan θ θ tanθ tan θ + tanθ tan θ tan θ tanθ tanθ tan θ tanθ tan θ + tanθ tan θ tan θ tanθ tan θ tan θ
Subject & Code: Basic Maths (70) Page No: / ) e) Prove that cos + cos cos 5 65 A cos B cos 5 cos A cos B 5 cos A + B cos Acos B sin Asin B 5 5 5 65 A + B cos 65 cos + cos cos 5 65 f) If 5 tan, tan y, show that 6 π + y 5 tan, tan y 6 5 tan, y tan 6 5 + y tan + tan 6 tan tan π 5 + 6 5 6
Subject & Code: Basic Maths (70) Page No: / ) 5 tan, tan y 6 tan + tan y tan ( + y) tan tan y 5 + 6 5 6 π + y tan 5) Attempt any FOUR of the following. a) Without using calculator prove that cos 0º cos 0º cos 60º cos80º 6 cos 0º cos 0º cos 60º cos 80º cos 0º cos 0º cos 80º ( cos 0º cos 0º ) cos 80º ( cos 60º + cos 0º ) cos 80º + cos 0º cos 80º cos 80º + cos 80º cos 0º cos 80º + cos 80º cos 0º cos 80º + ( c ) os00º + cos 60º cos 80º cos00º 8 + + cos 90º cos ( 0º ) 8 + 0 8 + 6
Subject & Code: Basic Maths (70) Page No: 5/ 5) Note The above problem may also be solved by making various combinations of cosine ratios. Consequently the solutions vary in accordance with the combinations. Please give the appropriate marks in accordance with the scheme of marking. For the sake of convenience one of the solutions is illustrated hereunder. cos 0º cos 0º cos60º cos80º cos 0º cos 0º cos80º ( cos 0º cos80º ) cos 0º ( cos0º + cos 0º ) cos 0º ( cos ( 90º + 0º ) + cos 0º ) cos 0º ( sin 0º + cos 0º ) cos 0º + cos 0º cos 0º cos 0º + cos 0º cos 0º cos 0º + cos 0º cos 0º cos 0º cos 60º cos ( 0º ) + + cos 0º cos 0º 8 + + 8 6
Subject & Code: Basic Maths (70) Page No: 6/ 5) b) Prove that sin + sin 5 + sin 6 tan 5 cos + cos5 + cos6 sin + sin 5 + sin 6 sin + sin 6 + sin 5 cos + cos5 + cos 6 cos + cos 6 + cos5 sin 5 cos + sin 5 cos5 cos + cos5 sin 5 cos + cos5 cos + tan 5 + c) Prove that + y y tan + tan y tan, > 0, y > 0, y < Put tan A and tan y B tan A and y tan B tan A + tan B tan ( A + B) tan A tan B + y y + y y A + B tan + y y tan + tan y tan d) Find the equation of a straight line passing through (, 5) and the point of intersection of the lines + y 0, y 9. + y 0 y 9 9 y Point of intersection,
Subject & Code: Basic Maths (70) Page No: 7/ 5) equation is, y y y y y 5 5 8 + y 0 ( ) Point of intersection, y y Slope m equation is, y y m y 5 8 5 8 8 + y 0 e) Find the equation of the straight line passing through (-, 0) and sum of their intercepts is 8. Let int a y int b a + b 8 equation is y y + or + a b a 8 a b + ay ab ( 8 a) ay a( 8 a) ( ) ( 8 a) 0a a ( 8 a) + But passing through, 0 + + a + 0a 8a a 5a 0 + a a, 8 y y + or + 5 8 6 + +
Subject & Code: Basic Maths (70) Page No: 8/ 5) f) Find the acute angle between the lines + y, 5y + 7 0 For + y, a slope m b For 5y + 7 0, a slope m b 5 5 m m tanθ + m m 5 + 5 6 or.55 6 θ tan or tan (.55) 6) Attempt any FOUR of the following. a) Find the equation of straight line passing through (5, 6) and making an angle 50º with -ais. Given θ 50º slope m tanθ tan50º equation is ( ) y y m y 6 5 y 6 + 5 + y 6 5 0
Subject & Code: Basic Maths (70) Page No: 9/ 6) equation is θ ( ) y y tan y 6 tan50º 5 y 6 5 ( ) y 6 + 5 + y 6 5 0 b) If the length of perpendicular from (5, ) on the straight line + y + k 0 is 5 units. Find the value of k. p 5 5 a + by + c a 5 + + k + k + 5 + b 5 5 + k 0 + k 0 + k or 0 + k + 6 k or k + c) The scores of two batsmen A and B in ten innings during a certain season are as under: A 8 7 6 7 9 0 60 96 B 9 8 5 67 90 0 6 0 80 Find which of the two batsmen is more consisting in scoring (use coefficient of variance).
Subject & Code: Basic Maths (70) Page No: 0/ 6) For Batsman A: 60 6 0 i 7660 60 σ 5.95 0 0 5.95 CV ( A) 00 55. 6 i 0 8 78 7 09 6 969 7 50 9 5 0 00 60 600 96 96 96 60 7660 For Batsman B: 500 50 0 i 0968 500 σ.9 0 0.9 CV ( B) 00 8.858 50 i 9 6 96 8 0 5 809 67 89 90 800 0 00 6 8 0 600 80 600 500 0968
Subject & Code: Basic Maths (70) Page No: / 6) CV ( A) CV B < B is more consistent. d) Find the range and the coefficient of range for the following: 0-0- 0-50- 60-70- 80-90- 9 9 9 59 69 79 89 99 of Students 0 5 6 0 09 08 L 99 S 0 Difference between two sets D Range L S + D 99 0 + 80 L S + D Coeff. of Range L + S 99 0 + 99 + 0 80 or 0.67 9 Class Cont. Class 0-9 9.5-9.5 0-9 9.5-9.5 0-9 9.5-9.5 50-59 9.5-59.5 60-69 59.5-69.5 70-79 69.5-79.5 80-89 79.5-89.5 90-99 89.5-99.5 L 99.5 S 9.5 Range L S 99.5 9.5 80 L S Coeff. of Range L + S 99.5 9.5 99.5 + 9.5 80 or 9 0.67
Subject & Code: Basic Maths (70) Page No: / 6) e) Calculate the mean deviation for the following data: Class intervals 0-59 60-79 80-99 00-9 0-9 of families 50 00 500 00 60 Class i f i fi i Di i fid i 0-59 9.5 50 75 8.559 97.95 60-79 69.5 00 0850 8.559 5567.7 80-99 89.5 500 750. 70.5 00-9 09.5 00 900. 88. 0-9 9.5 60 7770. 86.6 0 9775 990.8 + fii 9775 88.059 N 0 fidi M. D. N 990.8 0.505 f) Find the variance and coefficient of variance for the following distribution: Class 0-5- 0-5- 0-5- 50-55- 60- Intervals 5 0 5 0 5 50 55 60 65 Frequency 5 0 50 90 75 60 5 5 5 Class i f i fi i i fi i 0-5.5 5 56.5 506.5 656. 5-0 7.5 0 85 756.5 687.5 0-5.5 50 65 056.5 58.5 5-0 7.5 90 75 06.5 656 0-5.5 75 87.5 806.5 569 5-50 7.5 60 850 56.5 575 50-55 5.5 5 87.5 756.5 9668.8 55-60 57.5 5 7.5 06.5 8656. 60-65 6.5 5 97.5 906.5 5859.8 05 667.5 78
Subject & Code: Basic Maths (70) Page No: / 6) fii 667.5.08 N 05 fii fii S. D. N N 78 667.5 05 05 9.9 Variance ( S. D. ) 9.9 98.87 S. D. Coeff. of Variance 00 9.9 00.08. fii fii Variance N N 78 667.5 05 05 98.87 variance Coeff. of Variance 00 98.87 00.08. Class i f i d i i i f d d 0-5.5 5 - -00 6 00 5-0 7.5 0 - -90 9 70 0-5.5 50 - -00 00 5-0 7.5 90 - -90 90 0-5.5 75 0 0 0 0 5-50 7.5 60 60 60 50-55 5.5 5 70 0 55-60 57.5 5 75 9 5 60-65 6.5 5 60 6 0 05-5 65 i f d i i
Subject & Code: Basic Maths (70) Page No: / 6) i A A.5, h 5, di h fidi A + h N 5.5 + 5 05.08 fidi fidi S. D. h N N 65 5 5 05 05 9.9 Variance ( S. D. ) 9.9 98.87 S. D. Coeff. of Variance 00 9.9 00.08. f d f d i i i i Variance h N N 65 5 5 05 05 98.87 variance Coeff. of Variance 00 98.87 00.08. Important Note In the solution of the question paper, wherever possible all the possible alternative methods of solution are given for the sake of convenience. Still student may follow a method other than the given herein. In such case, FIRST SEE whether the method falls within the scope of the curriculum, and THEN ONLY give appropriate marks in accordance with the scheme of marking.