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Page 1 Strictly Confidential — (For Internal and Restricted Use Only) Senior School Certificate Examination March 2016 Marking Scheme — Mathematics 65/1/S, 65/2/S, 65/3/S General Instructions: 1. The Marking Scheme provides general guidelines to reduce subjectivity in the marking. The answers given in the Marking Scheme are suggested answers. The content is thus indicative. If a student has given any other answer which is different from the one given in the Marking Scheme, but conveys the meaning, such answers should be given full weightage 2. Evaluation is to be done as per instructions provided in the marking scheme. It should not be done according to one’s own interpretation or any other consideration — Marking Scheme should be strictly adhered to and religiously followed. 3. Alternative methods are accepted. Proportional marks are to be awarded. 4. In question (s) on differential equations, constant of integration has to be written. 5. If a candidate has attempted an extra question, marks obtained in the question attempted first should be retained and the other answer should be scored out. 6. A full scale of marks  0 to 100 has to be used. Please do not hesitate to award full marks if the answer deserves it. 7. Separate Marking Scheme for all the three sets has been given. 8. As per orders of the Hon’ble Supreme Court. The candidates would now be permitted to obtain photocopy of the Answer book on request on payment of the prescribed fee. All examiners/Head Examiners are once again reminded that they must ensure that evaluation is carried out strictly as per value points for each answer as given in the Marking Scheme. Page 2 Strictly Confidential — (For Internal and Restricted Use Only) Senior School Certificate Examination March 2016 Marking Scheme — Mathematics 65/1/S, 65/2/S, 65/3/S General Instructions: 1. The Marking Scheme provides general guidelines to reduce subjectivity in the marking. The answers given in the Marking Scheme are suggested answers. The content is thus indicative. If a student has given any other answer which is different from the one given in the Marking Scheme, but conveys the meaning, such answers should be given full weightage 2. Evaluation is to be done as per instructions provided in the marking scheme. It should not be done according to one’s own interpretation or any other consideration — Marking Scheme should be strictly adhered to and religiously followed. 3. Alternative methods are accepted. Proportional marks are to be awarded. 4. In question (s) on differential equations, constant of integration has to be written. 5. If a candidate has attempted an extra question, marks obtained in the question attempted first should be retained and the other answer should be scored out. 6. A full scale of marks  0 to 100 has to be used. Please do not hesitate to award full marks if the answer deserves it. 7. Separate Marking Scheme for all the three sets has been given. 8. As per orders of the Hon’ble Supreme Court. The candidates would now be permitted to obtain photocopy of the Answer book on request on payment of the prescribed fee. All examiners/Head Examiners are once again reminded that they must ensure that evaluation is carried out strictly as per value points for each answer as given in the Marking Scheme. QUESTION PAPER CODE 65/1/S EXPECTED ANSWER/VALUE POINTS SECTION A 1. Getting sin ? = 1 1 3 1 4 2 . 2 3 = 1 2 Hence  .  a b = 1 4 3 . . 1 2 2 3 = 1 2 2. 2  2  1 a b  = ? 1 . . 2 a b = 1 2 ? Angle between a and b = 4 p 1 2 3. Writing or using, that given planes are parallel 1 2 d =  4 10  2 units 4 9 36 + = + + 1 2 4. T T 2  AA  A  A  A = = 1 2 = 25 1 2 5. Getting AB = 7 8 0 10  ? ? ? ?  ? ? 1 2 AB = –70 1 2 6. k(2) = –8 ? k = –4 1 2 –4(3) = 4a ? a = –3 1 2 SECTION B 7. 1 (sin 2 ) sin ( 3 ) x y x x u v  = + = + ? dy du dv dx dx dx = + 1 2 (sin 2 ) log logsin 2 x u x u x x = ? = 1 2 1 2 cot 2 log sin 2 = · + du x x x u dx 1 65/1/S (1) 65/1/S Page 3 Strictly Confidential — (For Internal and Restricted Use Only) Senior School Certificate Examination March 2016 Marking Scheme — Mathematics 65/1/S, 65/2/S, 65/3/S General Instructions: 1. The Marking Scheme provides general guidelines to reduce subjectivity in the marking. The answers given in the Marking Scheme are suggested answers. The content is thus indicative. If a student has given any other answer which is different from the one given in the Marking Scheme, but conveys the meaning, such answers should be given full weightage 2. Evaluation is to be done as per instructions provided in the marking scheme. It should not be done according to one’s own interpretation or any other consideration — Marking Scheme should be strictly adhered to and religiously followed. 3. Alternative methods are accepted. Proportional marks are to be awarded. 4. In question (s) on differential equations, constant of integration has to be written. 5. If a candidate has attempted an extra question, marks obtained in the question attempted first should be retained and the other answer should be scored out. 6. A full scale of marks  0 to 100 has to be used. Please do not hesitate to award full marks if the answer deserves it. 7. Separate Marking Scheme for all the three sets has been given. 8. As per orders of the Hon’ble Supreme Court. The candidates would now be permitted to obtain photocopy of the Answer book on request on payment of the prescribed fee. All examiners/Head Examiners are once again reminded that they must ensure that evaluation is carried out strictly as per value points for each answer as given in the Marking Scheme. QUESTION PAPER CODE 65/1/S EXPECTED ANSWER/VALUE POINTS SECTION A 1. Getting sin ? = 1 1 3 1 4 2 . 2 3 = 1 2 Hence  .  a b = 1 4 3 . . 1 2 2 3 = 1 2 2. 2  2  1 a b  = ? 1 . . 2 a b = 1 2 ? Angle between a and b = 4 p 1 2 3. Writing or using, that given planes are parallel 1 2 d =  4 10  2 units 4 9 36 + = + + 1 2 4. T T 2  AA  A  A  A = = 1 2 = 25 1 2 5. Getting AB = 7 8 0 10  ? ? ? ?  ? ? 1 2 AB = –70 1 2 6. k(2) = –8 ? k = –4 1 2 –4(3) = 4a ? a = –3 1 2 SECTION B 7. 1 (sin 2 ) sin ( 3 ) x y x x u v  = + = + ? dy du dv dx dx dx = + 1 2 (sin 2 ) log logsin 2 x u x u x x = ? = 1 2 1 2 cot 2 log sin 2 = · + du x x x u dx 1 65/1/S (1) 65/1/S ? (sin 2 ) [2 cot 2 log sin 2 ] x du x x x x dx = + 1 2 1 3 1 3 2 dv dx x x =  1 ? 3 (sin 2 ) [2 cot 2 logsin 2 ] 2 1 3 x dy x x x x dx x x = + +  1 2 OR Let y = 2 2 1 1 2 2 2 1 1 tan and cos 1 1 x x z x x x   ? ? +   ? ? = ? ? + +  ? ? z = cos –1 x 2 ? x 2 = cos z ? y = 1 1 cos 1 cos tan 1 cos 1 cos z z z z  ? ? +   ? ? ? ? + +  ? ? 1 ? y = 1 1 cos sin 1 tan 2 2 2 tan tan cos sin 1 tan 2 2 2 z z z z z z   ? ? ? ?   ? ? ? ? = ? ? ? ? ? ? ? ? + + ? ? ? ? 1 2 + 1 2 ? y = 1 tan tan 4 2 4 2 z z  ? p ? p ? ?  =  ? ? ? ? ? ? ? ? 1 2 + 1 2 ? dy dz = 1 2  1 8. LHL = 0 lim .sin ( 1) 2  ? p + = x k x k 1 RHL = 3 0 tan (1 cos ) lim x x x x + ?  1 = 2 0 tan sin /2 1 lim . 2 2 /2 2 x x x x x + ? ? ? = ? ? ? ? 1 ? k = 1 2 1 9. When x = am 2 , we get y = ± am 3 1 ay 2 = x 3 ? 2 2 3 2 3 2 dy dy x ay x dx dx ay = ? = 1 slope of normal = 3 2 4 2 2 3 3 a am m a m = ± ± 1 ? Equation of normal is y ± am 3 = 2 2 ( ) 3 x am m  ± 1 [Full marks may be given, if only one value for point, slope and equation is derived] 65/1/S (2) 65/1/S Page 4 Strictly Confidential — (For Internal and Restricted Use Only) Senior School Certificate Examination March 2016 Marking Scheme — Mathematics 65/1/S, 65/2/S, 65/3/S General Instructions: 1. The Marking Scheme provides general guidelines to reduce subjectivity in the marking. The answers given in the Marking Scheme are suggested answers. The content is thus indicative. If a student has given any other answer which is different from the one given in the Marking Scheme, but conveys the meaning, such answers should be given full weightage 2. Evaluation is to be done as per instructions provided in the marking scheme. It should not be done according to one’s own interpretation or any other consideration — Marking Scheme should be strictly adhered to and religiously followed. 3. Alternative methods are accepted. Proportional marks are to be awarded. 4. In question (s) on differential equations, constant of integration has to be written. 5. If a candidate has attempted an extra question, marks obtained in the question attempted first should be retained and the other answer should be scored out. 6. A full scale of marks  0 to 100 has to be used. Please do not hesitate to award full marks if the answer deserves it. 7. Separate Marking Scheme for all the three sets has been given. 8. As per orders of the Hon’ble Supreme Court. The candidates would now be permitted to obtain photocopy of the Answer book on request on payment of the prescribed fee. All examiners/Head Examiners are once again reminded that they must ensure that evaluation is carried out strictly as per value points for each answer as given in the Marking Scheme. QUESTION PAPER CODE 65/1/S EXPECTED ANSWER/VALUE POINTS SECTION A 1. Getting sin ? = 1 1 3 1 4 2 . 2 3 = 1 2 Hence  .  a b = 1 4 3 . . 1 2 2 3 = 1 2 2. 2  2  1 a b  = ? 1 . . 2 a b = 1 2 ? Angle between a and b = 4 p 1 2 3. Writing or using, that given planes are parallel 1 2 d =  4 10  2 units 4 9 36 + = + + 1 2 4. T T 2  AA  A  A  A = = 1 2 = 25 1 2 5. Getting AB = 7 8 0 10  ? ? ? ?  ? ? 1 2 AB = –70 1 2 6. k(2) = –8 ? k = –4 1 2 –4(3) = 4a ? a = –3 1 2 SECTION B 7. 1 (sin 2 ) sin ( 3 ) x y x x u v  = + = + ? dy du dv dx dx dx = + 1 2 (sin 2 ) log logsin 2 x u x u x x = ? = 1 2 1 2 cot 2 log sin 2 = · + du x x x u dx 1 65/1/S (1) 65/1/S ? (sin 2 ) [2 cot 2 log sin 2 ] x du x x x x dx = + 1 2 1 3 1 3 2 dv dx x x =  1 ? 3 (sin 2 ) [2 cot 2 logsin 2 ] 2 1 3 x dy x x x x dx x x = + +  1 2 OR Let y = 2 2 1 1 2 2 2 1 1 tan and cos 1 1 x x z x x x   ? ? +   ? ? = ? ? + +  ? ? z = cos –1 x 2 ? x 2 = cos z ? y = 1 1 cos 1 cos tan 1 cos 1 cos z z z z  ? ? +   ? ? ? ? + +  ? ? 1 ? y = 1 1 cos sin 1 tan 2 2 2 tan tan cos sin 1 tan 2 2 2 z z z z z z   ? ? ? ?   ? ? ? ? = ? ? ? ? ? ? ? ? + + ? ? ? ? 1 2 + 1 2 ? y = 1 tan tan 4 2 4 2 z z  ? p ? p ? ?  =  ? ? ? ? ? ? ? ? 1 2 + 1 2 ? dy dz = 1 2  1 8. LHL = 0 lim .sin ( 1) 2  ? p + = x k x k 1 RHL = 3 0 tan (1 cos ) lim x x x x + ?  1 = 2 0 tan sin /2 1 lim . 2 2 /2 2 x x x x x + ? ? ? = ? ? ? ? 1 ? k = 1 2 1 9. When x = am 2 , we get y = ± am 3 1 ay 2 = x 3 ? 2 2 3 2 3 2 dy dy x ay x dx dx ay = ? = 1 slope of normal = 3 2 4 2 2 3 3 a am m a m = ± ± 1 ? Equation of normal is y ± am 3 = 2 2 ( ) 3 x am m  ± 1 [Full marks may be given, if only one value for point, slope and equation is derived] 65/1/S (2) 65/1/S 10. Writing 1 sin (1 sin ) 2sin sin (1 sin ) sin (1 sin ) x x x dx dx x x x x  +  = + + ? ? 1 = 1 1 2 sin 1 sin dx dx x x  + ? ? 1 = 2 (1 sin ) cosec 2 cos x x dx dx x   ? ? 1 = 2 log  cosec cot  2 (sec sec tan ) x x x x x dx    ? 1 2 = log  cosec cot  2(tan sec ) C    + x x x x 1 2 11. I = 2 1 log(log ) (log ) x dx x ? ? + ? ? ? ? ? = 2 1 log(log ).1 (log ) x dx dx x + ? ? 1 = 1 1 log(log ) log x x x x ·  · x · 2 1 (log ) dx dx x + ? ? 2 = 2 2 1 1 1 1 log(log ) log (log ) (log ) ? ?  ·  ·  · · + ? ? ? ? ? ? x x x x dx dx x x x x 1 2 = log (log ) C log x x x x  + 1 2 12. I = 2 /2 0 sin sin cos x dx x x p + ? ...(i) I = 2 2 /2 /2 0 0 sin ( /2 ) cos sin ( /2 ) cos ( /2 ) cos sin x x dx dx x x x x p p p  = p  + p  + ? ? ...(ii) 1 2I = /2 0 1 sin cos dx x x p + ? 1 ? I = /2 /2 0 0 1 1 1 sec 1 1 4 2 2 2 2 cos sin 2 2 p p p ? ? =  ? ? ? ? + ? ? dx x dx x x 1 = /2 0 1 log sec tan 4 4 2 2 x x p ? ? p p ? ? ? ?  +  ? ? ? ? ? ? ? ? ? ? ? ? 1 2 = 1 2 1 1 log or log  2 1 2 2 2 1 2 + +  1 2 65/1/S (3) 65/1/S Page 5 Strictly Confidential — (For Internal and Restricted Use Only) Senior School Certificate Examination March 2016 Marking Scheme — Mathematics 65/1/S, 65/2/S, 65/3/S General Instructions: 1. The Marking Scheme provides general guidelines to reduce subjectivity in the marking. The answers given in the Marking Scheme are suggested answers. The content is thus indicative. If a student has given any other answer which is different from the one given in the Marking Scheme, but conveys the meaning, such answers should be given full weightage 2. Evaluation is to be done as per instructions provided in the marking scheme. It should not be done according to one’s own interpretation or any other consideration — Marking Scheme should be strictly adhered to and religiously followed. 3. Alternative methods are accepted. Proportional marks are to be awarded. 4. In question (s) on differential equations, constant of integration has to be written. 5. If a candidate has attempted an extra question, marks obtained in the question attempted first should be retained and the other answer should be scored out. 6. A full scale of marks  0 to 100 has to be used. Please do not hesitate to award full marks if the answer deserves it. 7. Separate Marking Scheme for all the three sets has been given. 8. As per orders of the Hon’ble Supreme Court. The candidates would now be permitted to obtain photocopy of the Answer book on request on payment of the prescribed fee. All examiners/Head Examiners are once again reminded that they must ensure that evaluation is carried out strictly as per value points for each answer as given in the Marking Scheme. QUESTION PAPER CODE 65/1/S EXPECTED ANSWER/VALUE POINTS SECTION A 1. Getting sin ? = 1 1 3 1 4 2 . 2 3 = 1 2 Hence  .  a b = 1 4 3 . . 1 2 2 3 = 1 2 2. 2  2  1 a b  = ? 1 . . 2 a b = 1 2 ? Angle between a and b = 4 p 1 2 3. Writing or using, that given planes are parallel 1 2 d =  4 10  2 units 4 9 36 + = + + 1 2 4. T T 2  AA  A  A  A = = 1 2 = 25 1 2 5. Getting AB = 7 8 0 10  ? ? ? ?  ? ? 1 2 AB = –70 1 2 6. k(2) = –8 ? k = –4 1 2 –4(3) = 4a ? a = –3 1 2 SECTION B 7. 1 (sin 2 ) sin ( 3 ) x y x x u v  = + = + ? dy du dv dx dx dx = + 1 2 (sin 2 ) log logsin 2 x u x u x x = ? = 1 2 1 2 cot 2 log sin 2 = · + du x x x u dx 1 65/1/S (1) 65/1/S ? (sin 2 ) [2 cot 2 log sin 2 ] x du x x x x dx = + 1 2 1 3 1 3 2 dv dx x x =  1 ? 3 (sin 2 ) [2 cot 2 logsin 2 ] 2 1 3 x dy x x x x dx x x = + +  1 2 OR Let y = 2 2 1 1 2 2 2 1 1 tan and cos 1 1 x x z x x x   ? ? +   ? ? = ? ? + +  ? ? z = cos –1 x 2 ? x 2 = cos z ? y = 1 1 cos 1 cos tan 1 cos 1 cos z z z z  ? ? +   ? ? ? ? + +  ? ? 1 ? y = 1 1 cos sin 1 tan 2 2 2 tan tan cos sin 1 tan 2 2 2 z z z z z z   ? ? ? ?   ? ? ? ? = ? ? ? ? ? ? ? ? + + ? ? ? ? 1 2 + 1 2 ? y = 1 tan tan 4 2 4 2 z z  ? p ? p ? ?  =  ? ? ? ? ? ? ? ? 1 2 + 1 2 ? dy dz = 1 2  1 8. LHL = 0 lim .sin ( 1) 2  ? p + = x k x k 1 RHL = 3 0 tan (1 cos ) lim x x x x + ?  1 = 2 0 tan sin /2 1 lim . 2 2 /2 2 x x x x x + ? ? ? = ? ? ? ? 1 ? k = 1 2 1 9. When x = am 2 , we get y = ± am 3 1 ay 2 = x 3 ? 2 2 3 2 3 2 dy dy x ay x dx dx ay = ? = 1 slope of normal = 3 2 4 2 2 3 3 a am m a m = ± ± 1 ? Equation of normal is y ± am 3 = 2 2 ( ) 3 x am m  ± 1 [Full marks may be given, if only one value for point, slope and equation is derived] 65/1/S (2) 65/1/S 10. Writing 1 sin (1 sin ) 2sin sin (1 sin ) sin (1 sin ) x x x dx dx x x x x  +  = + + ? ? 1 = 1 1 2 sin 1 sin dx dx x x  + ? ? 1 = 2 (1 sin ) cosec 2 cos x x dx dx x   ? ? 1 = 2 log  cosec cot  2 (sec sec tan ) x x x x x dx    ? 1 2 = log  cosec cot  2(tan sec ) C    + x x x x 1 2 11. I = 2 1 log(log ) (log ) x dx x ? ? + ? ? ? ? ? = 2 1 log(log ).1 (log ) x dx dx x + ? ? 1 = 1 1 log(log ) log x x x x ·  · x · 2 1 (log ) dx dx x + ? ? 2 = 2 2 1 1 1 1 log(log ) log (log ) (log ) ? ?  ·  ·  · · + ? ? ? ? ? ? x x x x dx dx x x x x 1 2 = log (log ) C log x x x x  + 1 2 12. I = 2 /2 0 sin sin cos x dx x x p + ? ...(i) I = 2 2 /2 /2 0 0 sin ( /2 ) cos sin ( /2 ) cos ( /2 ) cos sin x x dx dx x x x x p p p  = p  + p  + ? ? ...(ii) 1 2I = /2 0 1 sin cos dx x x p + ? 1 ? I = /2 /2 0 0 1 1 1 sec 1 1 4 2 2 2 2 cos sin 2 2 p p p ? ? =  ? ? ? ? + ? ? dx x dx x x 1 = /2 0 1 log sec tan 4 4 2 2 x x p ? ? p p ? ? ? ?  +  ? ? ? ? ? ? ? ? ? ? ? ? 1 2 = 1 2 1 1 log or log  2 1 2 2 2 1 2 + +  1 2 65/1/S (3) 65/1/S OR I = 1 1 1 2 1 2 0 0 1 cot (1 ) tan 1 x x dx dx x x   ? ?  + = ? ?  + ? ? ? ? 1 2 = 1 1 1 1 1 1 0 0 0 (1 ) tan tan tan (1 ) 1 (1 ) x x dx x dx x dx x x    ? ? +  = +  ? ?   ? ? ? ? ? 1 = 1 1 0 2 tan x dx  ? 1 2 = ( ) 1 1 1 2 0 0 2 tan . 1  ? ?  ? ? + ? ? ? x x x dx x 1 2 = 1 1 2 0 1 2 tan log 1  2 x x x  ? ?  + ? ? ? ? 1 = 1 2 log 2 or log 2 4 2 2 p p ? ?   ? ? ? ? 1 2 13. The given differential equation can be written as 1 1 dy y dx x  + = 2 3 ( 1) . x x e + 1 2 Here, integrating factor = 1 1 1 1  + ? = + dx x e x 1 ? Solution is 3 1 ( 1) 1 x y x e dx x = + + ? 1 ? 1 y x + = 3 3 ( 1) C 3 9 x x e e x +  + 1 1 2 or y = 2 3 1 1 ( 1) C( 1) 3 9 x x x e x + ? ? +  + + ? ? ? ? 14. From the given differential equation, we can write dx dy = / / / / 2 2 / 1 2 2 x y x y x y x y xe y x y e ye e   = 1 Putting x y = v ? dx dy = dv v y dy + 1 2 ? dv v y dy + = 2 1 2 v v ve e  ? dv y dy = 1 2 v e  1 ? 2 v e dv ? = dy y  ? 1 2 ? 2 log   v e y + = C ? / 2 log   C x y e y + = 1 65/1/S (4) 65/1/SRead More
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