TED (10)-1002 (REVISION-2010) Reg. No.. Signature. FIRST SEMESTER DIPLOMA EXAMINATION IN ENGINEERING/ TECHNOLIGY- OCTOBER, 2010 TECHNICAL MATHEMATICS- I (Common Except DCP and CABM) (Maximum marks: 100) [Time: 3 hours Marks PART A (Maximum marks: 10) (Answer all questions. Each question carries 2 marks) I. (a) Find the values of a, b, c and d given that, [ ][ ] Here 2a 4 > a4/22 a+3b11 >2+3b11 > 3b9 > b9/33 5-c 7 > c 5-7 -2 (b) Evaluate sina x sina - -cosa x cosa sin 2 A + cos 2 A 1 [ a d-b c (c) If n n what is the value of n?
nc 12 nc 13 > n 12 + 13 25 [nc r nc s > r + snor r s] (d) Prove that tana + cota 2cosecA tana + cota + (e) Find the angle of inclination of the line joins the point (5,3) & (-8,3) 0 PART B Answer any five questions. Each question carries 6 marks II. (a) If A [ ] and B [ ] find AB & BA and show that AB BA AB [ ] [ ] [ ] BA [ ] [ ]
[ ] Clearly AB BA (b) Solve the following equations using cramer s rule. 2x + 3y z 5 x 2y + 3z 6 3x y + 2z 7 Here x y z x 14/14 1 y 2 z 3 (c) Obtain the coefficient of x 12 in the expansion ( ) 10 T r+1 (-1) r nc r a n-r b r T r+1 (-1) r 10c r (x 2 ) 10-r (1/x 2 ) r (-1) r 10c r x 20-2r (1) r /x 2r (-1) r 10c r x 20-4r But we have 20 4r 12 > -4r -8 > r 2
Then T 3 (-1) 2 10c 2 x 12 So coefficient of x 12 is 10c 2 (d) Prove that 2 sin 2 A - cos 2 A sin 2 A- (1-sin 2 A) sin 2 A sin 2 A 1 2 sin 2 A-1 Also sin 2 A - cos 2 A (1-cos 2 A) -cos 2 A 1-cos 2 A-cos 2 A 1-2cos 2 A Hence the results are same. (e) If sina4/5 and sinb 12/13, A and B are acute angles, find cos(a B) A 5 sina4/5 cosa3/5 3 A B 4 C 12 13 sinb12/13 cosb15/13 B C 5 cos(a B) cosa.cosb + sina.sinb 3/5 5/13 + 4/5 12/13 15/65 + 48/65 63/65s (f) Prove that sin(a+b).sin(a-b) sin 2 A-sin 2 B
sin(a+b) sina.cosb + cosa.sinb sin(a-b) sina.cosb - cosa.sinb sin(a+b).sin(a-b) (sina.cosb + cosa.sinb) (sina.cosb - cosa.sinb) sin 2 A.cos 2 B sina.cosasinb.cosb + sinacosa.sinbcosb - cos 2 A.sin 2 B sin 2 A.cos 2 B - cos 2 A.sins 2 B sin 2 A(1 - sin 2 B) (1 - sin 2 A)sin 2 B sin 2 A-sin 2 A.sin 2 B -sin 2 B -sin 2 A.sin 2 B sin 2 A - sin 2 B (g) Express the equation 3x + 4y 12 0 in (i) slope-intercept form and (ii) intercept form. Hence find the slope and intercept made on the axes. 3x + 4y 12 1 slope intercept form ymx + c + intercept form + 1 + 1 2 4y 12 3x y + y + 3 3 Slope intercept form: y + 3 : + 1
PART C Answer four full questions. Each question carries 15 marks. III. (a) If A [ ] find A 2 3A + 5I A 2 [ ] [ ] [ ] 3A 3[ ] [ ] 5I 5[ ] [ ] A 2 3A + 5I [ ] - [ ] + [ ] [ ] (b) Express the matrix [ ] as the sum of a symmetric & skew symmetric matrices [ ] [ ] [ ] [ ] This is a symmetric matrix. [ ] [ ] [ ] [ ] This is a skew symmetric matrix.
+ [ ] + [ ] [ ] Clearly A + That A can be expressed as a sum of symmetric and skew symmetric matrices. (c) Solve the following system of equations using the inverse of coefficient matrix 3x - 2y + 3z 4 2x + y - z 2 4x 3y + 2z 3 AX B [ ] [ ] [ ] X A -1 B Calculation for A -1-1 8-10 5-6 -1-1 -9 7
Minor matrix [ ] Cofactor matrix of A [ ] Adjoint matrix [ ] 3 + 2 + 3 3x -1 + 2 x 8 + 3 x -10-3 + 16 30-17 So inverse matrix X A -1 B [ ] [ ][ ] [ ] [ ] x 1 y 1 z 1 IV. (a) If A [ ] and B [ ] verify that (A + B) T A T + B T A + B [ ] + [ ] [ ] (A + B) T [ ] 1 A T [ ] B T [ ] A T + B T [ ] + [ ] [ ] 2
Clearly from & 1 2 (A + B) T A T + B T (b) Solve for x if 0 0 (5x 16) 2(10 12) + 3( 8 3x ) 0 5x 16 + 4 + 24-9x 0-4x -12 x 3 (c) If P [ ] and Q [ ] verify that (PQ) -1 Q -1 P -1 P [ ] Q [ ] PQ [ ] [ ] [ ] To find (PQ) -1 PQ [ ] Minors m 11 49 m 21 11 m 12 70 m 22 16 Minor matrix [ ] Cofactor matrix [ ] Adj (PQ) [ ] 784 770
(PQ) -1 14 [ ] To find Q -1 Q [ ] Minors m 11 5 m 21 1 m 12 6 m 22 4 Minor matrix [ ] Cofactor matrix [ ] Adj Q [ ] 20 6 14 Q -1 [ ] To find P -1 P [ ] Minors m 11 9 m 21 2 m 12 4 m 22 1 Minor matrix [ ]
Cofactor matrix [ ] Adj P [ ] [ ] 1 P -1 [ ] Q -1 x P -1 (PQ) -1 Q -1 P -1 [ ] [ ] [ ] x V. (a) Find the middle terms in the expression of n 7 n + 1 8, even, T r+1 n a n-r b r 7 So th and ) th are the middle terms ie, 4 th & 5 th are middle terms T 4 7c 3 (x 2 ) 4 ( 3 7c 3 x 8 7c 3 8 5 280x 5 T 5 7c 4 (x 2 ) 3 ( 4 7c 4 x 6 7c 4 2 4 2 560x 5 (b) Expand (3x + 2y) 5 We know (a + b) n a n + nc 1 a n-1 b + nc 2 a n-2 b 2 +....... + nc n b n Thus (3x + 2y) 5 (3x) 5 + 5c 1( 3x) 4 (2y)+ 5c 2 (3x) 3 (2y) 2 + 5c 3 (3x) 2 (2y) 3 + 5c 4 (3x) 1 (2y) 4 + (2y) 5 243x 5 + 5 81 2x 4 y + 10 27x 3 4y 2 + 10 9x 2 8 y 3 + 5 3x 16y 4 + 32y 5 243x 5 +810x 4 y + 1080 x3y 2 +720x 2 y 3 + 240xy 4 + 32y 5
(c) Show that sin60 o cos30 o cos60 o sin30 o 1/2 So - ¾ - ¼ ½ VI. (a) Find the constant term in the expansion of ( x 3 + 3/x 2 ) 15 T r+1 nc r a n-r b r 15c r (x 3 ) 15-r (3/x 2 ) r 15c r x 45-3r 3 r /x 2r 15c r x 45-3r-2r 3 r 15c r x 45-5r 3 r Now 45 5r 0 45 5r > r 9 So T 10 15c 9 (x) 0 3 9 15c 9 3 9 So constant term 3 9 15c 9 98513415 (b) Evaluate tan 2 60 o + 3tan 2 45 o tan60 tan 2 60 2 3 tan 2 45 o (tan45) 2 (1) 2 1 so tan 2 60 o + 3tan 2 45 o 3 + 3 1 6 (c) If sin -4/5, A 5 4 B 6 C 3 rd quadrant. Find all other trigonometric functions of sin -4/5 BC 3 So cos -3/5 ( 3 rd quadrant) tan 4/3 cot 3/4 cosec -5/4 sec -5/3 VII. (a) Prove that sin + sin + sin + sin 4 cos. cos. sin sin + sin + sin + sin [sinc + sind 2. ] (sin + sin7 ) + (sin + sin ) 2sin4.cos3 + 2sin4.cos
2sin4 (cos3 + cos ) [cosc + cosd 2. ] 2sin4. 2(cos2. cos ) 2sin4. cos2. cos, hence the result. (b) Prove that 2tan10 o + tan40 o tan50 o Consider tan(50 40) tan10 o > 2tan10 o >2tan10 o + tan40 o tan50 o (c) In any ABC, prove that a 2 + bc b 2 + c 2,if A 60 o By cosine rule we have a 2 b 2 + c 2-2bc.cosA b 2 + c 2-2bc.cos60 o b 2 + c 2-2bc.1/2 b 2 + c 2 bc > a 2 + bc b 2 + c 2 VIII. (a) Prove that cos20 o.cos40 o. cos80 o 1/8 cos20 o.cos40 o. cos80 o cos20 (cos40.cos80) cos20 [ ½ (cos120 + cos(-40) ) ] ½. cos20 [ cos120 + cos40 ] ½. cos20 [ -½ + cos40 ] -¼ cos20 + ½ cos20.cos40 -¼ cos20 + ¼ cos60 +¼ cos20
¼ ½ 1/8 (b) If A + B 45 O, show that (1 + tana)(1 + tanb) 2 tan(a + B) tan(a+ tanb) tan45 o 1 > > + 1 Adding 1 on both sides 1 + + 2 >(1 + tana)(1 + tanb) 2 (c) State and prove sine rule A Statement In any ABC 2R Proof Consider the circumcircle of ABC. The perpendicular bisectors of the sides BC, CA, and AB intersect at O. Therefore O is the circumcentre such that OA OB OC R B F O D E C We have 2 2A So A In ODB, sin<bod sina BD/OB sina > a 2RsinA Similarly, sinb >b 2RsinB sicb >c 2RsinC
It is clear that 2R IX. (a) Solve the with a 2cm, b 3cm and c 4cm A (28.96) o 28 o 57 B 46 o 34 C 180 o (A + B) 180 o (28 o 57 + 46 o 34 ) 104 o 29 (b) Find the acute angle between the lines 2x y + 3 0 and 3x 3y + 4 0 2x y - 3 m 1 2 3x 3y - 4 m 2 1 tan tan -1 (1/3) 18 o 26 (c) If A( 1, -1), B( -2, 1 ) and C( 3, 5 ) are the vertices of a triangle, then find the equation ofmedian through B A( 1, -1 ) Slope of AB D( 4/2, 4/2 ) Mid-point 2/3, (m 1 ) B( -2, 1 ) C( 3, 5 ) Slope of AC 6/2 3, (m 2 ) Slope of BC 4/5, (m 3 )
Equation of BD ( -2, 1 ) ( 2, 2 ) 4(y 1) x + 2 4y 4 x + 2 x 4y -6 x 4y + 6 0 B D X. (a) Using Napier s formula, find the values of the angles A, B in ABC, if a 5cm, b 8cm and C 30 o tan( ) cot tan -1 [ cot ] tan -1 [ cot ] tan -1 [ cot 15 o ] tan -1 [ ] -40.736 A B -81.473 A + B 180 30 150 Solving 1 + 2 2 1 A 34.2635 34 O 16 B 150-34.2635 115 O 44 Now we have to find c
We have c sin30 o 4.44cm (b) Find the angles of the triangle having vertices ( 3, 2 ), ( 5, -4 ) and ( 1, -2 ) A( 3, 2 ) m 1 m 2 B( 5, -4 ) m 3 C( 1, -2 ) m 1 slope of AB -3 m 2 slope of AC 2 m 1 slope of BC ( )
tan 1 tan 1 tanb 1 > B tan -1 (1) 45 o tana 1 implies A tan -1 (1) 45 0 so C 180 ( A + B ) 180 o 90 o 90 o (c) Show that the three lines are concurrent 5x + 2y 4 0 2x 5y + 11 0 3x 4y -18 0 5 2 4 5(-90 + 44) 2(-36 33) 4(-8 15) 5 x -46 2 x -69 4 x 23-230 + 138 + 92-230 + 230 0 So the given lines are concurrent