FIRST SEMESTER DIPLOMA EXAMINATION IN ENGINEERING/ TECHNOLIGY- OCTOBER, 2013
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1 TED (10)-1002 (REVISION-2010) Reg. No.. Signature. FIRST SEMESTER DIPLOMA EXAMINATION IN ENGINEERING/ TECHNOLIGY- OCTOBER, 2013 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) Which of the following matrices is symmetric * +, * +, * + * + * + * +is symmetric. (b) Find the value of r, if n n r + s n or r s. r + r r r 18 r 9
2 (c) State the identities for tan(a B) tan(a B) ( ) (d) State projection formula. In any ABC, a b.cosc + c.cosb or b a.cosc + c.cosa or c a.cosb + b.cosa (e) Define slope of a straight line. If a straight line is inclined at an angle with the x axis, the slope of that straight line is given by tan. It is represented by m tan. PART B Answer any five questions. Each question carries 6 marks II. (a) Solve the equations: 3x + y z 3 -x + y + z 1 x + y + z 3 by find the inverse of the coefficient matrix. 3x + y z 3 -x + y + z 1 x + y + z 3 AX B [ ] * + [ ] Calculations for A -1 m 11 0 m 12-2 m 13-2
3 m 21 2 m 22 4 m 23 2 m 31 2 m 32 2 m 33 4 Minor matrix [ ] Cofactor of A [ ] Adj A [ ] (1 1) 1(-1 1) 1(-1 1) 4 A -1 X A -1 B [ ] * + [ ][ ] [ ] [ ]
4 x 1 y 1 z 1 (b) If A * + and B * + show that (AB) -1 B -1 A -1 AB * + * + * + 4 Cofactor matrix * + Adj. (AB) * + Inverse of AB * + [ ] A * + 4 Cofactor matrix of A * + Adj. (A) * + * + A -1 B * + 1 Cofactor matrix of B * + Adj. (B) * + * + B -1 B -1 A -1 * + [ ] [ ] (AB) -1 B -1 A -1
5 (c) Prove that nc r + nc r-1 (n+1)c r nc r ( ) nc r-1 ( ) ( ) nc r + nc r-1 ( ) + ( ) ( ) n![ ( ) ( )( ) ( ) ] n![ ( ) ( ) ( )( ) ( ) ] ( ) ( ) * ( ) + ( ) ( ) * ( ) + ( ) ( ) ( ) ( ) ( ) ( ) (n+1)c r (d) Prove that cot3x L.H.S ( ) ( ) cot3x
6 (e) State and prove sine rule. 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 similarly, sinb sicb > a 2RsinA >b 2RsinB >c 2RsinC it is clear that 2R (f) 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 [ ]
7 A B A + B Solving A O 16 B O 44 Now we have to find c We have c sin30 o 4.44cm (g) Find the equation to the line passing through (4, 5) which is (i) parallel (ii) Perpendicular to the line 2x + 3y 4 Case I Equation of a parallel line is ax + by + k 0 a 2 b 3 2x + 3y + k Passes through (4, 5) 1 1 >2 x x 5 + k k k 0 K -23 > 2x + 3y 23 0 Case II Equation of perpendicular line is bx ay + k 0 a 2, b 3 3x 2y + k 0 2
8 2 Passes through (4, 5) 2 2 > 3 x 4 2 x 5 + k k k 0 K -2 > 3x 2y 2 0 PART C (Maximum mark: 60) Answer four full questions. Each question carries 15 marks. III. (a) If A * + B * +and C * +verify that A (B C) AB AC A (B C) * + (* + * +) * + (* +) * + AB AC * + * + * + * + * +-* + * + Clearly A(B C) AB AC (b) If A * +, B * +show that (AB) T B T A T AB * + * + * +
9 (AB) T * + 1 B T A T * + * + * + 2 Clearly (AB) T B T A T (c) Show that the eliminant of lx + my + n 0, mx + ny and nx + ly + m 0 is l 3 + m 3 + n 3 3mn Eliminant 0 > 1(mn 1) m(m 2 n) + n(m n 2 ) 0 >mn 1 m 3 + mn + nm n 3 0 > m 3 + n mn > m 3 + n mn IV. (a) If A [ ] find A 2 8A 20I A 2 A.A [ ] [ ] [ ] 8A 8[ ] [ ] 20I [ ] A 2 8A 20I [ ] - [ ]- [ ] [ ]
10 (b) Express the matrix A [ ] as the sum of two matrices of which one issymmetric and the other is skew symmetric. A [ ] A T [ ] [ ] [ ], this is a symmetric matrix. [ ] [ ], this is a skew symmetric matrix. Clearly [ ] + [ ] [ ] A (c) Solve using determinants: x + 2y z -1 x + 2y z -1 3x y 2z 5 x y 3z 0 x ( ) ( ) ( ) ( ) ( ) ( ) 3x y 2z 5 x y 3z 0 2
11 Y ( ) ( ) ( ) -1 z ( ) ( ) ( ) 1 V. (a) Expand (x+ ) 7 using binomial theorem. (a + b) n a n + nc 1 a n-1 b + nc 2 a n-2 b nc n b n Now (x ) 7 x 7 +7c 1 x 6 ( ) + 7c 2 x 5 ( ) 2 + 7c 3 x 4 ( ) 3 + 7c 4 x 3 ( ) 4 +7c 5 x 2 ( ) 5 + 7c 6 x( ) 6 + 7c 7 ( ) 7 x 7 + 7x 5 21x x + 35x x x 5 + x -7 (b) If tanx 7/24& x is in the 3 rd quadrant. Find the value of3sinx 4cosx. sinx -7/25 [x 3 rd quadrant.]
12 cosx -24/25 3sinx 4cosx 3 x (-7/25) 4 x (-24/25) (-21/25) + (96/25) 75/25 3 (c) Draw the graph of y cos3x The graph of y cosx X : 0 90 o 180 o 270 o 360 o Y : Graph 1-1 VI. (a) Find the term independent of x in the expansion of (x + 3/x 3 ) 10 T r+1 nc r a n-r b r, T r+1 10c r (x) 10-r (3/x) r 10c r 3 r x 10-r -x -r 10c r 3 r x 10-2r 10 2r 0 > r 5 The term independent of x 10c x
13 (b) Write the sign of (i) Cot(7 ) (ii) tan(500) (iii)cosec(280). (i) Cot(7 ) cot(7 x 45 o ) cot(315) Sign is negative. cot(3 x ) -tan45-1 (ii) tan(500) tan(5 x ) -cot50 Sign is negative.. (iii)cosec(280) Sign is negative. cosec(3 x ) ( ) (c) Prove that ( ) 2 ( ) ( ) ( ) ( ) ( ) ( ) 2 VII. (a) Prove the formula for cos3a Sin3A cos (2A + A) cos2acosa sin2acosa (2cos 2 A 1)cosA 2sinA.cosA.sinA
14 2cos 3 A cosa 2sin 2 A.cosA 2cos 3 A cosa 2sin 2 A.cosA 2cos 3 A cosa 2(1 cos 2 A).cosA 2cos 3 A cosa 2cosA + 2cos 3 A 4cos 3 A 3cosA (b) If sin18 ( -1)/4, find cos36 and sin54 Put 18 o Cos36 o cos2 x (18) cos2 1 2sin 2 1 2sin * x Sin54 o sin 3 x 18 sin3 3sin 4sin 3 3sin18 4sin * + 4* OR sin54 o sin(90 36) cos (c) Prove that cos + cos3 + cos5 + cos7 0 cos + cos3 + cos5 + cos7 cos + cos7 + cos3 + cos5 2cos( ). cos( )+ 2cos( ).2cos( ) 2cos. cos( ) + 2cos. cos( )
15 0 [cos 0] VIII. (a) If cosa -12/13, cotb 24/7 and A is in the quadrant II and B is in quadrant I. find cos(a B ) cos(a B) cosa.cosb + sina.sinb Given cosa sina [ A 2 nd quadrant ] CotB sinb [ B 1 st quadrant ] cosb cos(a B) cosa.cosb + sina.sinb - x + x (b) Prove that cota cot2a cosec2a cota cot2a cota cot2a - ( ) cosec2a
16 (c) Show that ( )sin 2 c/2 cos( ) Question is wrong. IX. (a) Derive the equation of a straight line of the form + 1 The x intercept a and y intercept b are given, consider a straight line AB having x intercept a and y-intercept b Hence OA a and OB b Let (x, y) be any point on the line. From the figure it is clear that and are similar. Their corresponding sides are proportional. 1 MA OA OM a x MP y Substituting these in 1 we have >ay ab bx >bx + ay ab +
17 > + 1 (b) Find the slope and intercept of the line 5x 3y x 3y -15 Slope 5x 3y > + 1 Intercept form + 1 (c) Find the angle of triangle having vertices (3, 2), (5, -4) and (1, -2) Slope of AB m 1-3 Slope of AC m 2 2 Slope of BC m 3 Angle between AB & AC tan -1 tan -1 tan -1 tan -1 (1) 45 o
18 Angle between AC & BC tan -1 tan -1 tan -1 tan -1 ( ) 90 o Angle between AB & AC 180 ( ) 45 o X. (a) Find the values of P if the lines (2P + 1)x (5 P)y 8 and (5P 1)x (P + 1)y 3 are parallel. (2P + 1)x (5 P)y 8 (5P 1)x (P + 1)y Slope of 1 is ( ) ( ) Slope of 2 is ( ) ( ) Since the lines are parallel, we have ( ) ( ) ( ) ( ) (2P + 1) (P + 1) (5P 1) (5 P) 2P 2 + 2P + P P 5P P 7P 2 23P P 3, 4/14 3 or 2/7
19 (b) Find the foot of the perpendicular from(-2, 1) on the line x -2y 6 Slope of x 2y 6 is m 1 Slope of AD -2 Equation of AD: y y1 m(x x1) y 1-2(x + 2) y 1-2x 4 2x + y -3 Solving x -2y 6 & 2x + y -3 x 0 Now 2 x 0 + y -3 > y -3 Foot of perpendicular is(0, -3) (c) Straight line cuts off on the axes of coordinates positive intercept whose sum is 5. Given that the line passes through (-4, 9), find its equation. Intercept form of a line is given by Given a + b 5 > b 5 a Equation 1 passes through (-4, 9) + 1 > -4(5 a) + 9a a(5 a) > a + 9a 5a a 2
20 > a 2 + 8a 20 0 a -10, 2 a 2 b 3
FIRST SEMESTER DIPLOMA EXAMINATION IN ENGINEERING/ TECHNOLIGY- OCTOBER, TECHNICAL MATHEMATICS- I (Common Except DCP and CABM)
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)
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