FIRST SEMESTER DIPLOMA EXAMINATION IN ENGINEERING/ TECHNOLIGY- MARCH, 2013

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1 TED (10)-1002 (REVISION-2010) Reg. No.. Signature. FIRST SEMESTER DIPLOMA EXAMINATION IN ENGINEERING/ TECHNOLIGY- MARCH, 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) If 0 find the value of x. 0 > 24x 14 0 > 24x 14 > x (b) If A * + B * + find 2A 3B 2A 3B 2* * + * + - * + * + (c) find n > r s or r + s n Then n r + s

2 n (d) Evaluate cos and tan if sin ½ Sin 2 + cos 2 1 Cos 2 1 (1/2) 2 Cos 2 1 (1/4) cos tan (e) Find the slope of the line whose inclination to the x axis is 45 o Slope tan tan45 o 1 PART B Answer any five questions. Each question carries 6 marks II. (a) If A * + show that AA T is symmetric. A * + A T [ ] AA T * + [ ] * + Clearly (AA T ) T * + * + AA T is symmetric. (b) Solve using determinants: 3x + y z 3 -x + y + z 1 x + y + z 3 AX B

3 [ ] [ ] [ ] x ( ) ( ) ( ) ( ) ( ) ( ) 1 y ( ) ( ) ( ) 1 z ( ) ( ) ( ) 1 (c) Find the term independent of x in the expression of( ) 5 T r+1 nc r a n-r b r 5c r (3x) 5-r (-y 2 ) r We have to find T 4 5c 3 (3x) 2 (-y 2 ) 3 5c x 2 (-1) 3 y 6

4 -90x 2 y 6 (d) Prove that sin + sin + sin + sin 4 cos. cos. sin sin + sin + sin + sin [sinc + sind 2.sin ] (sin + sin7 ) + (sin + sin ) 2sin4.cos3 + 2sin4.cos 2sin4 (cos3 + cos ) [cosc + cosd 2cos ] 2sin4. 2(cos2. cos ) 2sin4. cos2. cos, hence the result. (e) Prove that + 4cos2 + + ( ) (f) Find the equation of the line passing through the point (2, -1) and (-6, 3). Also find the slope of the line. Two points of a line is given by, (x 1, y 1 ) (2, -1) (x 2, y 2 ) (-6, 3)

5 -2(y+1) x 2-2y 2 x 2 x + 2y 0 Slope of the line x + 2y 0 is m (g) Find the value of k so that the following lines are concurrent. 5x + 2y 4 0 2x + ky x 4y x + 2y 4 0 2x + ky x 4y 18 0 Since the lines are concurrent (-18k + 44) 2(-36 33) 4(-8 3k) 0-90k k 0-78k k K 5

6 III. (a) If A* PART C (Maximummark: 60) Answer four full questions. Each question carries 15 marks. +and I unit matrix of same order, then find A 3-3A 2 + 2A + I A 3 A 2.A A 3 A 2. A * + * +A * + * + * + A 3-3A 2 + 2A + I * + - * + + * ++* + * + (b) If A * +, 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) * +

7 * + A -1 B * + 1 Cofactor matrix of B * + Adj. (B) * + * + B -1 B -1 A -1 * + [ ] [ ] (AB) -1 B -1 A -1 (c) Find x if 2 - x + 3 2(1 2) - x( 4 )+ 3( 4 ) -2-4x x 3x 4 3x x > 7x 14 >x 2 IV. (a) If A * +, B * +, show that( ) + * + + * + * + ( ) * + 1

8 * + * + * + * + + * + 2 From 1 & 2 it is clear that ( ) + (b) Find k, if the following system of equation are consistent x + y + 1 0, x + 2y + 1 0, 2x+3y + k 0 If the system is consistent then, (2k 3) (k 2) + (3-4) 0 2k 3 k k 2 0 k 2 (c) Solve using inverse of the coefficient matrix x + y + z 1, 2x + 2y + 3z 6, x + 4y + 9z 3 AX B [ ] [ ] [ ] Calculation for A

9 Minor matrix [ ] Cofactor matrix [ ] Adjoint matrix [ ] So inverse matrix, A -1 [ ] [ ][ ] [ ] X A -1 B x 7 y -10 z 4 V. (a) If 20C r 20C r+2 find r nc r nc s > r s or r + s n

10 ie, r + r r r 18, r 9 (b) Find middle term in the expansion of (x 2 +3/x) 20 T r+1 nc r a n-r b r, n 20 n , odd. ( ) 22/2 11th term is the middle term T 11 20c 10 (x 2 ) 10 (3/x) 10 20c 10 x x c 10 x c x 20 (c) Prove that 2 - ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 - VI. (a) Expand (3x ) 4 binomially (a + b) n a n + nc 1 a n-1 b + nc 2 a n-2 b nc n b n (3x ) 4 (3x) 4 4c 1 (3x) 3 ( ) + 4c 2 (3x) 2 ( ) - 4c 3 (3x) 1 ( ) + 4c 4 ( ) 81x x 3 ( ) + 6 9x 2 ( ) 4 3 x ( ) + ( ) 81x 4 54x 3 y + ( )x 2 y 2 ( ) +

11 (b) Find the constant term in the expansion of ( x + )10 T r+1 nc r a n-r b r T r+1 10c r ( x) 10-r ( )r 10c r 2 r x -2r 10c r x -2r 2 r 10c r 2 r 10c r 2 r Then 0 > 10 5r 0 > r 2 Therefore T 3 10c x is the constant term. (c) Prove that VII. (a)prove that in ABC, a(sinb - sinc) 0

12 in ABC, we know that sinb (by sine rule) andsinc a(sinb - sinc) a ( ) ( a (b c)) [a (b c) + b (c a) +c (a b)] [ab ac + bc ba +ca cb] x 0 0 (b)prove that cos20.cos40.cos60.cos80 We have cos60 ie, cos20.cos40.cos80 cos20 [cos120 cos(-40)]. cos20 [ + cos40] cos20 + cos20.cos40. cos20 + x (cos60 + cos20). cos20 + cos60 + cos20. cos60 x (c )Show that sin120.cos330 + cos240.sin330 1 Sin120 sin(1 x ) cos30

13 Cos330 cos(360 30) cos(4 x 90 30) cos30 Cos240 cos(270 30) cos(3 x 90 30) -sin30 Sin330 sin(360 30) sin(4 x 90 30) -sin30 sin120.cos330 + cos240.sin330 x + x + 1 VIII. (a) Express 3cos + 4sin in the form of Rsin( ) where is acute. cosx + sinx R.sin( ) R.sinx.cos + Rcosx.sin Equating the similar terms on both sides, cosx Rsin.cos Sinx Rsin.cos > Rsin 1 > 1 Rcos 2 Squaring and adding 1 & R 2 sin 2 + cos 2 4 R 2 > R > > tan > tan -1 ( ) > 60 o (b) 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

14 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.sin2B sin 2 A - sin 2 B (c) In any ABC, show that (b + c)sina/2 a.cos( ) LHS (a + b).sin (2RsinA + 2RsinB)sin * + 2R(sinA + sinb)sin 2R.2.sin.cos.sin cos.4r.sin.sin cos.4r.sin (90 - ).sin cos.2r.(2cos.sin ) cos.2r.sinc cos.c RHS IX. (a) Solve ABC if a 5cm, B 30 o & c 8cm tan( ) cot

15 tan -1 [ cot ] tan -1 [ cot ] tan -1 [ cot 15 o ] tan -1 [ ] A B A + B Solving A O 16 B O 44 Now we have to find C We have c sin30 o 4.44cm (b) Find the slope and intercept of the line 3x + 4y 12 Slope of 3x + 4y 12 is Intercept form of a line is 1 3x + 4y 12 1 > 1 X intercept 4 Y intercept 3 (c) Find k so that the lines kx + 2y 10 0, 2x 4y are (i). Perpendicular to each other.

16 (ii). Parallel to each other. (i). m 1 x m 2-1 x -1-1 > 2k 8 > k 4 (ii). m 1 m 2 X. (a) Solve using Napier s formula, given a 87cm, b 53cm and C 70 o tan( ) cot A B 2tan -1 [ cot ] A B 2tan -1 [ cot35] 2tan -1 [ cot 15 o ] 2tan -1 [ ] 2 x 19 o o 16 A + B o 2A 148 o 16 /2 74 o 08 A B 110 B o o 52 Now we have to find c, we have

17 >c 84.99cm (b) Find the equation to the line passing through the point of intersection of x y and 2x 3y +2 0 and perpendicular to the line x + y 6 0 Given x y x 3y x y -1 2x 3y -2 A * + B * + x -1 Y 0 Point of intersection (-1, 0) Given the line is, x + y 6 0 a 1, b 1, c -6 Perpendicular line is, 1 1 bx - ay + k 0 x y + k 0 Passes through (-1, 0) 1 > x y > k 0 K 1 1

18 (c ) A line passes through (-6, 3 ). The X-intrecept of the line is 3 times its Y-intercept. Find the. equation of the line. a 3 b (given) The equation of a line is + 1 Ie, + 1.(1) (1) Pass through (-6, 3 ) (1) Implies, , b 1 (1) Implies + 1

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)