Tancet Question Papers -Engineering Mathematics

2011 PART 01 — ENGINEERING MATHEMATICS (Common to all candidates) (Answer ALL questions)

1.

2.

3.

4.

5.

6.

 1 1 −1  4 4 −3  If the rank of a matri trix  b 2 2  9 9 b value of b is 1) 3 2) 1 3) –6

8.

The The part partic icul ular ar int integra egrall of of (D (D2+D)y= x 2+2 x +4 +4 is

1) x 2+4 3)

2)

x 3

+2 x 3 1 4) ( x x 3+4 x ) 3

x 3 + 12 x

3

4) 4

10. In the equ equatio ation n x (t)+2y(t)=–sin t, y (t)–2 x (t)=cos (t)=cos t, given x (0)=0 (0)=0 and y(0)=1, If the the rank rank of non non-sq -squar uare e matr matrix ix A and rank rank of of if x =cos =cos 2t–sin 2t–cos t, then y is equal to the augmented matrix of system of linear equations are equal, then the system 1) cos cos 2t–s 2t–sin in 2t+s t+sin t 2) cos cos 2t+s 2t+sin in 2t–s t–sin t 3) sin 2t 2t–c os os 2t 2t –s –sin t 4) c os os 2t+sin 2t+sin t 1) is inconsistent 2) ha s no solution 3) is consistent 4) does not have solution 11. If minim minimum um value value of f( x )= )= x 2 +2 b x +2 +2 c2 i s ’

’

I f th t h e s y s t e m – 2 x +y+z=a, + y+z=a, x –2y+z=b, +y–2z=c, where a, b, c are constants, is x +y–2z=c, consistent, consistent , then it has infinite solutions only when 1) a+b+c=0 2) a–b+c=0 3) a+b–c=0 4) a +b+c≠0

greater than maximum value g( x )=– )=– x +b 2, then for x is real, x 2–2c x +b

 1 2 2  0 2 1  , then the algebraic and I f A=   −1 2 2    geometric multiplicity are respectively 1) 2, 2 2) 1, 2 3) 1, 1 4) 2, 1

12. Form Form the partial partial differe differentia ntiall equatio equation n by eliminating the arbitrary constants a and b from

3) 0

4)

’

1)

4) 1

 x 3 + y 3    , then If  x − y  yu xy +y 2u yy equals x 2u xx +2 x yu 1) 0 2) sin u c os 3u 3) sin 3u co c os u 4) 2sin u cos 3u

2 |c|>b

2) p+q= x p+yq p+yq 4) p + q = z ( 2 D 2 –

of of

’

1 x +2y x +2y e +2y 2) – e x +2y 2 2

+2y 3) x e x +2y

4)

x 2

2

+2y e x +2y

y 14 . If f= f= tan–1   , then div (grad f) is equal to  x  1) –1 

2) 1 



3) 2

4) 0

∫∫



15. If F = a x i + by j + czk , then

I f u = x yz, v= x 2 + y 2 + z 2 , w = x +y+z, then ∂ ( x , y, z ) is equal to ∂ ( u, v, w ) 1) –2( x –y)(y–z)(z– x ) 2) ( x –y)(y–z)(z– x ) x –y)(y–z)(z– x –y)(y–z)(z– 1 3) 2 x − y y − z z − x 4) x yz ( )( )( )

e/tancet-MBA-MCA-2005-08/maths-tancet-2008-8p/ts 1

3) |c |> 2 |b|

13 . T h e pa pa r t i cu c u l ar ar i n te te g r a l x 2 +2y + 2y 3DD +D )z=e is

u = t a n–1

SAKTHI

2) no real real value value of a

1) x p = y q 3) y p = x q

 x 2  If u=l og  y  , then x u x +yu y is equal to   2) u

1) 0
of

 b ( y − 1)   as z=a log  1 − x   

The The sig signa natu ture re of quad quadra rati tic c for form m 2 x y+2yz+2z y+2yz+2z x is 1) 3 2) –1 3) 2 4) 1

1) 2u 7.

9.

0  1 2  is 3, then  3





F.dS , where S is

S

the surface of a unit sphere, is 4π 1) 3 ( a + b + c )

3)

1

4π (a+b+c) 3

2)

4 π(a+b+c)2 3

4) 0

TANCET - ENGINEE RING MATHEMATICS-2011

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∫

16. The val value ue of ( y − sin x ) dx + cosydy  , where C C

is the plane triangle enclosed by the lines y=0, 2 π y= and y= x , is π 2 8 1 1 1) 2) − (π2+8) 3) (π2+4) 4) π2+2 π 4π 8π 17. If f(z)=u+i f(z)=u+iv v is is analytic, analytic, then its first first derivativ derivative e equals ∂u ∂v ∂ v ∂u ∂u ∂u ∂u ∂v +i −i −i −i 1) 2) 3) 4) ∂ x ∂y ∂y ∂x ∂ x ∂y ∂ x ∂x

3z + 4 18. The The val valu ue of of 2z + 1 dz , where C is the circle C |z|=1, is π 1) πi 2) 3πi 3) 2πi 4) 3

∫

4z 2 + z + 5 dz , where C is the 19. The The va value lue of of z−4

∫

C

2

 3 x  ellipse   +y2=32, is  2  1) 3

2 3) 3

2) 0

1 is cosz − sinz π 2) 3) π 3

4) –1

20. The The pole pole of 1)

π 2

4)

π 4

∞

1 −t 2 21. The The val valu ue of of t ( e sin t ) dt is

∫

23. 23. The Fourier ourier sine sine trans transform form of e 1) t an–1 (s/a)

− a x

is

x

2) t an –1 (s/2a) 1 4) tan–(s/a) 2

3) tanh-1(s/a) 2z2 + 3z + 4

24. If Z(un )=

( z − 3)

3

of u3 is equal to 1) 21 21 2) 193

, |z|>3, then the value

3) 46 46

4) 139

25. As soon soon as a new value value of a variable variable is is found, found, it it is used immediately in the equations, such method is known as 1) Gauss-Jo Gauss-Jordan rdan method method 2) Gauss Gauss-Jac -Jacobi obi’s ’s method method 3) Gauss Gauss Elimin Elimination ation method method 4) Gauss Gauss-Se -Seida idall method method x for the data ( 0, 1), (1, 3), (2, 26. The The val value ue of x (2, 9), (3, x ) and (4, 81) is 1) 31 2) 18 18 3) 27 4) 36 36

27. 27. If y(0)=2 y(0)=2,, y(1)=4, y(1)=4, y(2)=8 y(2)=8 and and y(4)=32 y(4)=32,, then then y(3) is equal to 1) 12 12 2) 16.5 3) 18 18 4) 20 28. The joint joint probabi probability lity density density func function tion of of a r a n d o m v a r i a b l e ( x , y) is given by 2 2 f( x , y)=k x ye ye–( x +y ), where x , y>0. Then the value valu e of k is 1) 1 2) 3 3) 4 4) 2

0

1)

1 log 2 5

2)

1 log 5 4

3) log 3

4) 0

22. 22. The The sol solut utio ion n of (D2+9)y=cos 2t, y(0)=1 and y( π /2)=1 is given by 1) 2) 3) 4)

29. 29. The two two lines lines of regressi regression on are perpen perpendicu dicular lar to each other if the co-efficient of correlation equals 1) 0 2) 1 3) –1 4) ±1 30. Let the the random random variable variable X have the probab probabilit ility y density function

1 y= (cos 3t+4 sin 3t+4 cos 2t) 5 1 y= (2 cos 2t+sin 3t+cos 3t) 5 1 y= (cos 2t+4 sin 3t+4 cos 3t) 5 −1 y= (cos 2t–4 sin 3t+4 cos 3t) 5

 xe − x for x > 0 f ( x ) =  otherwise  0 Then the moment genera genera ting function functi on is 1)

1 1 − 2t

2)

1 1− t

3)

1 1+ t

4)

2 2− t

ME – ENGINEERING MATHS : ANSWERS

1 ......... 3 1 1. 1 ....... 3 2 1 .. ....... 2

2 .. ........ 3 1 2 .. ........ 2 2 2 . ......... *

3 ....... 1 1 3 ....... 2 2 3 ....... 1

4 ....... 1 1 4 .. . ...... 4 2 4 .. . ...... 3

5 .. . ....... 2 1 5 .. ........ 3 2 5 .. ........ 4

6 .. . ...... 4 1 6 ....... * 2 6 ....... 3

7 ....... 4 1 7 .. . ...... 4 2 7 .. . ...... 2

8 ........ 3 1 8 ........ 3 2 8 ........ 3

9 ....... 3 1 9 .. . ...... 2 2 9 .. . ...... 1

1 0 ........ 2 2 0. 0 ........ 4 3 0. 0 ......... *





1 −2 1 b 0 −3 3 a + 2b 2b ~ 0 3 −3 c − b

PART 01 — ENGINEERING — ENGINEERING MATHEMATICS

DETAILED SOLUTIONS

1.

1 −2 1 b 0 −3 3 a + 2b 2b ~ 0 0 0 a + b+c

(3 )

 1 1 −1  4 4 −3  b 2 2  9 9 b

0 1  2  3

Since the system has infinite solutions implies implies rank is less than 3. ∴ a+b+c = 0

0 0 1 4 0 1  ~ b 2 − b 2 + b  0 9+ b 9

0 1  2  3

4.

Algebraic Algeb raic multipli m ultipli city = 2 Geometric multiplicity = 2 5.

0 1 4 1  ~ b 2 + b  9 9 + b

(1)

0 0  1 0  2 2 − b  3 0 

(2)

0 1 1  1 0 1   A =   1 1 0  |A| = –1[0–1 0–1]+1[1– [1–0]

0 0 0  1 4 1 0 0   −b ~ b 2 + b 2 − b   0   9 9 + b −6 − b

= 1+1 = 2 D1 = |a 11| = 0 0 1 D2 = 1 0 = 0–1 = –1

Since the rank is 3 any determinant of order 4=0 1 0 0 2+b 2−b −b =0 ∴ 1× 9 + b −6 − b 0

D3 = |A| = 2 Difference between positive square terms and non positive square terms

⇒ 1(2–b)(6+b) = 0

= 1 –2 = – 1

∴ b = –6 (or) b = 2

2.

3.

(3 ) If ρ(A) = ρ(A, B) then the given system is consistent.

∴ Signature = –1

6.

(4)

 x 2  u = log  y   

(1 ) −2

[A , B ] =

1 1

1 1 a −2 1 b 1 −2 c

1 −2 1 b 2 1 1 a ~ − R1 ↔ R2 1 1 −2 c

u x =

∂u 1 2 x = . ∂ x  x 2  y    y 

=

y 2 x 2 . = x 2 y x





∂u 1  − x 2  .  uy = = ∂y  x 2   y 2     y 

∂ w ∂w ∂w =1; =1; =1 ∂ x ∂y ∂z

Now

y  − x 2 −1  = 2 2 = y  x  y 

∂u ∂u ∂ x ∂y ∂ ( u, v, w ) ∂ v ∂v = ∂ ( x, y, z) ∂x ∂y ∂ w ∂w ∂ x ∂y

 −1   2 + y    x   y 

∴ x u x +yuy = x 

= 2–1 = 1

7.

(4 ) L e t f ( u) = z = t a n u =

x 3 + y 3

=

x − y

yz x z 2 x 2y 1

xy

1

∂u ∂z ∂v ∂z ∂w ∂z

1 y

2z = 2 x 1 yz xz

1

1 z xy

Clearly z is a homogeneous function of degree 2. g(u) =

=

  ∵ = 2( 2 ( x – y)(y–z)(z– x )  x –y)(y–z)(z–  

n f (u) f ' (u) 2 × tan u se c 2 u

Now Now

sinu = 2× ×cos2u cosu = 2 sin sinu u cos cosu u = sin 2u

∂ ( x, y, z ) ∂ ( u, v, w )

=

Formula : 2 ∂ 2u ∂ 2u 2 ∂ u + 2 xy +y x ∂ x ∂y ∂ x 2 ∂y 2

=

2

= g(u)[g )[g (u)–1] ’

= sin sin 2u( 2u(2co 2coss 2u– 2u–1) 1) = 2sin 2sin 2u 2u cos cos 2u–si 2u–sin n 2u = sin sin 4u–s 4u–sin in 2u = 2sin 2sin u cos cos 3u

8.

(3 ) u = x yz yz ∂u ∂u ∂u ∴ = yz ; = xz ; = xy ∂ x ∂y ∂z 2

2

v = x +y +z

2

∂ v ∂v ∂v = 2 x ; = 2y ; = 2z ∂ x ∂y ∂z

w = x +y+z + y+z

     = ( a − b ) ( b − c ) ( c − a ) 

1 1 1 a b c bc ca ab

9.

1 ∂ ( u, v, w ) ∂ ( x, y, z )

1 2 ( x − y ) ( y − z ) ( z − x )

(3) Auxilla ry equati on is m2+m = m(m+1) = m = C. F =

0 0 0, m = –1 Ae0+Be – x = A+Be– x x 3 + 12 x If P.I. = , then 3

Solu Solutio tion n = y = A+ A+Be Be– x +

x 3

3

+4 x

dy = –Be– x + x 2+4 d x d2y = Be– x +2 x d x 2 d 2 y dy = x 2+2 x +4 +4 ∴ (D2+D)y = 2 + d x d x ∴ Correct option is (3) then





⇒ (1– x )p = a

10 . ( 2 ) –sin t x (t)+2y(t) = –s

. . . (1) . . . (2)

’

y (t)–2 x (t) = cos t cos 2t–si 2t–sin n 2t–cos 2t–cos t x = cos –2sin 2t–2 2t–2 cos 2t+s 2t+sin in t x (t) = –2sin ∴ (1) ⇒ –2sin 2t–2cos 2t+sin t+2 y(t) = – sin t 2y(t)) = 2sin 2sin 2t+2 2t+2 cos2 cos2t–2 t–2 sin sin t ∴ 2y(t y(t) = sin sin 2t+cos 2t+cos 2t–si 2t–sin nt ∴ y(t) ’

’

11 . ( 3 ) x ) = x 2+2b x +2c f( x +2c2 f ( x +2b x ) = 2 x +2b f ( x x ) = 2 ’

. . . (1 )

a b × = b ( y − 1) (1 − x ) 1 − x a (1 − x ) b a = b y − 1 × 1 − x = y − 1 ( ) ( ) ∴ (y – 1)q = a From (1) and (2) (1– x )p = (y–1)q ⇒ p– x p= yq–q ∂z q= ∂y

. . . (2 )

p+yq ∴ p+q = x p+yq 13. (2)

”

e x + 2 y PI = 2D2 − 3DD '+ D '2

f ( x +2b=0 x ) = 0 ⇒ 2 x +2b=0 ’

⇒ x = –b f (–b) = 2>0 ”

=

gives mini minimum mum ∴ x = –b gives Minim inimu um valu valuee = (–b) –b)2+2b(–b)+2c2

2

4 D − 3D '

14. (4) 2

g( x +b x ) = – x –2c x +b

Formula :

x ) = –2 x –2c g ( x –2c

div div (gr (grad ad f) = ∇.∇f

’

= ∇2f

g ( x x ) = –2 ”

g ( x –2c = 0 x ) = 0 ⇒ –2 x –2c

∂ 2 f ∂ 2 f + = ∂ x 2 ∂y 2 y f = ta n–1    x 

’

∴ x = –c

Now

= 4 1 −3 2 ( ) ( )

− xe x+2 y = 2

= b2–2b2+2c2 = –b2+2c2

xe x +2 y

x e x +2 y

g (–c) = –2<0 ”

gives maxi maximu mum m ∴ x = –c gives

Now

Maxim aximu um val value ue = –(–c –(–c))2–2c(–c)+b2 = –c 2+2c 2+b2 = c2+b2

1 ∂f = 1+  y   x  ∂ x  

Minimum value of f( x )>Maximum )>Maximum value of g( x )

=

⇒ –b2+2c 2>c 2+b2 ⇒ c 2>2b 2 ∴ |c|> 2 |b|

12. (2)

 b ( y − 1)  z = a log  1 − x    ∂z − b ( y − 1) a p= = × × −1 ∂ x b ( y − 1) (1 − x ) 2 1 − x a (1 − x ) b ( y − 1) a = 2 = b ( y − 1) × (1 − x ) (1 − x )

∂ 2 f ∂ x 2

=

x 2 x 2 + y 2

2

×

−y x 2

−y

×

x 2

−y = x 2 + y 2 ( )

( x 2 + y 2 ) 0 + y.2x 2 ( x 2 + y 2 ) 2 xy

= ∂f = ∂y

=

( x 2 + y 2 ) 1 2

2

.

1

 y  x 1+    x  x 2 2

2

.

1

x + y x

=

x 2

x + y 2





( x 2 + y 2 ) .0 − x ( 2 y ) 2 ( x 2 + y 2 )

∂ 2 f = ∂ y2

⇒ f (z) =

∂u ∂v +i ∂ x ∂x

=

∂u ∂u −i ∂ x ∂y

’

−2 xy

= x 2 + y (

2 2

)

2

∂u ∂v   equati tion onss =−  ∵ By CR equa ∂y ∂x  

(grad d f) = ∇ f ∴ div (gra ∂ 2f

=

+

∂ x 2

∂ 2 f ∂y 2

2 xy

=

( x

2

+y

2 2

)

−

18. (3)

2 xy

( x

2

+y

y

2 2

)

= 0 15 . ( 3 )     ∂  ∂  ∂  + + + + i j k . a i b y j cz c z k x ( ) =   ∇.F ∂z   ∂ x ∂y 

x

z=–1

= a+b+c By Gauss divergence theorem

∫∫



ˆ F.nds

=

S

∫∫∫ ∫∫∫



∇.Fdv

V

=

∫∫∫ ∫∫∫ ( a + b + c ) dv V

= (a+b+ (a+b+c) c)

∫∫∫ ∫∫∫ dv

3z + 4 3z + 4 Let f(z) (z) = 2z + 1 =  1 2 z +  2  ∴z =–

V

= (a+b+c) volume volume of the unit unit sphere 4π 3 (1) 3 4 π (a + b + c ) = 3 = (a+b+ (a+b+c)× c)×

Res idue a t z = –

2

=

y=0

y=0, y=

1 2

π

2 x π y =

O

1 is a simple pole 2

  1   3z + 4 = li m  z −  −   1  2   2 z + 1  z →−  2   2 

16 . ( * ) y=

z=1

1 Z=– 2

2 x π , y= π 2

will not form for m a tria ngle ∴ The data given in the problem are not correct.

Now

∫ C

lim

1 z →− 2

3z + 4 2

 1 3 −  + 4 −3 + 8 5 =  2 = = 2 4 4 3z + 4 dz 2z + 1 =

∫ f ( z ) dz C

= 2πi (sum of residues of poles within C) [By Cauchy’s Residue Theorem]

17 . ( 4 ) f(z) = u+iv

= 2πi ×

5 5πi = 4 2







19 . ( 2 )

∞

 sin 2 t  1 1 S  − 2 ds L  =  2  S S + 4   t  S f (t )  ∵ If t has a limit as t →0 and L(f(t))=F(s), then 

∫

2

 3 x  2 2  2  +y = 3   x 2

y2 ⇒ 2+ 2 2 3

= 1

∞   f ( t)   L = F s d s ( )    t  S

∫

∞

1   2 = log s − log (s + 4 )  2  S

z=3i

∞

1 s2  log  = 4 2  s + 4 s

z=2 z= 2 z= z=4 4

O

1 s2   = 4 0 − log 2 s +4  s2 −1 log = 4 s2 + 4

z = 4 lies outside of the ellipse ∴ f( z) =

4z2 + z + 5 is analytic inside C z−4

∴ By Cauchy’s theorem

∫ f ( z ) dzdz = 0 C

⇒

∫ C

2

4z + z + 5 dz = 0 z−4

 sint  s2 −1 L log  = 4  t  s2 + 4 ∞

∫

 sint  s2 −1 log  =  t  4 s2 + 4

∴ e− st  0

Put s=1 ∞

20 . ( 4 )

 sint  1 −1 ⇒ e− t  log  =  t  4 5 0

∫

To final pole of

1 is put cos z–sin z =0 cos z − sin z

⇒ cos z = sin z ⇒ ta n z = 1 π ∴z = 4 π ∴ Pole is z = 4

=

1 log5 4

22. ( * ) Auxillary Auxilla ry equation is given by m2+9 = 0 m2 = 9 m = ± 3i





y(0)=1 ⇒

y(3) = 2 3+1 = 24

Now 1 = A +

1 1 4 ⇒ A = 1− = 5 5 5

π y   =1 ⇒  2 1 = −B−

1 1 ⇒ B + = −1 5 5

−1 −6 B = −1 = 5 5

4 −6 sin 3t cos 2t cos3t + ∴ y(t) = 5 5 5

= 16 ≈ 16.5 If we use any interpolation method we get the value near to 16.5

28. (3) ∞∞

∫ ∫ k x ye

(

− x 2 + y 2

0 0 ∞

∫

ye i.e., k ye

− y2

0

1 = (4 cos 3t–6 sin 3t+cos 2t) 5

23 . ( 1 ) Fourier sine transform of

− a x e x

s is tan–1   a

25 . ( 4 ) Required method – Gauss siedal method.

) dx dy = 0

i.e.,

∞

2

∫

dy x e − x dx = 1 0

k =1 4

∴k=4

29. (1) If the two regression lines are perpendicular to each other, then the coefficient of correlation is equal

26 . ( 3 )

to 0. (0, 1) = (0, 3o) (1, 3) = (1, 31) (2, 9) = (2, 32) ( 4, 81) = (4, 34) 3

∴ (3, x ) = (3, 3 )

30. ( * ) Moment generating function is ∞

∫ 0

∞

∫

x ( t −1)

e t x ( x e− x ) dx = x e

dx

0

3

∴ x = 3 = 27

∞

∞  e x ( t − 1)  1 x ( t −1) dx =  x t − 1  − t − 1 e )  0 ( )0  (

∫

27 . ( 2 ) y(0) = 2 = 21 y(1) = 4 = 22 y(2) = 8 = 23 y(4) = 32 = 25 +1 ∴ y( x x ) = 2 x +1

x 1  e ( t −1)  = 0 − t − 1  t − 1  ) (

1

∞

0

Tancet Question Papers -Engineering Mathematics

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