2011 PART 01 — ENGINEERING MATHEMATICS (Common to all candidates) (Answer ALL questions)
1.
2.
3.
4.
5.
6.
1 4 If the rank of a matrix b 9 value of b is 1) 3 2) 1 3)
8.
–6
3)
4) 4
3) |c|> 2 |b|
4)
2 |c|>b
b ( y − 1) z=a log 1 − x as
1) xp = yq 3) yp = xq
2) p+q=xp+yq 4) p+q = z
13. Th e pa r t i c ul ar i nt e gr a l 3DD’+D’2)z=ex+2y is
1)
4) 1
x 3 + y3 If u=tan –1 x − y , then x 2u xx +2xyu xy +y 2u yy equals 1) 0 2) sin u cos 3u 3) sin 3u cos u 4) 2sin u cos 3u
1 x+2y x e 2) – ex+2y 2 2
3) xex+2y
(2 D 2 –
of
4)
x 2 x+2y e 2
y 14. If f=tan–1 , then div (grad f) is equal to x 1) –1
2) 1
3) 2
∫∫
15. If F = ax i + byj + czk , then
I f u= xyz , v= x 2 + y 2 +z 2 , w= x+ y+z , t he n ∂ ( x, y, z ) ∂ u, v, w is equal to
4) 0 F.dS , where S is
S
the surface of a unit sphere, is 4π 1) 3 ( a + b + c )
)
1) –2(x–y)(y–z)(z–x) 2) (x–y)(y–z)(z–x) 1 3) 2 x − y y − z z − x 4) xyz ( )( )( )
e/tancet-MBA-MCA-2005-08/maths-tancet-2008-8p/ts 1
2) no real value of a
eliminating the arbitrary constants a and b from
x2 If u=log y , then xux+yuy is equal to
SAKTHI
1) 0
12. For m the par tial differential equation by
The signature of quadratic form 2xy+2yz+2zx is 1) 3 2) –1 3) 2 4) 1
(
x 3 + 12 x 3
11. If minimum value of f(x)=x 2 +2bx+2c 2 is gr e at e r t han max imum va l ue of g(x)=–x2–2cx+b2, then for x is real,
1 2 2 0 2 1 If A= , then the algebraic and −1 2 2 geometric multiplicity are respectively 1) 2, 2 2) 1, 2 3) 1, 1 4) 2, 1
3) 0
x3 +2x 3 1 4) (x3+4x) 3
2)
10. In the equation x’ (t)+2y(t)=–sin t, y’(t)–2x(t)=cos t, given x(0)=0 and y(0)=1, if x=cos 2t–sin 2t–cos t, then y is equal to 1) cos 2t–sin 2t+sin t 2) cos 2t+sin 2t–sin t 3) sin 2t–cos 2t–sin t 4) cos 2t+sin 2t+sin t
I f t he s yst em –2x+y+z=a, x–2y+z=b, x+y–2z=c, where a, b, c are constants, is 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
2) u
The particular integral of (D2+D)y=x2+2x+4 is
1) x2+4
If the rank of non-square matrix A and rank of the augmented matrix of system of linear equations are equal, then the system 1) is inconsistent 2) has no solution 3) is consistent 4) does not have solution
1) 2u 7.
9.
1 −1 0 4 −3 1 2 2 2 is 3, then 9 b 3
3)
1
4π (a+b+c) 3
2)
4 π(a+b+c)2 3
4) 0
TANCET - ENGINEERING MATHEMATICS-2011
16. The value of
−ax
∫ ( y − sinx ) dx + cosydy , where C
23. The Fourier sine transform of e x is 1) tan–1 (s/a) 2) tan–1 (s/2a) 1 3) tanh-1(s/a) 4) tan–(s/a) 2
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
2z 2 + 3z + 4
24. If Z(un )=
17. If f(z)=u+iv is analytic, then its first derivative equals
of u3 is equal to 1) 21 2) 193
∂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
1) πi
2) 3πi
∫
19. The value of
C
3) 2πi
4)
π 3
26. The value of x for the data (0, 1), (1, 3), (2, 9), (3, x) and (4, 81) is 1) 31 2) 18 3) 27 4) 36
2
3x ellipse +y2=32, is 2 1) 3
2) 0
1 is cosz − sinz π 2) 3) π 3
27. If y(0)=2, y(1)=4, y(2)=8 and y(4)=32, then y(3) is equal to 1) 12 2) 16.5 3) 18 4) 20
4) –1
20. The pole of
1)
π 2
∞
21. The value of
1
∫ t (e
−t
4)
4) 139
3) Gauss Elimination method 4) Gauss-Seidal method
4z 2 + z + 5 dz , where C is the z−4
2 3) 3
3) 46
1) Gauss-Jordan method 2) Gauss-Jacobi’s method
C
|z|=1, is
, |z|>3, then the value
25. As soon as a new value of a variable is found, it is used immediately in the equations, such method is known as
∫ 2z + 1 dz , where C is the circle
18. The value of
( z − 3 )3
28. The joint probability density function of a r a nd om va ria bl e ( x , y) is gi v en by 2 2 f(x, y)=kx ye–(x +y ), where x , y>0. Then the value of k is 1) 1 2) 3 3) 4 4) 2
π 4
)
sin 2 t dt is
0
1)
1 log 2 5
2)
1 log 5 4
3) log 3
29. The two lines of regression are perpendicular to each other if the co-efficient of correlation
4) 0
equals 1) 0
22. The solution of (D2+9)y=cos 2t, y(0)=1 and y(π/2)=1 is given by
2) 1
3) –1
4) ±1
30. Let the random variable X have the probability density function
1 (cos 3t+4 sin 3t+4 cos 2t) 5 1 2) y= (2 cos 2t+sin 3t+cos 3t) 5 1 3) y= (cos 2t+4 sin 3t+4 cos 3t) 5 −1 4) y= (cos 2t–4 sin 3t+4 cos 3t) 5
1) y=
−x for x > 0 xe f (x) = otherwise 0 Then the moment generating function 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 ....... 3 2 1 ....... 2
2 ........ 3 12 ........ 2 22 ........ *
3 ....... 1 13 ....... 2 23 ....... 1
4 ....... 1 1 4 ....... 4 2 4 ....... 3
SAKTHI
e/tancet-MBA-MCA-2005-08/maths-tancet-2008-8p/ts 2
5 ........ 2 1 5 ........ 3 2 5 ........ 4
6 ....... 4 16 ....... * 26 ....... 3 2
7 ....... 4 1 7 ....... 4 2 7 ....... 2
8 ........ 3 18 ........ 3 28 ........ 3
9 ....... 3 1 9 ....... 2 2 9 ....... 1
1 0 ........ 2 2 0 ........ 4 3 0 ......... *
TANCET - ENGINEERING MATHEMATICS-2011
1 −2 1 b 0 −3 3 a + 2b ~ 0 3 −3 c − b
PART 01 — ENGINEERING MATHEMATICS DETAILED SOLUTIONS
1.
1 −2 1 b 0 −3 3 a + 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 rank is less than 3. ∴ a+b+c = 0
0 0 1 4 0 1 ~ b 2−b 2+ b 0 9+b 9
4.
0 1 2 3
Algebraic multiplicity = 2 Geometric multiplicity = 2 5.
0 1 4 1 ~ b 2 + b 9 9 + b
(2)
0 1 0 2 2 − b 3 0 0
0 1 1 1 0 1 A = 1 1 0 |A| = –1[0–1]+1[1–0]
0 0 0 1 4 1 0 0 ~ b 2 + b −b 2 − b 0 9 9 + b −6 − b
= 1+1 = 2 D1 = |a11| = 0 D2 =
Since the rank is 3 any determinant of order 4=0
∴ 1×
(1)
1 2+ b
0 −b
9 + b −6 − b
0 2−b 0
D3 = |A| = 2 Difference between positive square terms and non
=0
positive square terms
⇒ 1(2–b)(6+b) = 0
= 1–2 = –1
∴ b = –6 (or) b = 2 2.
3.
0 1 = 0–1 = –1 1 0
∴ Signature = –1
(3) If ρ(A) = ρ(A, B) then the given system is consistent.
6.
(4)
x2 u = log y
(1)
−2
1
1 1
−2 1 b 1 −2 c
1
−2
1
−2 ~ 1
1 1
1 a R1 ↔ R2 −2 c
[A, B] =
1
a ux =
∂u 1 2x = . ∂x x2 y y
b
SAKTHI
e/tancet-MBA-MCA-2005-08/maths-tancet-2008-8p/ts 3
=
3
y 2x 2 . = x2 y x
TANCET - ENGINEERING MATHEMATICS-2011
uy =
∂u 1 −x2 . = ∂y x2 y2 y
∂w ∂w ∂w =1; =1; =1 ∂x ∂y ∂z Now
y − x 2 −1 = 2 2 = y x y
∂u ∂x ∂ ( u, v, w ) ∂v = ∂ ( x, y, z ) ∂x ∂w ∂x
−1 2 ∴ xux+yuy = x + y x y
= 2–1 = 1
7.
(4) Let f(u)=z = tan u 3
=
x +y x−y
∂u ∂y ∂v ∂y ∂w ∂y
∂u ∂z ∂v ∂z ∂w ∂z 1
1
2x
x z xy 1 2y 2z = 2 x
y
z
1
1
xz
xy
yz 3
=
1
yz
Clearly z is a homogeneous function of degree 2. g(u) =
=
∵ = 2( x –y)(y–z)(z– x )
n f ( u) f ' ( u) 2 × tan u sec 2 u
Now
sin u ×cos2u cos u = 2 sinu cosu = sin 2u = 2×
∂ ( x, y, z ) ∂ ( u, v, w )
=
Formula : x2
2 ∂ 2u ∂2u 2 ∂ u + 2 xy + y ∂x ∂y ∂x 2 ∂y 2
=
= g(u)[g’(u)–1]
9.
= sin 2u(2cos 2u–1) = 2sin 2u cos 2u–sin 2u = 2sin u cos 3u
(3) u = xyz
∴
∂u ∂u ∂u = yz ; = xz ; = xy ∂x ∂y ∂z
v = x 2+y 2+z 2
∂v ∂v ∂v = 2x ; = 2y ; = 2z ∂x ∂y ∂z w = x+y+z
SAKTHI
e/tancet-MBA-MCA-2005-08/maths-tancet-2008-8p/ts 4
1 ∂ ( u, v, w ) ∂ ( x, y, z )
1 2 ( x − y )( y − z )( z − x )
(3) Auxillary equation is m2+m = m(m+1) = m = C.F =
0 0 0, m = –1 Ae 0+Be–x = A+Be–x x 3 + 12 x If P.I. = , then 3 x3 +4x Solution = y = A+Be–x+ 3 dy then = –Be–x+x2+4 dx d2y = Be–x+2x dx 2 d 2 y dy ∴ (D2+D)y = + = x2+2x+4 dx 2 dx ∴ Correct option is (3)
= sin 4u–sin 2u
8.
= ( a − b )( b − c )( c − a )
1 1 1 a b c bc ca ab
4
TANCET - ENGINEERING MATHEMATICS-2011
⇒ (1–x )p = a
10. (2)
x’(t)+2y(t) = –sin t
... (1)
y’(t)–2x(t) = cos t x = cos 2t–sin 2t–cos t
... (2)
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)
∂z q= ∂y
x’(t) = –2sin 2t–2 cos 2t+sin t ∴ (1) ⇒ –2sin 2t–2cos 2t+sin t+2 y(t) = – sin t ∴ 2y(t) = 2sin 2t+2 cos2t–2 sin t ∴ y(t) = sin 2t+cos 2t–sin t f(x) = x2+2bx+2c2 f’(x) = 2x+2b f”(x) = 2
13. (2)
f’(x) = 0 ⇒ 2x +2b=0
PI =
⇒ x = –b f ”(–b) = 2>0 ∴ x = –b gives minimum = b –2b +2c
2
= –b2+2c2 2
g(x) = –x –2cx+b
ex +2 y 2D 2 − 3DD'+ D '2
=
xe x+2 y x e x+2 y = 4 1 −3 2 ( ) ( ) 4D − 3D '
=
− xe x+2 y 2
Minimum value = (–b)2+2b(–b)+2c2 2
14. (4) Formula :
2
g’(x) = –2x–2c
div (grad f) = ∇.∇f = ∇2f
g”(x) = –2 g’(x) = 0 ⇒ –2x –2c = 0
∂2f ∂ 2f + = ∂x 2 ∂y 2 y f = tan–1 x
∴ x = –c g”(–c) = –2<0
Now
Now
∴ x = –c gives maximum Maximum value = –(–c)2–2c(–c)+b 2 2
2
= –c +2c +b
1 −y × 2 2 ∂f x = 1+ y ∂x x
2
= c2+b2
x2 −y × 2 = = 2 2 x +y x
Minimum value of f(x)>Maximum value of g( x) 2
2
2
⇒ –b +2c >c +b 2
⇒ c >2b
... (2)
(1–x)p = (y–1)q ⇒ p–x p= yq–q ∴ p+q = xp+yq
11. (3)
2
... (1)
2
2
∂ 2f
∴ |c|> 2 |b|
∂x
=
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 )
e/tancet-MBA-MCA-2005-08/maths-tancet-2008-8p/ts 5
=
2
(
)
)
+ y 2 0 + y.2 x
(x
2
+ y2
)
2
2 xy
12. (2)
SAKTHI
2
(x
−y
x 2 + y2
(x
)
2
+ y2 ∂f 1 1 . = 2 ∂y y x 1+ x
=
5
2
x2 1 x . = 2 2 2 x + y x x + y2
TANCET - ENGINEERING MATHEMATICS-2011
(x
∂2f = ∂ y2
2
)
+ y 2 .0 − x ( 2 y )
(x
2
+ y2
)
2
⇒ f ’ (z) =
∂u ∂v +i ∂x ∂x
=
∂u ∂u −i ∂x ∂y
−2 xy
=
(x
2
+y
2 2
)
2
∂u ∂v ∵ By CR equations ∂y = − ∂x
∴ div (grad f) = ∇ f =
∂ 2f ∂2f + ∂x 2 ∂y 2
=
2 xy
(x
2
+ y2
2
) (x
18. (3)
2 xy
−
2
+ y2
)
y
2
= 0 15. (3) ∂ ∂ ∂ ∇.F = ∂x i + ∂y j + ∂z k . ax i + byj + cz k
(
)
z=–1
Z=–
= a+b+c By Gauss divergence theorem ˆ F.nds ∇.Fdv =
∫∫
∫∫∫
S
V
=
∫∫∫ V
∫∫∫ dv ∴z =–
V
= (a+b+c) volume of the unit sphere 4π (1)3 3 4π (a + b + c ) = 3 = (a+b+c)×
Residue at z = –
2
2 π x
=
y=
y=0, y=
1 2
π
Now
y=0
O
1 is a simple pole 2
1 3z + 4 = lim z − − 1 2 2 z + 1 z →− 2 2
16. (*) y=
x
3z + 4 3z + 4 Let f(z) = 2z + 1 = 1 2 z + 2
( a + b + c ) dv
= (a+b+c)
z=1
1 2
∫ C
2x π , y= π 2
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
will not form a triangle ∴ The data given in the problem are not correct.
= 2πi (sum of residues of poles within C) [By Cauchy’s Residue Theorem]
17. (4)
= 2πi ×
f(z) = u+iv SAKTHI
e/tancet-MBA-MCA-2005-08/maths-tancet-2008-8p/ts 6
6
5 5 πi = 4 2
TANCET - ENGINEERING MATHEMATICS-2011
19. (2)
∞
sin 2 t 1 1 S L − 2 ds = 2 S S + 4 t S f (t) ∵ If t has a limit as t→0 and L(f(t))=F(s), then
∫
2
3x 2 2 +y = 3 2 ⇒
x 2 y2 + 22 3 2
= 1
∞ f (t) L = F ( s ) ds t S
∫
∞
1 2 = log s − log s + 4 2 S
(
z=3i
)
∞
1 s2 log = 4 s2 + 4 s
z=2 z=4
O
1 s2 = 4 0 − log 2 s + 4 z = 4 lies outside of the ellipse ∴ f(z) =
=
4z 2 + z + 5 is analytic inside C z−4
∴ By Cauchy’s theorem
sin t −1 s2 L log = 4 t s2 + 4
∫ f ( z ) dz = 0
∞
C
⇒
∫ C
−1 s2 log 2 4 s +4
sin t −1 s2 ∴ e− st log 2 = t 4 s +4 0
∫
2
4z + z + 5 dz = 0 z−4
Put s=1 ∞
sin t −1 1 ⇒ e− t log t = 4 5 0
20. (4)
∫
To final pole of
1 is put cos z–sin z =0 cos z − sin z
=
⇒ cos z = sin z ⇒ tan z = 1 π ∴z = 4 π ∴ Pole is z = 4
22. (*) Auxillary equation is given by m2+9 = 0 m2 = 9
m = ± 3i C.F = A cos 3t+B sin 3t
21. (2)
P.I. =
1 − cos 2t L(sin2t) = L 2 1 = [L(1)–L(cos 2t)] 2
=
e/tancet-MBA-MCA-2005-08/maths-tancet-2008-8p/ts 7
1 cos 2t D +9 2
1 cos 2t cos 2t = −4 + 9 5
y(t) = A cos 3t+B sin 3t+
1 1 S − 2 = 2 S S + 4 SAKTHI
1 log 5 4
7
cos 2t 5
TANCET - ENGINEERING MATHEMATICS-2011
y(0)=1 ⇒
y(3) = 23+1 = 24
Now 1 = A+
1 1 4 ⇒ A = 1− = 5 5 5
= 16 ≈ 16.5 If we use any interpolation method we get the value near to 16.5
π y =1 ⇒ 2
1 = −B−
28. (3)
1 1 ⇒ B + = −1 5 5
∞∞
∫∫ kx ye
−1 −6 B = −1 = 5 5
∴ y(t) =
(
− x 2 + y2
) dx dy = 0
0 0 ∞
4 −6 sin 3t cos 2t cos 3t + 5 5 5
2
∫
∞
2
∫
−y −x i.e., k ye dy x e dx = 1 0
1 = (4 cos 3t–6 sin 3t+cos 2t) 5
i.e.,
23. (1)
0
k =1 4
∴k=4
Fourier sine transform of
−ax e x
is tan–1
s a 29. (1)
25. (4) Required method – Gauss siedal method.
If the two regression lines are perpendicular to each other, then the coefficient of correlation is equal
26. (3)
to 0. o
(0, 1) = (0, 3 )
30. (*)
(1, 3) = (1, 31)
Moment generating function is
(2, 9) = (2, 32) (4, 81) = (4, 34)
∞
∫ (
∞
)
e tx x e − x d x =
∴ (3, x) = (3, 33)
0
∫xe
x ( t −1)
dx
0
∴ x = 33 = 27 ∞
∞ e x ( t − 1) 1 x t −1 x − e ( ) dx = t − 1 t − 1 ( ) ( ) 0 0
∫
27. (2)
y(0) = 2 = 21 y(1) = 4 = 22
x t −1 1 e ( ) − 0 = t − 1 ( t − 1)
y(2) = 8 = 23 y(4) = 32 = 25 ∴ y(x) = 2x+1
SAKTHI
e/tancet-MBA-MCA-2005-08/maths-tancet-2008-8p/ts 8
∞
0
1 = t −1 2 ( )
8
TANCET - ENGINEERING MATHEMATICS-2011
