aieee 2013 maths

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AIEEE/2013/MATHS 1

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S. No Questions Solutions

Q.1 The circle passing through (1, 2) and touching the axis of x at (3, 0) also passesthrough the point(a) (2, 5) (b) (5, 2)(c) (2, 5) (d) (5, 2)

Sol: 1 (b)

(x 3)2+ y2+ y = 0The circle passes through (1, 2)

4 + 4 2= 0 = 4(x 3)2+ y2+ 4y = 0 Clearly (5, 2) satisfies.

Q.2

ABCD is a trapezium such that AB and CD are parallel and BC

CD. If

ADB =

, BC = p and CD = q, then AB is equal to(a)

p2+q2cos p cos +qsin (b)

p2+q2

p2 cos +q2 sin (c)

(p2+q2)sin(pcos +qsin )2 (d)

(p2+q2)sin p cos +qsin

Sol: 2 (d)Using sine rule in triangle ABD

ABsin= BDsin (+a)AB =

p2+q2 sin sin cos +cos sin =

p2+q2 sin sin .qp 2+q 2+

cos .pp 2+q 2

AB =p2+q2 sin

(pcos +qsin ).Q.3 Given : A circle, 2x2+ 2y2= 5 and a parabola, y2= 4 5x.

Statement I : An equation of a common tangent to these curves is y = x + 5.Statement II : If the line,y= mx

5m

(m0) is their common tangent, then msatisfies m4 3m2+ 2 = 0.(1) Statement - I is True; Statement -II is true; Statement-II is not a

correct explanation for Statement-I(2) Statement -I is True; Statement -II is False.(3) Statement -I is False; Statement -II is True(4) Statement -I is True; Statement -II is True; Statement-II is a

correct explanation for Statement-I

Sol: 3 (a)

Let the tangent to the parabola be y = mx +5m

(m 0).Now, its distance from the centre of the circle must be equal to the radius of the circle.

So, 5

m = 5

21 + m2(1+m2)m2 = 2m4 + m2 2 = 0.(m2 1) m2 + 2 = 0 m = + 1So, the common tangents are y = x + 5and y = - x - 5.

Q.4

A ray of light along x + 3y= 3gets reflected upon reading xaxis, theequation of the reflected rays is

(a) 3y= x - 3 (b) y = 3x3(c)

3y= x1 (d) y = x +

3

Sol: 4 (a)

Slope of the incident ray is -1

3.So, the slope of the reflected ray must be

1

3.The point of incidence is (

3,0). So, the equation of reflected ray is y =

1

3(x-

3).

Q.5 All the students of a class performed poorly in Mathematics. The teacher decidedto give grace marks of 10 to each of the students. Which of the followingstatistical measures will not change even after the gracemarks were given ?(a) median (b) mode(c) variance (d) mean

Sol: 5 (c)

Variance is not change by the change of origin.

Alternate Solution:

=xx2n

for y = x + 10y= x +10

1=y+10y102n = yy2n =

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Q.6 If x, y, z are in A.P. and tan1x, tan1y and tan1z are also in A.P., then(a) 2x = 3y = 6z (b) 6x = 3y = 2z(c) 6x = 4y = 3z (d) x = y = z

Sol: 6 (d)If x, y, z are in A.P.2y = x + z

and tan1x, tan1y, tan1zare in A.P.2 tan1y = tan1x + tan1z x = y = z.Note: If y = 0, then none of the options is appropriate.

Q.7 If f(x) dx =(x), then x5f(x3)dx is equal to(a)

1

3x3 (x3)3 x3 (x3) dx+C

(b) 13

x3 (x3) x2(x3) dx+C(c)

1

3x3 x3 x3(x3) dx+C

(d)1

3x3 (x3) x2 (x3) dx+C

Sol: 7 (b) f (x)dx =(x)Let x3= t

3x2dx = dtthen x5f(x3)dx= 1

3 tf(t)dt

=1

3t ftdt {1. ftdt}dt= 1

3x3 (x3)- x2(x3)dx+C.

Q.8 The equation of the circle passing through the foci of the ellipsex2

16+

y2

9= 1, and

having centre (0,3) is

(a) x2+ y2- 6y + 7 = 0 (b) x2+ y2- 6y5 = 0(c) x2+ y2- 6y + 5 = 0 (d) x2+ y2- 6y7 = 0

Sol: 8 (d)

foci (+ae,0)We have a2e2 = a2 b2= 7Equation of circle (x0)2+ (y3)2= (7 0)2+ (03)2x2 +y6y7 = 0.

Q.9 The x-coordinate of the incentre of the triangle that has the coordinates of mid

points of its sides as (0, 1) (1, 1) and (1, 0) is

(a) 2 - 2 (b) 1+2 (c) 1 - 2 (d) 2 + 2Sol: 9 (a)

xcoordinate = ax 1+bx2+cx 3a+b+c

=2 2+22 0+20

2+2+22 =

4

4+22=2

2+2= 2 - 2Alternate Solution:

xcoordinate = r = (sa) tan A/2

=4+222

22tan 4= 2 - 2.

Q.10 The intercepts on xaxis made by tangents to the curve, y = tx0

dt, x R,which are parallel to the line y = 2x, are equal to

(a) +2 (b) +3 (c) +4 (d) +1

Sol: 10 (d)dy

dx=

x

= 2x = +2

y =

t

2

0dt = 2for x = 2

and y = t20 dt = - 2 for x = - 2tangents are y2 = 2 (x2)y = 2x2and y + 2 = 2 (x + 2) y = 2x + 2Putting y = 0 , we get x = 1 andl.

Q.11 The sum of first 20 terms of the sequence 0.7, 0.77, 0.777, .. , is

(a)7

9(99 - 1020) (b) 7

81(179 + 1020)

(c)7

9(99 + 1020) (d) 7

81(179 - 1020)

Sol: 11 (b)tr = 0.77r times

= 7 (101 + 102+ 103+ + 10r )=

7

9( 110r)

S20= tr20r=1 = 79 20 10r20r=1 = 7920 19 (1 1020 = 781

(179+1020)

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Q.12 Consider :

Statement I: (p ~ q) (~ p q) is a fallacy.Statement II: (pq)(~ q~ p) is a tautology.(1) Statement - I is True; Statement -II is true; Statement-II is not a

correct explanation for Statement-I(2) Statement -I is True; Statement -II is False.(3) Statement -I is False; Statement -II is True(4) Statement -I is True; Statement -II is True; Statement-II is a correct

explanation for Statement-I

Sol: 12 (a)S1:

p q ~ p ~ q p ~ q ~ p q (p ~ q)(~ p qTTFF

TFTF

FFTT

FTFT

FTFF

FFTF

FFFF

S2:p q ~ p ~ q pq ~ p~ q (pq)(~ p ~qTTFF

TFTF

FFTT

FTFT

TFTT

TFTT

TTTT

S2 is not an explanation of S1

Q.13 The area (in square units) bounded by the curves y = x,2yx + 3 = 0, xaxis,and lying in the first quadrant is

(a) 36 (b) 18 (c)27

4 (d) 9

Sol: 13 (d)

2x= x34x = x2- 6x + 9

x2

10x + 9x = 9, x = 1

2 y + 3 y230

]dy

y2 + 3y y33

0

3

= 9 + 9 9 = 9Q.14 The expression

t an A

1c ot A+c ot A

1t an Acan be written as(a) secA cosecA + 1 (b) tanA + cotA(c) secA + cosecA (d) sinA cosA + 1

Sol: 14 (a)1

cotA (1cotA) -co t2A

(1cotA )=1cot3 A

cotA (1cotA)=c os e c2A+cot A

cot A= 1 + sec A cosec A

Q.15 The real number k for which the equation 2x3+ 3x + k = 0 has two distinct realroots in [0,1](a) lies between 2 and 3 (b) lies between -1 and 0

(c) does not exist (d) lies between 1 and 0

Sol: 15 (c)

If 2x3 + 3 x + k = 0 has 2distinct real roots in [0,1], then f (x) will change sign but f (x) =6x2+ 3 > 0

So no value of k exists.

Q.16 Sol: 16 (c)

Q.17 Let Tn be the number of all possible triangles formed by joining vertices of an n

sided regular polygon. If Tn+1 Tn = 10, then the value of n is(a) 5 (b) 10 (c) 8 (d) 7Sol: 17 (a)

n+1 C3n C3 = 10 n C2 = 10 n = 5.

Fallacy

Tautology

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Q.18 At present, a firm is manufacturing 2000 items. It is estimated that the rate of

change of production P w.r.t. additional number of workers x is given bydP

dx=

10012 x.If the firm employs 25 more workers, then the new level ofproduction of items is(a) 3000 (b) 3500 (c) 4500 (d) 2500

Sol: 18 (b) dPP2000

= 100 12xdx250

(P2000) = 25 10012 2

3(25)3/2

P = 3500.

Q.19 StatementI: The value of the integral dx1+t an x

/3/6 is equal to 6 .

StatementII: f xdx = fa + b xdx.ba

b

a

(a) StatementI is True; StatementII i s true; StatementII is not acorrect explanation for statementI(b) StatementI is True; StatementII is False.(c) StatementI False; StatementII is True(d) StatementI True; StatementII is True; StatementII is a correct

explanation for StatementI

Sol: 19 (c)

I = dx1+t an x

/3/6

I = t an x

1+t an x/3

/6 dx2I =

6

I =

12.

Q.20If P = 1 31 3 3

2 4 4

is the adjoint of a 3 3 matrix A and A= 4, then is equalto(a) 11 (b) 5 (c) 0 (d) 4

Sol: 20 (a)

P = 1 a 31 3 32 4 4

Adj A= A2Adj A= 161(1212)

(46) + 3(46) = 16.

2 - 6 = 16.= 11.Q.21 The number of values of k, for which they system of equations

(k + 1)x + 8y =4kkx + (k + 3)y = 3k -1

has no solution, is(a) 1 (b) 2 (c) 3 (d) infinite

Sol: 21 (a)For no solutionk+1

k=

8

k + 3 4k

3k1 (1)(k + 1) (k + 3)8k = 0or k2 4k + 3 = 0 k = 1,3But for k = 1, equation (1) is not satisfiedHence k = 3.

Q.22 If y = sec(tan1x), then dydx

at x = 1is equal to

(a)1

2 (b) 1 (c) 2 (d) 12

Sol: 22 (d)

y = sec (tan1x)dydx

= sec (tan1x)tan (tan1x). 11+x2

dy

dxx=1

= 2 1 1

2=

1

2.Q.23 If the lines

x21

=y3

1=

z4kand

x1k

=y4

2=

z51

are coplanar, then k can have

(a) exactly one value (b) exactly two values(c) exactly three values (d) any value

Sol: 23 (b)

1 1 11 1 kk 2 1

= 01(1 + 2k) + 1(1 + k2)1 (2k) = 0 k2+ 1 + 2k + 12 + k = 0

k2+ 3k = 0(k) (k + 3) = 0 2 values of k.

Q.24 Let A and B be two sets containing 2 elements and 4 elements respectively. The

number of subsets of A B having 3 or more elements is(a) 220 (b) 219 (c) 211 (d) 256

Sol: 24 (b)

A B will have 8 elements.288C08C18C2= 256 1 8 28 = 219.

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