KARNATAKA CET - MATHEMATICS-2010 SOLVED PAPER · PDF fileKARNATAKA CET - MATHEMATICS-2010...

KARNATAKA CET - MATHEMATICS-2010 SOLVED PAPER

1. If A= [3 21 1] , then A2 +xA+yI=0 for

(x,y)= 1)(-1,3) 2)(-4,1) 3)(1,3) 4)(4,-1)

sol: ∣3− 21 1−∣ =0 ⇒

3+λ2 -4λ-2=0; λ2 -4λ+1=0 Ans is (2)

2. The constant term of the polynomial

∣x3 x 3x2x x1 x−1

x2 2x 3x1∣ is

1)2 2)0 3)1 4)-1By taking x=0, The given determinant= -1 Ans : 4

3. If a , b and c are nonzero coplanar vectors, then [ 2a−b 3b−c 4c−a ]is 1)0 2)25 3)9 4)27sol: Since a , b and c are coplanar, Given expression =0 Ans : (1)

4. A space vector makes the angle 1500

, 600 with +ve x axis and y-axis. The angle made by the vector with +ve z axis is 1)600 2) 900 3)1200 4)1800 sol: By cos2 1500 +cos2 60 + cos2 γ=1

341

4cos2 =1 ⇒ cos2γ=0;

γ=900 Ans : 2

5. If a b and c are unit vectors such that abc = 0 then

3a .b2b .cc .a =1)1 2)-1 3)3 4)-3since abc = 0 , it is equilateral triangle , and

G.E=3 cos 23 +2 cos

23

+cos23 =

−32

−22−1

2 =-3

Ans : 4

6. If a>b>0, sec-1 aba−b =2 sin-1 x,

then x 1) bab

2) - bab

3) aab

4) - aab

sol: 2 -sin-1 a−bab =2 sin-1 x

sin(2 -2 sin-1 x)= a−bab

cos(2 sin-1x)= a−bab 1-2x2 = a−bab ⇒ x= b

abAns: (1)

7. If x#nπ, x#(2n+1)2 , n∈Z ,

then sin−1cosx cos−1 sinxtan−1cotxcot−1tanx

=

1)6 2)

2 3)

3 4)

4

sol: for x=4 G.E=

4

44

4

=1

Note: Here Any of the options given is not correct.

8. The general solution of

1+sin2x=3sinx cosx, tanx#12 is

1)2nπ-4 n∈Z

2)2nπ+4 n∈Z

3)nπ+4 , n∈Z 4)nπ-

4

n∈ZCase (1) and (4) can't be choice as

when n=0, -4 ;

case (2) abd (3) satisfy when n=0,

when n=1, (2) ⇒ 2π+4 clearly

satisfies and (3) ⇒ π+4

satisfies.(3) must be the choice as (2) is in (3) Ans : (3)

1 pucpcmb.wordpress.com

9. The least +ve integer n, for which 1i n

1−i n−2 is +ve is

1) 4 2)3 3)2 4)1

sol: G.E=1i n

1−i n.1−i 2

=in (-2i) =2

when n=1 ; Ans : 4

comments: There is no concept of +ve and -ve complex number, so the question must be modified

10. If x+iy= (-1+i 3 )2010 , then x 1)22010 2) -22010 3)-1 4)1 sol: (x+iy) =(2ω)2010 =22010 ω2010 =22010

Ans: (1)11.

(sinθ+cosθ). (tanθ + cotθ) 1) 1 2)sinθ. cosθ 3) secθ cosecθ 4) secθ+cosecθ

Ans : 112. The sides of traingle are 6+

12 , 48 , 24 . The tangent of the smallest angle of the traingle is

1) 2−1 2)13

3)1 4) 3

sol: cos C=486122−24

261218

= 32

tan 30= 13

Ans : (2)

Note : It requires calculation and lengthy question.

13. A simple graph contains 24 edges, degree of each vertex is 3, the number of vertices is 1)12 2)8 3)16 4)21Ans: 3n=48 ⇒ n=16(using sum of degrees of all vertices in a graph = 2 x no of edges)Ans is (3)

14. limn∞ [n sin 2

3ncos 2

3n ] =

1)23 2)

6 3)

3 4)1

sol: limn∞ [n sin 2

3ncos 2

3n ]

= limn∞

2nsin[2 23n ]

23

23

Ans : 1

15. The function f(x) =[x] where [x] denotes the greatest integer not greater than x, is 1)continuous only at +ve integral values of x 2) continuous for all non integral values of x3)continuous only at rational values of x4)continuous for all real values of xAns : 2

16. The greatest value of x satisfying 21≡385(mod x) and 587≡167(mod x) is1)32 2)156 3)56 4)28Ans : 4

17. The number (492 -4)(493 -49) is divisible by 1) 9! 2)7! 3)5! 4)6!sol: (492 -22 )49 (492 -1)=(49+2)(49-2)49(49+1)(49-1) is divisible by 5! or 7! or 6!, Ans is 2 or 3 or 4Note : It has multi choice

18. The least +ve integer x satisfying 22010 ≡3x (mod 5) 1)4 2)3 3)2 4)1sol: 22 -1(mod 5)≡22010 (-1)≡ 1005 (mod 5) -1(mod 5)≡ -1 3x(mod 5)≡Ans : (2)

19. If A and Bare two square matrices of the same order such that AB=B and BA=A then A2 +B2 is always equal to 1)A+B 2)I 3)2BA 4)2ABsol: AA +BB=A(BA)+(AB)B=(AB)A+(BA)B=BA+AB=A+BAns : 1

20. If A is a 3x3 non singular matrix if |A|=3 then |(2A)-1 |=1)3 2)24 3)1/24 4)1/3

|(2A)-1 |=1

|2A | =1

8 |A | =1

24Ans is (3)


21. If a, -a, b are the roots of x3 – 5x2 – x + 5 = 0, then b is a root of …………………..1) x2 + 3x – 20 = 0 2) x2 – 5x + 10 = 0

3) x2 – 3x – 10 = 0

4) x2 + 5x – 30 = 0sol: a + (-a) + b = 5 clearly b = 5, which is a root of x2 – 3x – 10 = 0Ans is 4

22. In the binomial expansion of (1 + x)15, the coefficients of xr and xr – 3 are equal. Then r is ……….. 1) 8 2) 7 3) 4 4) 6sol:

15Cr xr =

15Cr+3, xr+3

15Cr = 15Cr+3

⇒ r=15-(r+s) ⇒ r=6 Ans : 4

23. The nth term of the series 1 + 3 + 7 + 13 + 21 + ………. Is 9901. The value of n is ……a) 90 b) 100 c) 99 d) 900

Sol: By the method of differences, tn = 1 + (n – 1) n

Given 1 + n (n – 1) = 9901n (n – 1) = 9900 which is satisfied by n = 100 Ans is (2)

24.If 1

3−5x23x

=A

3−5x +B

23x , then A:B is

1)5:3 2)2:3 3)3:2 4)3:5sol: put x=3/5 A=5/19put x=-2/3 B=5/19A:B = 5:3 Ans: 1

25. Which of the following is NOT true?1) {(p→q) ^ (q→r)} →(p→r) is a tautology.2)(p ^ ~q)↔(p→q) is a tautology.3)~(p↔ q)≡(p^~q) v(~p^q)4)p→( q^r)≡( p → q)^(p →r)Ans: 2

26. If i, j, k are unit vectors along the positive direction of X, Y and Z-axes, then a FALSE statement in the following is ……………..1) ∑ i× j×k = 0

2) ∑ i× jk = 0 3) ∑ i . jk = 0

4) ∑ i . j×k = 0sol: We have, i . (j x k) = i . i = 1

∑ i . j×k = 3 but #0 Ans : 4

27. In P (X), the power set of a nonempty set X, an binary operation * is defined byA * B =A∪B A, B ∈P (X). Under *, a TRUE statement is …………….1)inverse law is not satisfied2)identity law is not satisfied3)Associative law is not satisfied4)Commutative law is not satisfiedsol: Under the binary operation,

A * B =A∪B A, B ∈P (X)

Inverse of A doesn’t exists because A*B#∅ for any B∈P (X), as ∅ is identity element in P(x) Ans : (1)

28.The inverse of 2010 in the group of Q+

of all +ve Q+ under the binary

operation, * is defined by a*b=ab

2010∀ a ,b∈Q¿ is

1)2011 2)2009 3)2010 4)1Ans : (3)

29.If the 3 functions f(x), g(x) and h(x) are such that h(x)=f(x).g(x) and f'(x), g'(x) where c is a constant, then f ' ' x f x +

g ' ' x g x +

2cf x . g x

is

1)hx h ' ' x 2)h'(x). h''(x)

3)h x h ' x 4)

h ' ' xhx

sol: G.E=g x f ' ' x f x g ' ' x2c

f x g x

=h ' ' xhx Ans : 4

30.The derivative of eax cosbx w.r.t x is reax

cos(bx+tan-1 ba ) where a>0, b>0

the value of r is

1)1

ab 2) a2b2 3)a+b 4)ab

sol: put a=b=1 G.E ⇒ ex cosx-exsinx

=ex (12

cosx -12

sinx) 2

= 2 ex cos (x+4 )

Ans is (2)


31. The chord of the circle x2 + y2 – 4x = 0 which is bisected at (1, 0) is perpendicular to the line

……..a) y = x b) x + y = 0c) x = 1 d) y = 1sol: x2 + y2 – 4x = 0

centre = (2, 0)

y = 1 is perpendicular to the chord, because it is parallel to radius.

32.In traingle ABC, if a=2, B=tan-1 1/2 and C=tan-1 1/3, then (A, b)

1)(4, ,

225

) 2) 34,

25

3) 4,25 4) 3

4,225

sol: A= 1800 -tan-1(1/2) - tan-1 (1/3)

= 1800 - (tan-1

121

3

1−16

)=1800 -4,

=34 and using

bsinA

= asinA

b=225

Ans is (4)

33. The straight line 2x + 3y – k = 0, k > 0 cuts the X- and Y-axes at A and B. The area of OAB, where O is the origin, is 12 sq. units. The equation of the circle having AB as diameter is……………1) x2 + y2 + 4x – 6y = 0 2) x2 + y2 – 6x – 4y = 03) x2 + y2 – 4x – 6y = 04) x2 + y2 – 6x + 4y = 0

sol: Area of traingle OAB= k 2

12=12

k=12 Circle with diameter ends (0,4) and (6,0) is x2 + y2 – 6x – 4y = 0

34.Let P (x, y) be the midpoint of the line joining (1, 0) to a point on the curve

y2

= ∣x1 x2x3 x5∣

Then locus of P is symmetrical about …………..a) Y-axis b) X-axisc) x = 1 d) y = 1sol: y2 =x-1(by evaluting determinant) which is parabola. parametric representation of these is

t 2

4 1, t2Midpoint of t 2

4 1, t2 and (1,0) is

P(x,y) = t 2

81, t

4 which is similar

to (at2 +h, 2at+k)Note: Question is lenghty

35.The function f (x) = |x – 2| + x is ……………….1) differentiable at x = 2 but not at x = 0. 2)differentiable at both x = 2 and x =0

3) continuous at both x = 2 and x = 0.4) continuous at x = 2 but not at x = 0Ans: is 3

36.Let R be an equivalence relation defined on a set containing 6 elements. The minimum number of ordered pairs that R should contain is ……………….

1) 6 2)12 3) 36 4) 64

Ans: 137.The line joining A (2, -7) and B (6, 5) is

divided into 4 equal parts by the points P, Q and R such that AQ = RP = QB. The midpoint of PR is …………………..1)(-8, 1) 2)(4, 12) 3)(8, -2) 4)(4, -1)sol: Midpoint of PR is mid point of AB (draw the figure) Ans: 4

38.Let P(-1,0), Q(0,0) , R(3, 3 3 ) be three points. The equation of the bisector of the PQR is 1) 3 x-y=0 2)x- 3 y=0 3)

3 x+y=0 4)x+ 3 y=0

Ans is (3) from below figure

tanθ= 3 =θ=3


(1,0)(2,0)

39. If m is the slope of one of the lines represented by ax2 + 2hxy + by2

= 0, then (h + bm)2 =……..

1) (a - b)2 2) (a + b)2

3)h2 – ab 4)h2 + abAns: substitute y=mx in ax2 +2hmx2

+bm2x2 =0 a+2hm+bm2 =0..........(1)(h+bm)2 =h2 +b2m2 +2bmh =h2-ab(by using (1))Ans is (3)

40. Cot 120 Cot 1020 + Cot 1020 Cot 660 + Cot 660 Cot 120 = ……………

1)1 2)-2 3) 2 4) -1Ans: cot (102o) = - tan 12o

cot 12o (-tan 12o) + cot 66o [-tan 12o + cot 12o]

= -1 + cot 66o [ 1−tan2120

tan 120 ] = -1 + cot 66o x cot 24o x2 =-1 + cot 66o (tan 66o)x2

= -1 + 2 = 1 Ans: (1)41.A wire of length 20 cm is bent in the

form of a sector of a circle. The maximum area that can be enclosed by the wire is …………….

1) 25 sq. cm 2) 20 sq. cm

3) 30 sq. cm 4) 10 sq. cm

Ans: A=12 r2 θ, 2r+s=20=P

A=12 r2

sr =

12 r(P-2r)

dAdr =

12 (20-4r) form maximum

area, dAdr =0 gives r=5

Amax = 25 , Ans : 1

42. Two circles centered at (2, 3) and (5, 6) intersect each other. If the radii are equal, the equation of the common chord is …………….1) x – y + 1 = 0 2) x + y + 1 = 03) x -y – 8 = 4)x+y-8=0Ans: Radical axis is perpendicular to line joining centres and since radie are equal midpoint must be on radical axis(7/2, 9/2) which satisfies x+y-8=0 Ans is (4)

43. Equation of the circle centered at (4, 3) touching the circle x2 + y2 = 1 externally, is …………1) x2 + y2 + 8x + 6y + 9 = 02)x2 + y2 – 8x– 6y + 9 = 03) x2 + y2 + 8x – 6y + 9 = 04)x2 + y2 – 8x + 6y + 9 = 0Ans: (4,3) is the centre (2) onlyAns is (2)

44. The points (1, 0), (0, 1), (0, 0) and (2k, 3k),( k 0) are concyclic if k =1) -1/5 2)1/5 3)5/13 4)-5/13 Ans: The points (1, 0), (0, 1), (0, 0) gives the circle x2 +y2 -x-y=0Substituting (2k, 3k) into circle , k=0 or k= 5/13Ans : (3)

45. The locus of the point of intersection of the tangents drawn at the ends of a focal chord of the parabola x2 = -8y is ………….

1) x = -2 2) x = 2

3) y = -2 4) y = 2

Ans : The locus is the directrix i.e y=2

Ans is (4)46. The condition for the line y = mx + c to be a normal to the parabola y2 = 4ax is …………….1) c= - a/m 2) c= -2am-am3 3)3)c=2am + am3 4) c= a/mAns: consider y2 =4x if y=x+c is a tangent then c=a/m=1 y=x+1Point of contact is (a/m, 2a/m)=(1, 2)Equation of normal is y= -x+C2=-1+c, c=3, y= -x+3, m =-1, c=3 a=1, Ans is (2)

47. The eccentric angle of the point

(2 3 ) lying on x2

16y2

4=1

1)2 2)

4 3)

6 4)

3


P(-1,0) (0,0)

R(3,3√3)

3

32π/3

Q

Ans: (a cosθ, bsinθ) =(2 3 )

cosθ= 24=1

2 , sinθ= 32

θ=3 Ans is (3)

48. The distance of the focus of x2 – y2

= 4, from the directrix which is nearer to it, is 1)8 2 2) 4 2 3) 2 4)4 2

Ans: ae-a/e=2 2 -22

= 2

Ans is (3)

49. If ∫ f x sinx cosx dx =

12b2−a2

log f(x) +c where C is the

constant of integration then f(x) is

1)2

b2−a2cos2x2)

2abcos 2x

3)2

b2−a2 sin2x 4)

2absin2x

Ans: f x sin2x

2 =1

2b2−a2f ' x f x

[f(x)]2 sin2x=f ' xb2−a2

option (1) ⇒4

b2−a2cos2 2xsin 2x

1b2−a2 f'(x) ⇒

2b2−a22

2sec 2x tan 2x

=4sin 2x

b2−a22 cos2 2x option (1) is the

choice option (2) and (4) doesnot hold as it doesn't contains b2 -a2 and option (3) contains sin 2x which can't be satisfied . Ans is (1)Note: Questions is lengthy

50. If ∫ xx x1

dx = k tan-1 x, then

(k, m) is 11) (1,x) 2)(2, x) 3)(2, x ) 4) (1, x )

Ans: put x =t ⇒1

2 xdx =dt

G.E= ∫ 2t21

dt = 2 tan-1 t

⇒ 2 tan-1 x +C . Ans is (3)

51. ∫0

4sinxcosx3sin2x

dx =

1) log 3 2)14 log 3

3) 2 log 3 4) 12 log 3

Ans: G.E= ∫0

4

sinxcosx4−1−sin2x

dx =

∫0

4 d sinx−cosx

4−sinx−cosx 2dx =

12.2 log

[ 2 sinx−cosx 2− sinx−cosx ] between 0 to

4

= 1/4 log 3 Ans is (2)Note: Question is lengthy

52. ∫0

1

x 1−x 32 dx =

1)4/35 2)-2/35 3)-8/35 4)24/35

Using Short cut: ∫0

1

x 1−x ndx

1n1n2 Ans is 4/35

Ans is (1)53. The area bounded by the curve

y=x2 ; x0x x≥0

and the line y=4 is

1)8/3 2)32/3 3)16/3 4)40/3

Ans: A1 = ∫0

4

x dy = ∫0

4

y dy = 16/3

A2 = ∫0

4

y dy =8

Required area 16/3 +8=40/3Ans is 4Note: Question is lengthy

54. The order and degree of Differential

equation y=dpdx x + a2 p2b2 where

p=dydx (here a & b are arbitrary

constant) respective1)1,1 2)2,2 3)2,1 4)1,2Ans is (2)


55. The general solution of Differential

equation 2x dydx -y=3 is a family of

1)parabola 2)Hyperbola 3)circle 4)straight line

Ans: dy

3 y=dx

2x ⇒ log(3+y)=1/2 log

2x +logc ⇒ y+3= x c ⇒ (y+3)2 =xcAns is (1)56. If x= acos3 θ, y= asin3 θ then dydx =

1) 3 xy 2) 3 yx 3)-3 3 yx 4) −3 xydydx =

−3a sin2cos3a cos2 sin

= - tanθ

Ans is (3)57. If y = tan-1 x2−1 then the ratio

d 2 ydx2 : dy

dx

1)1−2x2

x x2−12)

x x2−112x2 3)

x x211−2x2

4)12x2

x x21

Ans: dydx =

1x x2−1

d 2 ydx2 =

1−2x2

x2x2−1 x2−1d 2 ydx2 : dy

dx=

1−2x2

x x2−1 Ans is (1)

Note: Question is lengthy and time consuming.58. P is the point of contact of the tangent from the origin to the curve y = Logex. The length of the perpendicular drawn from the origin to the normal at P is …………….

1)1e 2)

12e 3) e21 4)2 e21

Ans: y=mx be the tangent ; dydx =

1x =m; y=

1x . x =1

x=eLength of perpendicular = e21Ans is (3)

59. For the curve 4x5 = 5y4, the ratio of the cube of the subtangent at a point on the curve to the square of the subnormal at the same point is ………………….

1)y 54

4

2)x 45

4

3) 54

4

4) 45

4

sol: y' =54 yx

S.T=yy ' and SN=yy'

ST 3

SN 2 =y

y ' 5

= 45

4

(by substituting y' and

simplifying) Ans is(4)Ans: Questions is lenghty

60. The set of real values of x for which

f(x)=x

logx is increasing is

1)empty 2){x: xe} 3){1} 4){x:x<e}Ans: Increasing ⇒ f'(x)>0 logex>1 ⇒ x>e

Ans is (2) Note: Here answer given is x2 but x=2 is stationary point. it must be x>e

Note: This is one of the tough CET question paper and many problems are lengthy and requires calculations.


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