Post on 22-Mar-2018
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)
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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)
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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)
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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
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(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
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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)
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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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