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KCET-2020 (Code-C3)
MATHEMATICS
Subject No. Of
Questions Max. Marks
Maximum Time For Answering
Mathematics 60 60 70 Minutes
01. If ( ) nP n : 2 n!
1. Then the smallest positive integer for which
P(n) is true if
(A) 4 (B) 5
(C) 2 (D) 3
Sol. Answer (A)
n2 n!
By hit & trial
n 4= & 5 are to be true
So we require small n 4 =
2. If z x iy= + , then the equation z 1 z 1+ = −
represents
(A) x-axis (B) y-axis
(C) a circle (D) a parabola
Sol. Answer (B)
z 1 z 1+ = −
Locus of z is perpendicular bisector of line
segment joining -1 and 1
3. The value of 16 16 16 16
9 10 6 7C C C C+ − − is
(A) 17
10C (B) 17
3C
(C) 0 (D) 1
Sol. Answer (C)
+ − − =16 16 16 16
9 10 6 7C C C C 0
−
=n n
r n rC C
4. The number of terms in the expansion of
( )10
x y z+ + is
(A) 11 (B) 110
(C) 66 (D) 142
Sol. Answer (C)
( )10
x y z+ +
no. of terms in the expansion (x1+x2+....xr)n is
n r 1
r 1C+ −
−
10 3 1
3 1C+ −
−=
12
2
12 11C 66
2
= = =
5. If the sum of n terms of an A.P. is given by
2
nS n n,= + then the common difference of
the A.P. is
(A) 2 (B) 6
(C) 4 (D) 1
Sol. Answer (A)
2
nS n n= +
= + =1 1S 1 1 T
2
2 1 2S T T 2 2 6= + = + =
1T 2 a= =
( )+ = + + =1 2T T a a d 6
2a d 6 + = 4 d 6 d 2 + = =
Choose the correct answer:
30/07/2020
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6. The two lines lx my n+ = and l x m y n + =
are perpendicular if
(A) lm l m 0 + = (B) lm ml 0 + =
(C) ll mm 0 + = (D) lm ml =
Sol. Answer (C)
lx my x+ =
l x m y n + =
Product of their slopes is -1
− −
= −
l l1
m m
ll mm ll mm 0 = − + =
7. If the parabola 2x 4ay= passes through the
point (2, 1), then the length of the latusrectum
is
(A) 2 (B) 8
(C) 1 (D) 4
Sol. Answer (D)
2x 4ay=
It passes through (2, 1)
22 4.a.1 =
a 1 =
So parabola 2x 4y=
length of latusrectum is 4
8. x 0
tanxlim
2x 4 2→
+ − is equal to
(A) 4 (B) 6
(C) 2 (D) 3
Sol. Answer: (C)
x 0
tanxlim
2x 4 2→
+ −
( )
( )( )x 0
2x 4 2tanxlim
2x 4 2 2x 4 2→
+ +
+ − + +
( )
→
+ +=
+ −x 0
tanx 2x 4 2lim
2x 4 4
( )x 0
tanxlim 2x 4 2
2x→= + +
0
tanlim 1→
=
( )1 14 2 4 2
2 2= + = =
9. The negation of the statement “For all real
numbers x and y, x y y x+ = + is
(A) for some real numbers x and y,
+ +x y y x
(B) for some real numbers x and y,
x y y x− = −
(C) for all real numbers x and y, x y y x+ +
(D) for some real numbers x and y,
x y y x+ = +
Sol. Answer (A)
For some real numbers x and y, x y y x+ +
10. The standard deviation of the data 6,7,8,9,10
is
(A) 2 (B) 10
(C) 2 (D) 10
Sol. Answer (C)
6,7,8,9,10
6 7 8 9 10
X 85
+ + + += =
( )
2
ix x
N
− =
( ) ( )
2 2 2 22 1 0 1 2
5
− + − + + +=
10
5=
2=
11. If A, B, C are three mutually exclusive and
exhaustive events of an experiment such that
P(A)=2 P(B)=3 P(C), then P(B) is equal to
(A) 3
11 (B)
4
11
(C) 1
11 (D)
2
11
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Sol. Answer (A)
Let ( ) ( ) ( )P A 2.P B 3.P C x= = =
Events are mutually exclusive and exhaustive
( ) ( ) ( )P A P B P C 1+ + =
x x
x 12 3
+ + =
6x 3x 2x
16
+ + =
11x 6
1 x6 11
= =
( )x 3
P B2 11
= =
12. If a relation R on the set 1,2,3 be defined
by ( ) R 1,1= , then R is
(A) Symmetric and transitive
(B) Only symmetric
(C) Reflexive and symmetric
(D) Reflexive and transitive
Sol. Answer (A)
( ) R 1,1=
It is only symmetric & transitive
13. Let ) →f : 2, R be the function defined by
( ) 2f x x 4x 5,= − + then the range of f is
(A) ( )1, (B) )5,
(C) ( ),− (D) )1,
Sol. Answer (D)
( ) 2f x x 4x 5= − +
( ) ( )2
f x x 2 1= − +
( ) )f x 1,
14. If A a,b,c= , then the number of binary
operations on A is
(A) 33 (B)
93
(C) 3 (D) 63
Sol. Answer (B)
A a,b,c=
no. of binary operation 2nn of set is having in
elements = 93
15. The domain of the function defined by
( ) 1f x cos x 1−= − is
(A) 1,1− (B) 0,1
(C) 1,2 (D) 0,2
Sol. Answer (C)
( ) 1f x cos x 1−= −
( )x 1 0 x 1 1− →
1 x 1 1− −
0 x 1 1 −
x 1 1−
x 1 1−
( )x 2 2 →
From (1) & (2)
x 1,2
16. The value of 1 1cos sin cos3 3
− − +
is
(A) -1 (B) Does not exist
(C) 0 (D) 1
Sol. Answer (B)
For arc sin x, x should lie between -1 and 1
17. If
0 0 1
A 0 1 0
1 0 0
=
, then 4A is equal to
(A) I (B) 4A
(C) A (D) 2A
Sol. Answer (A)
0 0 1
A 0 1 0
1 0 0
=
2
0 0 1 0 0 1
A 0 1 0 0 1 0
1 0 0 1 0 0
=
2
1 0 0
A 0 1 0 I
0 0 1
= =
=2A I, 4 2 2A A .A I.I I= = =
KCET-2020 Mathematics (Code-C3)
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18. If A and B are square matrices of same order
and B is a skew symmetric matrix, then A BA
is
(A) Diagonal matrix
(B) Skew symmetric matrix
(C) Symmetric matrix
(D) Null matrix
Sol. Answer (B)
Let TP A BA=
( )T
T TP A BA=
( )T
T T TA B A=
( )T T T TPQR R Q P =
( )TA B A= −
T TP A B A= = −
TP P= −
Skew symmetric
19. If 2 1 1 0
A ,3 2 0 1
=
then the matrix A is
(A) 2 1
3 2
−
− (B)
2 1
3 2
−
(C) 2 1
3 2
(D) 2 1
3 2
− −
Sol. Answer (D)
2 1 1 0
A3 2 0 1
=
1
1 2 1A
3 2
−
− =
2 11
3 24 3
− =
−−
1 2 1A
3 2
−−
= −
20. If ( )
− + +
= − − +
− −
3
2
x x a x b x
f x x a x x c x ,
x b x c 0
then
(A) ( )f 0 0= (B) ( )f 1 0− =
(C) ( )f 1 0= (D) ( )f 2 0=
Sol. Answer (A)
( )
− + +
= − − +
− −
3
2
x x a x b x
f x x a x x c x
x b x c 0
( ) = − =
− −
0 a b
f 0 a 0 c 0
b c 0
Determinant of skew symmetric matrix of odd
order = 0
21. If 1 2 3 9a ,a ,a ,......,a are in A.P, then the value
of
1 2 3
4 5 6
7 8 9
a a a
a a a
a a a
is
(A) ( )e elog log e (B) 1
(C) ( )1 9
9a a
2+ (D) 1 9a a+
Sol. Answer (A)
Apply 1 1 3 2C C C 2C→ + −
22. If A is a square matrix of order 3 and |A|=5,
then A adjA is
(A) 25 (B) 625
(C) 5 (D) 125
Sol. Answer (D)
= nA.adjA A .I
3A.adjA A .I=
3 3A 5= =
23. If ( )
1 cosKx, if x 0
xsinxf x
1, if x 0
2
−
= =
is continuous at x 0,= then the value of K is
(A) 2 (B) 1
(C) 1
2 (D) 0
Sol. Answer (B)
( )
−
= =
1 coskx; x 0
xsinxf x
1; x 0
2
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x 0
1 coskxlim
x.sinx→
−
→
2
x 0
kx2sin
2limx.sin x
→
2
x 02
kxsin
kx 12lim2 . .kx 2 sinx
x2 x
2
2 2
2x 0
kxsin
2.k x 12lim . .kx sinx4.x
2 x
→
=
2k
2=
Since ( )f x is continuous at x=0
So ( )f 0 LHL RHL= =
2
21 kk 1 k 1
2 2= = =
24. The right hand and left hand limit of the
function ( )
1/ x
1/ x
e 1, if x 0
f x e 1
0 , if x 0
−
= + =
are
respectively
(A) -1 and -1 (B) -1 and 1
(C) 1 and 1 (D) 1 and -1
Sol. Answer (D)
RHL
Put x 0 u= +
1
u
1u 0u
e 1lim
e 1→
−
+
Take 1
ue common
1
u
1u 0u
1 elim 1
1 e
−
→ −
−=
+
LHL, Put x 0 u= −
1
u
1u 0 1u
e 1lim 1
e
−
→ − +
−= −
25. If x y x y2 2 2 ,++ = then dy
dx is
(A) x y2 − (B) y
x
2 1
2 1
−
−
(C) y x2 − (D) y x2 −−
Sol. Answer (D)
x y x y2 2 2 ++ =
x y x y2 2 2 .2+ =
x y2 2 1− − + =
Differentiating with respect to ‘x’
( ) ( )x y dy2 .log2. 1 2 .log2. 1 0
dx
− −− + − =
y
x
dy 2
dx 2
−=
26. If ( ) 1
2
2xf x sin
1 x
− =
+ , then ( )f 3 is
(A) 1
3 (B)
1
3−
(C) 1
2− (D)
1
2
Sol. Answer (C)
1
1 1
2
1
2tan x ; x 12x
sin 2tan x ; 1 x 11 x
2tan x; x 1
−
− −
−
−
= − +
− − −
( )
− +
= − +
− −
+
2
2
2
2;x 1
1 x
2f x ; 1 x 1
1 x
2;x 1
1 x
( ) − = = −
+
2 1f 3
1 3 2
27. If ( )y xxe e ,= then
dy
dx is
(A) ( )
logx
1 logx+ (B)
( )
xe
x y 1−
(C) ( )
2
logx
1 logx+ (D)
( )2
1
1 logx+
Sol. Answer (C)
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( )y xxe e=
Take log on both sides
( )e ey.ln xe x.ln e=
( )ey. ln x lne x+ =
x
yln x 1
=+
Differentiating with respect to ‘x’
( )
( ) ( )2 2
11. lnx 1 .x
lnxxylnx 1 lnx 1
+ − = =
+ +
28. If n 1
n
3y 2x ,
x
+= + then 2
2
2
d yx
dx is
(A) dy
x ydx
+ (B) y
(C) ( )6n n 1 y+ (D) ( )n n 1 y+
Sol. Answer (D)
+= +n 1
n
3y 2.x
x
( ) − − = + −n n 1y 2. n 1 .x 3nx
( ) ( )− − − = + + +n 1 n 2y 2.n n 1 .x 3n n 1 .x
Now according to question, multiply 2x on
both sides
( ) ( )− − − = + + +2 2 n 1 2 n 2x .y 2.n n 1 .x .x 3n n 1 .x .x
( ) ( )n 1
n
12.n n 1 .x 3n n 1 .
x
+= + + +
( )( ) = +2y .x n n 1 y
29. If the curves 22x y= and 2xy K= intersect
perpendicularly, then the value of 2K is
(A) 2 (B) 8
(C) 4 (D) 2 2
Sol. Answer (B)
Curves are ⊥ , so product of the slopes = -1
2
1
12x y 2 2y.y y m
y = = = =
k
2xy k y2x
= =
−
= = 22
ky m
2x
1 2m m 1= −
2
1 k. 1
y 2x
− = −
2k 2x y =
Put 2xy k= in 2k 2x y=
22xy 2x y=
( )xy x 1 0 − =
x 0,y 0= = or x 1=
Put x 1= in equation (1), to get y
22x y= = = 2y 2 y 2
Now =2xy k
(Squaring on both sides)
2 2 2k 4x y 4.1.2 8= = =
30. The maximum value of elog x,
x if x 0 is
(A) 1
e (B)
1
e−
(C) e (D) 1
Sol. Answer (A)
elog xy
x=
Differentiating with respect to ‘x’
e
2
1x. log x
xyx
− =
e
2
1 log xy
x
− =
Maxima occurs at x = e as dy
dx in the NBD
changes from +ve to –ve
( ) elog e 1f e
e e= =
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31. If the side of a cube is increased by 5%, then
the surface area of a cube is increased by
(A) 6% (B) 20%
(C) 10% (D) 60%
Sol. Answer (C)
dx
100 5x =
ds
100 ?s =
2s 6x=
logs log6 2.logx= +
1 dx
.ds 2s x
=
ds dx
2.s x
=
ds dx
100 2. 100 2 5s x = = =10%
32. The value of 4
6
1 xdx
1 x
+
+ is
(A) 1 1 31tan x tan x C
3
− −− +
(B) 1 1 21tan x tan x C
3
− −+ +
(C) 1 1 3tan x tan x C− −+ +
(D) 1 1 31tan x tan x C
3
− −+ +
Sol. Answer (D)
( )
4
6
1 xdx
1 x
+
+
( )( )( )( )
4 2
6 2
1 x 1 xdx
1 x 1 x
+ +
+ +
( ) ( )( )( )
6 2 2
6 2
1 x x 1 xdx
1 x 1 x
+ + +
+ +
2
2 6
dx x dx
1 x 1 x+
+ +
( )
2
2 23
dx 1 3xdx
31 x 1 x+
+ +
1 1 31tan x .tan x C
3
− −= + +
33. If ( )( )( )
3x 1dx
x 1 x 2 x 3
+
− − −
= − + − + − +Alog x 1 Blog x 2 Clog x 3 Constant ,
then the values of A, B and C are respectively
(A) 5,-7,5 (B) 2,-7,5
(C) 5,-7,-5 (D) 2,-7,-5
Sol. Answer (B)
( )( )( )
3x 1dx
x 1 x 2 x 3
+
− − −
( )( )( )
3x 1 A B C
x 1 x 2 x 3 x 1 x 2 x 3
+= + +
− − − − − −
Take LCM
( )( ) ( )( )+ = − − + − − +3x 1 A x 2 x 3 B x 1 x 3
( )( )C x 1 x 2− −
Put x 1=
( )( )4 A 1 2 A 2 = − − =
Put x=2
( )( )7 B 1 1 B 7= − = −
Put x=3
9 1 C.2.1 C 5+ = =
( )( )( )
3x 1 dx dx dxdx 2 7 5
x 1 x 2 x 3 x 1 x 2 x 3
+= − +
− − − − − −
2ln x 1 7.ln x 2 5.ln x 3 C= − − − + − +
A=2, B=-7, C=5
34. The value of sinxe sin2xdx is
(A) ( )sinx2e cosx 1 C+ +
(B) ( )sinx2e cosx 1 C− +
(C) ( )sinx2e sinx 1 C− +
(D) ( )sinx2e sinx 1 C+ +
Sol. Answer (C)
sinxe .2.sinx.cosx.dx
Put sinx = t
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=cos x dx dt
t2 e .t dt
−
t tdt
2 t e dt e dtdt
= − + t t2 e .t e C
( )t2e t 1 C= − +
Replace ‘t’ by sinx
( )sinx2e sinx 1 C= − +
35. The value of −
−
1
21
1
2
cos xdx is
(A) 1 (B) 2
2
(C) (D) 2
Sol. Answer (D)
( )1
121
2
I cos xdx 1−
−= →
( ) ( )b b
a af x dx f a b x dx= + −
( )1
121
2
I cos x dx−
−= −
( ) ( )1
121
2
I cos x dx 2−
−= − →
Add both (1) and (2)
( )− −
−= + −
11 12
1
2
2I cos x cos x dx
1
21
2
2I dx−
=
1 1
2I .2 2
= +
I2
=
36. The value of
−
+2
x
2
cosxdx
1 eis
(A) 1 (B) -2
(C) 2 (D) 0
Sol. Answer (A)
( )/ 2
x/ 2
cosxI dx 1
1 e
−= →
+
Use ( ) ( )b b
a af x dx f a b x dx= + −
x
cosxI dx
1 e−=
+
( )x
x
e .cos xI dx 2
1 e= →
+
Add (1) and (2)
2
2
2I cosxdx
−
=
( )/ 2
/ 22I sin x
−=
2I 1 1 I 1= + =
37. The value of ( )+
+1
2
0
log 1 xdx
1 x is
(A) 1
2 (B) log2
8
(C) log22
(D) log2
4
Sol. Answer (B)
( )1
20
log 1 xdx
1 x
+
+
Put x tan=
( )
+ =/ 4
0log 1 tan d log2
8
38. The area of the region bounded by the curve
2y 8x= and the line y 2x= is
(A) 3
4sq. units (B)
8
3 sq. units
(C) 16
3 sq. units (D)
4sq.units
3
Sol. Answer (D)
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2y 8x=
y=2x
substitute ( )2
2x 8x =
24x 8x=
x 0,2=
( )−2
08x 2x dx
1
2 22
0 02 2 x dx 2 xdx−
−
23
22
0
2 2.x 2.x
3 2
2
3
224 2
.2 23
= −
24.2 4
43 3
= − =
39. The area of the region bounded by the line
y 2x 1,= + x-axis and the ordinates x 1= −
and x 1= is
(A) 5
2 (B) 5
(C) 9
4 (D) 2
Sol. Answer (A)
y=2x+1
( )1,0−
1,0
2
−
( )1,0
= ( ) ( )
−
−−
− + + +
1
12
11
2
2x 1 dx 2x 1 dx
=1
12 22
11
2
x x x x−
− − − + + +
( ) ( )1 1 1 1
1 1 1 14 2 4 2
= − − − − + + − −
1 1 5
24 4 2
= + + =
40. The order of the differential equation obtained
by eliminating arbitrary constants in the family
of curves ( ) 4x c
1 2 3c y c c e += + is
(A) 3 (B) 4
(C) 1 (D) 2
Sol. Answer (C)
( ) 4x C
1 2 3C y C C .e += +
( ) 4C x
1 2 3C y C C .e .e= +
( )
42 3 C x
1
C Cy .e .e
C
+=
xy C.e=
Only 1 arbitrary constant
Order = 1
41. The general solution of the differential
equation − =2 4x dy 2xydx x cosxdx is
(A) 2y sinx cx= +
(B) 2y cosx cx= +
(C) 2 2y x sinx cx= +
(D) 2y x sinx c= +
Sol. Answer (C)
2 4x dy 2xy.dx x .cosxdx− =
2dy 2.y x .cosx
dx x− =
I.F −
2dx
xe
2.lnx
2
1e
x
−= =
Now solution
2
2 2
1 1y. x .cos x. dx
x x=
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= 2
ycosx dx
x
2
ysinx C
x= +
42. The curve passing through the point (1, 2)
given that the slope of the tangent at any
point ( )x,y is 2x
y represents
(A) Ellipse (B) Hyperbola
(C) Circle
(D) Parabola
Sol. Answer (B)
dy 2x
dx y=
ydy 2xdx=
Variable separable
2
2yx C
2= +
Passes through (1, 2)
4
1 C C 12= + =
2
2yx 1
2 = +
2
2 yx 1
2 − = − , Which is hyperbola
43.. The two vectors i j k+ + and i 3j 5k+ +
represent the two sides AB and AC
respectively of a ABC. The length of the
median through A is
(A) 7
(B) 14
(C) 14
2 (D) 14
Sol. Answer (B)
Let A be origin
D is midpoint of B & C
( ) ( )+ + + + +
=i j k i 3j 5k
AD2
= + +AD i 2j 3k
= + + =AD 1 4 9 14
44. If a and b are unit vectors and is the
angle between a and b , then sin2
is
(A) a b
2
− (B) a b−
(C) a b+
(D) a b
2
+
Sol. Answer (A)
− = + − 2 2
a b a b 2 a b .cos
1 1 2cos= + −
( )2 1 cos= −
a b 2.sin2
− =
45. If the vectors − + + −2i 3j 4k, 2i j k and
i j 2k − + are coplanar, then the value of
is
(A) -6 (B) 5
(C) 6 (D) -5
Sol. Answer (C)
−
− =
−
2 3 4
2 1 1 0
1 2
=6
46. If 2 2
a b a b 144 + = and =a 6 , then b
is equal to
(A) 2
(B) 4
(C) 6
(D) 3
Sol. Answer (A)
+ =2 2 2 2
a b a b a b , we get =b 2
Mathematics (Code-C3) KCET-2020
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47. The point (1, -3, 4) lies in the octant
(A) Fourth (B) Eighth
(C) Second (D) Third
Sol. Answer (A)
Conceptual
48. If a line makes an angle of 3
with each of x
and y-axis then the acute angle made by z-
axis is
(A) 3
(B)
2
(C) 4
(D)
6
Sol. Answer (C)
, , be angle with axes
2 2 2cos cos cos 1+ + =
2 2
2cos cos cos 13 3
+ + =
21 1cos 1
4 4+ + =
2 1cos 1
2 = −
1
cos2
=
=4
49. The distance of the point (1, 2, -4) from the
line x 3 y 3 z 5
2 3 6
− − += = is
(A) 293
49 (B)
293
49
(C) 293
7 (D)
293
7
Sol. Answer (D)
( )P 1,2, 4−
Q
− − +
= = = x 3 y 3 z 5
2 3 6
General point on line is ( )2 3,3 3,6 5 + + −
Direction ratio of PQ ( ) + + −2 2,3 1,6 1
Since line & PQ are perpendicular
So, product of their direction ratios will be 0.
So ( ) ( ) ( )2 2 .2 3 1 3 6 1 .6 0 + + + + − =
4 4 9 3 36 6 0 + + + + − =
1
49 1 049
+ = = −
Direction ratio PQ ( ) + + −2 2,3 1,6 1
− − −
+ + −
1 1 12 2, 3 1, 6 1
49 49 49
− =
96 46 55, ,
49 49 49
Distance
2 2 296 46 55 14537
49 49 49 492
− = + + =
293 293
49 7=
50. The sine of the angle between the straight
line − − −
= =−
x 2 3 y z 4
3 4 5 and the plane
2x 2y z 5− + = is
(A) 4
5 2 (B)
2
10
(C) 3
50 (D)
3
50
Sol. Answer (B)
x 2 y 3 z 4
3 4 5
− − −= = & 2x 2y z 5− + =
Angle between line & plane
( ) ( )
( )22 2 2 2 2
3.2 4 2 5 1sin
3 4 5 2 2 1
+ − + =
+ + + − +
6 8 5
sin5 2.3
− + =
3 1 2
103.5 2 5 2= = =
KCET-2020 Mathematics (Code-C3)
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51. The feasible region of an LPP is shown in the
figure. If Z = 11x + 7y, then the maximum
value of Z occurs at
Y
X
x y 5+ =
x 3y 9+ =
0
(A) (5, 0) (B) (3, 2)
(C) (0, 5) (D) (3, 3)
Sol. Answer (B)
Solve
+ =
+ =
− = −
x y 5
x 3y 9
2y 4
y 2 x 3= =
x 3y 9+ = meets y axis at (0, 3)
x y 5+ = meets at (0, 5)
Now check z 11x 7y= +
At (0,3) z 0 7.3 21 = + =
At (0,5) z 0 7.5 35 = + =
At (3,2) z 11.3 7.2 47 = + =
So, Z is maximum at (3,2)
52. Corner points of the feasible region
determined by the system of linear
constraints are (0, 3), (1, 1) and (3, 0). Let z =
px + qy, where p, q > 0. Condition on p and q
so that the minimum of z occurs at (3, 0) and
(1, 1) is
(A) p 3q= (B) p q=
(C) p 2q= (D) q
p2
=
Sol. Answer (D)
= +z px qy
Minimum occurs at (3,0)
z 3p=
Also at z p q = +
So 3p= p + q 2p q =
53. If A and B are two events such that
( ) ( )1 1
P A ,P B3 2
= = and ( )1
P A B ,6
= then
( )P A /B is
(A) 1
2 (B)
1
12
(C) 2
3 (D)
1
3
Sol. Answer (C)
( )1
P A3
=
( )1
P B2
=
( )1
P A B6
=
Since ( ) ( ) ( )P A B P A .P B =
So events are independent
Now ( )A
P P AB
=
1 2
13 3
= − =
54. A die is thrown 10 times, the probability that
an odd number will come up atleast one time
is
(A) 11
1024 (B)
1013
1024
(C) 1
1024 (D)
1023
1024
Sol. Answer (D)
1 – P(no odd number comes) 10
10 10
3 1 10231 1
10246 2= − − =
55. The probability of solving a problem by three
persons A, B and C independently is 1 1
,2 4
and 1
3 respectively. Then the probability of
the problem is solved by any two of them is
(A) 1
24 (B)
1
8
(C) 1
12 (D)
1
4
Sol. Answer (D)
Mathematics (Code-C3) KCET-2020
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Solved by two implies one person should fail
( ) ( ) ( )P A B C P A B C P A B C= + +
Events are independent
1 1 2 1 3 1 1 1 1
2 4 3 2 4 3 2 4 3= + +
6 1
24 4= =
56. Events 1E and 2E form a partition of the
sample space S. A is any event such that
( ) ( )1 2
1P E P E ,
2= = ( )2
1P E / A
2= and
( )2
2P A /E
3= , then ( )1P E / A is
(A) 1 (B) 1
4
(C) 1
2 (D)
2
3
Sol. Answer (C)
Using Baye’s theorem
( )
( ) ( )
( ) ( ) ( ) ( )=
+
2 2
2
1 1 2 2
P E .P A /EP E / A
P E P A /E P E .P A /E
`
1 2.
1 2 31 1 22
.x .2 2 3
=
+
1 2/3 2 4
x22 3 3
x3
= + =
+
( )= = 1
2x P A /E
3
Now use again Baye’s theorem
( )( ) ( )
( ) ( ) ( ) ( )=
+
2 2
2
1 1 2 2
P E .P A /EP E / A
P E P A /E P E .P A /E
1 2.
2 31 2 1 2
. .2 3 2 3
=
+
( ) =1
1P E / A
2
57. If A 1,2,3,4,5,6= , then the number of
subsets of A which contain atleast two
elements is
(A) 57 (B) 58
(C) 64 (D) 63
Sol. Answer (A)
6 6 6 6 6
2 3 4 5 6C C C C C + + + +
6 6 6
0 12 C C= − −
64 1 6 57= − − =
58. If n(A)=2 and total number of possible
relations from set A to set B is 1024, then
n(B) is
(A) 10 (B) 5
(C) 512 (D) 20
Sol. Answer (B)
No.of relations (maximum) = mn2
mn 102 1024 2= =
( ) 2.m 10n A 2 2 2 m 5= = =
59. The value of 2 0 2 0sin 51 sin 39+ is
(A) 0sin12 (B)
0cos12
(C) 1 (D) 0
Sol. Answer (C)
2 2sin cos 1+ =
60. If tan A + cot A = 2, then the value of
4 4tan A cot A+ =
(A) 4 (B) 5
(C) 2 (D) 1
Sol. Answer (C)
tan A cot A 2+ =
1
tanA 2tanA
+ =
2tan A 2tanA 1 0 + + =
( )2
tanA 1 0 tanA 1− = =
cot A 1=
So, 4 4tan A cot A 1 1 2+ = + =
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