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CAREER POINT
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JEE Main Online Exam 2019
Questions & Solutions 12th April 2019 | Shift - I
MATHEMATICS
Q.1 The integral xx1x2
4
3 dx is equal to :
(Here C is a constant of integration)
(1) Cx
1xlog 2
3
e
(2) Cx
1xlog
21
2
3
e
(3) Cx
1xlog3
e (4) C
x)1x(log
21
3
23
e
Ans. [3]
Sol. xx1x2
4
3 dx
xx)2x2()1x4(
4
33dx
xx1x4
4
3 dx – x
12 dx
x4 + x = t (4x3 + 1) dx = dt
tdt – dx
x12
n|t| – 2nx + C
n 2
4
xxx + C n
x1x3 + C
Q.2 For x (0, 3/2), let f(x) = )x(g,x = tan x and h(x) = 2
2
x1x1
If (x) = ((hof)og)(x), then
3 is equal to :
(1) 127tan (2)
1211tan (3)
12tan (4)
125tan
Ans. [2] Sol. (x) = [(hof)og](x)
hof(x) = x1x1
(hof)g(x) = xtan1xtan1
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3 =
3tan1
3tan1
3 = tan
34= tan
1211
Q.3 Let P be the point of intersection of the common tangents to the parabola y2 = 12x and the hyperbola
8x2 – y2 = 8. If S and S' denote the foci of the hyperbola where S lies on the positive x-axis then P divides SS' in a ratio :
(1) 14 : 13 (2) 13 : 11 (3) 5 : 4 (4) 2 : 1 Ans. [3]
Sol. genttancommon1
8yx:C
x12y:C2
22
21
sameBoth8)m(1mxy:C
m3mxy:CofTangent
22
1
2
m3
= m2 – 8
9 = m2 (m2–8) let m2 = t
t (t–8) = 9 t2 –8t – 9 = 0
(t–9) (t +1) = 0 m2 = 9
m = –3, 3 common tangent y = 3x + 1 y = –3x –1
P intersection of tangents P
0,31
foci of hyperbola a = 1, b = 2 2 8 = 1 (e2–1) e = 3 is S(3, 0), S(–3, 0)
x
k 1
S(3, 0) P
0,31 S(–3, 0)
31
= 1k
3k3
k = 5 : 4
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Q.4 If
20
),n(mdxecxcosxcot
xcot then mn is equal to
(1) – 1 (2) 1 (3) 21
(4) 21
Ans. [1]
Sol. I =
2/
0
dxecxcosxcot
xcot
I =
2/
0xcos1
xcos = 2/
02
2
2xcos2
12xcos2
I =
2/xsec211 2
2/
0
dx
I = 2/
0)2/xtan(
22x
12
= 2
2
m = 21 , n = –2
mn = –1
Q.5 If and are the roots of the equation 375 x2 – 25x – 2 = 0, then
n
1r
rn
1r nr
nlimlim is equal to :
(1) 116
7 (2) 35829 (3)
121 (4)
34621
Ans. [3]
Sol. + = 37525
= 375
2
& (–1, 1), then
n
lim
n
1r
rr )(
+ 2 ……….infinite + + 2........ infinite
11
)1)(1(
)1()1( =
121
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Q.6 The number of solutions of the equation 1 + sin4 x = cos2 3x, x
25,
25 is :
(1) 5 (2) 3 (3) 7 (4) 4 Ans. [1]
Sol. 1 + sin4x = cos2 3x ; x
25,
25
L.H.S. ≥ 1 R.H.S. ≤ 1 Both satisfy when L.H.S. = R.H.S. = 1 sin4x = 0 ; cos23x= 1 x = –2, –, 0, , 2 total 5 solution Q.7 The equation |z – i| = |z – 1|, i = ,1 represents :
(1) a circle of radius 1. (2) the line through the origin with slope – 1
(3) a circle of radius .2
1 (4) the line through the origin with slope 1.
Ans. [4] Sol. |z – i| = | z – 1| z = x + iy |x + i(y – 1)| = |(x – 1) + iy| x2 + (y – 1)2 = (x – 1)2 + y2 y = x straight line having slope (+)ve 1, pass origin (0, 0)
Q.8 If ey + xy = e, the ordered pair
,dxyd
,dxdy
2
2at x = 0 is equal to :
(1)
2e
1,e1 (2)
2e1,
e1 (3)
2e1,
e1 (4)
2e1,
e1
Ans. [1] Sol. ey + xy = e at x = 0; ey + 0 = e y = 1 y(0) = 1 diff. w.r.t. x ey. y1 + xy1 + y = 0 e(1).y1 + 0 + 1 = 0 y1 = –1/e at x = 0 diff again w.r.t. to x ey.y2 + y1 ey.y1 + xy2 + y1 + y1 = 0 at x = 0
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e(1) (y2) + 2
e1
e1 + 0.y2 + 2
e1 = 0
y2 = 1/e2 x = 0
2e1,
e1
Q.9 The number of ways of choosing 10 objects out of 31 objects of which 10 are identical and the remaining 21
are distinct, is : (1) 220 – 1 (2) 220 (3) 221 (4) 220 + 1
Ans. [2] Sol. total 31
10identical 10 & distinct 21
S = 21C10 + 21C9(1) + 21C8(1) + 21C7 + 21C6 …… 21C0(1) S = 21C11 + 21C12 + 21C13 …… 21C21 2S = 221 S = 220 Q.10 For x R, let [x] denote the greatest integer < x, then the sum of the series
10099
31.....
1002
31
1001
31
31 is :
(1) – 153 (2) – 135 (3) – 133 (4) – 131 Ans. [3]
Sol.
)times67(110066
31.....
1001
31
31
+
)times33(210099
31.....
10068
31
10067
31
= – 67 – 66 = – 133 Q.11 If m is the minimum value of k for which the function f(x) = 2xkxx is increasing in the interval [0,3]
and M is the maximum value of f in [0, 3] when k = m, then the ordered pair (m, M) is equal to :
(1) (5, 63 ) (2) (4, 33 ) (3) (4, 23 ) (4) (3, 33 )
Ans. [2] Sol. f(x) : f (x) ≥ 0 ; & f(x1) ≤ f(x2), x[0, 3]
f(0) < f(3) f (x) = 2xkx2
)x2k(x
+ 2xkx
0 < 9k33 = 2
2
xkx2
)xkx(2)x2k(x
≥ 0
k > 3 = 2
2
xkx2
x4kx3
≥ 0
& kx – x2 > 0
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3kx – 4x2 ≥ 0 x (3k – 4x) ≥ 0 x[0,3] x (4x – 3x) ≤ 0 k > 4 k = 4 minimum value of k m = 4
f(3) = M = 9)3(43 = 33
M = 33
(m, M) = (4, 3 3 ) Q.12 If the angle of intersection at a point where the two circles with radii 5 cm and 12 cm intersect is 90°, then
the length (in cm) of their common chord is :
(1) 5
13 (2) 1360 (3)
13120 (4)
213
Ans. [3] Sol.
D 12 5
C2 13 C1
A
B
AC1C2
tan = 5
12 sin = 1312 …..(i)
ACD:
sin = 5
2/AB …..(ii)
(i) & (ii)
5.2
AB = 1312
AB =13
120
Q.13 The coefficient of x18 in the product (1 + x) (1 – x)10 (1 + x + x2)9 is :
(1) 126 (2) – 84 (3) – 126 (4) 84 Ans. [4] Sol. coefficient of x18 (1 + x) (1 – x)10 (1 + x + x2)9 (1 + x) (1 – x) (1 – x)9 (1 + x + x2)9 (1 – x2) (1 – x3)9
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(1 – x2) [9Cr (–1)r (x3r)]
9Cr (–1)r x3r – 9Cr (–1)r x3r+2
for x18
3r = 18 3r+2=18
r = 6 r not possible
then coefficient of x18 is 9C6 (–1)6 = 84
Q.14 Consider the differential equation, y2dx +
y1x dy = 0, If value of y is 1 when x = 1, then the value of x
for which y = 2, is :
(1) e
123 (2) e
23 (3)
e1
21 (4)
e1
25
Ans. [1]
Sol. dydx + 2y
x = 3y1
I.F. = dyy/1 2e = e–1/y
sol. y/1xe = y/1e (1/y3)dy
y1
= t
2y1 dy = dt
xet = dt).t(et
x.et = – tt ee.t + c
x = –t + 1 + c.e–t
x = y1 +1 + c. y/1e
give y(1) = 1
1 = 1 +1 + c.e1/1
c = –e–1
x = y1 + 1 + (–)e–1. e1/y
at y = 2
x = 23 –
e1
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Q.15 If the area (in sq. units) of the region {(x, y) : y2 4x, x + y 1, x 0, y 0} is a 2 + b, then a – b is equal
to :
(1) 38 (2) 6 (3)
310 (4)
32
Ans. [2] Sol. C1 : y2 ≤ 4x C2 : x + y ≤ 1 x ≥ 0 y ≥ 0
(0,1) P A(2 2 -2)
(1, 0) B (0, 0) O
y2 = 4x ; y2 = 4 (1–y) y2 + 4y + 4 = 0
y = 2 2 –2 , – 2 2 –2
Area : shaded region of curve OAB
A = Area of OBP – Area of region OAP
OBP = 21 × 1 × 1 =
21
Area of OAP = 222
0
2
4y dy +
1
222
)y1( dy
= 121 222
03y
+ 1
222
2
2yy
= 121 [(2 2 –2)3] +
2)222()222(
211
2
= 623 – 2
38
A = 21 –
623 +
328
a = 38 , b = –
620
a – b = 6
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Q.16 If the data x1, x2, …… x10 is such that the mean of first four of these is 11, the mean of the remaining six is 16 and the sum of squares of all of these is 2,000 ; then the standard deviation of this data is :
(1) 2 (2) 2 (3) 22 (4) 4 Ans. [2] Sol. x1 + x2 + x3 + x4 = 44 x5 + x6 ….. x10 = 96 xi = 140
40
140x = 14
2x = 2000
SD = 22
nx
nx
=2
10140
102000
= 196200 = 2
Q.17 If A is a symmetric matrix and B is a skew-symmetric matrix such that A + B =
1532
, then AB is equal
to :
(1)
4124
(2)
4124
(3)
4124
(4)
4124
Ans. [4] Sol. A is symmetric AT = A B is skew 8 symmetry BT = – B
A + B =
1532
… (i)
Transpose
AT + BT =
1352
A – B =
1352
….(ii)
From (i) + (ii)
A =
1442
From (i) – (ii)
B =
0110
then AB =
4124
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Q.18 If the normal to the ellipse 3x2 + 4y2 = 12 at a point P on it is parallel to the line, 2x + y = 4 and the tangent to the ellipse at P passes through Q(4,4) then PQ is equal to :
(1) 261 (2)
2221 (3)
2157 (4)
255
Ans. [4]
Sol. Ellipse : 9y
4x 22
= 1
Normal at P : (2cos, 3 sin) is 2xsec – 3 cosec y = 4 = 3 = 1 Normal at P: 2x sec– 3 y cosec
Slope of normal =
eccos3sec2 =
32
cossin
Normal parallel to 2x + y = 4
then 3
2 tan = –2; tan = – 3
32
Point P
23,1 , Q(4, 4)
PQ = 2
2234)14(
= 4
125 = 255
Q.19 Let a = k2j2i3 and b
k2j2i be two vectors. If a vector perpendicular to both the vectors
ba
and ba
has the magnitude 12 then one such vector is : (1) )kj2i2(4 (2) )kj2i2(4 (3) )kj2i2(4 (4) )kj2i2(4
Ans. [1] Sol. k2j2i3a
k2j2ib
j4i4ba
k4i2ba
A vector r perpendicular to )ba(&)ba(
& magnitude 12 n.12r
n = |)ba()ba(|
)ba()ba(
)ba()ba(
= 402044kji
= )8(k)16(j)16(i = ]kj2i2[8 | )ba()ba(
| = 8.3
n = 3
)kj2i2(
r = 4. )kj2i2(
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Q.20 If B =
13120125
is the inverse of a 3 × 3 matrix A, then the sum of all values of for which
det(A) + 1 = 0, is : (1) 2 (2) – 1 (3) 0 (4) 1
Ans. [4] Sol. If B is inverse of A then
AB = I det(AB) = det(I)
det(A).det(B) = 1 Given det (A) = –1 then det (B) = –1
13
120125
= –1
5(–2 –3) – 2[0 – ] + 1 [–2] = –1 22 – 2 – 25 = – 1 22 – 2 – 24 = 0 ( – 4)( + 3) = 0 ; = 4, – 3 Sum of value of = 4 – 3 = 1 Q.21 A 2 m ladder leans against a vertical wall. If the top of the ladder begins to slide down the wall at the rate
25 cm/sec, then the rate (in cm/sec.) at which the bottom of the ladder slides away from the wall on the horizontal ground when the top of the ladder is 1 m above the ground is :
(1) 3
25 (2) 25 (3) 325 (4) 3
25
Ans. [4] Sol. x2 + y2 = 4 Diff. w.r.t. t
2x dtdx + 2y
dtdy = 0
2m
x
dtdx =2
s/cm25dtdy
y
At y = 1; x = 3 then
sec/cm3
25dtdx
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Q.22 The equation y = sinx sin (x + 2) – sin2 (x + 1) represents a straight line lying in : (1) first, second and fourth quadrants (2) first, third and fourth quadrants (3) second and third quadrants only (4) third and fourth quadrants only
Ans. [4] Sol. y = sinx sin(x + 2) – sin2(x + 1)
y = )]2xsin(xsin2[21
– )]1x(sin2[21 2
y = )]2x2cos()2[cos(21
– )]1x(2cos1[21
y = )]12[cos21
y = (–) 1sin)2(21 2
y = – sin21
y = – sin2l Graph IIIrd & IVth Q.23 If the truth value of the statement p (~q r) is false (F), then the truth values of the statements p, q, r are
respectively : (1) T, F, T (2) F, T, T (3) T, T, F (4) T, F, F
Ans. [3] Sol. p (~ q r) is false It is true when p T & (~ q r) = false It will true : ~ q false & r false ~q F | r F q T Truth value of p, q, r T, T, F
Q.24 Let a random variable X have a binomial distribution with mean 8 and variance 4. If P(X < 2) = ,2k16 then k
is equal to : (1) 17 (2) 1 (3) 137 (4) 121
Ans. [3] Sol. Given : np = 8 ; nqp = 4
q = 21 p =
21 n = 16
P(x 2) = P(x = 0) + P(x = 1) + (x = 2)
= 16C0 016
21
21
+ 16C1
15
21
1
21
+ 16C2
14
21
2
21
=
16
21
[1 + 16 + 120] = 162
137
k = 137
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Q.25 Let Sn denote the sum of the first n terms of an A.P. If S4 = 16 and S6 = – 48, then S10 is equal to : (1) – 320 (2) – 380 (3) – 410 (4) – 260
Ans. [1] Sol. Sn = Sum of n terms of an A.P. S4 = 16 = a + 3d …..(i) S6 = –48 = a + 5d …..(ii) From (i) & (ii) d = – 32 a = 112
S10 = 2
10 [2.(112) + (10 – 1)(–32)]= 5[–64]
S10 = – 320
Q.26 Let f : R R be a continuously differentiable function such that f(2) = 6 and f(2) = .481 If
f(x)
63 ),x(g)2x(dtt4 then )x(gLim
2xis equal to :
(1) 18 (2) 36 (3) 12 (4) 24 Ans. [1]
Sol. )x(f
6
3 dt.t4 (x – 2)g(x) ; f(2) = 6 ; f '(2) = 481
g(x) = )2x(
dtt4)x(f
6
3
2x
lim
)2x(
dtt4)x(f
6
3
at x = 2
00 form
L-hospital rule
2x
lim
1
)x('f.)]x(f[4 3
At x = 2, 4[f(2)]3.f '(2)
4(6)3
481 = 18
Q.27 The value of sin–1
1312
53sin 1 is equal to :
(1)
6563sin 1 (2)
659cos
21 (3)
6533cos 1 (4)
6556sin
21
Ans. [4]
Sol. sin–1
1312 – sin–1
53
= sin–1
1691441
53
2591
1312
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= sin–1
6533
= cos–1
6556
= 2 – sin–1
6556
Q.28 If the volume of parallelopiped formed by the vectors kj,kji and ki is minimum, then is
equal to :
(1) 3
1 (2) 3 (3)
31 (4) 3
Ans. [3] Sol. Volume of parallelopiped
10
1011
= 1[1 – 0] – [– 2] + 1 [–]
V = |1 + 3 – |
13ddV
0ddV
= 3
1 , –3
1
6d
Vd2
2 for minimum
2
2
dVd
> 0 at = 3
1
Q.29 If three of the six vertices of a regular hexagon are chosen at random, then the probability that the triangle
formed with these chosen vertices is equilateral is :
(1) 101 (2)
103 (3)
203 (4)
51
Ans. [1] Sol. Number of total triangle = 6C3
Equilateral = 2
Prob. = 3
6C2 =
101
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Q.30 If the line 11z
21y
32x
intersects the plane 2x + 3y – z+13 = 0 at a point P and the plane
3x + y + 4z = 16 at a point Q, then PQ is equal to :
(1) 72 (2) 14 (3) 142 (4) 14
Ans. [3]
Sol.
11z
g1y
32x
General point (3 + 2, 2 – 1, – + 1) = intersect plane 2x + 3y – z + 13 = 0 at P then 2(3+ 2) + 3 (2 – 1) – (– + 1) + 13 = 0 = – 1 P(– 1, – 3, 2) Line intersect plane : 3x + y + 4z = 16 at Q then 3(3 + 2) + 2 – 1 + 4(– +1) = 16 = 1
Q(5, 1, 0) then PQ = 222 )20()31()15(
PQ = 2 14
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