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JEE Main Online Exam 2019
Questions & Solutions 12th April 2019 | Shift - II
MATHEMATICS
Q.1 Let a
2,0 be fixed. If the integral
tan–xtantanxtan dx = A(x) cos 2 + B(x) sin 2 + C, where C is a
constant of integration, then the functions A(x) and B(x) are respectively : (1) x – and loge|cos(x – )| (2) x + and loge|sin(x – )| (3) x + and loge|sin(x + )| (4) x – and loge|sin(x – )|
Ans. [4]
Sol. dx
sinxcos–cosxsinsinxcoscosxsin
= dx
)–xsin()xsin(
= dx)–xsin(
)2–xsin(
= dx)–xsin(
2cos)–xsin(
+ dx)–xsin(
2sin)–xcos(
= (x – ) cos2 + sin 2 log |sin(x – )| + C Q.2 The general solution of the differential equation (y2 – x3)dx – xydy = 0 (x 0) is : (where c is a constant of integration)
(1) y2 + 2x2 + cx3 = 0 (2) y2 + 2x3 + cx2 = 0 (3) y2 – 2x + cx3 = 0 (4) y2 – 2x3 + cx2 = 0
Ans. [2] Sol. (y2 – x3)dx – xydy = 0 (x 0)
y2 – x3 – xydxdy = 0
xydxdy – y2 = –x3
ydxdy –
x1 y2 = –x2 … (i)
Let y2 = v 2ydxdy =
dxdv
put in eqn (i)
21
dxdv – v
x1 = –x2
dxdv + v
x2–
= –2x2 …(ii)
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I.F. = dxx2–
e = xn2–e = 2x1
solution of eqn (ii)
v × 2x1 =
dx
xx2– 2
2 – C
2xv = – 2x – C
y2 = –2x3 – cx2
y2 + 2x3 + cx2 = 0 Q.3 The equation of common tangent to the curves y2 = 16x and xy = –4, is :
(1) x – y + 4 = 0 (2) x + y + 4 = 0 (3) x – 2y + 16 = 0 (4) 2x – y + 2 = 0 Ans. [1]
Sol. y = mx + m4 is always tangent to y2 = 16x … (i)
if it is tangent to the xy = –4
x
m4mx = –4
m2x2 + 4x = –4m m2x2 + 4x + 4m = 0 for tangent D = 0 16 – 16m3 = 0 m = 1 put in eqn (i) y = x + 4 Q.4 A plane which bisects the angle between the two given planes 2x – y + 2z – 4 = 0 and x + 2y + 2z – 2 = 0,
passes through the point : (1) (1, –4, 1) (2) (1, 4, –1) (3) (2, 4, 1) (4) (2, –4, 1)
Ans. [4] Sol. Eqn of angle bisectors are
222 2)1(–2
4–z2y–x2
= ±
222 z21
2–z2y2x … (i)
Case I : take positive sign 2x – y + 2z – 4 = x + 2y + 2z – 2 x – 3y – 2 = 0 … (ii) Case-II : take negative sign
2x – y + 2z – 4 = –(x + 2y + 2z – 2)
2x – y + 2z – 4 = –x – 2y – 2z + 2
3x + y + 4z – 6 = 0 … (iii)
option (4) satisfy eqn (iii)
(2, –4, 1)
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Q.5 A triangle has a vertex at (1, 2) and the mid points of the two sides through it are (–1, 1) and (2, 3). Then the
centroid of this triangle is :
(1)
2,
31 (2)
35,
31 (3)
37,1 (4)
1,
31
Ans. [1] Sol.
A(1, 2)
F(2, 3)E(–1, 1)
B (x2, y2)
C (x3, y3)
21x2 = –1,
22y2 = 1
21x3 = 2 &
22y3 = 3
x2 = –3, y2 = 0 x3 = 3, y3 = 4
B(–3, 0) C(3, 4)
centroid
3
yyy,
3xxx 321321
3402,
333–1 =
2,
31
Q.6 The Boolean expression ~(p (~q)) is equivalent to :
(1) p q (2) (~p) q (3) q ~p (4) p q
Ans. [1]
Sol. – (p ~q) = p q
Q.7 Let A, B and C be sets such that A B C. Then which of the following statements is not true ?
(1) If (A – B) C, then A C (2) B C
(3) (C A) (C B) = C (4) If (A – C) B, then A B
Ans. [4] Sol. Let A = {1, 2, 3, 4} B = {3, 4, 5, 6} C = {1, 2, 3, 4, 7, 8}
Here A B = {3, 4} C
A – C = B
but A B
So not true (wrong) statement is 4th
If A – C B then A B
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Q.8 If the area (in sq. units) bounded by the parabola y2 = 4x and the line y = x, > 0, is 91 , then is equal to :
(1) 34 (2) 62 (3) 48 (4) 24 Ans. [4] Sol.
y
x
y2 = 4x & y = x 2x2 = 4x
x = 0 & x = 4
Area =
/4
0
dx)x–x4( = 91
2 ×
/4
0
2/3
23
x –
/4
0
2
2x =
91
34
× 2/3
2/32 x)2(
– 2 × 2
16
= 91
3
32 – 8 =
91
38 =
91 = 24
Q.9 If , and are three consecutive terms of a non-constant G.P. such that the equations x2 + 2x + = 0 and
x2 + x – 1 = 0 have a common root, then ( + ) is equal to : (1) (2) 0 (3) (4)
Ans. [4] Sol. ,, are in G.P. x2 + 2x + = 0 & x2 + x – 1 = 0 have a common roots. Both roots will be common
1 =
12 =
1– =
= , B = 2 , = –
( + ) =
–2
= 2
– 2 =
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Q.10 If 20C1 + (22) 20C2 + (32) 20C3 + ..... + (202)20C20 = A(2), then the ordered pair (A, ) is equal to : (1) (420, 19) (2) (420, 18) (3) (380, 18) (4) (380, 19)
Ans. [2] Sol. (1 + x)20 = 20C0 + 20C1x + 20C2x2 + ........ + 20C20x20 … (i) different eqn (i) w.r.t. x 20(1 + x)19 = 20C1.1 + 2.20C2x + ......... + 2020C20x19 … (ii) Multiply eqn (ii) by x 20x(1 + x)19 = 20C1.x+ 2.20C2x2 + ........ + 2020C20x20 … (iii) diff. eqn (iii) w.r.t. x 20[(1 + x)19 + 19x(1 + x)18] = 1.20C1 + 22.20C2x + ........ + (202)20C20x19 … (iv) put x = 1 in eqn (iv) 20(219 + 19.218) = 12.20C1 + 22.20C2 + ........ + (202) 20C20
= 20 × 218(2 + 19) = 20 × 21 × 218 = 420 × 218 A = 420, = 18
Q.11 The length of the perpendicular drawn from the point (2, 1, 4) to the plane containing the lines
r = ( ji ) + ( k–j2i ) and
r = ( ji ) + µ(– k2–ji ) is :
(1) 31 (2)
31 (3) 3 (4) 3
Ans. [4] Sol. Equation of plane containing both lines is
2–11–1–21
z1–y1–x = 0
(x – 1) (–4 + 1) + (y – 1) (1 + 2) + z (1 + 2) = 0 –3(x – 1) + 3(y – 1) + 3z = 0 –x + 1 + y – 1 + z = 0 –x + y + z = 0 distance from point (2, 1, 4) is
222 111
412–
= 3
Q.12 A circle touching the x-axis at (3, 0) and making an intercept of length 8 on the y-axis passes through the
point : (1) (1, 5) (2) ( 2, 3) (3) (3, 5) (4) (3, 10)
Ans. [4] Sol. Equation of required circle will be (x – 3)2 + (y ± r)2 = r2
x2 – 6x + 9 + y2 ± 2ry + r2 = r2 x2 + y2 – 6x ± 2ry + 9 = 0 …(i)
Length of y intercept = 2 c–f 2 f = ±r
8 = 9–r2 2 16 = r2 – 9 r = 5
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So eqn of required circle will be x2 + y2 – 6x ± 10y + 9 = 0
two circles
x2 + y2 – 6x + 10y + 9 = 0 … (ii)
x2 + y2 – 6x – 10y + 9 = 0 … (iii)
option 4th (3, 10) satisfy eqn (iii)
Q.13 A person throws two fair dice. He wins Rs. 15 for throwing a doublet (same numbers on the two dice), wins Rs. 12 when the throw results in the sum of 9, and loses Rs. 6 for any other outcome on the throw. Then the expected gain/loss (in Rs.) of the person is :
(1) 41 loss (2)
21 gain (3)
21 loss (4) 2 gain
Ans. [3] Sol.
win +15 +12 –6
Prob. 366
364
3626
Probability of doublet = 366
Probability of sum of 9 = 364
Other probability = 3626
Expected gain/loss = 15 × 366 + 12 ×
364 – 6 ×
3626
= 3690 +
3648 –
36156 =
21–
21–
So, 21 loss
Q.14 Let R and the three vectors a = k3ji ,
b = k–ji2 and
c = k3j2–i . Then the set
S = { : a ,
b and
c are coplanar}
(1) contains exactly two numbers only one of which is positive (2) is singleton (3) contains exactly two positive numbers (4) is empty
Ans. [4]
Sol. [a
b
c ] = 0
32–
–1231
= 0
(3 – 2) + 1 (– 2 – 6) + 3(– 4 – ) = 0 3 – 22 – 2 – 6 – 12 – 3 = 0 –32 – 18 = 0 2 + 6 = 0 not possible for real S is empty set
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Q.15 The tangents to the curve y = (x – 2)2 – 1 at its points of intersection with the line x – y = 3, intersect at the point :
(1)
1–,
25 (2)
1–,
25– (3)
1,
25 (4)
1,
25–
Ans. [1] Sol. x – y – 3 = 0 … (i) will be chord of contact of parabola y = x2 – 4x + 3 Let the required point is P(x1, y1) chord of contact for point P is
2
yy 1 = xx1 – 42
)xx( 1 + 3
y + y1 = 2x1x – 4x – 4x1 + 6 (2x1 – 4)x – y + (–4x1 – y1 + 6) = 0 … (ii) eqn (i) & (ii) are same line
1
4–x2 1 =1–1– =
3–6y–x4– 11
2x1 – 4 = 1 – 4x1 – y1 + 6 = –3
x1 = 25 –10 – y1 + 9 = 0
y1 = –1
Ans.
1–,
25
Q.16 If [x] denotes the greatest integer x, then the system of linear equations [sin ]x + [–cos]y = 0, [cot]x + y = 0
(1) has a unique solution if
32,
2 and have infinitely many solutions if
67,
(2) have infinitely many solutions if
32,
2 and has a unique solution if
67,
(3) have infinitely many solutions if
32,
2
67,
(4) has a unique solution if
32,
2
67,
Ans. [2] Sol. [sin ]x + [–cos]y = 0 … (i) [cot]x + y = 0 … (ii) Case-I :
When
32,
2 sin
1,
23
cos
0,
21– – cos
21,0
cot
0,
31–
[sin ] = 0, [– cos] = 0, [cot] = –1 eqn (i) & (ii) will
0yx–0y0x0
system will have infinitely many solution
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Case-II :
When
67, sin
0,
21–
cos
23–,1–
cot ),3(
[sin] = –1, [cos] = –1 [cot] = {1,2,3,.....} –x – y = 0 Ix + y = 0 I = {1, 2, .....} It will have unique solution is all cases x = 0, y = 0 Q.17 A group of students comprises of 5 boys and n girls. If the number of ways, in which a team of 3 students can
randomly be selected from this group such that there is at least one boy and at least one girl in each team, is 1750, then n is equal to :
(1) 24 (2) 25 (3) 27 (4) 28 Ans. [2] Sol. Given 5 boys and n girls total ways of forming team of 3 member under given condition = 5C1. nC2 + 5C2. nC1 5C1 . nC2 + 5C2 . nC1 = 1750
2
)1–n(n5 + 10n = 1750
2
)1–n(n + 2n = 350
n2 + 3n = 700 n2 + 3n – 700 = 0 n = 25 Q.18 An ellipse, with foci at (0, 2) and (0, –2) and minor axis of length 4, passes through which of the following
points ?
(1) (2, 2 ) (2) (2, 22 ) (3) ( 2 , 2) (4) (1, 22 ) Ans. [3] Sol. Given 2a = 4 and 2be = 4 a = 2, be = 2 b2e2 = 4 b2 – a2 = 4 b2 = 8 equation of ellipse
4
x2 +
8y2
= 1
Clearly option (3) satisfy the given curve.
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Q.19 For an initial screening of an admission test, a candidate is given fifty problems to solve. If the probability
that the candidate solve any problem is 54 , then the probability that he is unable to solve less than two
problems is :
(1) 48
51
25164
(2)
48
54
25316
(3)
49
51
5201
(4)
49
54
554
Ans. [4] Sol. Total problems = 50
P(Solving) = 54
P(Not solving) = 51
P(unable to solve less than two problems) = P(not solving one problem) + P(not solving zero problem)
= 50C0 0
51
50
54
+ 50C1
1
51
49
54
= 50
50
54 + 50. 49
49
5.54
= 50
54
+ 10.
49
54
= 49
54
+
10
54
= 5
54 . 49
54
Q.20 A straight line L at a distance of 4 units from the origin makes positive intercepts on the coordinate axes and
the perpendicular from the origin to this line makes an angle of 60º with the line x + y = 0. Then an equation of the line L is :
(1) x + 3 y = 8 (2) 3 x + y = 8
(3) ( 3 + 1)x + ( 3 – 1)y = 28 (4) ( 3 – 1)x + ( 3 + 1)y = 28 Ans. [3,4] Sol.
A
P
O
B
4
OP = 4 Given OP makes 60º with x + y = 0
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let slope of OP = m
tan60º = m–11m
1–m1m = 3 or – 3
m + 1= m3 – 3 or m + 1 = 3 – 3 m
m( 3 – 1) = 3 + 1 or m( 1 + 3 ) = 3 – 1
m = 1–313 or m =
131–3
tan = 1–313 or tan =
131–3
eqn of line x cos + y sin = P ( 3 + 1)x + ( 3 – 1)y = 28 or ( 3 – 1)x + ( 3 + 1)y = 28
Q.21 A value of such that
1
)1x)(x(dx = loge
89 is :
(1) 2 (2) –2 (3) 21 (4)
21–
Ans. [2]
Sol.
1
)1x)(x(dx = loge
89
1
dx)1x)(x()x(–)1x( = loge
89
1
xdx –
1
1xdx = loge
89
loge
1
1xx
= loge
89
loge
2212 – log
122 = loge
89
log
212
2212 = loge
89
)1(4
)12( 2
=
89
8[42 + 4 + 1] = 9[42 + 4] 322 + 32 + 8 = 362 + 36 42 + 4 – 8 = 0 2 + – 2 = 0 ( + 2) ( – 1) = 0 = 1, –2
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Q.22 1x–xsin–1xsin2x
xsin2xlim220x
is :
(1) 6 (2) 1 (3) 3 (4) 2 Ans. [4]
Sol. 1x–xsin–1xsin2x
xsin2xlim220x
= 1–xxsin–1xsin2x
xsin2xlim 220x
× )1x–xsin1xsin2x( 22
= xxsin–xsin2x
xsin2xlim 220x
× (2)
Applying L’H Rule
= 1xcosxsin2–xcos2x2
)xcos21(2lim0x
= 12)3(2
= 2
Q.23 The derivative of tan–1
xcosxsinxcos–xsin , with respect to
2x , where
2
,0x is :
(1) 32 (2) 1 (3) 2 (4)
21
Ans. [3]
Sol. Given y = tan–1
xcosxsinxcos–xsin
y = tan–1
1xtan1–xtan
y = –tan–1
xtan1xtan–1
y = –tan–1
x–
4tan
0 < x < 2
–2 < –x < 0
–4 <
4 – x < 0
y = –
x–
4
2
,2
–xxxtantan 1–
y = x –4
)2/x(ddy =
)2/1(1 = 2
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Q.24 Let S be the set of all R such that the equation, cos2x + sinx = 2– 7 has a solution. Then S is equal to : (1) [2, 6] (2) [3, 7] (3) [1, 4] (4) R
Ans. [1] Sol. Given cos2x + sinx = 2 – 7 1 – 2sin2x + sinx = 2 – 7 2sin2x – sinx + 2 – 8 = 0
sin x =4
)8–2(8–2
sin x =4
)8–(
sin x = 4
8– , 4
8–
sin x = 2 (Not possible) for solution
–1 4
8–2 1
–4 2a – 8 4 4 2 12 [2, 6]
Q.25 Let z C with Im(z) = 10 and it satisfies nz2n–z2
= 2i – 1 for some natural number n. Then :
(1) n = 20 and Re(z) = –10 (2) n = 40 and Re(z) = 10 (3) n = 40 and Re(z) = –10 (4) n = 20 and Re(z) = 10
Ans. [3] Sol. Let z = x + 10i
given nz2n–z2
= 2i – 1
n)i10x(2n–)i10x(2
= 2i – 1
(2x – n) + 20i = (2i – 1) [(2x + n) + 20i] Comparing real and imaginary part 2x – n = 2(–20) – (2x + n) and 20 = 2(2x + n) – 20 2x – n = –40 – 2x – n and 20 = 4x + 2n – 20 4x = –40 and 4x + 2n = 40 x = –10 and –40 + 2n = 40 n = + 40 n = 40 and Re(z) = –10 Q.26 Let f(x) = 5 – |x – 2| and g(x) = |x + 1|, x R. If f(x) attains minimum value at , then
–xlim
8x6–x)6x5–x)(1–x(
2
2
is equal to :
(1) 21 (2)
21– (3)
23 (4) –
23
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Ans. [1] Sol. f(x) = 5 – |x – 2| f(x) attains maximum value when |x – 2| = 0 x = 2 = g(x) = |x + 1| g(x) attains minimum value of x = –1 =
–x
lim8x6–x
)6x5–x)(1–x(2
2
= )4–x)(2–x(
)3–x)(2–x)(1–x(lim2x
= )4–2(
)3–2)(1–2( = 21
Q.27 The angle of elevation of the top of a vertical tower standing on a horizontal plane is observed to be 45º from
a point A on the plane. Let B be the point 30 m vertically above the point A. If the angle of elevation of the top of the tower from B be 30º, then the distance (in m) of the foot of the tower from the point A is :
(1) 15(1 + 3 ) (2) 15(3 – 3 ) (3) 15(3 + 3 ) (4) 15(5 – 3 ) Ans. [3] Sol.
P
A45º
30º B
M
N
AB = 30 m = NP In ANM
tan45º = ANMN = 1
MN = AN PM = MN – 30 = AN – 30 In BPM
tan30º = PBPM =
AN30–AN
3
1 = AN
30–AN
AN = 3 AN – 330
AN = 1–3
330 = 2
)13(330 = 15 ( 33 )
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Q.28 A value of
3,0 , for which
6cos41sincos6cos4sin1cos6cos4sincos1
22
22
22
= 0, is :
(1) 18 (2)
9 (3)
247 (4)
367
Ans. [2]
Sol.
3,0
6cos41sincos6cos4sin1cos6cos4sincos1
22
22
22
= 0
R2 R2 – R1, R3 R3 – R1
101–011–
6cos4sincos1 22 = 0
C1 C1 + C2
101–010
6cos4sin2 2 = 0
Expanding along first column 2[1 – 0] – 1[– 4cos6] = 0 2 + 4cos6 = 0
cos6 = –21
6 = 3
2 = 9
Q.29 If a1, a2, a3, ..... are in A.P. such that a1 + a7 + a16 = 40, then the sum of the first 15 terms of this A.P. is :
(1) 120 (2) 150 (3) 280 (4) 200 Ans. [4] Sol. a1, a2, a3, ......, an are in A.P. a1 + a7 + a16 = 40 a + a + 6d + a + 15d = 40 3a + 21d = 40
a + 7d = 3
40
S15 =2
15 [2a + 14d]
= 15[a + 7d]
= 15 × 3
40
= 200
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Q.30 The term independent of x in the expansion of
81x–
601 8
.6
22
x3–x2
is equal to :
(1) –36 (2) –108 (3) 36 (4) –72 Ans. [1]
Sol.
81x–
601 8
.6
22
x3–x2
term independent of x will be
601 × term independent of x in
6
22
x3–x2
–
81 × term of z–8 in
6
22
x3–x2
Tr+1 in 6
22
x3–x2
will be
Tr+1 = 6Cr(2x2)6 – rr
2x3–
= 6Cr 26 – r (–1)r × 3r × x12 – 2r – 2r Case-I : For term independent of x is 12 – 4r = 0 r = 3 T4 = – 6C3 × 23 × 33x6 = –20 × 23 × 33 Case-II : For term of x–8 12 – 4r = –8 4r = 20 r = 5 T6 = 6C5. 21. (–1). 35 . x–8
Required ans. =601 × (–20)23 × 33 –
811 × 6 × 2 × (–1) × 35
= –72 + 36 = –36
