Allen: Test Paper of Jee (Main) Examination - 2019
Allen: Test Paper of Jee (Main) Examination - 2019
Allen: Test Paper of Jee (Main) Examination - 2019
Ans (3)
A = {–2, 2, 3} Sol. DR's of line are 2, 1, –2
B = {x Î Z : –3 < 2x – 1 < 9}
normal vector of plane is ˆi – 2ˆj – kkˆ
B = {0, 1, 2, 3, 4}
2.
A × B has is 15 elements so number of subsets
of A × B is 215.
If sin a + 4cos b + 2 = 4 2 sin a cos b ;
4 4
EN sin a =
(2iˆ + ˆj – 2k).(i
ˆ ˆ – 2ˆj – kk)
3 1 + 4 + k2
ˆ
2k
a, b Î [0, p ] , then cos(a + b) – cos(a – b) is sin a = .......(1)
equal to :
3 k2 + 5
given that sin4 a + 4 cos4 b + 2 = 4 2 sina cosb the coordinate axes is bisected at P, then its
Þ A.M. = G.M. Þ sin4 a = 1 = 4 cos4 b equation is :
(1) x – y + 7 = 0
1
sin a = 1, cos b = ± (2) 3x – 4y + 25 = 0
2
(3) 4x + 3y = 0
1 (4) 4x – 3y + 24 = 0
Þ sin b = as b Î [0, p]
2 Ans (4)
cos (a + b) – cos (a – b) = –2 sin a sin b
=– 2
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JEE ( Main ) Examination–2019/Evening Session/12-01-2019
6. There are m men and two women participating in
a chess tournament. Each participant plays two
B(0, b) games with every other participant. If the number
P(–3, 4)
of games played by the men between themselves
Sol. A(a, 0) exceeds the number of games played between the
men and the women by 84, then the value of m
is :
(1) 9 (2) 11 (3) 12 (4) 7
x y Ans (3)
Let the line be =1
a b Sol. Let m-men, 2-women
mC × 2 = mC1 2C1 . 2 + 84
a b 2
(–3, 4) = ,
2 2 m2 – 5m – 84 = 0 (m – 12) (m + 7) = 0
a = –6, b = 8 m = 12
equation of line is 4x – 3y + 24 = 0 7. If the function f given by f(x) = x3 –3(a – 2)x2 +
3ax + 7, for some a R is increasing in (0, 1] and
3x13 2x11 decreasing in [1, 5), then a root of the equation,
5. The integral 4
dx is equal to :
2x 4 3x 2 1 f (x) 14
2
0(x 1) is :
(where C is a constant of integration) x 1
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JEE ( Main ) Examination–2019/Evening Session/12-01-2019
9. The tangent to the curve y = x2 – 5x + 5, parallel 11. If a circle of radius R passes through the origin O
to the line 2y = 4x + 1, also passes through the point. and intersects the coordinate axes at A and B, then
the locus of the foot of perpendicular from O on
1 7 7 1 AB is :
(1) , (2) ,
4 2 2 4 (1) (x2 + y2)2 = 4Rx2y2
(2) (x2 + y2)(x + y) = R2xy
1 1 (3) (x2 + y2)3 = 4R2x2y2
(3) ,7 (4) , 7
8 8 (4) (x2 + y2)2 = 4R2x2y2
Ans (4) Ans (3)
y
Sol. y = x2 – 5x + 5
dy 7
= 2x – 5 = 2 x= B
dx 2
Sol. P(h, k)
7 –1 x
at x = ,y= A
2 4
7 –1 29 –h
Equation of tangent at , is 2x – y – =0 Slope of AB =
2 4 4 k
Equation of AB is hx + ky = h2 + k2
Now check options
h2 k2 h2 k2
1 A ,0 , B 0,
x = , y = –7 h k
8
10. Let S be the set of all real values of such that AB = 2R
a plane passing through the points (– 2, 1, 1), (h2 + k2)3 = 4R2h2k2
(1, – 2, 1) and (1, 1, – 2) also passes through the (x2 + y2)3 = 4R2x2y2
point (–1, –1, 1). Then S is equal to : 12. The equation of a tangent to the parabola,
x2 = 8y, which makes an angle with the positive
(1) 3 (2) 3 3 direction of x-axis, is :
(1) x = ycot + 2tan (2) x = ycot – 2tan
(3) {1 , –1} (4) {3, –3} (3) y = xtan – 2cot (4) y = xtan + 2cot
Ans (2) Ans (1)
Sol. All four points are coplaner so Sol. x2 = 8y
2 dy x
1– 2 0 = tan
2 – 2
1 0 0 dx 4
2 x1 = 4tan
2 2 – –1
y1 = 2 tan2
( 2 + 1)2 (3 – 2) =0 Equation of tangent :-
y – 2tan2 = tan (x – 4tan )
=± 3
x = y cot + 2 tan
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JEE ( Main ) Examination–2019/Evening Session/12-01-2019
13. If the angle of elevation of a cloud from a point
n n n 1
P which is 25 m above a lake be 30º and the angle 15. lim 2 2 2 2 2 2
....
n n 1 n 2 n 3 5n
of depression of reflection of the cloud in the lake
from P be 60º, then the height of the cloud (in is equal to :
meters) from the surface of the lake is :
(1) 42 (2) 50 (3) 45 (4) 60 (1) (2) tan–1(2)
4
Ans (2)
cloud
(3) tan–1(3) (4)
x 2
P 30° Ans. (2)
60° y
25m 25m 2n
n
lim 2
Sol. surface
x
r 1 n r2
x + 25m 2n
1 2
lim dx
x 2
tan 1 2
r 1
n 1 r2 =
1 x2
0
x n
tan 30° = y= 3x .....(i)
y 16. The set of all values of for which the system of
25 x 25 linear equations.
tan 60° = x – 2y – 2z = x
y
x + 2y + z = y
3 y = 50 + x
–x – y = z
3x = 50 + x
has a non-trivial solution.
x = 25 m
(1) contains more than two elements
Height of cloud from surface = 25 + 25 = 50m
(2) is a singleton
e 2x x
x e (3) is an empty set
14. The integral log e xdx is equal
1
e x (4) contains exactly two elements
to :
Ans. (2)
1 1 3 1 1
(1) e 2 (2)
2 e 2 e 2e2 –1 2 2
1 1 1 3 1 1 2– 1 0 ( – 1)3 0 =1
(3) (4) e 1 1 1
2 e 2e 2 2 2e 2
Ans. (4)
17. If nC4, nC5 and nC6 are in A.P., then n can be:
e 2x e
x e (1) 14 (2) 11 (3) 9 (4) 12
log e x.dx log e x.dx
1 e 1 x Ans. (1)
x
x
2x
e 2.nC5 = nC4 + nC6
Let t, v
e x n n n
1
1 1 2. 5 n 5 4n 4 6n 6
= dt dv
2 1
2 e
e 2 1 1 1
.
1 1 3 1 5 n 5 (n 4)(n 5) 30
= 1 2 (1 e) – –e
2 e 2 2e 2
n = 14 satisfying equation.
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JEE ( Main ) Examination–2019/Evening Session/12-01-2019
lim 2 1 2
1 sin 1 · =
0 2
19. If A sin 1 sin ; then for all 2 sin
2
–1 sin 1
21. The expression ~( p q) is logically equvalent to :
3 5 (1) ~ p ^ ~ q (2) p ^ q
, , det(A) lies in the interval :
4 4 (3) ~ p ^ q (4) p ^ ~ q
Ans. (1)
5 3
(1) ,4 (2) ,3
2 2 p q ~p ~p q ~(~p q) (~p ^ ~q)
T T F T F F
3 5
(3) 0, (4) 1, F T T T F F
2 2
T F F T F F
Ans (2) F F T F T T
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JEE ( Main ) Examination–2019/Evening Session/12-01-2019
23. The mean and the variance of five observation are Ans. (3)
4 and 5.20, respectively. If three of the mSB . mS'B = –1
observations are 3, 4 and 4; then then absolute
value of the difference of the other two
observations, is :
45º 45º
(1) 1 (2) 3 (3) 7 (4) 5 (–ae, 0) (ae,0) x
Ans. (3)
2
mean x = 4, = 5.2, n = 5,. x1 =3 x2 = 4 = x3
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JEE ( Main ) Examination–2019/Evening Session/12-01-2019
27. The number of integral values of m for which the 29. If a cuver passes through the point (1, –2) and has
quadratic expression. slope of the tangent at any point (x, y) on it as
(1 + 2m)x2 – 2(1 + 3m)x + 4(1 + m), x R, is
x 2 2y
always positive, is : , then the curve also passes through the
x
(1) 8 (2) 7 (3) 6 (4) 3
Ans. (2) point :
Exprsssion is always positve it (1) 2,1 (2) 3,0
1 (3) (–1, 2) (4) (3, 0)
2m + 1 > 0 m>–
2 Ans. (2)
&
dy x2 2y
D<0 m2 – 6m – 3 < 0 (Given)
dx x
3 12 m 3 12 .... (iii)
Common interval is
dy y
2 x
dx x
3– 12 < m < 3 12
2
dx
Intgral value of m {0,1,2,3,4,5,6} x
I.F = e = x2
28. In a game, a man wins Rs. 100 if he gets 5 of 6
on a throw of a fair die and loses Rs. 50 for getting
y.x2 = x.x 2 dx C
any other number on the die. If he decides to throw
the die either till he gets a five or a six or to a
maximum of three throws, then his expected gain/ x4
loss (in rupees) is : = C
y
400 400
(1) gain (2) loss 9
3 3 hence bpasses through (1, –2) C=
4
400
(3) 0 (4) loss x4 9
9 yx2 = –
4 4
Ans. (3)
Expected Gain/ Loss = Now check option(s) , Which is satisly by op-
tion (ii)
= w × 100 + Lw (–50 + 100) + L2w (–50 –50 +
30. Let Z1 and Z2 be two complex numbers satisfying
100) + L3 (–150)
|Z1| = 9 and |Z2–3–4i|=4 . Then the minimum value
2
1 2 1 2 1 of |Z1–Z2| is :
= 100 . 50 + 0 +
3 3 3 3 3 (1) 0 (2) 1 (3) 2 (4) 2
3
Ans. (1)
2
150 = 0 z1 9 , z2 3 4i 4
3
here w denotes probability that outcome 5 or 6 ( C1 (0, 0) radius r1 = 9
C2(3, 4), radius r2 = 4
2 1 C1C2 = |r1–r2 | = 5
w= )
6 3 Circle touches internally
here L denotes probability that outcome z1 z2 0
min
4 2
1,2,3,4 ( L = )
6 3
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