JEE Main 2019 Paper 2 B.Arch Attempt Shift - 1 (07th April, 2019)
JEE Main 2019 Paper 2 B.Arch Attempt Shift - 1 (07th April, 2019)
JEE Main 2019 Paper 2 B.Arch Attempt Shift - 1 (07th April, 2019)
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PART : MATHEMATICS
Straight Objective Type (lh/ks oLrqfu"B izdkj)
This section contains 30 multiple choice questions. Each question has 4 choices (1), (2), (3) and (4) for its
answer, out of which Only One is correct.
bl [k.M esa 30 cgq&fodYih iz'u gSaA izR;sd iz'u ds 4 fodYi (1), (2), (3) rFkk (4) gSa] ftuesa ls flQZ ,d lgh gSA
1. If a, b and c (all distance) are the sides of a triangle ABC opposite to the angles A, B and C,
c sin(A B) b sin(C A)
respectively, then is equal to :
a 2 b2 c 2 a2
(1) 1 (2) 2 (3) –1 (4) 0
Ans. (4)
C sin A cosB – cos A sinB b sinCcos A – cosC sin A
Sol. –
a2 – b2 c 2 – a2
a a2 c 2 – b2 b b2 c 2 – a2 C b2 c 2 – a2 a a2 b2 – c 2
– –
c 2ac 2bc
– b
2bc 2ab
=
2R a2 – b2 2R c 2 – a2
1 a c – b – b – c a b c – a – a – b c
2 2 2 2 2 2 2 2 2 2 2 2
1
= – = [2 – 2] = 0
4R a –b
2 2
c –a
2 2
4R
2. Let P be the point on the parabola y2 = 3x such that OP makes an angle of with the x-axis, where O
6
is the origin. A normal is drawn to the parabola at P intersecting the axis, of the parabola at Q. If S is the
focus of the parabola, then SQ is equal to :
39 39 41
(1) (2) (3) 9 (4)
4 2 4
Ans. (1)
3 3
P t2, t
4 2
Sol. 3
S ,0
4
3
4a=3 a=
4
3 3
Let P t 2 , t
4 2
3t 4 1
2 2 =
3t 3
2 1
= t= 2 3
t 3
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| JEE MAIN-2019 | DATE : - 07 04-2019 (SHIFT-1) | PAPER-2 | B.ARCH | MATHEMATICS
3 3
Point P 4 3, 2 3
4 2
P (9, 3 3 )
equation of normal is y–3 3 = –t(x – 9)
y – 3 3 = –2 3 (x – 9)
For Pt Q, Put y = 0
3 3 21 21 3 39
=x–9 x=9+ = SQ = – = .
2 2 2 2 4 4
tan 3 = =
1– 3 tan 2 1– 3 tan2
1 3 – 1
tan 3 =
1– 3
tan 3 = 1
tan23 = 1
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5. Let A be the set of all 3-digit natural number and B = {xA : H.C.F. (x, 15) = 1}. Then the number of
elements in B is :
(1) 420 (2) 480 (3) 360 (4) 240
Ans. (3)
Sol. A = {100, 101, 102, ………………999}
number divisibleby 3 = 102, 105, ………..999
total = 300
number divisible by 5 = 100, 105, 110, ……995
total = 180
number divisible by 15 = 105, 120, …………..990
total = 60
total number which are divisible by 3 or 5
= 300 + 180 – 60
= 420
n(B) = 900 – 420 = 480
6. Let A (1, 3) and C(5, 1) be two opposite vertices of a rectangle. The other two vertices B (a, b) and D(c,
d) lie on the line y = 2x + k for some k. Then the value of (a + b) (c + d) is :
(1) 8 (2) 24 (3) 32 (4) 16
Ans. (4)
Sol. (a, b) & (c, d) lies on circle
(x – 1) (x – 5) + (y – 3) (y – 1) = 0
(a, b) & (c, d) lies on x2 +y2 – 6x – 4y + 8 = 0
Point of interaction of diagonal is (3, 2)
(a, b) & (c, d) also lies on y = 2 x – 4
Now point of intersection of y = 2x – 4 and
x2 + y2 – 6x – 4y + 8 = 0 are (2, 0) and (4, 4)
a + b = 2 or c +d = 8
(a + b ) (c + d) = 16
3),….,(xn – 3) is equal to :
13 16
(1) 5 (2) (3) 8 (4)
3 3
Ans. (1)
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n
Sol. x
i 1
1 – 2n 180 ………………(i)
n
x
i 1
i – 7n 30 ………………(ii)
n x i
240
i 1
x i 240
30
i 1
=
30
8x
2 6 14
9. The sum of the infinite series 1 + 2 + + 2 + 4 +…….is :
3 3 3
9
(1) (2) 6 (3) 5 (4) 4
2
Ans. (3)
2 6 10 14
Sol. Let S = + 2 + 3 + 4 + ……………….
3 3 3 3
S 2 6 10
= 2 + 3 + 4 + …………
3 3 3 3
S 2 4 1 1
S– = + 2 1 2 ........
3 3 3 3 3
2S 2 4 1 2 2 4
= + × = + = S=2
3 3 9 1 3 3 3
9 –
3
Required equals to 1 + 2 + S = 1 + 2 +2 = 5
x 6 s in(x 3) 3
10. lim is equal to :
x 3 ( x 3) cos (x 3)
2 5 1 5
(1) (2) (3) (4)
3 6 6 6
Ans. (2)
Sol.
x 6 – 3 – sin x – 3
cos x – 3 x – 3
lim
x 3
x6 –3 ln x – 3
–
cos x – 3 x – 3 cos x – 3 x – 3
= lim lim
x 3 x 3
= lim
x – 3 –
1
x 3
x – 3 cos x – 3 x 6 3 cos 0
1 1 –5
– 1= –1
cos0
=
a 3 6 6
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11. Let the tangent drawn at any point P(x,y) on a curve intersect the x and y axes at two distinct points A
and B respectively. If AP : PB = 5 : 1, and the curve passes through the point (2, 2) then an equation of
the curve is :
(1) xy5 = 26 (2) x5y = 26 (3) x4y = 25 (4) xy4 = 25
Ans. (1)
1 p(x, y)
Sol. :
5
A(6x, 0)
dy dy y 0 dy –dx
Slope of AP is equal to
dx dx x – 6x y 5x
5dy dx
0 5ny + lnx = lnc xy5 = C
y x
Because it passes through (2, 2)
so 2(25) = C c = 26 equation of curve is xy5 = 26
12. In an increasing geometric series, the sum of the first and the sixth term is 66 and the product of the
second and the fifth terms is 128. Then the sum of the first 6 terms of this series is
(1) 129 (2) 126 (3) 127 (4) 128
Ans. (2)
Sol. Let terms of G.P are a, ar, ar2
Now a + ar5 = 66 & ar × ar4 = 128
128
a+ = 66 = 66 a = 2 or 64
a
1 26 – 1
If a = 2 , r = 2 & if a = 64, r a = 2 & r = 2 sum of 6 terms are = 2 = 2 × 63 = 126
2 2 –1
4 1
13. If for two event A and B, in a random experiment, P(A|B) = and P(B|A) = , then P (A|A B) is
5 4
equal to
5 16 11 5
(1) (2) (3) (4)
16 17 16 17
Ans. (2)
A 4 P A B 4
Sol. P =
B 5 P B 5
B P A B 1 PA 16
P = = =
A P
A 4 P B 5
But P(A) = 16x, P(B) = 5x & P(AB) = 4x
P (AB) = P(A) + P(B) – P (AB) = 17 x
A P A A B PA
16x 16
Now P = = = =
A B P A B P A B 17x 17
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3
14. Let f : [0, 5] R be continuous function such that f (x ) 3 for all x [0, 5]. The value of f ( t)dt can be :
0
(1) –4 (2) 10 (3*) 6 (4) 12
Ans. (3)
3 5 5 5
Sol. f t dt =
0
f t dt –
0
f t dt = 3 –
3
f t dt
3
5
Now –3 (5 –3) f t dt (5–3) 3
3
5 5 3
–6 – f t dt
3
6 –33– f t dt
3
9 – 3 f t dt 9
0
z(1 z)
15. Let z( –1) be any complex number such that z 1 . Then the imaginary part of is :
z(1 z)
(Here = arg z)
(1) tan s in (2) tan cos (3) tan s in (4) tan cos
2 2 2 2
Ans. (2)
Sol. Let z = (cos + i sin)
z 1– z = z –1
z zz – z
z 1 z z zz z
Now =
z 1
2
sin cos i
– sin 2 – sin 2 cos i 2 2
= Img = Img – tan
cos2 isin cos 2
cos sin i
2 2 2 2 2
i –
e 2 2 i 2 –
= Img – tan = –tan img e = – tan cos
2
2 2
e
i
2
16. Let f : R – {0} R be defined by f(x) = a loge x bx3 x 2 . If x = –1 and x = 1 are the critical points of
f(x), then :
(1) f''(1) – f''(–1) = 4
(2) both x = 1 and x = –1 are local minima of f(x)
(3) f''(1) + f''(–1) = 0
(4) x = 1 is a local minima and x = –1 is a local maxima of f(x)
Ans. (4)
a –a
Sol. f'(x) = + 3bx2 + 2x f''(x) = +2
x x2
Now f'(1) = a + 3b + 2 = 0
& f'(–1) = –a + 3b – 2 = 0 a = – 2 & b = 0 f'' (1) = 4 & f''(–1) = 4
2 x – 1 x 1
& f'(x) =
x
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17. Let the ellipse x2 + 16y2 = 16 be inscribed in a rectangle whose sides are parallel to the coordinate
axes. If the rectangle is inscribed in another ellipse that passes through the point (16, 0) then the
equation of the outer ellipse is
(1) x2 + 248y2 = 162 (2*) x2 + 240y2 = 162 (3) x2 + 232y2 = 162 (4) x2 +256y2 = 162
Ans. (2)
(4, 1)
(0,1) E1 E2
Sol. (16, 0)
(4, 0)
x2 y2
Let E1 : x2 + 16y2 = 16 & E2 : 2
+ 2 =1
16 b
16 1 16 x2 15y 2
Now (4, 1) lies on E2 + 2
= 1 b2 = E2 is + = 1 x2 + 240 y2 = 162
162 b 15 162 16
3
5x 4
18. The value of
–3
1 e –x
dx is :
35
(1) 34 (2) 35 (3) 2(35) (4)
5
Ans. (2)
5 –x
4
3 5x 4 3
–3 1 e– x dx
3
dx I = 35
3
Sol. Let I= I= 2I = 3 5x4 dx = x5
–3 1 e x –3
a sin2x – bcos x
, x
2
–x
2
19. Let f be a continuous function defined of f(x) = 4, x . Then the value of a + b is :
2
2bcos x
, x
2
–x
2
(1) 8 (2) 5 (3) 4 (4) 1
Ans. (2)
sin – x
2b cos x 2
Sol. lim = 2b lim = 2b = 4 b = 2
x
–
– x
x
2 2 2 – x
2
sin x sin – x sin x
a sin 2x – b cos x 2 2
lim = 2a lim
– b lim
x x
–x 2 x
–x x
2 2 2
2 2
= 2a – b = 4 a=3
a+b=5
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20. If three vectors V1 ˆi ˆj kˆ , V2 ˆi ˆj – 2k and V3 ˆi ˆj are coplanar and V1 and V3 are
perpendicular then the vector V1 V2 is :
(1) 2iˆ – 2ˆj kˆ (2) – ˆi ˆj 2kˆ (3) ˆi – ˆj 2kˆ (4) – ˆi ˆj
Ans. (3)
Sol. V1 V3 are V1 V3 0
+ 1 + 1 + 0 = 0 = –1
–1 1 1
V1 V2 ,V3 0 1 –2 0
1 1 0
–1 0 0
1 1 –1 0 +1+2=0 =–3
1 2 1
i j k
Now V1 V2 = –1 1 1 = î (–2 + 3) – ĵ (2–1) + k̂ (3 – 1) = î – j + 2k
1 –3 –2
21. The area (in square units) above the x-axis bounded by the parabola x – y2 – 1 = 0 and the line
x – y – 3 = 0 is :
13 10 8
(1) 4 (2) (3) (4)
3 3 3
Ans. (3)
1 3 5
Sol.
–3
y2 = x – 1 & x – y – 3 = 0
(x–3)2 – x + 1 = 0
x2 – 6x + 9 – x + 1 = 0
x2 – 7x + 10 = 0
x = 2, 5
y = –1, 2
y 3 – y
2
Area = 2
1 dy
0
2
2 y 2 y3 4 8
y – y
2 dy = – 2y – 4
2
= =
0 2 3 0 2 3
8 10
= 6– =
3 3
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x – 2m y z–4 x–2 y z – 2m
22. If the two lines and are parallel for some m R then the
2m 5 8m 2 m – 2 –1 1– 3m
distance between them is :
(1) 10 (2) 34 (3) 2 5 (4) 29
Ans. (1)
2m 5 8m 2
Sol. Lines are parallel = =
m–2 –1 1– 3m
–2 m – 5 = 8m2 – 16m & 8m – 24m2 = –2
8m2 – 14m + 5 = 0 12m2 – 4m – 1 = 0
8m2 – 4m – 10m + 5 = 0 12m2 – 6m + 2m – 1 = 0
(4m – 5) (2m – 1) = 0 (6m + 1) (2m – 1) = 0
5 1 –1 1
m= , m= ,
4 2 6 2
1
m=
2
x –1 y z–4
Lines are L1 = = =
6 4 2
x–2 y z –1
L2 = = =
–3 –1 1
–
2 2
Distance =
a 2
– a1 b
b
i j k
Now a2 – a2 × b = 1 0 –3 = 6i – 10j + 2k
3 2 1
36 100 4 140
Distance = = = 10
9 4 1 14
Ans. (1)
23. Let R be a relation defined on Z × Z by (a, b) R (c, d) a – d = b – c where Z is the set of all integers
then R is :
(1) Transitive but neither reflexive nor symmetric.
(2) Reflexive but neither symmetric nor transitive.
(3) Symmetric but neither reflexive nor transitive.
(4) Symmetric and transitive but not reflexive.
Ans. (3)
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24. Let the abscissae of two points A and B on a circle be the roots of x2 + 2x – 4 = 0 and the ordinates of A
and B be the roots of y2 + 4y – 16 = 0. If AB is a diameter of this circle then the radius of this circle is :
(1) 5 (2) 2 10 (3) 6 (4) 2 6
Ans. (1)
Sol. Let A (x1, y1) & B (x2, y2)
x1 + x2 = –2, x1x2 = –4
y1 + y2 = –4, y1 + y2 = –16
x2 – x1 y2 – y1
2 2
Now AB = 2r =
x1 x2 – 4x1x2 y1 y2 – 4y1y2
2 2
=
= 4 16 16 64 10
radius = 5
25. If r is the remainder obtained on dividing (98)5 by 12 then the coefficient of x3 in the binomial expansion
2x
of 1 x is :
2
91 55
(1) 70 (2) (3) 102 (4)
2 2
Ans. (1)
Sol. (98)5 = (96 +2)5
= 965 + 5C1 (96)4 + ………. + 5C4 96 + 25
= 96 K + 32 = 12 (8K + 2) + 8
r=8
2r 16
x
= 1 x
1 2
2
r
Tr +1 = 16 x
Cr
2
x3
r = 3, T4 = 16C3
8
1 16.15.14
coefficient of x3 =
16
C3 = 70
8 3.2.1 8
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sec x
26. sin x.cos5 x
dx is equal to : (Where C is a constant of integration)
2 1
(1) 2 tan x tan x C (2) 2 tan x tan x C
1/ 2 5/2 1/ 2 5/2
–
5 5
2 2
(3) 2 tan x tan x C (4) tan x tan x C
1/ 2 5/2 1/ 2 5/2
5 5
Ans. (3)
sec x
Sol. sin x cos5 x
dx
28. Let the planes x – 2y + kz = 0 and x + 5y – z = 0 be perpendicular. Then the plane through the point
(2, –2, –2) and perpendicular to the given planes also passes through the point :
(1) (1, 0, 7) (2) (0, 5, –8) (3) (–1, 0, –7) (4) (0, 5, 8)
Ans. (1)
Sol. x – 2y + kz = 0
and x + 5y – z = 0 perpendiculars
1 – 10 – k = 0
k = –9
Any plane through (2, –2, –2) is
a(x–2) + b(y +2) + c(z+2) = 0
& a – 2b – 9c = 0
a + 5b – c = 0
a b c
= =
2 45 –9 1 52
Plane is 47(x–2) –8(y+2) +7(z+2) = 0
47x – 8y + 7z – 96 = 0
Clearly (1, 0, 7) satisfying the plane
xy 2 1
dy
29. Let y be an implicit function of x defined by 1 xy 2 + 12y = 0. If y(0) = –1, then at
dx
1 2 xy
x = 0 is :
4 5 –1 1
(1) – (2) (3) (4)
5 4 2 2
Ans. (4)
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zy 2 1
Sol. 1 xy 2 + 12y = 0
1 2 xy
1 2 1
(x+y+3) 1 x y 2 + 12y = 0
1 2 xy
0 0 1– x – y
(x+y+3) 1 x y 2 + 12y = 0
1 2 xy
(x+y+3) (1–x–y) (2–x–y) + 12y = 0
Difference w.r.t. x
dy dy dy dy
1 dx (1–x–y) (2–x–y) + (x+y+3) –1– dx (2–x–y) + (x + y + z) (1–x–y) –1– dx + 12 =0
dx
dy dy dy
1 .0 + 4 –1– .0 + 4.0 + 12 =0
dx dx dx
dy dy dy 1
12 =4+4 =
dx dx dx 2
30. The set of all real values of for which the equation |x + 2| . |x – 2| = 2 – 2 has real solutions for x is :
(1) (–, 0] [2, 1 + 5 ] (2) [–1 – 5 , 1 – 5 ] [1 + 5 , )
(3) (–, 0] [2, ) (4) [1 – 5 , 0] [2, 1 + 5 ]
Ans. (3)
Sol. x 2 – 4 = 2 – 2
|x| 0 2 – 2 0
(–2) 0
(– , 0) [2, )
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APTITUDE TEST
1. Zaha Hadid was born in which country amongst the following?
(1) Iran (2) Iraq (3) Turkistan (4) Afghanistan
Ans. (2)
4. A small lift for carrying small loads only is known as which of the following?
(1) A deaf bearer (2) A push upper (3) A dumb waiter (4) A jockey boy
Ans. (3)
5. Which amongst the following is the city in Italy that is known for its leaning tower?
(1) Pisa (2) Venice (3) Florence (4) Rome
Ans. (1)
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11. The lotus temple is located in which one of the following city?
(1) New Delhi (2) Luckow (3) Kanpur (4) Nagpur
Ans. (1)
14. [ktqjkgks lewg ds eafnjksa esa lcls izfl} eafnj fuEufyf[kr esa ls dkSu lk gS?
(1) danfj;k egknso eafnj (2) d`".kk eafnj (3) x.ks'k eafnj (4) f'ko eafnj
Ans. (1)
15. Hindustan parryware in the indian market is known for which of the following product?
(1) Wooden tables (2) Sanitary ware (3) Wall tiles (4) Pipes
Ans. (2)
Directions : One of the following answer figures is hidden in the problem figure in the same size and
direction. Select the correct one.
16.
Directions : One of the following answer figures is hidden in the problem figure in the same size and
direction. Select the correct one.
17.
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Directions : One of the following answer figures is hidden in the problem figure in the same size and
direction. Select the correct one.
18.
Directions : One of the following answer figures is hidden in the problem figure in the same size and
direction. Select the correct one.
19.
Directions : One of the following answer figures is hidden in the problem figure in the same size and
direction. Select the correct one.
20.
Directions : Which one of the answer figure will complete the sequence of the three problems figures?
21.
Ans. (2)
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Directions : Which one of the answer figure will complete the sequence of the three problems figures?
22.
Ans. (1)
Directions : Which one of the answer figure will complete the sequence of the three problems figures?
23.
Directions : Which one of the answer figure will complete the sequence of the three problems figures?
24.
Ans. (2)
Directions : Which one of the answer figure will complete the sequence of the three problems figures?
25.
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Directions : The 3D figure shows the view of an object. Identify the correct front view looking in the direction
of the arrow from amongst the answer figures.
26.
Directions : The 3D figure shows the view of an object. Identify the correct front view looking in the direction
of the arrow from amongst the answer figures.
27.
Directions : The 3D figure shows the view of an object. Identify the correct front view looking in the direction
of the arrow from amongst the answer figures.
28.
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Directions : The 3D figure shows the view of an object. Identify the correct front view looking in the direction
of the arrow from amongst the answer figures.
29.
(1) (2)
(3) (4)
Ans. (3)
Directions : The 3D figure shows the view of an object. Identify the correct front view looking in the direction
of the arrow from amongst the answer figures.
30.
(1) (2)
(3) (4)
Ans. (3)
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Direction : Which one of the answer figures is the correct mirror image of the problem figure with respect
to X-X?
31.
(1) (2)
(3) (4)
Ans. (3)
Direction : Which one of the answer figures is the correct mirror image of the problem figure with respect
to X-X?
32.
(1) (2)
(3) (4)
Ans. (1)
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| JEE MAIN-2019 | DATE : - 07 04-2019 (SHIFT-1) | PAPER-2 | B.ARCH | MATHEMATICS
Direction : Which one of the answer figures is the correct mirror image of the problem figure with respect
to X-X?
33.
(1) (2)
(3) (4)
Ans. (2)
Direction : Which one of the answer figures is the correct mirror image of the problem figure with respect
to X-X?
34.
(1) (2)
(3) (4)
Ans. (4)
This solution was download from Resonance JEE MAIN 2019 Solution portal PAGE # 20
| JEE MAIN-2019 | DATE : - 07 04-2019 (SHIFT-1) | PAPER-2 | B.ARCH | MATHEMATICS
Direction : Which one of the answer figures is the correct mirror image of the problem figure with respect
to X-X?
35.
Ans. (1)
Directions : The problem figure showns the top view of an object. Identiy the correct elevation from amongst
the answer figures looking in the direction of the arrow.
36.
Ans. (2)
Directions : The problem figure showns the top view of an object. Identiy the correct elevation from
amongst the answer figures looking in the direction of the arrow.
37.
Ans. (3)
This solution was download from Resonance JEE MAIN 2019 Solution portal PAGE # 21
| JEE MAIN-2019 | DATE : - 07 04-2019 (SHIFT-1) | PAPER-2 | B.ARCH | MATHEMATICS
Directions : The problem figure showns the top view of an object. Identiy the correct elevation from amongst
the answer figures looking in the direction of the arrow.
38.
Ans. (3)
Directions : The problem figure showns the top view of an object. Identiy the correct elevation from amongst
the answer figures looking in the direction of the arrow.
38.
Ans. (2)
Directions : The problem figure showns the top view of an object. Identiy the correct elevation from amongst
the answer figures looking in the direction of the arrow.
39.
Ans. (1)
This solution was download from Resonance JEE MAIN 2019 Solution portal PAGE # 22
| JEE MAIN-2019 | DATE : - 07 04-2019 (SHIFT-1) | PAPER-2 | B.ARCH | MATHEMATICS
Directions : The problem figure showns the top view of an object. Identiy the correct elevation from amongst
the answer figures looking in the direction of the arrow.
40.
Ans. (2)
Directions : The 3D figure shown the view of an object. Identify the correct top view from amongst the answer
figure.
41.
Ans. (1)
Directions : The 3D figure shown the view of an object. Identify the correct top view from amongst the answer
figure.
42.
Ans. (2)
This solution was download from Resonance JEE MAIN 2019 Solution portal PAGE # 23
| JEE MAIN-2019 | DATE : - 07 04-2019 (SHIFT-1) | PAPER-2 | B.ARCH | MATHEMATICS
Directions : The 3D figure shown the view of an object. Identify the correct top view from amongst the answer
figure.
43.
Directions : The 3D figure shown the view of an object. Identify the correct top view from amongst the answer
figure.
44.
Ans. (2)
Directions : The 3D figure shown the view of an object. Identify the correct top view from amongst the answer
figure.
45.
This solution was download from Resonance JEE MAIN 2019 Solution portal PAGE # 24
| JEE MAIN-2019 | DATE : - 07 04-2019 (SHIFT-1) | PAPER-2 | B.ARCH | MATHEMATICS
Directions : The 3D figure shown the view of an object. Identify the correct top view from amongst the answer
figure.
46.
Ans. (3)
Directions : The 3D figure shown the view of an object. Identify the correct top view from amongst the answer
figure.
47.
Directions : The 3D figure shown the view of an object. Identify the correct top view from amongst the answer
figure.
48.
This solution was download from Resonance JEE MAIN 2019 Solution portal PAGE # 25
| JEE MAIN-2019 | DATE : - 07 04-2019 (SHIFT-1) | PAPER-2 | B.ARCH | MATHEMATICS
Directions : The 3D figure shown the view of an object. Identify the correct top view from amongst the answer
figure.
49.
Ans. (1)
Directions : The 3D figure shown the view of an object. Identify the correct top view from amongst the answer
figure.
50.
Ans. (3)
This solution was download from Resonance JEE MAIN 2019 Solution portal PAGE # 26
| JEE MAIN-2019 | DATE : - 07 04-2019 (SHIFT-1) | PAPER-2 | B.ARCH | MATHEMATICS
DRAWING
1. In the space provided in the answer sheet for this question, draw margin lines to form a frame. In this
frame create an aesthetic composition using only cubes. These can be of any size, and may be placed
separate, overlapping or within each other. The idea is to produce an aesthetic and visually exiting
composition of these shapes in the frame without making it represent any realistic form like house face
etc. These shapes and the other spaces should be filled with some colors of your choice so that the visual
quality of the composition is enhanced.
2. Copy the graphic image shown in the space provided for the answer of this question. Credit will be
given to the exactness of your answer.
3. In the space provided for the answer of this question attempt any ONE of the following:
Design and draw an appropriate pattern for a square table cloth. Color or shade it to enhance its visual
quality.
OR
Draw a picture of a classroom looking towards the teacher from behind the student.
OR
Draw from imagination a picture of an officer sitting in his office.
This solution was download from Resonance JEE MAIN 2019 Solution portal PAGE # 27