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JEE Main Online Exam 2019
Questions & Solutions 10th April 2019 | Shift - I
Mathematics Q.1 The region represented by |y–x| 2 and | x + y| 2 is bounded by a :
(1) rhombus of area 28 sq. units (2) square of side length 22 units (3) square of area 16 sq. units (4) rhombus of side length 2 units Ans. [2]
Sol. shown figure is square with side length 22 .
(0,2)
(2,0)
(0,–2)
(–2, 0)
y
x
Q.2 All the pairs (x, y) that satisfy the inequality 5xsin2–xsin 22
ysin 24
1 1 also satisfy the equation
(1) sin x = |ysin| (2) sin x = 2 sin y (3) 2 sin x = sin y (4) |xsin|2 = 3 sin y Ans. [1]
Sol. 5xsin2–xsin 22 ysin– 2
4 1
4)1–x(sin 2
2 ysin24
4)1–x(sin 2 0 2
2 sin2 y 2
this is possible only if sin x = 1 & |sin y| = 1
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Q.3 If and are the roots of the quadratic equation, x2 + x sin – 2 sin = 0,
2,0 , then
2412–12–
1212
)–( is equal to :
(1) 6
12
)8–(sin2
(2) 12
6
)8(sin2
(3) 12
12
)8(sin2
(4) 12
12
)4–(sin2
Ans. [3]
Sol. x2 + x sin – 2 sin = 0
+ = – sin
= – 2 sin
Now, 24
1212
1212
)–(11
= 24
12
)–()(
= 122
12
4–)(
)(
= 12
2 4–)(
= 12
2 sin8sinsin2–
= 12
12
)8(sin2
Q.4 If a > 0 and z = i–a)i1( 2 , has magnitude
52 , then z is equal to :
(1) – 51 +
53 i (2) i
53–
51– (3) i
53–
51 (4) i
51–
53–
Ans. [2]
Sol. z = i–a)i1( 2 =
1a)ia(i2
2
| z | = 1a
22
= 52 a = 3
z = 10
)i–3(i2–
5
i3–1–
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Q.5 If y = y(x) is the solution of the differential equation dxdy = (tan x – y) sec2 x, x
2,
2– , such that
y (0) = 0, then
4–y is equal to :
(1) 21 – e (2) e – 2 (3)
e12 (4) 2–
e1
Ans. [2]
Sol. dxdy = (tan x – y) sec2 x
dxdy + y sec2 x = tan x sec2 x
Let tan x = t sec2 x = dxdt
dtdy = (t – y)
dtdy + y = t (Linear differential equation)
After solving we get y et = et (t – 1) + c y = (tan x – 1) + ce – tan x y (0) = 0 c = 1 y = tan x –1 + e– tan x
So, y
4– = e – 2
Q.6 If the length of the perpendicular from the point (, 0, ) ( 0) to the line, 1x =
01–y =
1–1z is
23 , then
is equal to : (1) 2 (2) 1 (3) –2 (4) –1 Ans. [4] Sol.
B(, 0, )
A(0, 1, –1) C (, 1, – –1)
1x =
01–y =
1–1z =
A point on this line is A(0, 1, –1) AC BC = 0
we get = – 21
C
21–,1,
21–
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32BC
2
22
21)1(
21
=
32
= 0, –1 = – 1 ( 0)
Q.7 If 1 = x1cos1x–sin–
cossinx
and 2 =
x12cos1x–2sin–2cos2sinx
, x 0 ; then for all
2,0 :
(1) 1 – 2 = x (cos 2 – cos 4) (2) 1 + 2 = – 2x3
(3) 1 + 2 = – 2(x3 + x –1) (4) 1 – 2 = – 2x3 Ans. [2]
Sol. 1 = x1cos1x–sin–
cossinx
= x (– x2 – 1) – sin (– x sin – cos ) + cos (– sin + x cos ) – x3
2 = x12cos1x–2sin–2cos2sinx
– x3 1 + 2 = – 2x3 Q.8 Assume that each born child is equally likely to be a boy or a girl. If two families have two children each,
then the conditional probability that all children are girls given that at least two are girls is :
(1) 101 (2)
171 (3)
111 (4)
121
Ans. [3]
Sol. P (Boy) = P(girl) = 21
Required probability = girlstwoleastAt
girlsfourall
= 4
24
4
34
4
4
21C
21C
21
21
= 111
Q.9 Which one of the following Boolean expressions is a tautology ? (1) (p q) (~ p q) (2) (p q) (p q) (3) (p q) (p q) (4) (p q) (p q) Ans. [2]
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Sol. from options (p q) (~ p q) (p q) (p q) Not a tautology (p q) (p q) p (q q) tautology (p q) (p q) p (q q) Not a tautology (p q) (p q) p (q q) Not a tautology Q.10 Let f(x) = x2, x R. For any A R, define g (A) = { x R : f(x) A}. If S = [0,4], then which one of the
following statements is not true ? (1) g(f(S)) S (2) f(g(S)) S (3) f(g(S)) f(S) (4) g(f(S)) = g(S) Ans. [4] Sol. g(S) = [–2, 2] f(g(S)) = [0,4] = S f(S) = [0, 16] f(g(S)) f(S) g(f(S)) = [–4, 4] g(S) therefore , g(f(S)) S
Q.11 If 1x
lim
1–x1–x4
= kx
lim
22
33
k–xk–x , then k is :
(1) 23 (2)
38 (3)
34 (4)
83
Ans. [2]
Sol. 1x
lim
1–x1–x4
= 1x
lim
(x + 1) (x2 + 1) .....(i)
kx
lim
22
33
k–xk–x =
k2kkk 222 ....(ii)
(i) = (ii)
k = 38
Q.12 If a directrix of a hyperbola centred at the origin and passing through the point (4, –2 3 ) is 5x = 4 5 and
its eccentricity is e, then : (1) 4e4 – 24e2 + 27 = 0 (2) 4e4 – 24e2 + 35 = 0 (3) 4e4 – 12e2 – 27 = 0 (4) 4e4 + 8e2 – 35 = 0 Ans. [2]
Sol. Let hyperbola be 2
2
2
2
by–
ax = 1 & passes through (4, 32– ) therefore
22 b12–
a16 = 1 .....(i) b2 = a2 (e2 –1)
x = ea
554
a2 = 2e5
16 ....(ii)
one solving (i) & (ii) 4e4 – 24 e2 + 35 = 0
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Q.13 If f (x) =
0x,x
x–xx0x,q
0x,x
xsinx)1psin(
2/3
2
is continuous at x = 0, then the ordered pair (p, q) is equal to :
(1)
21,–
23– (2)
23,
21– (3)
21,
23– (4)
21,
25
Ans. [3]
Sol. RHL = 0x
lim 2/3
2
xx–xx
= 0x
lim x
1–x1 = 21
LHL = –0x
lim
x
xsinx)1psin(
= (p + 1) + 1 = p + 2 for function to be continuous LHL = RHL = f(0)
(p, q) =
21,
23–
Q.14 If the coefficients of x2 and x3 are both zero, in the expansion of the expression (1 + ax + bx2) (1 – 3x)15 in
powers of x, then the ordered pair (a,b) is equal to : (1) (28, 861) (2) (28, 315) (3) (–21, 714) (4) (–54, 315) Ans. [2]
Sol. coefficient of x2 = 215C × 9 – 3a )C( 1
15 + b = 0
215C × 9 – 45 a + b = 0 ...(i)
coefficient of x3 = –27 × 315C + 9a × 2
15C – 3b × 115C = 0
– 273 + 21a – b = 0 ...(ii) (i) + (ii) –24 a + 672 = 0 a = 28 b = 315 Q.15 If the circles x2 + y2 + 5Kx + 2y + K = 0 and 2(x2 + y2) + 2Kx + 3y –1 = 0, (KR), intersect at the points
P and Q, then the line 4x + 5y – K = 0 passes through P and Q, for : (1) exactly two values of K (2) no value of K. (3) exactly one value of K (4) infinitely many values of K Ans. [2] Sol. equation of common chord
4 kx + 21 y + k +
21 = 0 ...(i)
and given line 4x + 5y – k = 0 ...(ii)
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on comparing (i) & (ii) we get
k = k–
2/1k101
No. real value of k exist
Q.16 The sum 222
333
22
33
2
3
321)321(7
21)21(5
113
+......... upto 10th term is :
(1) 660 (2) 680 (3) 600 (4) 620 Ans. [1]
Sol. Tn = )n........21(
)n......21)(2)1–n(3(222
333
=23 n(n + 1)
Sn = nT
= )1n(n23
on solving Sn = 2
)2n)(1n(n S10 = 660
Q.17 Let f : R R be differentiable at c R and f(c) = 0. If g(x) = )x(f , then at x = c, g is :
(1) differentiable if f ' (c) = 0 (2) differentiable if f ' (c) 0 (3) not differentiable (4) not differentiable if f ' (c) 0 Ans. [1]
Sol. g'(c) = 0h
lim
h
|)c(f|–|)hc(f|
= 0h
lim
h
|)hc(f| ( f (c) = 0)
= 0h
lim
h
|h|h
)c(f–)hc(f
= 0h
lim
|f '(c)| h
|h| = 0 if f ' (c) = 0
i.e. g(x) is differentiable at x = c if f '(c) = 0 Q.18 If Q(0, –1 –3) is the image of the point P in the plane 3x – y + 4z = 2 and R is the point (3, –1, –2), then the
area (in sq. units) of PQR is :
(1) 265 (2) 132 (3)
291 (4)
491
Ans. [3]
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Sol.
P
M
R
Q
MQ = 1619
|2–12–1|
= 26
13
= 2
13
PM = 26
RQ = 1019
RM = 27
213–10
Ar (PQR) = 21 × 26 ×
27 =
291
Q.19 If 22 )10x2–x(dx = A
10x2–x)x(f
31–xtan 2
1– + C where C is a constant of integration then :
(1) A =541 and f(x) = 9(x–1)2 (2) A =
541 and f(x) = 3(x–1)
(3) A =811 and f(x) = 3(x–1) (4) A =
271 and f(x) = 9(x–1)
Ans. [2]
Sol. 22 )10x2–x(dx =
22 9)1–x(
dx
Let x – 1 = 3 tan dx = 3 sec2 d
dcos271 2 =
541 d2cos1 =
541
22sin
= 541 C
10x2–x)1–x(3
31–xtan 2
1–
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Q.20 The line x = y touches a circle at the point (1,1). If the circle also passes through the point (1, – 3), then its radius is :
(1) 3 (2) 2 (3) 2 2 (4) 23 Ans. [3] Sol.
(1,1)
Equation of circle is given as S + L = 0 (x – 1)2 + (y – 1)2 + (x – y) = 0 passes through(1, – 3) 16 + × 4 = 0 = – 4 (x – 1)2 + (y – 1)2 – 4 (x –y) = 0 r = 22
Q.21 The value of dx)]x3cos1(x2[sin2
0
, where [t] denotes the greatest integer function is :
(1) 2 (2) (3) –2 (4) – Ans. [4]
Sol. I = dx)]x3cos1(x2[sin2
0
2I =
2
0
dx]x3cosx2sin–x2sin[–)]x3cos1(x2[sin
2I = 2
0
dx–
2I =
0
dx–2
I =
0
dx– –
Q.22 Let A (3,0, –1), B(2, 10, 6) and C(1, 2, 1) be the vertices of a triangle and M be the midpoint of AC. If G divides BM in the ratio, 2 : 1, then cos (GOA (O being the origin) is equal to :
(1) 151 (2)
1061 (3)
301 (4)
1521
Ans. [1] Sol. G will be centroid of ABC G (2, 4, 2) OG = k2j4i2
OA = k–i3
cos (GOA) = |OA||OG|
OAOG = 151
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Q.23 If the system of linear equations x + y + z = 5 x + 2y + 2z = 6 x + 3y + z = , (, R), has infinitely many solutions, then the value of + is : (1) 10 (2) 9 (3) 12 (4) 7 Ans. [1] Sol. x + 3y + z – = a (x + y + z – 5) + b (x + 2y + 2z – 6) comparing coefficients we get a + b = 1 and a + 2b = 3 (a, b) (–1, 2) So, x + 3y + z – = x + 3y + 3z – = 7, = 3 Q.24 The number of 6 digit numbers that can be formed using the digits 0, 1, 2, 5, 7 and 9 which are divisible by
11 and no digit is repeated is : (1) 36 (2) 60 (3) 72 (4) 48 Ans. [2] Sol. Let the six digit number be abcdef for this number to be divisible by 11, |(a + c + e) – (b + d + f)| must be
multiple of 11 possibility is a + c + e = b + d + f = 12 Case : 1 {a, c, e} = {7, 5, 0} & {b, d, f} = {9, 2, 1} So, number of numbers = 2 × 2! × 3! = 24 Case : 2 {a, c, e} = {9, 2, 1} & {b, d, f} = {7, 5, 0} So, number of numbers = 3! × 3! = 36 total 24 + 36 = 60 Q.25 If for some x R, the frequency distribution of the marks obtained by 20 students in a test is :
Marks 2 3 5 7 Frequency (x + 1)2 2x – 5 x2 – 3x x
then the mean of the marks is (1) 3.0 (2) 2.8 (3) 2.5 (4) 3.2 Ans. [2]
Sol. Mean x =
i
ii
f
fx
if = (x + 1)2 + (2x –5) + (x2 – 3x) + x = 20 x = 3, – 4(rejected)
x =
i
ii
f
fx = 2.8
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Q.26 ABC is a triangular park with AB = AC = 100 metres. A vertical tower is situated at the mid-point of BC. If the angles of elevation of the top of the tower at A and B are cot–1 (3 2 ) and cosec–1 (2 2 ) respectively, then the height of the tower (in metres) is :
(1) 33
100 (2) 25 (3) 20 (4) 510
Ans. [3] Sol.
Q
h
P
100C A
100
B
x
cosec = 22 cot = 23
hx = 23 ...(i)
So 24 x–10
h = 7
1 .....(ii)
from (i) & (ii) h = 20 Q.27 If a1, a2, a3, ............... an are in A.P. and a1 + a4 + a7 + ........... + a16 = 114, then a1 + a6 + a11 + a16 is equal to : (1) 38 (2) 98 (3) 76 (4) 64 Ans. [3] Sol. a1 + a4 + a7 + a10 + a13 + a16 = 114
26 (a1 + a16) = 114
a1 + a16 = 38
So, a1 + a6 + a11 + a16 = 24 (a1 + a16)
= 2 × 38 76 Q.28 Let f(x) = ex – x and g(x) = x2 – x, x R. Then the set of all x R, where the function h(x) = (fog) (x) is
increasing, is :
(1) [0, ) (2)
21–,1–
,
21 (3) ),1[0,
21–
(4)
21,0 [1, )
Ans. [4]
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Sol. h(x) = f(g(x)) h'(x) = f '(g(x)) g'(x) and f '(x) = ex –1 h'(x) = (eg(x) – 1) g'(x)
h'(x) = 0)1–x2()1–e( x–x 2
Case :1 1e x–x2 and 2x –1 0
x
21,0 .....(i)
Case : 2 1e x–x 2 and 2x –1 0
x [1, ) ....(ii) from (i) & (ii)
x
21,0 [1,)
Q.29 n
lim
3/4
3/1
3/4
3/1
3/4
3/1
n)n2(.......
n)2n(
n)1n( is equal to :
(1) 34 (2)3/4 (2)
43 (2)4/3 –
43 (3)
34 (2)4/3 (4)
34–)2(
43 3/4
Ans. [2]
Sol. n
lim3/1n
1r nrn
n1
= 3/11
0
)x1( dx = 43 (24/3 –1)s
Q.30 If the line x – 2y = 12 is tangent to the ellipse 2
2
ax + 2
2
by = 1 at the point
29–,3 , then the length of the latus
rectum of the ellipse is :
(1) 8 3 (2) 212 (3) 5 (4) 9
Ans. [4] Sol. Tangent at (3, – 9/2)
2ax3 – 2b2
y9 = 1
comparing with x – 2y = 12
2a3 = 2b4
9 = 121
a = 6 & b = 33
length of latus rectum = ab2 2
= 9