JEE Main 2018 Maths paper January 10 Shift 1 solutions are available on this page. Students are recommended to revise these solutions so that they can increase their speed and accuracy. These solutions will help them to understand the difficulty level of questions from each chapter. Students can easily download them in PDF format for free.
1. If the tangent at (1, 7) to the curve x^{2} = y-6 touches the circle x^{2}+y^{2}+16x+12y+c = 0, then the value of c is :
Solution:
Equation of tangent at (1, 7) to x^{2} = y-6 is 2x-y = -5.
It touches circle x^{2}+y^{2}+16x+12y+c = 0.
Hence length of perpendicular from centre (-8, -6) to tangent equals radius of circle.
d =
radius =
√(100-c) = √5
So c = 95
Answer: (b)
2. If L_{1} is the line of intersection of the planes 2x-2y+3z-2 = 0, x-y+z+1 = 0 and L_{2} is the line of intersection of the planes x+2y-z-3 = 0, 3x-y+2z-1 = 0, then the distance of the origin from the plane, containing the lines L_{1} and L_{2} is :
Solution:
L_{1} =
= l+m (drs of line L_{1})
L_{2} =
= 3l-5m-7n (drs of line L_{2})
Normal plane containing line L_{1} and L_{2} =
= 7l-7m-8n
For one point of line L_{1}
2x-2y+3z-2 = 0
x-y+z+1 = 0
Put x = 0, and solving above equations we get (0, 5, 4)
So the equation of the plane is -7(x-0)+7(y-5)-8(z-4) = 0
7x-7y+8z+3 = 0
Distance =
= 3/√162
= 1/3√2
Answer: (d)
3. If α and β ∈ C are the distinct roots of the equation x^{2}-x+1 = 0, then α^{101}+β^{107} is equal to:
Solution:
x^{2}-x+1 = 0
x =
= (1±i√3)/2
= (1+i√3)/2 , (1-i√3)/2
= -(-1-i√3)/2, -(-1+i√3)/2
= -ω^{2}, -ω
α= -ω^{2}
β = -ω
α^{101}+β^{107} = (-ω^{2})^{101}+(-ω)^{107}
= -(ω^{202}+ ω^{107})
= -[(ω^{3})^{67} ω+ (ω^{3})^{35} ω^{2}]
= -[ω +ω^{2}]
= 1
Answer: (a)
4. Tangents are drawn to the hyperbola 4x^{2}- y^{2} = 36 at the points P and Q. If these tangents intersect at the point T(0, 3) then the area (in sq. units) of ar(PTQ) is :
Solution:
Given equation of hyperbola is 4x^{2}-y^{2} = 36
(x^{2}/9)-(y^{2}/36) = 1
a^{2} = 9
b^{2 }= 36
From T(0, 3) tangents are drawn to hyperbola at P and Q.
Hence equation of Chord of contact PQ is
(x(0)/9) - (y(3)/36) = 1
y = -12
(x^{2}/9)-(144/36) = 1
x^{2} = 45
x = ±3√5
Hence P = (3√5, -12) and Q = (-3√5, -12)
Hence ar(PQT) is
Answer: (c)
5. If the curves y^{2} = 6x, 9x^{2 }+ by^{2} = 16 intersect each other at right angles, then the value of b is :
Solution:
Given y^{2} = 6x
Differentiating
2y(dy/dx) = 6
(dy/dx) = 3/y
Given 9x^{2 }+ by^{2} = 16
Differentiating
18x+2by(dy/dx) = 0
(dy/dx) = -18x/2by
(dy/dx) = -9x/by
Let the intersection point be (x_{1}, y_{1})
m_{1 }= 3/y_{1}
m_{2 }= -9x_{1} /by_{1}
Since the curves intersect at right angles, m_{1}m_{2} = -1
(3/y_{1})( -9x_{1} /by_{1}) = -1
27x_{1} = by_{1}^{2 } [Since (x_{1}, y_{1}) lies on y^{2} = 6x y_{1}^{2} = 6x_{1}]
27x_{1} = b(6x_{1})
b = 9/2
Answer: (b)
6. If the system of linear equations :
x+ky+3z = 0
3x+ky-2z = 0
2x+4y-3z = 0
has a non-zero solution (x, y, z), then xz/y^{2} is equal to :
Solution:
Given system of equations have non zero solutions.
So, Δ =
1(-3k+8)-k(-9+4)+3(12-2k) = 0
-3k+8+5k+36-6k = 0
4k = 44
k = 44/4 = 11
Now the given equations become
x+11y+3z = 0
3x+11y-2z = 0
2x+4y-3z = 0
x/(-22-33) = y/(-(-2-9) = z/(11-33)
x/-55 = y/11 = x/-22
x/5 = y/-1 = z/2 = L (let)
xz/y^{2} = (5L)(2L)/(-L^{2}) = 10
Answer: (d)
7. Let S = {x ∈R: x ≥ 0 and 2|√x-3|+√x(√x-6)+6 = 0}. Then S:
Solution:
2|√x-3|+√x(√x-6)+6 = 0
2√x-6+x-6√x+6 = 0 if √x >3
x-4√x = 0
√x = 0 or 4
√x = 4
Also 2(3-√x)+√x(√x-6)+6 = 0 if √x<3
6-2√x+x-6√x+6 = 0
x-8√x+12 = 0
(√x)^{2}-6√x-2√x+12 = 0
(√x-2)(√x-6) = 0
√x = 2, 6
x = 4
Answer: (a)
8. If sum of all the solutions of the equation 8cos x(cos ((π/6)+x). cos ((π/6)-(x))-(1/2)) = 1 in [0, π] is kπ, then k is equal to :
Solution:
8cos x(cos ((π/6)+x). cos ((π/6)-(1/2))-(1/2)) = 1
8.cos x[(cos (π/3)+ cos 2x-2)/2] = 1
4 cosx(cos 2x-(1/2)) = 1
4 cosx(2cos^{2}x-(3/2)) = 1
8cos^{3}x-6cosx-1 = 0
2(4cos^{3}x-3cosx)-1 = 0
2cos3x-1 = 0
cos 3x = 1/2 = cos(π/3)
3x = 2nπ ±π/3
x = (2nπ /3) ±(π/9)
= 7π/9, 5π/9, π/9 in [0, π]
k = 13/9
Answer: (d)
9. A bag contains 4 red and 6 black balls. A ball is drawn at random from the bag, its colour is observed and this ball along with two additional balls of the same colour are returned to the bag. If now a ball is drawn at random from the bag, then the probability that this drawn ball is red, is :
Solution:
From total Probability Theorem
P (R) = (4/10)×(1/2)+(6/10) × (4/12)
= (1/5)+(1/5)
= 2/5
Answer: (d)
10. Let f(x) = x^{2}+(1/x^{2}) and g(x) = x-(1/x), x∈ R-{-1,0,1}. If h(x) = f(x)/g(x), then the local minimum value of h(x) is:
Solution:
f(x) = x^{2}+(1/x^{2})
= (x-(1/x))^{2}+2
g(x) = x-(1/x)
Put x-(1/x) = t
h(x) = f(x)/g(x)
Let h(x) = k(t)
= (t^{2}+2)/t
= t+2/t
Differentiating
k’(t) = 1-2/t^{2}
For minima or maxima, 1-2/t^{2} = 0
t^{2} = 2
t = ±√2
k’’(t) = 4/t^{3}
At t = √2, k’’(t) = 2/√2
At t = -√2, k’’(t) = -2/√2
At t = √2. k’’(t) is positive.
So it is a minima.
h(x) = t+2/t
= √2+2/√2
= 2√2
Answer: (b)
11. Two sets A and B are as under :
A = {(a,b) ∈R×R:|a-5| < 1 and |b-5| < 1}
B = {(a,b) ∈R×R: 4(a-6)^{2}+9(b-5)^{2 }≤ 36}. Then:
Solution:
A = {(a,b) ∈R×R:|a-5| < 1 and |b-5| < 1}
-1<a-5<1
4<a<6
4<b<6
B = {(a,b) ∈R×R
4(a-6)^{2}+9(b-5)^{2} ≤ 36
((a-6)^{2 }/9 )+((b-5)^{2} /4) ≤ 1
Answer: (d)
12. The Boolean expression : ~(p ˅q) ˅(~p˄q) is equivalent to:
Solution:
p (1) |
q (2) |
~p (3) |
(p ˅q) (4) |
~(p ˅q) (5) |
~p˄q (6) |
~(p ˅q) ˅(~p˄q) |
T |
T |
F |
T |
F |
F |
F |
T |
F |
F |
T |
F |
F |
F |
F |
T |
T |
T |
F |
T |
T |
F |
F |
T |
F |
T |
F |
T |
Entries in column (3) and (7) are identical.
Answer: (c)
13. Tangent and normal are drawn at P(16, 16) on the parabola y^{2} = 16x, which intersect the axis of the parabola at A and B, respectively. If C is the centre of the circle through the points P, A and B and ∠CPB = θ, then a value of tan θ is :
Solution:
Tangent and normal are drawn at P (16, 16) on y^{2} = 16x
y^{2} = 16x
Slope of tangent dy/dx = 16/2y
= 8/y
(dy/dx)_{(16,16)} = 1/2
Equation of the tangent is
y-16 = (1/2)(x-16)
2y-32 = x-16
x-2y+16 = 0 ..(1)
Equation of normal
y-16 = -2(x-16)
2x+y-48 = 0 ..(2)
Tangent & normal intersect the axis of parabola
A(-16,0)
B(24,0)
Slope of AP×slope of PB
[(16-0)/(16+16)][(16-0)/(16-24)] = -1
So AB is the diameter of circle with centre C(4,0)
m_{PC} = (16-0)/(16-4) = 16/12 = 4/3
m_{PB} = (16-0)/(16-24) = 16/-8 = -2
tan = m_{1 }-m_{2}/(1+m_{1}m_{2})
= (4/3)+2)/(1+(4/3)×2)
= 2
Answer: (d)
14. If
Solution:
= (5x-4)(-1)(x+4)(x-4-2x)
= -1(5x-4)(x+4)(-x-4)
= (5x-4)(x+4)(x+4)^{2}
= (-4+5x)(x-(-4))^{2}
Hence A = -4, B = 5
Answer: (a)
15. The sum of the coefficients of all odd degree terms in the expansion of (x+√(x^{3}-1))^{5}+(x-√(x^{3}-1))^{5}, (x>) is :
Solution:
(a+b)^{n}+(a-b)^{n} = 2(^{n}C_{0}a^{n} + ^{n}C_{2}a^{n-2}b^{2}+^{ n}C_{4}a^{n-4}b^{4}…)
(x+√(x^{3}-1))^{5}+(x-√(x^{3}-1))^{5}, (x>)
= 2(^{5}C_{0}x^{5}+^{5}C_{2}x^{3}(√(x^{3}-1))^{2}+^{5}C_{4}x(√(x^{3}-1))^{4})
= 2(x^{5}+10x^{6}-10x^{3}+5x^{7}-10x^{4}+5x)
Sum of the coefficient of odd term is given by
2(1-10+5+5)
= 2
Answer: (b)
16. Let a_{1}, a_{2}, a_{3},..a_{49} be in A.P such that Ʃ_{k=0}^{12} a_{4k+1} = 416 and a_{9}+a_{43} = 66. If a_{1}^{2}+a_{2}^{2}+…+a_{17}^{2} = 140m, then m is equal to:
Solution:
a_{1}, a_{2}, a_{3},..a_{49} are in A.P
Let A be the first term and D is the common difference.
a_{9}+a_{43} = 66
A+8D+A+42D = 66
2A+50D = 66
A+25D = 33
a_{26} = 33 ..(i)
Ʃ_{k=0}^{12} a_{4k+1}^{= 416}
a_{1}+a_{5}+a_{9}+…+a_{49} = 416
13A+312D = 416
A+24D = 23
a_{25} = 32 ..(ii)
From (i) and (ii), a_{26}-a_{25} = D = 1
Also A+25D = 33
So A = 8
a_{1}^{2}+a_{2}^{2}+…+a_{17}^{2} = 8^{2}+9^{2}+10^{2}+…+24^{2}
Ʃ_{r=1}^{24} r^{2}- Ʃ_{r=1}^{7} r^{2 }= 140m
(24×25×49/6)-(7× 8 ×15/6) = 140m
4900-140 = 140m
m = 34
Answer: (a)
17. A straight line through a fixed point (2, 3) intersects the coordinate axes at distinct points P and Q. If O is the origin and the rectangle OPRQ is completed, then the locus of R is :
Solution:
line y-3 = m(x-2)
y-3 = mx-2m
P = ((2m-3)/m, 0)
Q = (0, 3-2m)
Let R(α, β)
So α= ((2m-3)/m, 0) and β = 3-2m
m = 3/(2-α) and m = (3-β)/2
3/(2-α) = (3-β)/2
6 = 6-2β-3α+αβ
Locus of R (α,β) is 3x+2y = xy
Answer: (a)
18. The value of
Solution:
Answer: (b)
19. Let g(x) = cos x^{2}, f(x) = √x and α, β (α< β) be the roots of the quadratic equation 18x^{2}-9πx+π^{2 }= 0. Then the area in sq.units) bounded by the curve y = (gof)(x) and the lines x = αand x = β and y =0 is:
Solution:
18x^{2}-9πx+π^{2 }= 0
x = π/6, π/3 (so α= π/6, β= π/3)
y = (gof)(x)
= g(f(x))
= g(√x)
= cos x
A =
= (√3/2)-(1/2)
= (√3-1)/2
Answer: (c)
20. For each t ∈ R, let [t] be the greatest integer less than or equal to t. Then
Solution:
Answer: (a)
21. If ∑^{9}_{i=1} (x_{i}-5) = 9 and ∑^{9}_{i=1} (x_{i}-5)^{2} = 45, then the standard deviation of the 9 items x_{1}, x_{2},…x_{9} is:
Solution:
Variance = (1/n) Ʃ_{i=1}^{n} x_{i}^{2}-[(1/n) Ʃ_{i=1}^{n} x_{i}]^{2}
= (1/9)×45-((1/9)×9)^{2}
= 5-1
= 4
S.D = √4 = 2
Answer: (a)
22. The integral
where C is the constant of integration.
Solution:
Answer: (d)
23. Let S = {t∈R: f(x) = |x-π|(e^{|x|}-1)sin x is not differentiable at t}. Then the set S is equal to:
Solution:
f(x) = |x-π|(e^{|x|}-1) sinx
Obviously differentiable at x = 0
Check at x = π
So, differentiable at x = π.
Answer: (c)
24. Let y = y(x) be the solution of the differential equation sin x(dy/dx)+y cosx = 4x, x∈(0,π). If y(π/2) = 0, then y(π/6) is equal to:
Solution:
(dy/dx)+y cotx = 4x/sin x
IF = e^{∫cot xdx} = e^{log|sin x|} = sin x as x∈(0,π)
y sin x = c+∫4x dx
= c+2x^{2}
As y(π/2) = 0 0.sin(π/2) = c+2(π/2)^{2}
c = -(π^{2}/2)
So y sin x = 2x^{2}-(π^{2}/2)
Put x = π/6
y sin π/6 = 2(π/6)^{2}-(π^{2}/2)
y(1/2) = (π^{2}/18)- (π^{2}/2)
= (-8π^{2}/9)
= (-8/9)π^{2}
Answer: (a)
25. Let u be a vector coplanar with the vectors
and
Solution:
As coplanar
4u_{1}-2u_{2}+2u_{3} = 0
2u_{1}-u_{2}+u_{3} = 0 ..(1)
2u_{1}-3u_{2}-u_{3} = 0 ..(2)
u_{2}+u_{3}= 24 ..(3)
Solving u_{1} = -4, u_{2 }= 8, u_{3} = 16
Answer: (c)
26. The length of the projection of the line segment joining the points (5, -1, 4) and (4, -1, 3) on the plane, x+y+z = 7 is :
Solution:
Direction ratios of AB = (1,0,1)
Let be angle between line AB and normal of plane.
So cos = 1×1+0×1+1×1)/√2√3
= √2/√3
So projection of line AB =
= √2×√(1-2/3)
= √(2/3)
Answer: (b)
27. PQR is a triangular park with PQ = PR = 200 m. A T.V. tower stands at the mid-point of QR. If the angles of elevation of the top of the tower at P, Q and R are respectively 45^{0}, 30^{0} and 30^{0}, then the height of the tower (in m) is :
Solution:
Let TW = h
ΔPTW, tan 45^{0} = h/y_{1}
h = y_{1}
Similarly, h/y_{2} = tan 30^{0}
y_{2} = √3h
ΔPQT, PQ^{2} = QT^{2}+PT^{2}
40000 = 3h^{2}+h^{2}
4h^{2} = 40000
h = 100
Answer: (c)
28. From 6 different novels and 3 different dictionaries, 4 novels and 1 dictionary are to be selected and arranged in a row on a shelf so that the dictionary is always in the middle. The number of such arrangements is :
Solution:
We have 6 novels and 3 dictionaries. We can select 4 novels and 1 dictionary in
^{6}C_{4 }×^{3}C_{1} = 6!×3/(4!2!)
= 6×5×3/2
= 45 ways.
Now 4 novels and 1 dictionary are to be arranged so that dictionary is always in middle. So remaining 4 novels can be arranged in 4! ways.
Hence total arrangements possible are
45×24 = 1080 ways
Answer: (c)
29. Let A be the sum of the first 20 terms and B be the sum of the first 40 terms of the series.
1^{2 }+2.2^{2}+3^{2}+2.4^{2}+5^{2 }+ 2.6^{2}+..
If B-2A = 100λ, then λ is equal to :
Solution:
B = (1^{2}+3^{2}+5^{2}+…+39^{2})+2(2^{2}+4^{2}+…+40^{2})
= (1^{2}+2^{2}+3^{2}+…+40^{2})+(2^{2}+4^{2}+…+40^{2})
= (1^{2}+2^{2}+3^{2}+…+40^{2})+4(1^{2}+2^{2}+3^{2}+…+20^{2})
A = (1^{2}+2^{2}+3^{2}+…+20^{2})+4(1^{2}+2^{2}+3^{2}+…+10^{2})
Using 1^{2}+2^{2}+..n^{2} = n(n+1)(2n+1)/6
B-2A = 24800
Hence λ = 248
Answer: (d)
30. Let the orthocentre and centroid of a triangle be A(-3, 5) and B(3, 3) respectively. If C is the circumcentre of this triangle, then the radius of the circle having line segment AC as diameter, is:
Solution:
As we know centroid divides line joining circumcentre and orthocentre internally 1:2.
So, C(6,2)
AC = √[(6+3)^{3}+(2-5)^{2}]
= √90
= 3√10
r = AC/2
= 3√10/2
= 3√(5/2)
Answer: (a)