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.
January 10 Shift 1 - Maths
1. If the tangent at (1, 7) to the curve x2 = y-6 touches the circle x2+y2+16x+12y+c = 0, then the value of c is :
- a) 85
- b) 95
- c) 195
- d) 185
Solution:
-
Equation of tangent at (1, 7) to x2 = y-6 is 2x-y = -5.
It touches circle x2+y2+16x+12y+c = 0.
Hence length of perpendicular from centre (-8, -6) to tangent equals radius of circle.
d =
\(\left | \frac{-16+6+5}{\sqrt{2^{2}+(-1)^{2}}} \right |= \sqrt{5}\)radius =
\(\sqrt{64+36-c}\)√(100-c) = √5
So c = 95
Answer: (b)
2. If L1 is the line of intersection of the planes 2x-2y+3z-2 = 0, x-y+z+1 = 0 and L2 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 L1 and L2 is :
- a) 1/2√2
- b) 1/√2
- c) 1/4√2
- d) 1/3√2
Solution:
-
L1 =
\(\begin{vmatrix} l &m &n \\ 2& -2 & 3\\ 1 & -1 & 1 \end{vmatrix}\)= l+m (drs of line L1)
L2 =
\(\begin{vmatrix} l &m &n \\ 1& 2 & -1\\ 3 & -1 & 2 \end{vmatrix}\)= 3l-5m-7n (drs of line L2)
Normal plane containing line L1 and L2 =
\(\begin{vmatrix} l &m &n \\ 1& 1 & 0\\ 3 & 5 & -7 \end{vmatrix}\)= 7l-7m-8n
For one point of line L1
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 =
\(\left | \frac{7\times 0-7\times 0+8\times 0+3}{\sqrt{7^{2}+7^{2}+8^{2}}} \right |\)= 3/√162
= 1/3√2
Answer: (d)
3. If α and β ∈ C are the distinct roots of the equation x2-x+1 = 0, then α101+β107 is equal to:
- a) 1
- b) 2
- c) -1
- d) 0
Solution:
-
x2-x+1 = 0
x =
\(\frac{1\pm \sqrt{1-4\times 1\times 1}}{2\times 1}\)= (1±i√3)/2
= (1+i√3)/2 , (1-i√3)/2
= -(-1-i√3)/2, -(-1+i√3)/2
= -ω2, -ω
α= -ω2
β = -ω
α101+β107 = (-ω2)101+(-ω)107
= -(ω202+ ω107)
= -[(ω3)67 ω+ (ω3)35 ω2]
= -[ω +ω2]
= 1
Answer: (a)
4. Tangents are drawn to the hyperbola 4x2- y2 = 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 :
- a) 60√3
- b) 36√5
- c) 45√5
- d) 54√3
Solution:
-
Given equation of hyperbola is 4x2-y2 = 36
(x2/9)-(y2/36) = 1
a2 = 9
b2 = 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
(x2/9)-(144/36) = 1
x2 = 45
x = ±3√5
Hence P = (3√5, -12) and Q = (-3√5, -12)
Hence ar(PQT) is
\(\frac{1}{2}\begin{vmatrix} 3\sqrt{5} & -12 & 1\\ -3\sqrt{5} & -12&1 \\ 0& 3 & 1 \end{vmatrix} = 45\sqrt{5}\)Answer: (c)
5. If the curves y2 = 6x, 9x2 + by2 = 16 intersect each other at right angles, then the value of b is :
- a) 4
- b) 9/2
- c) 6
- d) 7/2
Solution:
-
Given y2 = 6x
Differentiating
2y(dy/dx) = 6
(dy/dx) = 3/y
Given 9x2 + by2 = 16
Differentiating
18x+2by(dy/dx) = 0
(dy/dx) = -18x/2by
(dy/dx) = -9x/by
Let the intersection point be (x1, y1)
m1 = 3/y1
m2 = -9x1 /by1
Since the curves intersect at right angles, m1m2 = -1
(3/y1)( -9x1 /by1) = -1
27x1 = by12 [Since (x1, y1) lies on y2 = 6x y12 = 6x1]
27x1 = b(6x1)
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/y2 is equal to :
- a) -30
- b) 30
- c) -10
- d) 10
Solution:
-
Given system of equations have non zero solutions.
So, Δ =
\(\begin{vmatrix} 1 & k &3 \\ 3 & k & -2\\ 2&4 & -3 \end{vmatrix}= 0\)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/y2 = (5L)(2L)/(-L2) = 10
Answer: (d)
7. Let S = {x ∈R: x ≥ 0 and 2|√x-3|+√x(√x-6)+6 = 0}. Then S:
- a) contains exactly two elements.
- b) contains exactly four elements.
- c) is an empty set.
- d) contains exactly one element.
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 :
- a) 8/9
- b) 20/9
- c) 2/3
- d) 13/9
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(2cos2x-(3/2)) = 1
8cos3x-6cosx-1 = 0
2(4cos3x-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 :
- a) 1/5
- b) 3/4
- c) 3/10
- d) 2/5
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) = x2+(1/x2) 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:
- a) -2√2
- b) 2√2
- c) 3
- d) -3
Solution:
-
f(x) = x2+(1/x2)
= (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)
= (t2+2)/t
= t+2/t
Differentiating
k’(t) = 1-2/t2
For minima or maxima, 1-2/t2 = 0
t2 = 2
t = ±√2
k’’(t) = 4/t3
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:
- a) A⋂B = phi(an empty set)
- b) neither A⊂B nor B⊂A
- c) B⊂A
- d) A⊂B
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:
- a) q
- b) ~q
- c) ~p
- d) p
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 y2 = 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 :
- a) 3
- b) 4/3
- c) 1/2
- d) 2
Solution:
-
Tangent and normal are drawn at P (16, 16) on y2 = 16x
y2 = 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)
mPC = (16-0)/(16-4) = 16/12 = 4/3
mPB = (16-0)/(16-24) = 16/-8 = -2
tan = m1 -m2/(1+m1m2)
= (4/3)+2)/(1+(4/3)×2)
= 2
Answer: (d)
14. If
- a) (-4, 5)
- b) (4, 5)
- c) (-4, -5)
- d) (-4, 3)
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+√(x3-1))5+(x-√(x3-1))5, (x>) is :
- a) 1
- b) 2
- c) -1
- d) 0
Solution:
-
(a+b)n+(a-b)n = 2(nC0an + nC2an-2b2+ nC4an-4b4…)
(x+√(x3-1))5+(x-√(x3-1))5, (x>)
= 2(5C0x5+5C2x3(√(x3-1))2+5C4x(√(x3-1))4)
= 2(x5+10x6-10x3+5x7-10x4+5x)
Sum of the coefficient of odd term is given by
2(1-10+5+5)
= 2
Answer: (b)
16. Let a1, a2, a3,..a49 be in A.P such that Ʃk=012 a4k+1 = 416 and a9+a43 = 66. If a12+a22+…+a172 = 140m, then m is equal to:
- a) 34
- b) 33
- c) 66
- d) 68
Solution:
-
a1, a2, a3,..a49 are in A.P
Let A be the first term and D is the common difference.
a9+a43 = 66
A+8D+A+42D = 66
2A+50D = 66
A+25D = 33
a26 = 33 ..(i)
Ʃk=012 a4k+1 = 416
a1+a5+a9+…+a49 = 416
13A+312D = 416
A+24D = 23
a25 = 32 ..(ii)
From (i) and (ii), a26-a25 = D = 1
Also A+25D = 33
So A = 8
a12+a22+…+a172 = 82+92+102+…+242
Ʃr=124 r2- Ʃr=17 r2 = 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 :
- a) 3x+2y = xy
- b) 3x+2y = 6xy
- c) 3x+2y = 6
- d) 2x+3y = xy
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
- a) 4π
- b) π/4
- c) π/8
- d) π/2
Solution:
-
Answer: (b)
19. Let g(x) = cos x2, f(x) = √x and α, β (α< β) be the roots of the quadratic equation 18x2-9πx+π2 = 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:
- a) (1/2)(√3-√2)
- b) (1/2)(√2-1)
- c) (1/2)(√3-1)
- d) (1/2)(√3+1)
Solution:
-
18x2-9πx+π2 = 0
x = π/6, π/3 (so α= π/6, β= π/3)
y = (gof)(x)
= g(f(x))
= g(√x)
= cos x
A =
\(\int_{\frac{\pi }{6}}^{\frac{\pi }{3}}\cos x dx = [\sin x]^{\frac{\pi }{3}}_{\frac{\pi }{6}}\)= (√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
- a) is equal to 120
- b) does not exist (in R)
- c) is equal to 0
- d) is equal to 15
Solution:
-
Answer: (a)
21. If ∑9i=1 (xi-5) = 9 and ∑9i=1 (xi-5)2 = 45, then the standard deviation of the 9 items x1, x2,…x9 is:
- a) 2
- b) 3
- c) 9
- d) 4
Solution:
-
Variance = (1/n) Ʃi=1n xi2-[(1/n) Ʃi=1n xi]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.
- a) 1/(1+cot3x) + C
- b) -1/(1+cot3x) + C
- c) 1/3(1+tan3x) + C
- d) -1/3(1+tan3x) + C
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:
- a) {π}
- b) {0,π}
- c) (an empty set)
- d) {0}
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:
- a) (-8/9)π2
- b) (-4/9)π2
- c) (4/9√3)π2
- d) (-8/9√3)π2
Solution:
-
(dy/dx)+y cotx = 4x/sin x
IF = e∫cot xdx = elog|sin x| = sin x as x∈(0,π)
y sin x = c+∫4x dx
= c+2x2
As y(π/2) = 0 0.sin(π/2) = c+2(π/2)2
c = -(π2/2)
So y sin x = 2x2-(π2/2)
Put x = π/6
y sin π/6 = 2(π/6)2-(π2/2)
y(1/2) = (π2/18)- (π2/2)
= (-8π2/9)
= (-8/9)π2
Answer: (a)
25. Let u be a vector coplanar with the vectors
and
- a) 256
- b) 84
- c) 336
- d) 315
Solution:
-
\(\vec{u}= u_{1}\hat{i}+u_{2}\hat{j}+u_{3}\hat{k}\)
As coplanar
\(\begin{vmatrix} u_{1} & u_{2} &u_{3} \\ 2& 3 & -1\\ 0&1 & 1 \end{vmatrix}=0\)4u1-2u2+2u3 = 0
2u1-u2+u3 = 0 ..(1)
\(\vec{u}.\vec{a}=0\)2u1-3u2-u3 = 0 ..(2)
\(\vec{u}.\vec{b}=0\)u2+u3= 24 ..(3)
Solving u1 = -4, u2 = 8, u3 = 16
\(|\vec{u^2}|=336\)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 :
- a) 1/3
- b) √(2/3)
- c) 2/√3
- d) 2/3
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 =
\(|\vec{AB}|sin \theta\)= √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 450, 300 and 300, then the height of the tower (in m) is :
- a) 100√3
- b) 50√2
- c) 100
- d) 50
Solution:
-
Let TW = h
ΔPTW, tan 450 = h/y1
h = y1
Similarly, h/y2 = tan 300
y2 = √3h
ΔPQT, PQ2 = QT2+PT2
40000 = 3h2+h2
4h2 = 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 :
- a) at least 500 but less than 750
- b) at least 750 but less than 1000
- c) at least 1000
- d) less than 500
Solution:
-
We have 6 novels and 3 dictionaries. We can select 4 novels and 1 dictionary in
6C4 ×3C1 = 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.
12 +2.22+32+2.42+52 + 2.62+..
If B-2A = 100λ, then λ is equal to :
- a) 464
- b) 496
- c) 232
- d) 248
Solution:
-
B = (12+32+52+…+392)+2(22+42+…+402)
= (12+22+32+…+402)+(22+42+…+402)
= (12+22+32+…+402)+4(12+22+32+…+202)
A = (12+22+32+…+202)+4(12+22+32+…+102)
Using 12+22+..n2 = 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:
- a) 3√(5/2)
- b) 3√5/2
- c) √10
- d) 2√10
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)
Video Lessons - January 10 Shift 1 Maths
JEE Main 2018 Maths Paper With Solutions Shift 1 January 10
