JEE Main 2018 Paper With Solutions Maths (Jan 10 Shift 1) - Download PDF

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

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 =

\(\begin{array}{l}\left | \frac{-16+6+5}{\sqrt{2^{2}+(-1)^{2}}} \right |= \sqrt{5}\end{array} \)

radius =

\(\begin{array}{l}\sqrt{64+36-c}\end{array} \)

√(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

L₁ =

\(\begin{array}{l}\begin{vmatrix} l &m &n \\ 2& -2 & 3\\ 1 & -1 & 1 \end{vmatrix}\end{array} \)

= l+m (drs of line L₁)

L₂ =

\(\begin{array}{l}\begin{vmatrix} l &m &n \\ 1& 2 & -1\\ 3 & -1 & 2 \end{vmatrix}\end{array} \)

= 3l-5m-7n (drs of line L₂)

Normal plane containing line L₁ and L₂ =

\(\begin{array}{l}\begin{vmatrix} l &m &n \\ 1& 1 & 0\\ 3 & 5 & -7 \end{vmatrix}\end{array} \)

= 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 =

\(\begin{array}{l}\left | \frac{7\times 0-7\times 0+8\times 0+3}{\sqrt{7^{2}+7^{2}+8^{2}}} \right |\end{array} \)

= 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

x²-x+1 = 0

x =

\(\begin{array}{l}\frac{1\pm \sqrt{1-4\times 1\times 1}}{2\times 1}\end{array} \)

= (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

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

\(\begin{array}{l}\frac{1}{2}\begin{vmatrix} 3\sqrt{5} & -12 & 1\\ -3\sqrt{5} & -12&1 \\ 0& 3 & 1 \end{vmatrix} = 45\sqrt{5}\end{array} \)

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

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

Given system of equations have non zero solutions.

So, Δ =

\(\begin{array}{l}\begin{vmatrix} 1 & k &3 \\ 3 & k & -2\\ 2&4 & -3 \end{vmatrix}= 0\end{array} \)

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.

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

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

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

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

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

JEE Main 2018 Papers With Solutions Jan 10 Maths Shift 1 Set B

Answer: (d)

12. The Boolean expression : ~(p ˅q) ˅(~p˄q) is equivalent to:
a) q
b) ~q
c) ~p
d) p

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

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)

m_PC = (16-0)/(16-4) = 16/12 = 4/3

m_PB = (16-0)/(16-24) = 16/-8 = -2

tan = m₁-m₂/(1+m₁m₂)

= (4/3)+2)/(1+(4/3)×2)

= 2

Answer: (d)

14. If

\(\begin{array}{l}\begin{vmatrix} x-4 & 2x &2x \\ 2x& x-4 & 2x\\ 2x& 2x & x-4 \end{vmatrix}\end{array} \)

= (A+Bx)(x-A)² then the ordered pair (A, B) is equal to :
a) (-4, 5)
b) (4, 5)
c) (-4, -5)
d) (-4, 3)

= (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

(a+b)ⁿ+(a-b)ⁿ = 2(ⁿC₀aⁿ + ⁿC₂a^n-2b²+ⁿC₄a^n-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 Ʃ_k=0¹² a_4k+1 = 416 and a₉+a₄₃ = 66. If a₁²+a₂²+…+a₁₇² = 140m, then m is equal to:
a) 34
b) 33
c) 66
d) 68

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)

Ʃ_k=0¹² a_4k+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

Ʃ_r=1²⁴ r²– Ʃ_r=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

JEE Main Solution 2018 Maths Papers Jan 10 Shift 1 Set B

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

\(\begin{array}{l}\int_{\frac{-\pi }{2}}^{\frac{\pi }{2}}\frac{\sin ^{2}x}{1+2^{x}}dx\end{array} \)

is:
a) 4π
b) π/4
c) π/8
d) π/2

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)

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 =

\(\begin{array}{l}\int_{\frac{\pi }{6}}^{\frac{\pi }{3}}\cos x dx = [\sin x]^{\frac{\pi }{3}}_{\frac{\pi }{6}}\end{array} \)

= (√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

\(\begin{array}{l}\lim_{x\to0+}x([\frac{1}{x}]+[\frac{2}{x}]+…[\frac{15}{x}])\end{array} \)

a) is equal to 120
b) does not exist (in R)
c) is equal to 0
d) is equal to 15

JEE Main Jan 10 Shift 1 2018 Solved Maths Papers Set B

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 x₁, x₂,…x₉ is:
a) 2
b) 3
c) 9
d) 4

Variance = (1/n) Ʃ_i=1ⁿ x_i²-[(1/n) Ʃ_i=1ⁿ x_i]²

= (1/9)×45-((1/9)×9)²

= 5-1

= 4

S.D = √4 = 2

Answer: (a)

22. The integral

\(\begin{array}{l}\int \frac{\sin ^{2}x\cos ^{2}x}{\sin ^{5}x+\cos ^{3}x\sin ^{2}x+\sin ^{3}x\cos ^{2}x+\cos ^{5}x}dx\end{array} \)

is equal to:

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

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}

f(x) = |x-π|(e^|x|-1) sinx

Obviously differentiable at x = 0

Check at x = π

Shift 1 Set B 2018 JEE Main Jan 10 Solved Maths Papers

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

(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+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

\(\begin{array}{l}\vec{a}= 2\hat{i}+3\hat{j}-\hat{k}\end{array} \)

and

\(\begin{array}{l}\vec{b}= \hat{j}+\hat{k}\end{array} \)

. If vector u is perpendicular to vector a and

\(\begin{array}{l}\vec{u}.\vec{b}= 24\end{array} \)

, then u² is equal to:
a) 256
b) 84
c) 336
d) 315

\(\begin{array}{l}\vec{u}= u_{1}\hat{i}+u_{2}\hat{j}+u_{3}\hat{k}\end{array} \)

As coplanar

\(\begin{array}{l}\begin{vmatrix} u_{1} & u_{2} &u_{3} \\ 2& 3 & -1\\ 0&1 & 1 \end{vmatrix}=0\end{array} \)

4u₁-2u₂+2u₃ = 0

2u₁-u₂+u₃ = 0 ..(1)

\(\begin{array}{l}\vec{u}.\vec{a}=0\end{array} \)

2u₁-3u₂-u₃ = 0 ..(2)

\(\begin{array}{l}\vec{u}.\vec{b}=0\end{array} \)

u₂+u₃= 24 ..(3)

Solving u₁ = -4, u₂= 8, u₃ = 16

\(\begin{array}{l}|\vec{u^2}|=336\end{array} \)

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

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 =

\(\begin{array}{l}|\vec{AB}|sin \theta\end{array} \)

= √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

Shift 1 Set B Jan 10 2018 JEE Main Solved Maths Papers

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

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

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

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

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