**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 :

a) 85

b) 95

c) 195

d) 185

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 :

a) 1/2√2

b) 1/√2

c) 1/4√2

d) 1/3√2

L_{1} =

= *l*+m (drs of line L_{1})

L_{2} =

= 3*l*-5m-7n (drs of line L_{2})

Normal plane containing line L_{1} and L_{2} =

= 7*l*-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:

a) 1

b) 2

c) -1

d) 0

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 :

a) 60√3

b) 36√5

c) 45√5

d) 54√3

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 :

a) 4

b) 9/2

c) 6

d) 7/2

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 :

a) -30

b) 30

c) -10

d) 10

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:

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(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 :

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) = 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:

a) -2√2

b) 2√2

c) 3

d) -3

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:

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

**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 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 :

a) 3

b) 4/3

c) 1/2

d) 2

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)] = -1So 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

^{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+√(x

^{3}-1))

^{5}+(x-√(x

^{3}-1))

^{5}, (x>) is :

a) 1

b) 2

c) -1

d) 0

(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:

a) 34

b) 33

c) 66

d) 68

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 :

a) 3x+2y = xy

b) 3x+2y = 6xy

c) 3x+2y = 6

d) 2x+3y = xy

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

**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:

a) (1/2)(√3-√2)

b) (1/2)(√2-1)

c) (1/2)(√3-1)

d) (1/2)(√3+1)

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

a) is equal to 120

b) does not exist (in R)

c) is equal to 0

d) is equal to 15

**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:

a) 2

b) 3

c) 9

d) 4

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.

a) 1/(1+cot^{3}x) + C

b) -1/(1+cot^{3}x) + C

c) 1/3(1+tan^{3}x) + C

d) -1/3(1+tan^{3}x) + 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

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)π

^{2}

b) (-4/9)π

^{2}

c) (4/9√3)π

^{2}

d) (-8/9√3)π

^{2}

(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

^{2}is equal to:

a) 256

b) 84

c) 336

d) 315

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 :

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 =

= √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 :

a) 100√3

b) 50√2

c) 100

d) 50

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 :

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

^{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 :

a) 464

b) 496

c) 232

d) 248

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:

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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