﻿ For each t ∈R, let [t] be the greatest integer less than or equal to t. Thenlimx→0+ x1x+2x+......+15x from Mathematics JEE Year 2018 Free Solved Previous Year Papers

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# JEE Mathematics Solved Question Paper 2018

#### Multiple Choice Questions

1.

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,
then a value of tan θ is

• 4/3

• 1/2

• 2

• 3

C.

2

y2 = 16x

Tangent at P (16,6) is 2y = x + 16..... (1)
Normal at P (16,16) is y = -2x + 48... (2)
i.e., A is (-16,0); B is (24,0)
Now Centre of circle is (4,0)

Now, mPC = 4/3
mPB = -2 # 2.For each t ∈R, let [t] be the greatest integer less than or equal to t. Thendoes not exist (in R) is equal to 0 is equal to 15 is equal to 120

D.

is equal to 120

3.

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

• At least 750 but less than 1000

• At least 1000

• Less then 500

• At least 500 but less than 750

B.

At least 1000

Number of ways of selecting 4 novels from 6 novels
= 6C4
Number of ways of selecting 1 dictionary from
3 dictionaries = 3C1
Required arrangements = 6C4 × 3C1 × 4! = 1080
=> At least 1000

4.

If α, β ∈ C are the distinct roots, of the equation x2 -x + 1 = 0, then α101 + β107 is equal to

• 2

• -1

• 0

• 1

D.

1

x2-x + 1 = 0

Roots are -ω, -ω2

Let α = -ω, β = -ω2

α101 + β107 = (-ω)101 + (-ω2)107

= -( ω101214)
= - (ω2 + ω)
= 1

5.

The Boolean expression ~ (p v q )v (-p ∧q ) is equivalent to

• ~q

• ~ p

• p

• q

B.

~ p

~ (p v q )v (p ∧q )

By property (-~ p ∧- q )v (-p ∧ q )

~p

6.

Two sets A and B are as under:

A = {(a-b)∈ RxR:|a-5|<1 and |b-5|<1}

B = {(a,b)∈ Rx R: 4(a-6)2 + 9 (b-5)2 ≤ 36},then

• Neither A ⊂ B nor B ⊂ A

• B ⊂ A

• A ⊂ B

C.

A ⊂ B

As, |a-5| < 1 and |b-5| < 1

Taking axes as a-axis and b-axis The set A represents square PQRS inside set B representing ellipse and hence A ⊂ B

7.

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

• 95

• 195

• 185

• 85

A.

95

Equation of tangent at (1,7) to curve x2 = y -6 is

2x - y +5 = 0 ... (i)

Centre of circle  = (-8,-6)

8.

The sum of the coefficients of all odd degree terms in the expansion of

• 2

• -1

• 0

• 1

A.

2

Sum of odd degree terms coefficients
= 2(5 + 1 – 10 + 5)
= 2

9.

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 45o, 30o and 30o, then the height of the tower (in m) is

• $50\sqrt{2}$

• 100

• 50

• $100\sqrt{3}$

B.

100 Let ST = h (height of tower)

PT = ST = h

10.

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 △PTQ is

• $36\sqrt{5}$

• $45\sqrt{5}$

• $54\sqrt{3}$

• $60\sqrt{3}$

B.

$45\sqrt{5}$ Clearly, PQ is a chord of contact,
i.e., the equation of PQ is T = 0
=> y = –12

Solving with curve, 4x2 - y2 = 36