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This mock test of Mathematics Test 3 - Entire XI Syllabus for JEE helps you for every JEE entrance exam.
This contains 30 Multiple Choice Questions for JEE Mathematics Test 3 - Entire XI Syllabus (mcq) to study with solutions a complete question bank.
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QUESTION: 1

Value of lim_{x → 0}(1+Sin(x))^{Cosec(x)}

Solution:

lim_{x → 0}(1+Sin(x))^{Cosec(x)}

Put sin(x) = t we get

lim_{t → 0}(1+t)^{(1⁄t)} = e.

QUESTION: 2

The number of straight lines that can be drawn out of 10 points of which 7 are collinear is

Solution:

1. One line passing through 7 collinear points

2. 7 lines passing from each collinear point to non collinear one (3 times as there are 3 non-collinear points)

3. 3 lines joining each pair of non-collinear points 1+(7*3)+3 1+21+3 =25

QUESTION: 3

If the correlation coefficient between two variables is 1, then the two least square lines of regression are

Solution:

QUESTION: 4

The real part of is

Solution:

First we consider theta as x

1/1-cosx+isinx=1/2cos^2 x/2 + 2isinx/2cosx/2

=sec(x/2)/2 {1/cosx/2 + isinx/2}

=sec(x/2)/2 (1/e^ix/2). (e^ix=cosx+isinx){eulers theorem)

=sec(x/2)/2 e^-ix/2

=sec(x/2)/2 [ cos(-x/2) + i sun(-x/2)]

=1/2secx/2.cosx/2.-1/2itanx/2

=1/2.-i(1/2tanx/2)

=Re(z)=1/2

QUESTION: 5

The value of log3 tan1º + log3 tan2º +....+log3 tan 89º is

Solution:

Loga+log=log(ab)

tan(90-A)=cotA

log(tab.tan2......tan45........cot2.cot1). tan45=1

tanA.cotA=1

=log(1)=0

QUESTION: 6

If the solution of quadratic equation x^{2}-11x+22 are x=3 and x=6, then the base of the number is

Solution:

QUESTION: 7

1^{2} + (1^{2} + 2^{2}) + (1^{2} + 2^{2} + 3^{2 })+..... upto 22^{nd} term is

Solution:

QUESTION: 8

In ΔABC, ∠B = 90^{o} and b a = 4. The area of the triangle is the maximum when ∠C is

Solution:

QUESTION: 9

Let E be the ellipse and C be the circle x^{2} y^{2} = 4. Let P and Q be the point (1,2) and (2,1) respectively. Then

Solution:

QUESTION: 10

The third term of a G.P. is 4. The products of the first five term is

Solution:

QUESTION: 11

The sum of the infinIte of terms of G.P. is 20, and the sum of their square is 100, then the first term of

Solution:

QUESTION: 12

If cos^{-1} x/2 + cos ^{-1} y/3 = θ Then 9x^{2} - 12xy cos θ + 4y^{2} is

Solution:

QUESTION: 13

The foci of the ellipse 25 (x+1)^{2} + 9(y+2)^{2} = 225 are at

Solution:

QUESTION: 14

If A_{1}, A_{2 }be two A.M’s and G_{1}, G_{2} be two G..M.’s between a and b, then ( A_{1}, A_{2 })/G_{1},G_{2} is equal to

Solution:

QUESTION: 15

A tree is broken by wind, its upper part touches the ground at a point 10 meters from the foot of the tree and makes an angle of 60º with the ground the entire lenght of the tree of

Solution:

QUESTION: 16

How many diagonals can be drawn in a polygon of 15 sides ?

Solution:

QUESTION: 17

9x^{2} - 16y^{2} = 144 represents

Solution:

QUESTION: 18

If then r is equal to

Solution:

QUESTION: 19

Coefficient of X^{4} in the expansion of is

Solution:

QUESTION: 20

The locus of the mid-point of the chords of a circle x^{2} + y^{2} =4, which subtended a right angle at the centre is

Solution:

QUESTION: 21

The real root of the equation 7 ^{log7 (x2 - 4x + 5)} = x-1 is

Solution:

QUESTION: 22

A lady gives a dinner party for six guests.the number of ways in which they may be selected from among ten friends,if two of the friends will not attend the party together is

Solution:
First keep aside those two friends who doesn't want to come together and select 6 members from rest of 8 members.=8c2= 28

Now take one of those two and select 5 from the other 8 members=8c5=56

Now do the same with other one of the two friends =8c5=56

Now add all =(56Ã—2) +28 =140

Now take one of those two and select 5 from the other 8 members=8c5=56

Now do the same with other one of the two friends =8c5=56

Now add all =(56Ã—2) +28 =140

QUESTION: 23

Find the equation of the circle through the intersection of the circles x^{2}+ y^{2} - x + 7y - 3 = 0 and x^{2} + y^{2} - 5x - y + 1 = 0, having its centre on the line x + y = 0.

Solution:

x^{2} + y^{2 }- x + 7y - 3 + λ(x^{2} + y^{2} - 5x - y + 1) = 0, (λ ≠1)

⇒(1 + λ) (x^{2} + y^{2}) - (1 + 5λ)x + (7 - λ)y - 3 + λ = 0

⇒ x^{2} + y^{2} - (1+5λ)/(1+λ)x +(7- λ)/(7+ λ)+ λ/(1+ λ) = 0 …………….(i)

Clearly, the co-ordinates of the centre of the circle (i) are [1+5λ/2(1+λ),( λ-7)/2(7+ λ) ]

By question, this point lies on the line x + y = 0.

Therefore, put coordinates of centre in the equation

⇒1 + 5λ + λ - 7 = 0

⇒ 6λ = 6

⇒ λ = 1

Therefore, the equation of the required circle is 2(x^{2} + y^{2}) - 6x + 6y - 2 = 0, [putting λ = 1 in (1)]

⇒ x^{2} + y^{2} - 3x + 3y - 1 = 0.

QUESTION: 24

A circle is the set of …… in a plane that are equidistant from a fixed point in the plane.

Solution:

QUESTION: 25

The number of non-negative integral solution x_{1} + x_{2} + x_{3} + x_{4} = 20 is

Solution:

To solve this question, you have to consider some cases.

case I : let x4=0

x1+x2+x3=20.

You have to distribute 20 identical objects into 3 groups. No. of solutions to this = (20+3-1)C(3-1)=**22C2**

case II : let x4=1

x1+x2+x3=16

You have to distribute 16 identical objects into 3 groups. No. of solutions to this = (16+3-1)C(3-1)=**18C2**

case III : let x4=2

x1+x2+x3=12

You have to distribute 12 identical objects into 3 groups. No. of solutions to this = (12+3-1)C(3-1)=**14C2**

case IV : let x4=3

x1+x2+x3=8

You have to distribute 8 identical objects into 3 groups. No. of solutions to this = (8+3-1)C(3-1)=**10C2**

case V : let x4=4

x1+x2+x3=4

You have to distribute 4 identical objects into 3 groups. No. of solutions to this = (4+3-1)C(3-1)=**6C2**

case VI : let x4=5

x1+x2+x3=0

You have to distribute 0 identical objects into 3 groups. No. of solutions to this = **2C2 = 1**

Hence total number of solutions = **1 +6C2 + 10C2 + 14C2 + 18C2 + 22C2**

= 536

QUESTION: 26

The value of x satisfying sin^{ -1}x + sin^{-1} (1-x) = cos^{-1} are

Solution:

QUESTION: 27

A line passes through (2,2) and is perpendicular to the line 3x+ y = 3 its y intercept is

Solution:

QUESTION: 28

A circle having area = 154 sq. units has two diameters 2x–3y–5 = 0 and 3x–4y–7 0, then the equation of the circle is

Solution:

QUESTION: 29

In x = 33^{n }, n is a positive integral value, then what is the probability that x will have 3 at its unit’s place?

Solution:

QUESTION: 30

equal to

Solution:

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