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JEE Main Practice Test- 13 - JEE MCQ


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30 Questions MCQ Test - JEE Main Practice Test- 13

JEE Main Practice Test- 13 for JEE 2024 is part of JEE preparation. The JEE Main Practice Test- 13 questions and answers have been prepared according to the JEE exam syllabus.The JEE Main Practice Test- 13 MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for JEE Main Practice Test- 13 below.
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JEE Main Practice Test- 13 - Question 1

JEE Main Practice Test- 13 - Question 2

If 60a = 3 and 60b = 5 then the value of  equals

Detailed Solution for JEE Main Practice Test- 13 - Question 2

60a = 3 ⇒ a = log603
60b = 5 ⇒ b = log605


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JEE Main Practice Test- 13 - Question 3

The set of values of x satisfying the inequality

Detailed Solution for JEE Main Practice Test- 13 - Question 3

Domain is (- 3, – 2) ∪ (- 1, ∞) ; for x > - 1, LHS is negative & RHS is positive and for - 3 < x < - 2 it is the other way
⇒ x > - 1 is the final answer

JEE Main Practice Test- 13 - Question 4

Given that x + sin y = 2008 and x + 2008 cos y = 2007 where 0 ≤ y ≤ π/2. The value of [x + y], is. (Here [x] denotes greatest integer function)

Detailed Solution for JEE Main Practice Test- 13 - Question 4

x + sin y = 2008

This is possible only if cos y = 0

JEE Main Practice Test- 13 - Question 5

Let p be the statement ' Ram races' and q be the statement 'Ram wins'. Then ~ (p v (~ q)) is

Detailed Solution for JEE Main Practice Test- 13 - Question 5

Given
p : Ram races
q : Ram wins
∴ The statement of given proposition ~ (p v (~ q)) = ~ p ∧ q is "Ram does not race and Ram wins." 

JEE Main Practice Test- 13 - Question 6

A vertical pole PS has two marks at Q and R such that portions PQ, PR and PS subtend angles α, β, γ respectively at a point on the ground which is at distance x from the bottom of pole P. If PQ = a, PR = b, PS = c and α + β + γ = 180º, then x2 is equal to

Detailed Solution for JEE Main Practice Test- 13 - Question 6

We have


tan α + tan β + tan γ = tan α tan β tan γ

JEE Main Practice Test- 13 - Question 7


Then the number of subsets of set A×(A ∩ B) which contains exactly 3 elements is

Detailed Solution for JEE Main Practice Test- 13 - Question 7

A = {1, 2, 3, 4, 5, 6} ⇒ n(A) = 6
B = {1, 2} ⇒ n(B) = 2
∴ A ∩ B = {1, 2} ⇒ n(A ∩ B) = 2
So, number of elements in A× (A ∩ B) = 12
∴ Number of subsets containing 3 elements = 12C3 = 220

JEE Main Practice Test- 13 - Question 8

The relation P defined from R to R as a P b ⇔ 1+ ab > 0, for all a,b ∈ R is

JEE Main Practice Test- 13 - Question 9

If x1, x2 & x3 are the three real solutions of the equation


where x1 > x2 > x3 , then

Detailed Solution for JEE Main Practice Test- 13 - Question 9

RHS when simplified is equal x.

JEE Main Practice Test- 13 - Question 10

The variance of 20 observations is 5. If each observation is multiplied by 2 then the new variance of the resulting observations, is

Detailed Solution for JEE Main Practice Test- 13 - Question 10


New, observations are,
2x1 , 2x2 , 2x3 , ......, 2x20
Theirmean,

Now, variance,

JEE Main Practice Test- 13 - Question 11

The last two digits of the number 3400 are

Detailed Solution for JEE Main Practice Test- 13 - Question 11

3400 = 81100 = (1 + 80)100 = 100C0
+ 100C1 80 + ....... + 100C100 80100
⇒ Last two digits are 01

JEE Main Practice Test- 13 - Question 12

There exist positive integers A, B and C with no common factors greater than 1, such that A log2005 + B log2002 = C. The sum (A + B + C) equals

Detailed Solution for JEE Main Practice Test- 13 - Question 12

A log2005 + B log2002 = C


A log 5 + B log 2 = C log 200 = C log(52 23)
= 2C log 5 + 3 C log 2
hence, A = 2C
B = 3C
for no common factor greater than 1, C = 1
∴ A = 2; B = 3 ⇒ A + B + C = 6

JEE Main Practice Test- 13 - Question 13

The expression 

Detailed Solution for JEE Main Practice Test- 13 - Question 13

Let
E = sin2α + sin2β + cos2(α + β) + 2 · sin α · sin β · cos(α + β)
= sin2α + sin2β + cos2 (α + β) + [cos (α - β) - cos (α + β)] · cos (α + β)
= sin2α + sin2β +(cos2α - sin2β) = 1. Ans.

Aliter: E = sin2α + sin2β + cos2 (α + β) + 2sin α sin β cos (α + β)
= sin2α - sin2 (α + β) + sin2β + 1 + 2sin α sin β cos (α + β)
= - sin (2α + β) · sin β + sin2 β + 1 + 2sin α sin β cos (α + β)
= 1 - sin β [sin (2α + β) - sin β] + 2 sin α sin β cos (α + β)
= 1 - sin β [2 cos (α + β) sin α] + 2 sin α sin β cos (α + β)
= 1 - 2 sin α sin β cos (α + β) + 2 sin α sin β cos (α + β)
∴ E = 1 ⇒ (A). Ans.

JEE Main Practice Test- 13 - Question 14

Smallest positive x satisfying the equation cos33x + cos35x = 8 cos34x · cos3x is

Detailed Solution for JEE Main Practice Test- 13 - Question 14

cos33x + cos35x = (2 cos 4x cos x)3 = (cos 5x + cos 3x)3
cos33x + cos35x = cos35x + cos33x + 3 cos 5x cos 3x (cos 5x + cos 3x)
⇒ (3 cos 3x · cos 5x) (2 cos 4x · cos x) = 0
⇒ cos x · cos 3x · cos 4x · cos 5x = 0

⇒  smallest + ve values of x is π/10 i.e. 18° Ans. 

JEE Main Practice Test- 13 - Question 15

If the quadratic equations 3x2 + ax + 1 = 0 and 2x2 + bx + 1 = 0 have a common root, then the value of the expression 5ab - 2a2 - 3b2 is

Detailed Solution for JEE Main Practice Test- 13 - Question 15

6x2 + 2ax + 2 = 0 and 6x2 + 3bx + 3 = 0 subtracting x (2a – 3b) – 1 = 0

(put in any equation)

2 + b (2a – 3b) + (2a – 3b)2 = 0
4a2 + 5b2 – 12ab + 2ab – 3b2 + 2 = 0
–10ab + 6b2 + 4a2 + 1 = 0
⇒ 5ab –3b2 – 2a2 = 1 ⇒ B 

JEE Main Practice Test- 13 - Question 16

Solution set of the inequality 

JEE Main Practice Test- 13 - Question 17

The sum to infinity of the series

Detailed Solution for JEE Main Practice Test- 13 - Question 17


JEE Main Practice Test- 13 - Question 18

Consider the sequence 8A + 2B, 6A + B, 4A, 2A – B, ........ Which term of this sequence will have a coefficient of A which is twice the coefficient of B?

Detailed Solution for JEE Main Practice Test- 13 - Question 18

coefficient of A in nth term = 8 + (n – 1)(– 2)
= 10 – 2n
coefficient of B in nth term = 2 + (n – 1)(– 1)
= 3 – n
10 – 2n = 2(3 – n) ⇒ 10 = 6
which is absurd ⇒ none

JEE Main Practice Test- 13 - Question 19


then the value of d is

Detailed Solution for JEE Main Practice Test- 13 - Question 19


JEE Main Practice Test- 13 - Question 20

Let k1, k2 (k1 < k2) be two values of k for which the expression x2 – y2 +kx +1 can be factorised into two real linear factors, then (k2 – k1) is equal to

Detailed Solution for JEE Main Practice Test- 13 - Question 20

x2 + kx + 1 – y2 = 0


for real linear factors 4y2 + 0y + k2 – 4 must be a perfect square.
Hence D = 0 ⇒ 0 – 16(k2 – 4) = 0
∴ k = 2, – 2 ⇒ k1 = – 2 and k2 = 2
k2 – k1 = 2 – (–2) = 4  Ans.

Aliternatively: Comparing x2 – y2 + kx + 1, with Ax2 + 2Hxy + By2 + 2Gx + 2Fy + C;
we get A = 1, B = – 1, H = 0, G = k/2 , 
F = 0, C = 1
Now, using condition,
ABC + 2FGH – AF2 – BG2 – CH2 = 0, we get

⇒ k1 = – 2, k2 = 2
Hence, (k2 – k1) = 2 – (– 2) = 4. Ans.

*Answer can only contain numeric values
JEE Main Practice Test- 13 - Question 21

(Instruction to attempt numerical value (integer) type question: If your answer is 100 write 100 only. Do not write 100.0)

Rolle’s theorem holds for the function f(x) = x3 + bx2 + cx, 1 ≤ x ≤ 2 at the point 4/3, then the value of 100c − 500b must be


Detailed Solution for JEE Main Practice Test- 13 - Question 21

*Answer can only contain numeric values
JEE Main Practice Test- 13 - Question 22

If p, q, r are three distinct real numbers, p ≠ 0 such that x2 + qx + pr = 0 and x2 + rx + pq = 0 have a common root, then the value of p + q + r is .....


Detailed Solution for JEE Main Practice Test- 13 - Question 22

*Answer can only contain numeric values
JEE Main Practice Test- 13 - Question 23

The number of solutions of the equation [2x] - [x + 1] = 2x must be equal to (where [.] denotes the greatest integer function)


Detailed Solution for JEE Main Practice Test- 13 - Question 23

*Answer can only contain numeric values
JEE Main Practice Test- 13 - Question 24

If λ = logx2  then the value of 3216λ must be


Detailed Solution for JEE Main Practice Test- 13 - Question 24

*Answer can only contain numeric values
JEE Main Practice Test- 13 - Question 25

The three angles of a quadrilateral are 60°, 60g and 5π/6, if fourth angle is λ, then the value of λ must be


Detailed Solution for JEE Main Practice Test- 13 - Question 25

JEE Main Practice Test- 13 - Question 26

A special kind of ruler can be used to measure reaction time. You position the 'zero' mark between your friend's first finger and thumb. Without warning, you let go of the the rular, and your friend has to grab it as soon as possible. You can read the reaction time from labelled marks on the ruler. The marks, are separated from each other by 0.01 s intervals of falling time. This means that

JEE Main Practice Test- 13 - Question 27

An object moves to the East across a frictionless surface with constant speed. A person then applies a constant force to the North on the object. What is the resulting path that the object takes?

Detailed Solution for JEE Main Practice Test- 13 - Question 27

Component of velocity along north will increase

JEE Main Practice Test- 13 - Question 28

An aeroplane is flying vertically upwards. When it is at a height of 1000m above the ground and moving at a speed of 367 m/s., a shot is fired at it with a speed of 567 m/s from a point directly below it. What should be the acceleration of aeroplane so that it may escape from being hit ?

Detailed Solution for JEE Main Practice Test- 13 - Question 28


0 = (200)2 – 2(arel) (1000)
or arel = 20 m/s2
So to avoit the hit,
arel > 20 m/s2
or ap > 10 m/s2

JEE Main Practice Test- 13 - Question 29

A jet plane flying at a constant velocity V at a height h = 8 km is being tracked by a radar R located at O directly below the line of flight. If the angle θ is decreasing at the rate of 0.025 rad/s., when θ = 60° the velocity of the plane is :

Detailed Solution for JEE Main Practice Test- 13 - Question 29


JEE Main Practice Test- 13 - Question 30

The figure Shows the acceleration of a particle versus time graph. The particle starts from rest at t = 0. Find the approximate velocity at t = 1s.

Detailed Solution for JEE Main Practice Test- 13 - Question 30

Counting the squares under the graph from t = 0 to t = 1. We get change in velocity.

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