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Test: Integrals- 1 - JEE MCQ


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25 Questions MCQ Test Mathematics (Maths) for JEE Main & Advanced - Test: Integrals- 1

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

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Test: Integrals- 1 - Question 2

Find the distance travelled by a car moving with acceleration given by a(t)=Sin(t), if it moves from t = 0 sec to t = π/2 sec, and velocity of the car at t=0sec is 10 km/hr.

Detailed Solution for Test: Integrals- 1 - Question 2

Acceleration is the derivative of velocity, so we integrate a(t) to get v(t):

v(t)=∫a(t)dt=∫sin(t)dt=−cos(t)+C

Now, we are given that the velocity at t=0 is 10 km/hr. We can use this information to find the constant C:

v(0)=−cos(0)+C=−1+C=10

Solving for C, we get C=11.

Now, we have the velocity function:

v(t)=−cos(t)+11

Finally, we integrate v(t) to get the displacement function s(t):

s(t)=∫v(t)dt=∫(−cos(t)+11)dt

s(t)=−sin(t)+11t+D

Now, we need to find the constant D. We are given that the car moves from t=0 to t=π/2, and we know that s(0)=0 (starting position). Plugging in these values, we can solve for D:

s(0)=−sin(0)+11(0)+D=0

D=0

So, the displacement function is:

s(t)=−sin(t)+11t

Now, to find the distance traveled, we evaluate s(t) over the given time interval:

Distance=s(π/2​)−s(0)

Distance=(−sin(π/2​)+11(π/2​))−(−sin(0)+11(0))

Distance=−1+11π​/2 = 16.27887 kilometers

Therefore, the distance traveled by the car from t=0 to t=π/2​ is 16.27887 kilometers​.

 

 

Test: Integrals- 1 - Question 3

Detailed Solution for Test: Integrals- 1 - Question 3

Using By parts,


2I = x|x|

I=x|x|/2

Test: Integrals- 1 - Question 4

Detailed Solution for Test: Integrals- 1 - Question 4

The greatest integer function ⌊x⌋ returns the largest integer less than or equal to x. For x2, this means that ⌊x2⌋ will be the greatest integer less than or equal to x2.

The integral becomes:

The function ⌊x2⌋ will be 0 on the interval (0,1)(0,1), 1 on the interval [1,√2​], and 4 on the interval [√2,2].

So, the integral is the sum of the areas of these intervals:

Evaluating these integrals:


8 - 1 + √2 - 4√2

7 - 3√2

So, the value of the integral is 7-3√2.

Test: Integrals- 1 - Question 5

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Test: Integrals- 1 - Question 11

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Test: Integrals- 1 - Question 12

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Test: Integrals- 1 - Question 13

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Test: Integrals- 1 - Question 15

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Test: Integrals- 1 - Question 16

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The correct option is d : 26/3

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Test: Integrals- 1 - Question 18

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Test: Integrals- 1 - Question 19

dx can be evaluated by the substitution

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Test: Integrals- 1 - Question 20

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Test: Integrals- 1 - Question 21

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Since g(x) and h(x) are integrals of the same function , therefore ; g(x) – h(x) is constant.

Test: Integrals- 1 - Question 22

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Test: Integrals- 1 - Question 23

then the value of the integral

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Test: Integrals- 1 - Question 24

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Test: Integrals- 1 - Question 25

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