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JEE Exam  >  JEE Tests  >  Mathematics (Maths) for JEE Main & Advanced  >  Test: Limit Of A Sum - JEE MCQ

Test: Limit Of A Sum - JEE MCQ


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10 Questions MCQ Test Mathematics (Maths) for JEE Main & Advanced - Test: Limit Of A Sum

Test: Limit Of A Sum for JEE 2024 is part of Mathematics (Maths) for JEE Main & Advanced preparation. The Test: Limit Of A Sum questions and answers have been prepared according to the JEE exam syllabus.The Test: Limit Of A Sum MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Limit Of A Sum below.
Solutions of Test: Limit Of A Sum questions in English are available as part of our Mathematics (Maths) for JEE Main & Advanced for JEE & Test: Limit Of A Sum solutions in Hindi for Mathematics (Maths) for JEE Main & Advanced course. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free. Attempt Test: Limit Of A Sum | 10 questions in 10 minutes | Mock test for JEE preparation | Free important questions MCQ to study Mathematics (Maths) for JEE Main & Advanced for JEE Exam | Download free PDF with solutions
Test: Limit Of A Sum - Question 1
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In the definite integral  , the variable of integration is called​

Test: Limit Of A Sum - Question 2
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Express the shaded area in the form of an integral.

Detailed Solution for Test: Limit Of A Sum - Question 2

As the curve goes from c to d and the equation is x = f(y)
So the shaded area is ∫(c to d)f(y)dy

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Test: Limit Of A Sum - Question 3
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Evaluate as limit of  sum 

Detailed Solution for Test: Limit Of A Sum - Question 3

 ∫(0 to 2)(x2 + x + 1)dx
= (0 to 2) [x3/3 + x2/2 + x]½
= [8/3 + 4/2 + 2]
 = 40/6
= 20/3

Test: Limit Of A Sum - Question 4
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Evaluate as limit of sum 

Detailed Solution for Test: Limit Of A Sum - Question 4




Test: Limit Of A Sum - Question 5
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The value of definite integral depends on
Test: Limit Of A Sum - Question 6
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Find   
Detailed Solution for Test: Limit Of A Sum - Question 6

Using trigonometric identities, we have
cos2x=cos2x-sin2x  -(1) and cos2x+sin2x =1 -(2)
cos2x=1-sin2x , substituting this in equation (1) we get 
cos2x=1-sin2x-sin2x=1-2sin2x
So,cos2x=1-2sin2x
2sin2x=1-cos2x


 

Test: Limit Of A Sum - Question 7
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Evaluate as limit of  sum  ​
Test: Limit Of A Sum - Question 8
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The value of    is:​
Test: Limit Of A Sum - Question 9
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Evaluate as limit of sum 
Detailed Solution for Test: Limit Of A Sum - Question 9

 ∫(0 to 4)3x dx
= [3x2/2] (0 to 4)
[3(4)2] / 2
= 24 sq unit

Test: Limit Of A Sum - Question 10
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The value of   is:
Detailed Solution for Test: Limit Of A Sum - Question 10

∫(0 to 3)1/[(3)2 - (x)^2]½
∫1/[(a)2 - (x)2] = sin-1(x/a)
= [sin-1(x/3)](0 to 3)
= sin-1[3/3] - sin-1[0/3]
= sin-1[1]
= π/2

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