JEE Exam  >  JEE Tests  >  Daily Test for JEE Preparation  >  Test: Definite Integration: Leibnitz Theorem (3 Oct) - JEE MCQ

Test: Definite Integration: Leibnitz Theorem (3 Oct) - JEE MCQ


Test Description

10 Questions MCQ Test Daily Test for JEE Preparation - Test: Definite Integration: Leibnitz Theorem (3 Oct)

Test: Definite Integration: Leibnitz Theorem (3 Oct) for JEE 2024 is part of Daily Test for JEE Preparation preparation. The Test: Definite Integration: Leibnitz Theorem (3 Oct) questions and answers have been prepared according to the JEE exam syllabus.The Test: Definite Integration: Leibnitz Theorem (3 Oct) MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Definite Integration: Leibnitz Theorem (3 Oct) below.
Solutions of Test: Definite Integration: Leibnitz Theorem (3 Oct) questions in English are available as part of our Daily Test for JEE Preparation for JEE & Test: Definite Integration: Leibnitz Theorem (3 Oct) solutions in Hindi for Daily Test for JEE Preparation course. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free. Attempt Test: Definite Integration: Leibnitz Theorem (3 Oct) | 10 questions in 20 minutes | Mock test for JEE preparation | Free important questions MCQ to study Daily Test for JEE Preparation for JEE Exam | Download free PDF with solutions
Test: Definite Integration: Leibnitz Theorem (3 Oct) - Question 1

Let f(x) = sin(x)/1+x2. Let y(n) denote the nth derivative of f(x) at x = 0 then the value of y(100) + 9900y(98) is

Detailed Solution for Test: Definite Integration: Leibnitz Theorem (3 Oct) - Question 1

The key here is a simple manipulation and application of the Leibniz rule.
Rewriting the given function as
y(1 + x2) = sin(x)…….(1)
The Leibniz rule for two functions is given by

Differentiating expression (1) in accordance to Leibniz rule (upto the hundredth derivative) we have

(y(1+x2)) = (sin(x))(100) = sin(x)
Now substituting gives us
y(100)+9900y(98) = sin(0) = 0
Hence, Option 0 is the required answer.

Test: Definite Integration: Leibnitz Theorem (3 Oct) - Question 2

Let f(x) = ln(x)/x+1 and let y(n) denote the nth derivative of f(x) at x = 1 then the value of 2y(100) + 100y(99)

Detailed Solution for Test: Definite Integration: Leibnitz Theorem (3 Oct) - Question 2

Assume f(x) = y
Rewrite the function as
y(x + 1 ) = ln(x)
Now differentiate both sides up to hundredth derivative in accordance to the Leibniz rule we have

Using the nth derivative for ln(x+a) as 

we have the right hand side as

Now substituting x = 1 yields

= -(99)!

1 Crore+ students have signed up on EduRev. Have you? Download the App
Test: Definite Integration: Leibnitz Theorem (3 Oct) - Question 3

Let f(x) =  and let y(n) denote the nth derivative of f(x) at x = 0 then the value of 6y (1) y(2) + 2y(3) is

Detailed Solution for Test: Definite Integration: Leibnitz Theorem (3 Oct) - Question 3

Explanation:Assume f(x) = y
Rewriting the function as
y2 = 1 – x2
Differentiating both sides of the equation up to the third derivative using leibniz rule we have

(1-x2)(3)=0
Now substituting x = 0 in both the equations and equating them yields
2y (3) y + 6y (2) y(1) = 0.

Test: Definite Integration: Leibnitz Theorem (3 Oct) - Question 4

Let f(x) = tan(x) and let y(n) denote the nth derivative of f(x) then the value of y(9998879879789776) is

Detailed Solution for Test: Definite Integration: Leibnitz Theorem (3 Oct) - Question 4

Explanation:Assume y = f(x) and we also know that tan(x)= 
Rewriting the function as y(cos(x))=sin(x)
Now differentiating on both sides upto nnt derivative we have
(y(cos(x))(n)
Now observe that y(0)=tan(0)=0….(1)
Now consider the second derivative at x=0 on both sides

Using (1) and the above equation one can conclude that
y(2) = 0
This gives the value of second derivative to be zero
Similarly for any even value of n all the odd derivatives of y in the expression would have sin(x) as their coefficients and as the values of y(0) and y(2) are zero. Every even derivative of the tan(x) function has to be zero.
Thus, we have
y(9998879879789776) = 0.

Test: Definite Integration: Leibnitz Theorem (3 Oct) - Question 5

If the first and second derivatives at x = 0 of the function f(x)=  were 2 and 3 then the value of the third derivative is

Detailed Solution for Test: Definite Integration: Leibnitz Theorem (3 Oct) - Question 5

Explanation: Assume
f(x) = y
Write the given function as
y(x2 – x + 1) = cos(x)
Now applying Leibniz rule up to the third derivative we get

Equating both sides we have

Now in the question it is assumed that the y(1)=2 and y(2)=3
Substituting these values in (1) we have

Substituting x = 0 gives
sin(0) = y(3) + 9 -12
y(3) = 12 – 9 = 3.

Test: Definite Integration: Leibnitz Theorem (3 Oct) - Question 6

For the given function f(x)=  the values of first and second derivative at x = 1 are assumed as 0 and 1 respectively. Then the value of the third derivative could be

Detailed Solution for Test: Definite Integration: Leibnitz Theorem (3 Oct) - Question 6

Explanation:Rewriting the given function as
y2 = x3 + x7
Now applying the Leibniz rule up to the third derivative we have
(y2)(3)=(x3+x7)(3)

Equating both sides and substituting x = 1 we get
y(1) = 0
Now assumed in the question are the values y(1) = 0 and y(2) = 1
We also know y(1) = √2
Putting them in equation (1) we get
2√2 y(3) = 3! + 210 = 216
y(3) = 54√2.

Test: Definite Integration: Leibnitz Theorem (3 Oct) - Question 7

Let f(x)=  and let the nth derivative at x = 0 be given by y(n) Then the value of the expression for y(n) is given by

Detailed Solution for Test: Definite Integration: Leibnitz Theorem (3 Oct) - Question 7

Expanding sin(x)/x into Taylor series we have

Now Taking the nth derivative of function using Leibniz rule we have

Now substituting x = 0 we have

Hence, Option  is the right answer.

Test: Definite Integration: Leibnitz Theorem (3 Oct) - Question 8

Let f(x) = ex sinh(x) / x, let y(n) denote the nth derivative of f(x) at x = 0 then the expression for y(n) is given by

Detailed Solution for Test: Definite Integration: Leibnitz Theorem (3 Oct) - Question 8

The key here is to find a separate Taylor expansion for sinh(x) / x which is

Now consider   applying the Leibniz rule for nth derivative we have
 

Now substituting x = 0 yields 

Test: Definite Integration: Leibnitz Theorem (3 Oct) - Question 9


Detailed Solution for Test: Definite Integration: Leibnitz Theorem (3 Oct) - Question 9


Test: Definite Integration: Leibnitz Theorem (3 Oct) - Question 10


Detailed Solution for Test: Definite Integration: Leibnitz Theorem (3 Oct) - Question 10

360 tests
Information about Test: Definite Integration: Leibnitz Theorem (3 Oct) Page
In this test you can find the Exam questions for Test: Definite Integration: Leibnitz Theorem (3 Oct) solved & explained in the simplest way possible. Besides giving Questions and answers for Test: Definite Integration: Leibnitz Theorem (3 Oct), EduRev gives you an ample number of Online tests for practice

Top Courses for JEE

Download as PDF

Top Courses for JEE