JEE Main Differentiation Important Questions - Are you preparing to take the JEE Main 2024 Exam ? Do you want to know how to pass the JEE Main Differentiation Section? We've got your back. By preparing and answering the Differentiation section of the JEE Main Mathematics Differentiation Questions, you may ace the JEE Main 2024 exam. To tackle the differentiation section of the question paper, one must have a thorough comprehension of the concepts of differentiation as well as effective problem-solving skills.
The National Testing Agency or NTA has released the JEE Main 2024 Exam Dates on the official website. JEE Main 2024 Exam date session 1 is January 24 to February 1, 2024, and the JEE Main session 2 exam date is April 1 to 15, 2024. Candidates can check the JEE Main Differentiation Important Questions below.
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JEE Main Differentiation Important Questions
Solving questions related to differentiation in the JEE Mains mathematics portion is a difficult challenge. Some essential questions, along with their step-by-step solutions, are provided in this article for students to prepare and ace their JEE Main. These questions will also aid you in time management whilst taking the JEE Main 2024 Exam.
Question 1: If f (x) = sin [cos-1 (1 – 22x) / (1 + 22x)] and its first derivative with respect to x is (- b / a) log 2 when x = 1, where a and b are integers, then the minimum value of |a2 – b2| is:
Solution:
Given that f (x) = sin [cos-1 (1 – 22x) / (1 + 22x)]
cos-1 [1 – 4x] / [1 + 4x]
Let 2x = t > 0
cos-1 [(1 – t2) / (1 + t2)], t > 0 and t = tan θ
cos-1 (cos 2θ) = 2θ ∈ (0, π), θ ∈ π / 2, 2θ ∈ (0, π)
⇒ 2θ
sin {cos-1 [[1 – 4x] / [1 + 4x]]} = sin2θ
So y = [2 tanθ / (1 + tan2 θ)]
= 2t / (1 + t2)
= [2 × 2x] / [1 + 4x]
dy/dx = [20 ln 2 – 32 ln 2] / 25
= – 12 ln 2 / 25
a = 25, b = 12
|a2 – b2|min = |252 – 122| = 481
Hence, the minimum value of |a2 – b2| is 481.
Question 2 : Find y(π) if y(0) = 7 and dy/dx = 2(y – 2 sin x – 10)x + 2 cos x.
Solution:
Given that dy/dx = 2(y – 2 sin x – 10)x + 2 cos x
dy/dx – 2 cos x = 2(y – 2 sin x – 10)x
(d/dx (y – 2 sin x – 10))/((y – 2 sin x – 10) ) = 2x
⇒∫d(y – 2 sin x – 10)/((y – 2 sin x – 10) ) = ∫2x dx
⇒ log |y – 2 sin x – 10| = x2 + C
When x = 0, y = 7
⇒ log |7 – 0 – 10| = 0 + C
So C = log 3
When x = π
⇒ log |y – 2 sin π – 10| = π2+ log 3
⇒ log ((y – 10)/3) = π
= y( π ) = 3e
Question 3: d/dx ( log e x) ( log a x) =
(a) (1/x) log a x
(b) (1/x) log x x
(d) (2/x) log a x
(d) (2/x) log x
Solution:
Let y = ( loge x) ( log a x)
= (log x/log e) ( log x/ log a)
= (log x)2/log a
Differentiating with respect to x, we get;
dy/dx = 2 log x (1/x)/log a
= (2/x) log a x
Hence, option (c) is the answer.
Question 4: If f (x) = sin [cos-1 (1 – 22x) / (1 + 22x)] and its first derivative with respect to x is (- b / a) loge 2 when x = 1, where a and b are integers, then the minimum value of |a2 – b2| is:
Solution:
Given that f (x) = sin [cos-1 (1 – 22x) / (1 + 22x)]
cos-1 [1 – 4x] / [1 + 4x]
Let 2x = t > 0
cos-1 [(1 – t2) / (1 + t2)], t > 0 and t = tan θ
cos-1 (cos 2θ) = 2θ ∈ (0, π), θ ∈ π / 2, 2θ ∈ (0, π)
⇒ 2θ
sin {cos-1 [[1 – 4x] / [1 + 4x]]} = sin2θ
So y = [2 tanθ / (1 + tan2 θ)]
= 2t / (1 + t2)
= [2 × 2x] / [1 + 4x]
dy/dx = [20 ln 2 – 32 ln 2] / 25
= – 12 ln 2 / 25
a = 25, b = 12
|a2 – b2|min = |252 – 122| = 481
Hence, the minimum value of |a2 – b2| is 481.
Question 5 : If ey + xy = e, then the value of d2y/dx2 for x = 0 is
(a) 1/e
(b) 1/e2
(c) 1/e3
(d) none of these
Solution:
Given that e y + xy = e
When x = 0, we get y = 1
Differentiate w.r.t.x
eydy/dx + x dy/dx + y = 0 …(i)
Put x = 0 and y = 1, we get
dy/dx = -y/(x+ ey) = -1/e
Again differentiate (i) w.r.t.x
ey(dy/dx)2 + eyd2y/dx2 + dy/dx + x d2y/dx2 + dy/dx = 0
d2y/dx2 [ ey + x] = -2dy/dx – ey(dy/dx)2 …(ii)
Put x = 0 and y = 1 and dy/dx = -1/e in (ii), we get;
d2y/dx2 [ e + 0] = 2/e – e/e2
d2y/dx2 e = 1/e
d2y/dx2 = 1/e2
Hence, option (b) is the answer.
Question 6 : If loge (x+y) = 4xy, find (d2y)/(dx2) at x = 0.
Solution:
Given that loge (x+y) = 4xy
Differentiating with respect to x, we get-
(1/(x + y)) [1 + (dy/dx)] = 4[x (dy/dx) + y]
1 + (dy/dx) = 4(x + y) [x (dy/dx) + y]⋯(i)
If x = 0, then y = 1.
From (i), we get
1 + dy/dx = 4
⇒ dy/dx = 3
Again differentiate (i) w.r.t. x.
d2y/dx2 = 4(x + y)[x (d2y)/(dx2) + 2 (dy/dx)] + 4[x (dy/dx) + y](1 + (dy/dx))
At x = 0, y = 1, dy/dx = 3
d2y/dx2 = 4(0 + 1)[0 + 2x3]+4[0 + 1](1 + 3)
= 40
So, d2y/dx2 = 40.
Also, check
| JEE Main 2024 Mathematics subject-wise weightage |
|---|
| JEE Main 2024 Chemistry subject-wise weightage |
| JEE Main 2024 Physics subject-wise weightage |
About Differentiation
In calculus, the process of determining a function's derivatives is known as differentiation. A derivative is the pace at which a function changes in relation to another quantity. Sir Isaac Newton established the laws of Differential Calculus. Limit and derivative ideas are applied in many scientific areas. Calculus' main principles are differentiation and integration.
Differentiation determines the highest or lowest value of a function, the velocity and acceleration of moving objects, and the tangent of a curve. If y = f(x) and x is differentiable, the differentiation is denoted by f'(x) or dy/dx.
JEE Main Mathematics Question Paper
JEE Main Differentiation Preparation Books
Candidates can find the best JEE Main Differentiation Section Books from the table below.Differential Calculus by Amit M Agarwal (Arihant Publications) | Complete mathematics for JEE Main TMH |
|---|---|
Integral Calculus by Amit M Agarwal (Arihant Publications) | Objective Mathematics by R D Sharma |
| Play with Graphs by Amit M Agarwal (Arihant Publications) | - |
Download JEE Main Mathematics Previous Year Question Papers
Source: Aakash BYJU's
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FAQs
Differentiation in JEE Main Mathematics is the process of finding the derivative of a function.
The quotient rule of differentiation is - Quotient rule: (d/dx)(u/v) = (v (du/dx) – u (dv/dx))/v2.
The product rule of differentiation is - Product rule: (d/dx) (uv) = u (dv/dx) + v (du/dx).
(fg)' = f'g + g'f, for all functions f and g. (4) Quotient Rule: (f/g)' = (f'g – fg')/g2, for all functions f and g such that g ≠ 0.
250 marks is considered a good score in JEE Mains. Candidates scoring 250+ marks will get a percentile between 85 -95.
JEE Main Previous Year Question Paper
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