Differential Calculus MCQ Level - 2 - Physics MCQ

# Differential Calculus MCQ Level - 2 - Physics MCQ

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## 10 Questions MCQ Test Topic wise Tests for IIT JAM Physics - Differential Calculus MCQ Level - 2

Differential Calculus MCQ Level - 2 for Physics 2024 is part of Topic wise Tests for IIT JAM Physics preparation. The Differential Calculus MCQ Level - 2 questions and answers have been prepared according to the Physics exam syllabus.The Differential Calculus MCQ Level - 2 MCQs are made for Physics 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Differential Calculus MCQ Level - 2 below.
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Differential Calculus MCQ Level - 2 - Question 1

### The maximum value of x2 log (1/x) is :

Detailed Solution for Differential Calculus MCQ Level - 2 - Question 1

x2(log 1 – log x)
= –x2  log x

or x e–1/2

Hence, y is maximum at x = 1/√e

Differential Calculus MCQ Level - 2 - Question 2

### The critical point of the function f(x) = (x – 1)2 + 1 from the following graph is :

Detailed Solution for Differential Calculus MCQ Level - 2 - Question 2

Turning point is (1, 1) as on one side of (1, 1), the curve is decreasing while the other, is increasing.

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Differential Calculus MCQ Level - 2 - Question 3

### If for the twice differentiable function, f'(x) = f ''(x) = 0, x is

Detailed Solution for Differential Calculus MCQ Level - 2 - Question 3

Explanation : If f'(x) = f''(x) = 0,

but we don't know the state of higher derivative. Therefore the answer is cant say.

Differential Calculus MCQ Level - 2 - Question 4

The function  at the point (3, 8) has :

Detailed Solution for Differential Calculus MCQ Level - 2 - Question 4

As we can see that the denominator in f'(x) will always be positive.

Also, we notice that  i.e., f is a decreasing function and  i.e., f is an increasing function.

Hence, we conclude that there must be minima at x = 3.

Differential Calculus MCQ Level - 2 - Question 5

The function sin x(1 + cos x) at x = π/3  is :

Detailed Solution for Differential Calculus MCQ Level - 2 - Question 5

Let  f(x) = sin x(1 + cosx)

Now, f''(x) = –sin x – 2 sin 2x

= –1 < 0
Hence, f(x) will be maximum at x = π/3.

Differential Calculus MCQ Level - 2 - Question 6

The critical points of the function f(x) = (x – 1)(x – 2)2 in the interval [1, 2) :

Detailed Solution for Differential Calculus MCQ Level - 2 - Question 6

Explanation : Let f(x) = (x – 1)(x – 2)2

f(x)'= (x – 2)2 + 2(x – 1)(x – 2)

= (x – 2)(3x – 4)

Putting f'(x) = (x – 2)(3x – 4) = 0

We get, x = 2, 4/3

Differential Calculus MCQ Level - 2 - Question 7

The turning points of the function y = x3 – 3x + 2 are :

Detailed Solution for Differential Calculus MCQ Level - 2 - Question 7

f(x) = x3 – 3x + 2
Putting f'(x) = 3x2 – 3

We get   as the critical or turning points.
The correct answer is: –1 and +1

Differential Calculus MCQ Level - 2 - Question 8

The absolute maximum and absolute minimum value of the function f(x) = x3 – 3x2 + 1 for  are :

Detailed Solution for Differential Calculus MCQ Level - 2 - Question 8

f(x) = x3 – 3x2 + 1
⇒
Putting f'(x) = 0,
i.e., 3x2 – 6x = 0
3x(x – 2) = 0
⇒ x = 0, 2
Now, f''(x) = 6x – 6
f''(0) = –6 < 0
and f''(0) = 6 > 0

f(4) = 17
f(0) = 1
f(2) = –3
Hence, absolute maxima will be 17 (at x = 4)
and hence absolute minima will be –3 (at x = 2).

The correct answer is: 17, –3

Differential Calculus MCQ Level - 2 - Question 9

The critical points of f(x) = x3/5(4 – x) are :

Detailed Solution for Differential Calculus MCQ Level - 2 - Question 9

f(x)=x3/5(4 – x)
= 4x3/5 + x8/5

Putting f'(x) = 0,

Now, either, 3 – 2x = 0  ⇒ x = 3/2 is the critical point or x = 0 (at x = 0, f'(x) doesn’t exist)

The correct answer is:  3/2 and 0

Differential Calculus MCQ Level - 2 - Question 10

Two numbers whose sum is a, if the product of one by the cube of the other is to be maximum. Then the numbers are :

Detailed Solution for Differential Calculus MCQ Level - 2 - Question 10

Let the two numbers be x and y
⇒ x + y = a
or      x = a – y
We need to maximise, z = xy3
or  z = (a – y)y3
ay3 – y4

Putting dz/dy = 0, i.e
3ay2 – 4y3 = 0
or y(3a – 4y) = 0

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