Test: Calculus- 2 - Civil Engineering (CE) MCQ

# Test: Calculus- 2 - Civil Engineering (CE) MCQ

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## 15 Questions MCQ Test Engineering Mathematics - Test: Calculus- 2

Test: Calculus- 2 for Civil Engineering (CE) 2024 is part of Engineering Mathematics preparation. The Test: Calculus- 2 questions and answers have been prepared according to the Civil Engineering (CE) exam syllabus.The Test: Calculus- 2 MCQs are made for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Calculus- 2 below.
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Test: Calculus- 2 - Question 1

### What is the maximum value of the function f(x) = 2x2 - 2x + 6 in the interval [0, 2]?

Detailed Solution for Test: Calculus- 2 - Question 1

We need absolute maximum of
f(x) = 2x2 - 2x + 6
in the interval [0, 2]
First find local maximum if any by putting f'{x) = 0

is a point of local minimum. So there is no point of local maximum.
Now tabulate the values of f at end point of interval and at local maximum if any (in this case no point of local maximum).

Clearly the absolute maxima is at x = 2 and absolute maximum value is 10.

Test: Calculus- 2 - Question 2

### The following definite integral evaluates to

Detailed Solution for Test: Calculus- 2 - Question 2

Comparing with area under the standared normal curve from -∝ to 0.
We get

So, the required integral

Test: Calculus- 2 - Question 3

### If f(x) is defined as follows, what is the minimum value of f{x) for x ∈ (0, 2]?

Detailed Solution for Test: Calculus- 2 - Question 3

For the function 25/8x the minimum value will come when x is maximum since it is a decreasing function.
The maximum value of x is 3/2.
the function has the value

But since for this function  this function we get the minimum of this function which is
Now comparing the minimum value 2.0833 of the first function with minimum value 2.166 of the second function, we get the overall minimum of this function to be 2.0833   which is option (b).

Test: Calculus- 2 - Question 4

A point on a curve is said to be an extremum if it is a local minimum or a local maximum. The number of distinct extrema for the curve is

Detailed Solution for Test: Calculus- 2 - Question 4

Now
x = 0

∴ f(x) has a minimum at x = 0

So x = 2 is a saddle point (point of inflection)
∴ f(x) has no extremum at x = 2. So f(x) has only one point of extremum (at x = 0).

Test: Calculus- 2 - Question 5

The value of

Detailed Solution for Test: Calculus- 2 - Question 5

⇒ y = 1

Test: Calculus- 2 - Question 6

If   at x=0 ais equal to

Detailed Solution for Test: Calculus- 2 - Question 6

= 1

Test: Calculus- 2 - Question 7

If   then S has the value

Detailed Solution for Test: Calculus- 2 - Question 7

Test: Calculus- 2 - Question 8

Consider function f(x) = (x2- 4)2 where x is a real number. Then the function has

Detailed Solution for Test: Calculus- 2 - Question 8

∴ There is only one maxima and only two minima for this function.

Test: Calculus- 2 - Question 9

The value of the quantity P, where   is equal to

Detailed Solution for Test: Calculus- 2 - Question 9

Test: Calculus- 2 - Question 10

The function f(x) = 2x - x2 + 3 has

Detailed Solution for Test: Calculus- 2 - Question 10

So at x = 1 we have a relative maxima.

Test: Calculus- 2 - Question 11

The maximum value of  in the interval [1, 6] is

Detailed Solution for Test: Calculus- 2 - Question 11

We need absolute maximum of
f(x) = x3 - 9x2 + 24x + 5 in the interval [1,6]
First find local maximum if any by putting f'(x) = 0.

Now tabulate the values of f at end pt. of interval and at local maximum pt., to find absolute maximum in given range, as shown below:

Clearly the absolute m axim a is at x = 6 and absolute maximum value is 41.

Test: Calculus- 2 - Question 12

Let f(x) = x e-x. The maximum value of the funntion in the interval (0, ∝) is

Detailed Solution for Test: Calculus- 2 - Question 12

Hence f(n) have maximum value at n = 1

Let,

Test: Calculus- 2 - Question 13

Minimum of the real valued function f(x) = (x-1)2/3 occurs at  x equal to

Detailed Solution for Test: Calculus- 2 - Question 13

As f(x) is square of hence its minimum value be 0 where at x = 1.

Test: Calculus- 2 - Question 14

The minimum value of the function f(x) = x3-3x2 - 24x + 100 in the interval [-3, 3] is

Detailed Solution for Test: Calculus- 2 - Question 14

Hence f(x) has minimum value at x = 3 which is 28.

Test: Calculus- 2 - Question 15

If a continuous function f(x) does not have a root in the interval [a, b], then which one of the following statements is TRUE?

Detailed Solution for Test: Calculus- 2 - Question 15

Intermediate value theorem states that if a function is continuous and f(a) • f(b) < 0, then surely there is a root in (a, b). The contrapositive of this theorem is that if a function is continuous and has no root in (a, b) then surely f(a) . f{b) ≥ 0. But since it is given that there is no root in the closed interval [a ,b] it means f(a) . f(b) ≠ 0.
So surely f(a) . f(b) > 0.

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## Engineering Mathematics

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