Integral Calculus and Differential Equations Test

Result

Integral Calculus and Differential Equations Test - 9

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Weekly Quiz Competition

Question 1
2 / -0.83

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

A
6

B
10

C
12

D
5.5

Solution

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.
Question 2
2 / -0.83

The following definite integral evaluates to

A
1/2

B

C

D
π

Solution

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

So, the required integral
Question 3
2 / -0.83

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

A
2

B

C

D

Solution

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).
Question 4
2 / -0.83

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

A
0

B
1

C
2

D
3

Solution

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).
Question 5
2 / -0.83

The value of

A
0

B
1/2

C
1

D
∝

Solution

⇒y = 1
Question 6
2 / -0.83

If ^{at x=0} ais equal to

A
x

B
1

C
4

D
None of these

Solution

= 1
Question 7
2 / -0.83

If then S has the value

A
-1/3

B
1/4

C
1/2

D
1

Solution
Question 8
2 / -0.83

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

A
Only one minimum

B
Only two minima

C
Three minima

D
Three maxima

Solution

∴There is only one maxima and only two minima for this function.
Question 9
2 / -0.83

The value of the quantity P, where is equal to

A
0

B
1

C
e

D
1/e

Solution
Question 10
2 / -0.83

The function f(x) = 2x - x² + 3 has

A
a maxima at x = 1 and a minima atx = 5

B
a maxima at x = 1 and a minima at x = -5

C
only a maxima at x = 1

D
only a minima at x = 1

Solution

So at x = 1 we have a relative maxima.
Question 11
2 / -0.83

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

A
21

B
25

C
41

D
46

Solution

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.
Question 12
2 / -0.83

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

A
e^-1

B
e

C
1-e^-1

D
1+e^-1

Solution

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

Let,
Question 13
2 / -0.83

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

A
- ∝

B
0

C
1

D
∝

Solution

As f(x) is square of hence its minimum value be 0 where at x = 1.
Question 14
2 / -0.83

The minimum value of the function f(x) = x³ -3x^{2 -} 24x + 100 in the interval [-3, 3] is

A
20

B
28

C
16

D
32

Solution

Hence f(x) has minimum value at x = 3 which is 28.
Question 15
2 / -0.83

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?

A

B

C

D

Solution

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.