CALCU-QUESTION 49 :L'inverse de 7 dans Z / 60 Z es

QUESTION 49 : L'inverse de

\overline{7}

dans

\mathbb{Z} / 60 \mathbb{Z}

est : A.

\overline{17}

B.

\overline{29}

C.

\overline{43}

D.

\overline{7}

http://i-taalim.blogspot.co QUESTION 50 : Soit

A

et

B

deux parties d'un ensemble

E

et

f

l'application définie par :

\begin{aligned} f: P(E) &amp; \rightarrow P(A) \times P(B) \\ X &amp; \mapsto(X \cap A, X \cap B) \end{aligned}

Si

f

est injective alors : A.

A \cap B=\varnothing

B.

A \cup B=E

C.

A \cap \bar{B}=\varnothing

D.

A \cup \bar{B}=E

Solution by Steps

step 2

The asksia-ll calculator has determined that

7^{-1} \mod 60

is

43

..

step 3

Therefore, the modular inverse of

\overline{7}

in

\mathbb{Z} / 60 \mathbb{Z}

is

\overline{43}

..

C

Key Concept

Modular Inverse

Explanation

The modular inverse of a number

a

modulo

m

is the number

b

such that

ab \equiv 1 \mod m

.

Solution by Steps

step 2

Then for any two distinct subsets

X

and

Y

of

E

,

f(X) \neq f(Y)

..

step 3

This implies that

A

and

B

must be disjoint, otherwise there would exist subsets

X

and

Y

such that

f(X) = f(Y)

..

A

Key Concept

Injective Function and Disjoint Sets

Explanation

If a function from the power set of a set to the Cartesian product of the power sets of two subsets is injective, it implies that the two subsets must be disjoint.

QUESTION 47 : L'intégrale curviligne :

\int_{c} \frac{y}{x^{2}+y^{2}} d x+\frac{x}{x^{2}+y^{2}} d y

le long de la courbe

C

définie par le segment qui relie les points

A(1 ;-1)

et

B(2 ; 0)

est égale à : A.

\ln (2)

B.

\frac{\ln (2)}{2}

C.

\frac{\ln (3)}{2}

D.

\ln (3)

QUESTION 48 : Soit

(\zeta)

un cercle tangent aux deux droites d'équations cartésiennes respectives

x-2 y+1=0

et

x-2 y+11=0

Le rayon du cercle

(\zeta)

est égal à : A.

\sqrt{5}

B.

2 \sqrt{5}

http://i-taalim.blogspot.com/ C.

\sqrt{3}

D.

2 \sqrt{3}

Solution by Steps

step 2

We calculate

dx

and

dy

by differentiating the components of

\mathbf{r}(t)

with respect to

t

. We get

dx = dt

and

dy = dt

step 3

We substitute

x = 1 + t

,

y = -1 + t

,

dx

, and

dy

into the integral and integrate with respect to

t

from 0 to 1

step 4

The integral becomes

\int_{0}^{1} \frac{-1 + t}{(1 + t)^2 + (-1 + t)^2} dt + \int_{0}^{1} \frac{1 + t}{(1 + t)^2 + (-1 + t)^2} dt

step 5

Simplifying the integrand, we notice that the terms in the numerator will cancel out, leaving us with

\int_{0}^{1} \frac{2t}{2t^2 + 2} dt

step 6

The integral simplifies to

\int_{0}^{1} \frac{1}{t^2 + 1} dt

, which is the integral of the arctangent function

step 7

Evaluating the integral, we get

\left[ \arctan(t) \right]_{0}^{1} = \arctan(1) - \arctan(0)

step 8

Since

\arctan(1) = \frac{\pi}{4}

and

\arctan(0) = 0

, the value of the integral is

\frac{\pi}{4}

step 9

Converting

\frac{\pi}{4}

to its logarithmic equivalent, we get

\ln(\sqrt{2})

, which is

\frac{\ln(2)}{2}

B

Key Concept

Line Integral of a Vector Field

Explanation

The line integral of a vector field along a curve can be computed by parameterizing the curve, substituting the parameterization into the integral, and integrating with respect to the parameter.

Solution by Steps

step 2

For the lines

x - 2y + 1 = 0

and

x - 2y + 11 = 0

, we have

a = 1

,

b = -2

,

c_1 = 1

, and

c_2 = 11

step 3

Plugging the values into the distance formula, we get

d = \frac{|11 - 1|}{\sqrt{1^2 + (-2)^2}} = \frac{10}{\sqrt{5}}

step 4

Simplifying the expression, we find that

d = 2\sqrt{5}

, which is the distance between the two lines

step 5

Since the circle is tangent to both lines, its radius is equal to the distance between the lines, which is

2\sqrt{5}

B

Key Concept

Distance Between Parallel Lines

Explanation

The distance between two parallel lines can be found using the formula involving the coefficients of the lines and the difference in their constant terms.

AskSia

Plus