Math Problem Statement

حل هذه التمارين

حل هذه التمارين

Solution

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التمرين الأول (6 نقاط):

1. الحساب:

Given: log40+log25\log 40 + \log 25, we use the property of logarithms:

log40+log25=log(40×25)=log1000=3\log 40 + \log 25 = \log (40 \times 25) = \log 1000 = 3

Thus, the result is 33.

2. حل المعادلة:

Solve ln(2x)=ln10\ln(2x) = \ln 10. Using the fact that lna=lnba=b\ln a = \ln b \Rightarrow a = b:

2x=10x=102=52x = 10 \quad \Rightarrow \quad x = \frac{10}{2} = 5

3. حل المعادلة:

Given e3x=e2e^{3x} = e^2. Since the exponential function is one-to-one, we can equate the exponents:

3x=2x=233x = 2 \quad \Rightarrow \quad x = \frac{2}{3}

4. النهاية:

limx0ex1x\lim_{x \to 0} \frac{e^x - 1}{x}. Using L'Hôpital's rule:

limx0ex1x=limx0ex1=1\lim_{x \to 0} \frac{e^x - 1}{x} = \lim_{x \to 0} \frac{e^x}{1} = 1

التمرين الثاني (6 نقاط):

Given the sequence un=4n3u_n = 4n - 3, we proceed as follows:

1. تحقق أن:

To prove un=4n3u_n = 4n - 3, it's already given. The formula defines the sequence explicitly.

2. المتتالية:

The first term u0u_0, second term u1u_1, and sum of the first 18 terms:

u0=4(0)3=3,u1=4(1)3=1u_0 = 4(0) - 3 = -3, \quad u_1 = 4(1) - 3 = 1

To find the sum S=u0+u1++u17S = u_0 + u_1 + \dots + u_{17}, we use the sum formula for an arithmetic sequence:

Sn=n2×(a1+an)S_n = \frac{n}{2} \times (a_1 + a_n)

Where a1=u0=3a_1 = u_0 = -3, an=u17=4(17)3=65a_n = u_{17} = 4(17) - 3 = 65, and n=18n = 18:

S18=182×(3+65)=9×62=558S_{18} = \frac{18}{2} \times (-3 + 65) = 9 \times 62 = 558

التمرين الثالث (6 نقاط):

The function f(x)=2exf(x) = 2 - e^{-x} is defined. Now:

1. تحقق أن:

Check f(ln2)=0f(\ln 2) = 0:

f(ln2)=2eln2=212=32(not zero, there seems to be a mistake)f(\ln 2) = 2 - e^{-\ln 2} = 2 - \frac{1}{2} = \frac{3}{2} \quad \text{(not zero, there seems to be a mistake)}

Similarly for the limits:

limx+f(x)=2,limxf(x)=\lim_{x \to +\infty} f(x) = 2, \quad \lim_{x \to -\infty} f(x) = -\infty

2. اشتقاق:

Differentiate f(x)f(x):

f(x)=ddx(2ex)=exf'(x) = \frac{d}{dx} \left(2 - e^{-x}\right) = e^{-x}

التمرين الرابع (6 نقاط):

This problem involves probability with colored balls (3 white, 5 blue).

1. احسب عدد التوزيعات الممكنة:

Total number of ways to distribute 6 balls (3 white, 5 blue) in a sequence without distinction:

(63)=6!3!3!=20\binom{6}{3} = \frac{6!}{3!3!} = 20

2. احتمال الحصول:

Probability of obtaining three blue balls when choosing 3 out of 5:

P=(33)(63)=120P = \frac{\binom{3}{3}}{\binom{6}{3}} = \frac{1}{20}


Would you like to dive into the details of any of these steps? Here are some follow-up questions you might want to explore:

  1. Can we apply another method for solving logarithmic equations?
  2. How does L'Hôpital's rule work in more complex limits?
  3. What are other properties of arithmetic sequences that can be useful?
  4. Can you derive the differentiation of composite functions like f(x)f(x)?
  5. How is probability affected if we add more balls of different colors?

Tip:

In probability, breaking down problems into distinct cases often helps clarify the solution path!

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Math Problem Analysis

Mathematical Concepts

Logarithms
Exponential Equations
Limits
Arithmetic Sequences
Differentiation
Probability

Formulas

log(a) + log(b) = log(ab)
ln(a) = ln(b) implies a = b
Limit formula: lim(x→0) (e^x - 1)/x
Sum formula for arithmetic sequence: S_n = n/2 * (a_1 + a_n)

Theorems

L'Hôpital's Rule
Properties of Exponential Functions
Basic Properties of Logarithms

Suitable Grade Level

Grades 10-12