Natural Exponetial Function

Question 1: Stock Price Growth (from file image_c6353c.png)

The Question: Suppose that a stock’s price continues to increase at the rate of 5 % per year. If the value of one share of this stock is ₹1000 at present, then find the value(in ₹) of one share of this stock two years from now. (Use the formula $S = a(1+r)^n$)

Detailed Solution:

1. Identify the variables from the problem and the formula $S = a(1+r)^n$:
   • Initial value (present value), $a = 1000$.
   • Increase rate per year, $r = 5% = 0.05$.
   • Number of years, $n = 2$.
   • Final value, $S$, is what we need to find.
2. Substitute the values into the formula: $$S = 1000(1 + 0.05)^2$$
3. Calculate the result: $$S = 1000(1.05)^2$$$$S = 1000(1.1025)$$$$S = 1102.5$$

Final Answer: The value of one share two years from now will be ₹1102.5.

Question 2: Which of the following options yield the value ’e’? (from file image_c6322d.png)

The Question: Which of the following options yield the value ’e’? (Multiple Select Question)

Detailed Solution:

Let’s analyze each option based on the properties of $e$.

• “The slope of the tangent line of the curve $f(x) = e^x$ at x = 1”

  • A unique property of the function $f(x)=e^x$ is that its derivative (which gives the slope of the tangent line) is also $e^x$.
  • So, the slope at any point $x$ is $f’(x) = e^x$.
  • At $x=1$, the slope is $f’(1) = e^1 = e$. This statement is TRUE.

• “The slope of the tangent line of the curve $f(x) = x^2 + ex$ at x = 1”

  • The derivative is $f’(x) = 2x + e$.
  • At $x=1$, the slope is $f’(1) = 2(1) + e = 2+e$. This is not $e$. This statement is FALSE.

• “The area under the curve $f(x) = 2e^x$, from $(-\infty, 1)$”

  • This requires integration: $\int_{-\infty}^{1} 2e^x ,dx = [2e^x]{-\infty}^{1} = 2e^1 - \lim{a \to -\infty} 2e^a = 2e - 0 = 2e$. This is not $e$. This statement is FALSE.

• “The area under the curve $f(x) = e^x$ from $(-\infty, 1)$”

  • This requires integration: $\int_{-\infty}^{1} e^x ,dx = [e^x]{-\infty}^{1} = e^1 - \lim{a \to -\infty} e^a = e - 0 = e$. This statement is TRUE.

Final Answer:

• The slope of the tangent line of the curve $f(x) = e^x$ at x = 1
• The area under the curve $f(x) = e^x$ from $(-\infty, 1)$

Question 3: Correct Statements about ’e’ (from file image_c6322d.png)

The Question: Which of the following statements is(are) CORRECT?

Detailed Solution:

• “If $n \to \infty$, then $(1+\frac{1}{n})^n \to \infty$”: FALSE. This is the famous limit that defines the number $e$. It does not go to infinity.
• “If $n \to \infty$, then $(1+\frac{1}{n})^n \to e$”: TRUE. This is the formal definition of the natural number $e$.
• “Every $f(x) = a^x, a < 1$ has same properties as $g(x) = e^x$”: FALSE. The base $e \approx 2.718$, which is greater than 1, so $g(x)=e^x$ is an increasing (growth) function. For $0<a<1$, $f(x)=a^x$ is a decreasing (decay) function. They have opposite properties.
• “Every $f(x) = a^x, a > 1$ has same properties as $g(x) = e^x$”: TRUE. Since $e > 1$, $g(x)=e^x$ is an exponential growth function. Any function $f(x)=a^x$ where the base $a$ is also greater than 1 will have the same fundamental properties (it will be increasing, have a domain of $\mathbb{R}$, a range of $(0, \infty)$, etc.).

Final Answer: The correct statements are:

• If $n \to \infty$, then $(1+\frac{1}{n})^n \to e$
• Every $f(x) = a^x, a > 1$ has same properties as $g(x) = e^x$

Question 4: Blue Whale Population (Future Prediction) (from file image_c6322d.png)

The Question: In 1990, the population of blue whales in the world is approximately 3200. Let $F(t)$ be the approximate population… after $t$ years since 1990, defined as $F(t) = 3200e^{-0.15t}$. What is the approximate population of blue whales in the world in the year 2020?

Detailed Solution:

1. Find the value of $t$: The model starts in 1990 ($t=0$). We need to find the population in 2020.
   • $t = 2020 - 1990 = 30$ years.
2. Substitute $t=30$ into the function:
   • $F(30) = 3200e^{-0.15 \times 30}$
   • $F(30) = 3200e^{-4.5}$
3. Calculate the value:
   • Using a calculator, $e^{-4.5} \approx 0.011109$.
   • $F(30) \approx 3200 \times 0.011109 \approx 35.55$.
4. Approximate the answer: The closest whole number is 36.

Final Answer: The approximate population is 36.

Question 5: Blue Whale Population (Finding Time) (from file image_c63196.png)

The Question: …using the same function $F(t) = 3200e^{-0.15t}$, in which year will the population of blue whales approximately be 160?

Detailed Solution:

1. Set the function equal to the target population:
   • $160 = 3200e^{-0.15t}$
2. Isolate the exponential term:
   • Divide both sides by 3200:
   • $\frac{160}{3200} = e^{-0.15t}$
   • $\frac{1}{20} = e^{-0.15t} \implies 0.05 = e^{-0.15t}$
3. Use the natural logarithm ($\ln$) to solve for the exponent:
   • Take the natural log of both sides: $\ln(0.05) = \ln(e^{-0.15t})$
   • Using the property $\ln(e^x) = x$, we get: $\ln(0.05) = -0.15t$
4. Calculate and solve for $t$:
   • Using a calculator, $\ln(0.05) \approx -2.9957$.
   • $-2.9957 = -0.15t$
   • $t = \frac{-2.9957}{-0.15} \approx 19.97$ years.
5. Find the year:
   • We can round $t$ to approximately 20 years.
   • The year is $1990 + t = 1990 + 20 = 2010$.

Final Answer: The population will be approximately 160 in the year 2010.

Question 6: Bacterial Growth Model (from file image_c63196.png)

The Question: Suppose the population of bacteria in a culture is growing exponentially. At today 2:00 pm, 100 bacteria were present and by 5:00 pm, 448 bacteria were present. Find an exponential function $y = ae^{kt}$ that models this growth…

Detailed Solution:

Let $t$ be the number of hours since 2:00 pm.

1. Use the first data point to find ‘a’ (the initial amount):

   • At 2:00 pm, $t=0$ and the population $y=100$.
   • Substitute into the model: $y = ae^{kt}$
   • $100 = ae^{k \cdot 0} = ae^0 = a \cdot 1$
   • So, the initial amount $a = 100$.
   • Our model is now $y = 100e^{kt}$.

2. Use the second data point to find ‘k’ (the growth constant):

   • At 5:00 pm, $t = 5 - 2 = 3$ hours have passed. The population is $y=448$.
   • The problem provides a hint to “Check from the given options” rather than solving for k directly. Let’s test the options where $a=100$.
   • Option: $y=100e^{0.5t}$. Let’s check if it works for our second point ($t=3, y=448$).
     • $y = 100e^{0.5 \times 3} = 100e^{1.5}$
     • Using a calculator, $e^{1.5} \approx 4.4817$.
     • $y \approx 100 \times 4.4817 = 448.17$. This is a very close match to 448.
   • Option: $y=100e^{2t}$. Let’s check this one.
     • $y = 100e^{2 \times 3} = 100e^6$.
     • Using a calculator, $e^6 \approx 403.4$.
     • $y \approx 100 \times 403.4 = 40340$. This is not close to 448.

Final Answer: The function that models the growth is $y = 100e^{0.5t}$.