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Punjab Board Class 11 Mathematics

Exercise 12.3 Solutions — Limit and Continuity

Complete worked solutions for Exercise 12.3 from Chapter 12 (Limit and Continuity), Punjab Board Class 11 Mathematics. Each question shows every step and names the formula used.

Mathematics 11 (Punjab Board)

Exercise 12.3 Solutions

Worked solutions for Punjab Board Class 11 Mathematics, Chapter 12: Limit and Continuity.

Question 1: A substance decays exponentially as \(A(t) = A_0 e^{-0.1t}\), where \(A_0\) is the initial amount. Find \(\lim_{t \to \infty} A(t)\).

Decay model mein exponent negative hai. \(t \to \infty\) par \(e^{-0.1t} \to 0\), is liye poora amount zero ki taraf jata hai.

\[\lim_{t \to \infty} A(t) = \lim_{t \to \infty} A_0 e^{-0.1t} = A_0 \cdot 0 = 0\]
Answer: \(0\) — the substance decays completely over time.

Question 2: A town's population is modelled by \(P(t) = \frac{100{,}000}{1 + 9e^{-0.5t}}\). Find the long-term population as \(t \to \infty\).

Logistic model hai. \(t \to \infty\) par \(e^{-0.5t} \to 0\), to denominator \(1\) reh jata hai aur population carrying capacity tak pahunch jati hai.

\[\lim_{t \to \infty} P(t) = \lim_{t \to \infty} \frac{100{,}000}{1 + 9e^{-0.5t}} = \frac{100{,}000}{1 + 9(0)} = 100{,}000\]
Answer: \(100{,}000\) — the population stabilizes at its carrying capacity.

Question 3: Weekly sales (in thousands) follow \(S(t) = \frac{500t}{t + 10}\). Find \(\lim_{t \to \infty} S(t)\) and interpret it.

Numerator aur denominator dono ko \(t\) se divide karenge. \(\frac{10}{t} \to 0\), to limit leading coefficients ka ratio hogi.

\[\lim_{t \to \infty} S(t) = \lim_{t \to \infty} \frac{500t}{t+10} = \lim_{t \to \infty} \frac{500}{1 + \frac{10}{t}} = \frac{500}{1+0} = 500\]
Answer: \(500\) — long-term weekly sales stabilize at \(500{,}000\).

Question 4: Signal strength at distance \(d\) from a tower is \(S(d) = \frac{1000}{d^2}\). (i) Find the signal strength at \(d = 10\). (ii) What happens to the signal strength as \(d \to \infty\)?

(i) mein seedha value put karenge. (ii) mein denominator infinity ki taraf jata hai, to fraction zero ki taraf jata hai.

(i)

\[S(10) = \frac{1000}{10^2} = \frac{1000}{100} = 10\]

(ii)

\[\lim_{d \to \infty} S(d) = \lim_{d \to \infty} \frac{1000}{d^2} = 0\]
Answer: (i) \(10\), (ii) \(0\) — the signal fades away at large distances.

Question 5: A stock price grows as \(P(t) = 50e^{0.05t}\). (i) Find \(\lim_{t \to \infty} P(t)\). (ii) Calculate the price after 10 years.

(i) mein positive exponent ki wajah se \(e^{0.05t} \to \infty\). (ii) mein \(t = 10\) put karke \(e^{0.5} \approx 1.6487\) use karenge.

(i)

\[\lim_{t \to \infty} P(t) = \lim_{t \to \infty} 50e^{0.05t} = 50 \cdot \infty = \infty\]

(ii)

\[P(10) = 50e^{0.05(10)} = 50e^{0.5} \approx 50(1.6487) = 82.435\]
Answer: (i) \(\infty\); (ii) approximately \(82.44\).

Question 6: The factory's cost function is \(C(x) = \begin{cases} 10x + 500 & \text{if } x \le 100 \\ 12x + 300 & \text{if } x > 100 \end{cases}\). Is the cost function continuous at \(x = 100\)?

Continuity ki teen shartein check karenge: \(C(100)\), LHL aur RHL. Teenon barabar hon to continuous.

\[\lim_{x \to 100^-} (10x+500) = 1000 + 500 = 1500, \qquad \lim_{x \to 100^+} (12x+300) = 1200 + 300 = 1500\]
\[C(100) = 10(100) + 500 = 1500\]

Since LHL = RHL = \(C(100) = 1500\), all three conditions hold.

Answer: Yes — the cost function is continuous at \(x = 100\).

Question 7: Inflation is modelled by \(I(t) = I_0 e^{0.03t}\), where \(I_0\) is the initial price index. (i) Find the inflation rate after 5 years if \(I_0 = 100\). (ii) Find the expected price index after 10 years.

(i) mein \(t = 5\) par index nikal kar percentage increase calculate karenge. (ii) mein \(t = 10\) put karenge. (\(e^{0.15} \approx 1.1618\), \(e^{0.3} \approx 1.3499\).)

(i)

\[I(5) = 100e^{0.03(5)} = 100e^{0.15} \approx 100(1.1618) = 116.18\]
\[\text{Inflation rate} = \frac{116.18 - 100}{100} \times 100\% = 16.18\%\]

(ii)

\[I(10) = 100e^{0.03(10)} = 100e^{0.3} \approx 100(1.3499) = 134.99\]
Answer: (i) \(16.18\%\); (ii) \(134.99\).

Question 8: The cost to produce \(x\) units is \(C(x) = \begin{cases} 5x + 20 & \text{if } x \le 10 \\ 6x + 10 & \text{if } x > 10 \end{cases}\). Is the cost function continuous at \(x = 10\)?

Pichle sawal ki tarah LHL, RHL aur \(C(10)\) compare karenge.

\[\lim_{x \to 10^-} (5x+20) = 50 + 20 = 70, \qquad \lim_{x \to 10^+} (6x+10) = 60 + 10 = 70\]
\[C(10) = 5(10) + 20 = 70\]

Since LHL = RHL = \(C(10) = 70\), all three conditions hold.

Answer: Yes — the cost function is continuous at \(x = 10\).