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Punjab Board Class 11 Mathematics

Exercise 9.2 Solutions — Division of Polynomials

Complete worked solutions for Exercise 9.2 from Chapter 9 (Division of Polynomials), Punjab Board Class 11 Mathematics. Each question shows every step and names the formula used.

Mathematics 11 (Punjab Board)

Exercise 9.2 Solutions

Worked solutions for Punjab Board Class 11 Mathematics, Chapter 9: Division of Polynomials. This exercise applies the remainder and factor theorems to polynomial regression models and signal processing systems.

Question 1: A polynomial regression model \(P(x) = x^3+2x^2+x-3\) is fitted to monthly sales data. If the observed sales in the 5th month are 240 units, find the percentage error.

Pehle \(x = 5\) rakh kar predicted sales \(P(5)\) nikalenge (yehi remainder theorem jaisa evaluation hai). Phir percentage error \(= \left|\frac{\text{observed}-\text{predicted}}{\text{observed}}\right|\times 100\).

\[ P(5) = (5)^3 + 2(5)^2 + 5 - 3 = 125 + 50 + 5 - 3 = 177 \]
\[ \text{Percentage error} = \left|\frac{240-177}{240}\right|\times 100 = \frac{63}{240}\times 100 = 26.25\% \]
Answer: \(26.25\%\).

Question 2: A retailer models weekly demand as \(D(w) = w^3-2w^2+5w-4\). Use the remainder theorem to predict demand for the 3rd week. If the observed demand is 22 units, calculate the prediction error.

Remainder theorem ke mutabiq \(D(3)\) hi predicted demand hai. Observed (22) mein se predicted subtract karke error nikalenge.

\[ D(3) = (3)^3 - 2(3)^2 + 5(3) - 4 = 27 - 18 + 15 - 4 = 20 \]
\[ \text{Prediction error} = 22 - 20 = 2 \text{ units} \quad \left(\frac{2}{22}\times 100 \approx 9.09\%\right) \]
Answer: Predicted demand \(= 20\) units; prediction error \(= 2\) units (about \(9.09\%\)).

Question 3: A digital signal processing system has a transfer function with numerator \(B(z) = z^2-z-2\). Use the factor theorem to find the zeros of the system.

Zeros wahan hote hain jahan numerator zero ho. \(B(z)\) ko factorize karke zeros nikalenge, phir factor theorem se verify karenge ke \(B(2) = 0\) aur \(B(-1) = 0\).

\[ B(z) = z^2-z-2 = (z-2)(z+1) \]

Check: \(B(2) = 4-2-2 = 0\) and \(B(-1) = 1+1-2 = 0\). By the factor theorem, \((z-2)\) and \((z+1)\) are factors.

Answer: The zeros are \(z = 2\) and \(z = -1\).

Question 4: A signal processing system has transfer function \(H(z) = \frac{z^2+3z+2}{z^2-0.2z+0.9}\). Find the zero(s) of the transfer function using the factor theorem.

Transfer function ke zeros numerator ke zeros hote hain. \(N(z) = z^2+3z+2\) ko factorize karenge.

\[ N(z) = z^2+3z+2 = (z+1)(z+2) \]

Check: \(N(-1) = 1-3+2 = 0\) and \(N(-2) = 4-6+2 = 0\). By the factor theorem, \((z+1)\) and \((z+2)\) are factors of the numerator.

Answer: The zeros are \(z = -1\) and \(z = -2\).

Question 5: A signal processing system has transfer function \(H(z) = \frac{z^2-0.5z-0.5}{z^2+1}\). Find the zero(s) of the transfer function using the factor theorem.

Numerator \(N(z) = z^2-0.5z-0.5\) ko factorize karenge. \(z = 1\) try karenge: \(1-0.5-0.5 = 0\) ✓, to \((z-1)\) aik factor hai.

\[ N(z) = z^2-0.5z-0.5 = (z-1)(z+0.5) \]

Check: \(N(1) = 1-0.5-0.5 = 0\) and \(N(-0.5) = 0.25+0.25-0.5 = 0\). By the factor theorem, \((z-1)\) and \((z+0.5)\) are factors.

Answer: The zeros are \(z = 1\) and \(z = -0.5\).

Question 6: A signal processing system has a transfer function with denominator \(A(z) = z^2-0.3z-0.4\). Use the factor theorem to find the poles of the system and determine if the system is stable.

Poles denominator ke zeros hote hain. Quadratic formula se \(z\) nikalenge (10 se multiply karke asaan bana lenge). Stability ke liye har pole ka \(|z| \lt 1\) hona chahiye.

\[ z^2-0.3z-0.4 = 0 \Rightarrow 10z^2-3z-4 = 0 \]
\[ z = \frac{3\pm\sqrt{9+160}}{20} = \frac{3\pm 13}{20} \Rightarrow z = 0.8 \text{ or } z = -0.5 \]

Check with the factor theorem: \(A(0.8) = 0.64-0.24-0.4 = 0\) ✓ and \(A(-0.5) = 0.25+0.15-0.4 = 0\) ✓. Since \(|0.8| \lt 1\) and \(|-0.5| \lt 1\), both poles lie inside the unit circle.

Answer: Poles at \(z = 0.8\) and \(z = -0.5\); the system is stable.

Question 7: The denominator of a signal processing system's transfer function is \(A(z) = z^2+1.2z+0.35\). Use the factor theorem to determine the location of the poles and assess the stability of the system.

Quadratic formula se poles nikalenge, phir \(A(z)\) mein put karke verify karenge, aur \(|z| \lt 1\) check karke stability batayenge.

\[ z = \frac{-1.2\pm\sqrt{1.44-1.40}}{2} = \frac{-1.2\pm 0.2}{2} \Rightarrow z = -0.5 \text{ or } z = -0.7 \]

Check: \(A(-0.5) = 0.25-0.6+0.35 = 0\) ✓ and \(A(-0.7) = 0.49-0.84+0.35 = 0\) ✓. Since \(|-0.5| \lt 1\) and \(|-0.7| \lt 1\), both poles lie inside the unit circle.

Answer: Poles at \(z = -0.5\) and \(z = -0.7\); the system is stable.