CBSE Class 9 Maths Ganita Manjari Part 1 Ch 02 Introduction to Linear Polynomials Long Answer and Case Study Questions with Answers

5 Mark Questions

Question 1. A taxi service charges a fixed amount of ₹50 and ₹15 per km travelled. Let the distance travelled be x km and the total fare be represented by F(x).
(a) Form a polynomial F(x) representing the total fare.
(b) State its degree, coefficient of x, and constant term.
(c) For what distance will the fare be ₹200?
(d) Explain why the polynomial obtained in this situation is a linear polynomial.
Answer:
(a) \( F(x) = 15x + 50 \)
(b) Degree = 1; coefficient of x = 15; constant term = 50.
(c) \( 15x + 50 = 200 \Rightarrow 15x = 150 \Rightarrow x = 10 \). The fare is ₹200 for 10 km.
(d) The fare increases by the same amount (₹15) for every extra kilometre, and x appears only to the power 1. So the highest power of x is 1, which makes F(x) a linear polynomial.
In simple words: The fare is ₹50 plus ₹15 for each km, and ₹200 means 10 km.

Exam Tip: A constant rate of change in a real situation always leads to a linear polynomial.

Question 2. Two linear polynomials are given by \( p(x) = 3x + 5 \) and \( q(x) = 5x - 7 \).
(a) Find p(2) and q(2).
(b) Find the value of x for which p(x) = q(x).
(c) Determine the zero of each polynomial.
Answer:
(a) \( p(2) = 3(2) + 5 = 11 \); \( q(2) = 5(2) - 7 = 3 \)
(b) \( 3x + 5 = 5x - 7 \Rightarrow 12 = 2x \Rightarrow x = 6 \). Check: p(6) = 23 and q(6) = 23 ✓
(c) \( 3x + 5 = 0 \Rightarrow x = -\frac{5}{3} \); \( 5x - 7 = 0 \Rightarrow x = \frac{7}{5} \)
Zeros: \( -\frac{5}{3} \) for p(x) and \( \frac{7}{5} \) for q(x).
In simple words: Substitute to find values, set them equal to find where they match, and set each to zero to find its zero.

Exam Tip: In part (b), collect x-terms on one side and numbers on the other before dividing.

Question 3. A linear polynomial is of the form \( p(x) = ax + b \), where a and b are real numbers. It is known that \( p(2) = 7 \) and \( p(5) = 16 \).
(a) Form two equations in a and b using the given information.
(b) Find the values of a and b.
(c) Hence, find the polynomial p(x).
(d) Find the zero of p(x).
Answer:
(a) \( 2a + b = 7 \) … (1) and \( 5a + b = 16 \) … (2)
(b) (2) − (1): \( 3a = 9 \Rightarrow a = 3 \). From (1): \( 6 + b = 7 \Rightarrow b = 1 \).
(c) \( p(x) = 3x + 1 \). Check: p(5) = 15 + 1 = 16 ✓
(d) \( 3x + 1 = 0 \Rightarrow x = -\frac{1}{3} \)
In simple words: Two given values give two equations; subtracting them removes b and gives a.

Exam Tip: Always check the final polynomial with both given values.

Question 4. A student claims: "If \( p(x) = ax + b \) is a linear polynomial and \( p(3) = p(7) \), then \( a = 0 \)."
(a) Determine whether the student's claim is correct or incorrect.
(b) Give a mathematical justification for your answer.
(c) Is such a polynomial a linear polynomial? Give a reason based on the definition of the degree of a polynomial.
(d) What condition on a is necessary for \( ax + b \) to be a linear polynomial?
Answer:
(a) The algebra in the claim is correct: p(3) = p(7) does force a = 0.
(b) \( 3a + b = 7a + b \Rightarrow 0 = 4a \Rightarrow a = 0 \)
(c) No. If a = 0, then p(x) = b, which has no x-term. Its degree is 0 (or it is the zero polynomial if b = 0), so it is a constant polynomial, not a linear one. This means a truly linear polynomial can never have p(3) = p(7).
(d) For \( ax + b \) to be linear, we need \( a \neq 0 \).
In simple words: Equal values at two different x-values mean the x-term has vanished, so the polynomial is really just a constant.

Exam Tip: A linear polynomial takes different values at different x; if two values match, it is not linear.

Case Study Based Questions

Question 5. The Robot That Refuses Stage 0
A robot builds towers according to \( H(n) = 5n - 2 \) centimetres, where n = 1, 2, 3, … is the tower number. A student says the constant term −2 means the first tower has negative height.
(a) Explain the error in the student's interpretation.
(b) Find H(1), H(2) and H(10).
(c) Algebraically H(0) = −2. Why need this not describe a real tower?
(d) Can a tower in this sequence have height 48 cm?
Answer:
(a) The constant term −2 is the value of H at n = 0, not at n = 1. The first tower is n = 1, and \( H(1) = 3 \) cm, which is positive.
(b) \( H(1) = 5 - 2 = 3 \) cm; \( H(2) = 10 - 2 = 8 \) cm; \( H(10) = 50 - 2 = 48 \) cm
(c) The rule is only meant for n = 1, 2, 3, … . n = 0 is outside this range, so the value −2 is just an algebraic result and does not stand for any real tower.
(d) \( 5n - 2 = 48 \Rightarrow 5n = 50 \Rightarrow n = 10 \). Since 10 is an allowed tower number, yes, the 10th tower is 48 cm tall.
In simple words: The −2 belongs to a 'tower 0' that does not exist; real towers start at 3 cm and the 10th one is 48 cm.

Exam Tip: Always check whether a value of the variable is allowed in the real situation before interpreting it.

Question 6. The Reverse Barcode
A barcode machine prints a value \( B(x) = ax + 9 \). It prints 29 when x = 4.
(a) Find a.
(b) Find the zero of B(x).
(c) A worker says the zero must be positive because all printed values seen so far were positive. Is that valid?
(d) If x is restricted to whole-number package counts, does the zero necessarily correspond to a real package count?
Answer:
(a) \( 4a + 9 = 29 \Rightarrow 4a = 20 \Rightarrow a = 5 \). So \( B(x) = 5x + 9 \).
(b) \( 5x + 9 = 0 \Rightarrow x = -\frac{9}{5} = -1.8 \)
(c) No. Positive outputs at some inputs tell us nothing about where the polynomial becomes zero. Here the zero is −1.8, which is negative.
(d) No. −1.8 is neither a whole number nor non-negative, so it cannot be a package count. It has algebraic meaning only.
In simple words: The machine rule is 5x + 9, and it becomes zero at −1.8, which is not a possible number of packages.

Exam Tip: The zero of a polynomial can be negative or a fraction even when all real-life values are positive.

Question 7. The Discount That Becomes Impossible
A coupon model gives final price \( F(n) = 500 - 35n \), where n is the number of discount tokens used.
(a) Is F(n) increasing or decreasing? Why?
(b) The algebraic rule eventually gives negative prices. Does that mean a shop would pay the customer?
(c) Find the greatest whole number n for which F(n) is still non-negative.
Answer:
(a) Decreasing, because the coefficient of n is −35: each extra token lowers the price by ₹35.
(b) No. The model only makes sense for a limited range of n. In real life a shop would set a limit on tokens or a minimum price; the formula simply stops being valid beyond that point.
(c) \( 500 - 35n \ge 0 \Rightarrow 35n \le 500 \Rightarrow n \le 14.28… \)
The greatest whole number is n = 14, giving \( F(14) = 500 - 490 = ₹10 \). (For n = 15, F = −25, which is not possible.)
In simple words: Each token cuts ₹35, so after 14 tokens the price is ₹10 and one more token would make it negative.

Exam Tip: When a question asks for the greatest whole number, round down, not to the nearest whole number.