HOTS Questions on Class 9 Maths Chapter 2 Introduction to Linear Polynomials

HOTS Questions on Class 9 Maths Chapter 2 Introduction to Linear Polynomials is an easy way for students to learn about algebra. It will also help them understand variables,coefficients,constants, and linear polynomials in an easy way. This chapter builds a strong base for solving equations and improving math skills. A downloadable PDF is also available for easy revision and practice.

HOTS Questions Based on Identifying Linear Polynomials

Q1. Which of the following are linear polynomials? Justify each answer.

(a) 3x + 7 (b) x² − 2x + 1 (c) 5 (d) 4x + 0 (e) √x + 3 (f) 2/x + 1 (g) 0·x + 9 (h) x − x + 1

Ans (a) 3x + 7: degree 1, coefficient of x is 3 ≠ 0, Linear polynomial

Ans (d) 4x + 0 = 4x: degree 1, coefficient 4 ≠ 0, Linear polynomial 

Q2. A student says that any expression with exactly two terms is a linear polynomial. Is this always true? Give two counterexamples.

Ans. False. A two-term (binomial) expression is not always linear.

Counterexample 1: x² + 3 has two terms but is quadratic (degree 2).

Counterexample 2: x³ − 5 has two terms but is cubic (degree 3).

A linear polynomial must have the highest degree equal to exactly 1. Having two terms is a necessary condition for the standard form ax + b, but it is not sufficient the degree must be verified.

Q3. If p(x) = (k − 2)x + 5 is a linear polynomial, what condition must k satisfy? What happens when k = 2?

Ans. For p(x) to be linear, the coefficient of x must not be zero:

k − 2 ≠ 0, k ≠ 2

When k = 2: p(x) = (2 − 2)x + 5 = 0·x + 5 = 5

This is a constant polynomial of degree 0, not a linear polynomial. So k = 2 is not allowed.

Q4. Two polynomials p(x) = ax + b and q(x) = cx + d are added. Under what conditions is their sum also a linear polynomial?

Ans. Sum = (ax + b) + (cx + d) = (a + c)x + (b + d)

For this to be a linear polynomial, the coefficient of x must not be zero:

(a + c) ≠ 0

If a + c = 0, the x terms cancel and the sum reduces to a constant polynomial (b + d), which is not linear.

Q5. Can the product of two linear polynomials ever be a linear polynomial? Justify mathematically.

Ans. No. Let p(x) = ax + b and q(x) = cx + d, where a ≠ 0 and c ≠ 0.

Product = (ax + b)(cx + d) = acx² + (ad + bc)x + bd

Since a ≠ 0 and c ≠ 0, the product ac ≠ 0, so the x² term does not vanish.

The product is always a degree-2 (quadratic) polynomial — never linear.

HOTS Questions Based on Zeroes of Linear Polynomials

Q1. The zero of the linear polynomial p(x) = 3x − k is 4. Find k and write the full polynomial.

Ans. At x = 4, p(4) = 0:

3(4) − k = 0, 12 − k = 0, k = 12

Full polynomial: p(x) = 3x − 12

Q2. The zero of ax + 5 is −5/3. Find a. Then find the zero of 2ax + 10 and compare.

Ans. At x = −5/3:

a(−5/3) + 5 = 0, −5a/3 = −5, a = 3

New polynomial: 2(3)x + 10 = 6x + 10

Zero of 6x + 10: 6x + 10 = 0, x = −10/6 = −5/3

Observation: The zero is unchanged. Multiplying a linear polynomial by a non-zero constant (here, 2) does not change its zero it only scales the output values.

Q3. The zero of (m + 1)x − (2m − 1) is 3. Find m and verify your answer.

Ans. At x = 3:

(m + 1)(3) − (2m − 1) = 0

3m + 3 − 2m + 1 = 0

m + 4 = 0, m = −4

Substitute m = −4: p(x)

= (−4 + 1)x − (2(−4) − 1)

= −3x − (−9)

= −3x + 9

Verify at x = 3: −3(3) + 9 = −9 + 9 = 0

Q4. Two linear polynomials p(x) = 2x + a and q(x) = 3x + b have the same zero. Prove that 3a = 2b.

Ans. Let the common zero be α.

p(α) = 0: 2α + a = 0, α = −a/2

q(α) = 0: 3α + b = 0, α = −b/3

Since both equal α:

−a/2 = −b/3

Cross multiply: −3a = −2b, 3a = 2b (proved)

Q5. A linear polynomial has its zero at x = −7. Write three different linear polynomials that could have this zero and state the general form.

Ans. If zero = −7, then p(−7) = 0. Any polynomial with this zero has the form:

p(x) = a(x + 7) for any non-zero real a

Three examples:

• p(x) = x + 7 (when a = 1)

• p(x) = 2x + 14 (when a = 2)

• p(x) = −3x − 21 (when a = −3)

All give zero at x = −7. The general form is p(x) = a(x + 7), a ≠ 0.

Q6. Is it possible for a linear polynomial to have no zero? Justify your answer.

Ans. No. A linear polynomial p(x) = ax + b where a ≠ 0 always has exactly one zero:

Setting ax + b = 0: x = −b/a

Since a ≠ 0, this division is always valid and gives a unique real number as the zero.

It is impossible for a linear polynomial to have no zero over the real numbers. This is one of the key properties that distinguishes linear polynomials from quadratic polynomials, which can have no real roots.

Q7. If the zero of p(x) = (2k − 1)x + (k + 3) is 0, find k. What does a zero at x = 0 tell you about the polynomial?

Ans. p(0) = 0:

(2k − 1)(0) + (k + 3) = 0 → k + 3 = 0 → k = −3

Substitute: p(x) = (2(−3) − 1)x + 0 = −7x

Interpretation: A zero at x = 0 means the constant term b = 0. The polynomial becomes p(x) = ax, which passes through the origin. Its graph is a straight line through (0, 0). Every polynomial of the form p(x) = ax has zero at x = 0.

HOTS Questions Based on Algebraic Relationships

Q1. Observe: x − 1 has zero 1; x − 2 has zero 2; x − n has zero n. Write the polynomial whose zero is n² − 1 using the same pattern.

Ans. Following the pattern x − (zero) = polynomial:

The polynomial whose zero is n² − 1 is:

p(x) = x − (n² − 1) = x − n² + 1

This can also be written as x − (n + 1)(n − 1), which links to the difference of squares identity.

Q2. If p(x) = 5x + 10 and q(x) = 2p(x) − 3, is q(x) linear? Find its zero.

Ans. q(x) = 2(5x + 10) − 3 = 10x + 20 − 3 = 10x + 17

Degree = 1, coefficient of x = 10 ≠ 0 → Yes, q(x) is a linear polynomial.

Zero: 10x + 17 = 0, x = −17/10

Q3. The zeroes of p(x) = ax + b and r(x) = bx + a are added. Find their sum.

Ans. Zero of p(x) = ax + b is α₁ = −b/a

Zero of r(x) = bx + a is α₂ = −a/b

Sum = −b/a + (−a/b) = −b/a − a/b

= (−b² − a²) / ab = −(a² + b²) / ab

Q4. A student claims that if p(x) has zero α, then 2p(x) has zero 2α. Is this correct? Give a proof or a counterexample.

Ans. The claim is incorrect.

Let p(x) = ax + b with zero α = −b/a.

Then 2p(x) = 2ax + 2b.

Zero of 2p(x): 2ax + 2b = 0, x = −2b/2a = −b/a = α

The zero of 2p(x) is still α, not 2α.

Multiplying a polynomial by a non-zero constant scales all output values by that constant but does not change the input value (x) where the output is zero.

Q5. If the zero of p(x) = kx + (k² − 4) is positive, find all possible values of k (k ≠ 0).

Ans. Zero = −(k² − 4)/k = (4 − k²)/k

For the zero to be positive: (4 − k²)/k > 0

Case 1: 4 − k² > 0 and k > 0

k² < 4 and k > 0

−2 < k < 2 and k > 0, 0 < k < 2

Case 2: 4 − k² < 0 and k < 0

k² > 4 and k < 0, k < −2, k < −2

Answer: k ∈ (0, 2) ∪ (−∞, −2)

Q6. Two linear polynomials have zeroes that are additive inverses of each other. What condition must their coefficients satisfy?

Ans. Let zeroes be α and −α.

p(x) = ax + b has zero α = −b/a

q(x) = cx + d has zero −α = b/a

So −d/c = b/a, −ad = bc, ad + bc = 0

This is the required condition on the coefficients.

HOTS Questions Based on Real-Life Applications of Polynomials

Q1. A shopkeeper's profit is given by p(x) = 15x − 300, where x is the number of items sold. At how many items does the shopkeeper break even?

Ans. At breakeven, profit = 0:

15x − 300 = 0, 15x = 300, x = 20 items

The shopkeeper breaks even at 20 items. This is the zero of the profit polynomial. Below 20 items, the shopkeeper makes a loss; above 20 items, a profit.

Q2. A taxi fare is modelled by f(x) = 12x + 30, where x is distance in km. At what distance is the fare zero? Does this have practical meaning?

Ans. 12x + 30 = 0, x = −30/12 = −2.5 km

The mathematical zero is −2.5 km, which has no practical meaning since distance cannot be negative.

This illustrates an important principle: a mathematical solution must always be interpreted within its real-world context. The zero of the fare polynomial lies outside the physically meaningful domain (x ≥ 0).

Q3. The temperature in a city at time t hours after midnight is T(t) = 2t − 10 (°C). When does the temperature cross 0°C?

Ans. 2t − 10 = 0, 2t = 10, t = 5 hours after midnight

The temperature crosses zero degrees at 5:00 AM.

Q4. A school's fee collection is modelled by F(n) = 500n − 2000, where n is the number of students. Below how many students does the school run at a loss?

Ans. Loss when F(n) < 0:

500n − 2000 < 0, n < 4

The school runs at a loss when fewer than 4 students are enrolled. The zero (breakeven) is at n = 4.

Q5. Riya earns ₹(3x + 200) per day and her colleague earns ₹(4x + 100), where x is extra hours worked. On what value of x do they earn the same? Express this as a zero of a polynomial.

Ans. Set equal: 3x + 200 = 4x + 100, 100 = x, x = 100

The polynomial whose zero gives this answer is the difference:

(4x + 100) − (3x + 200) = x − 100

Zero of x − 100 is x = 100

The answer to "when are two linear expressions equal" is always the zero of their difference.

HOTS Questions Involving Graphical Interpretation

Q1. The graph of a linear polynomial p(x) passes through the points (1, 3) and (3, 7). Find the polynomial.

Ans. Let p(x) = mx + c (slope m, y-intercept c).

From (1, 3): m + c = 3. From (3, 7): 3m + c = 7.

Subtract: 2m = 4, m = 2. Then c = 3 − 2 = 1.

p(x) = 2x + 1.

Q2. A linear polynomial p(x) crosses the x-axis at x = −4 and the y-axis at y = 8. Find p(x) and verify both intercepts.

Ans. x-intercept (zero) at −4 means p(x) = a(x + 4) for some a ≠ 0.

y-intercept at y = 8 means p(0) = 8: a(0 + 4) = 8, 4a = 8, a = 2.

p(x) = 2(x + 4) = 2x + 8.

Verify: p(−4) = 2(0) = 0.

p(0) = 8.

Q3. Look at the graph above. The line crosses the x-axis to the right of the origin. What does this tell you about the sign of the zero and the relationship between a and b in p(x) = ax + b?

Ans. The x-intercept (zero) is at a positive value. Zero = −b/a > 0. This means b and a have opposite signs: if a > 0, then b < 0; if a < 0, then b > 0.

Q4. If a linear polynomial has a positive slope and crosses the x-axis at a positive value, what can you say about the y-intercept?

Ans. Positive slope means a > 0. Zero = −b/a is positive, so −b > 0, meaning b < 0. A negative y-intercept. Geometrically: the line goes upward left to right and crosses the y-axis below the x-axis, then crosses the x-axis at a positive value.

Q5. The graph of p(x) = ax + b passes through the origin. What is the zero of this polynomial, and what is the value of b?

Ans. If the graph passes through the origin, then p(0) = 0: a(0) + b = 0, b = 0. So the polynomial is p(x) = ax, and the zero is x = 0. The graph of any polynomial of the form p(x) = ax passes through the origin.

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