Polynomials CBQs — CBSE Class 10 (2027) with step-wise solutions

Sample Polynomials CBQs

1. Question 1MCQ1 mark
   A decorative arch is built over a garden path. Taking the ground as the x-axis and a lamp post as the origin, the height of the arch (in metres) at a distance of x metres from the lamp post is h(x) = −x² + 11x − 24. The arch stands on the ground at the two points where h(x) = 0. How wide is the arch at ground level?
   • 5 m
   • 11 m
   • 8 m
   • 24 m
   Show step-wise solution

   1. [1 mark] The feet of the arch are the zeroes of h(x): −x² + 11x − 24 = 0 ⇒ x² − 11x + 24 = 0 ⇒ (x − 3)(x − 8) = 0 ⇒ x = 3 or 8; width at ground = 8 − 3 = 5 m.

   Answer: 5 m

2. Question 2Assertion-Reason1 mark
   A stone is thrown upward from the edge of a 15 m high cliff. Its height above the ground after t seconds is h(t) = −5t² + 10t + 15 metres. Assertion (A): The stone hits the ground 3 seconds after it is thrown. Reason (R): The time at which the stone hits the ground is the positive zero of h(t).
   • (A) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.
   • (B) Both Assertion and Reason are true but Reason is NOT the correct explanation of Assertion.
   • (C) Assertion is true but Reason is false.
   • (D) Assertion is false but Reason is true.
   Show step-wise solution

   1. [1 mark] (A): h(t) = 0 ⇒ t² − 2t − 3 = 0 ⇒ (t − 3)(t + 1) = 0 ⇒ t = 3 (t = −1 rejected). A is true, R is true, and A follows from R.

   Answer: (A)

3. Question 3Short answer2 marks
   (x² - 3√2x + 4)/(x - √2) ; x ≠ √2 At how many points does the graph of the above expression intersect the x-axis? Show your work.
   Show step-wise solution

   1. [1 mark] Factorises the numerator to write the given expression as (x - √2)(x - 2√2)/(x - √2).
   2. [1 mark] Writes that the above expression simplifies to x - 2√2, whose graph intersects the x-axis at exactly one point.

   Answer: Exactly one point — the expression simplifies to x - 2√2.

4. Question 4Case study4 marks
   In a city park, a fountain nozzle at ground level throws a jet of water. The height of the jet (in metres) at a horizontal distance x metres from the nozzle is h(x) = −x² + 5x. The jet falls back to the ground where h(x) = 0. Based on the above information, answer the following questions: (i) How far from the nozzle does the jet land? [1 mark] (ii) Find the height of the jet at a horizontal distance of 2 m from the nozzle. [1 mark] (iii) The gardener adjusts the nozzle. The new jet is again described by a quadratic polynomial: it leaves the nozzle at x = 0, lands at x = 7 m, and is 6 m high at x = 1 m. Find the polynomial for the new jet. [2 marks]
   Show step-wise solution

   1. [1 mark] (i) −x² + 5x = −x(x − 5) = 0 ⇒ x = 0 (the nozzle) or x = 5, so the jet lands 5 m from the nozzle.
   2. [1 mark] (ii) h(2) = −4 + 10 = 6, so the jet is 6 m high.
   3. [1 mark] (iii) The new jet has zeroes 0 and 7, so its polynomial is p(x) = kx(x − 7), where k ≠ 0.
   4. [1 mark] (iii) p(1) = k × 1 × (−6) = −6k = 6 ⇒ k = −1, so p(x) = −x² + 7x.

   Answer: (i) 5 m (ii) 6 m (iii) p(x) = −x² + 7x

5. Question 5Long answer5 marks
   At a school science fair, a team launches a small stone with a catapult from the edge of a cliff 40 m high. The height of the stone above the ground t seconds after launch is h(t) = -5t² + 10t + 40 metres. The path of the stone is symmetric about the time at which it is highest. Based on the above information, answer the following questions: (a) Find the zeroes of h(t) and hence the time after which the stone hits the ground. [2 marks] (b) Using the symmetry of the zeroes, find the time at which the stone is highest and its greatest height. [2 marks] (c) Verify the relationship between the zeroes and the coefficients of h(t). [1 mark]
   Show step-wise solution

   1. [1 mark] (a) h(t) = 0 gives -5(t² - 2t - 8) = 0, i.e. t² - 2t - 8 = 0.
   2. [1 mark] (t - 4)(t + 2) = 0, so the zeroes are 4 and -2; time cannot be negative, so the stone hits the ground after 4 s.
   3. [1 mark] (b) The highest point lies midway between the zeroes: t = (4 + (-2))/2 = 1 s.
   4. [1 mark] Greatest height = h(1) = -5 + 10 + 40 = 45 m.
   5. [1 mark] (c) Sum of zeroes = 4 + (-2) = 2 = -b/a = -10/(-5); product of zeroes = 4 × (-2) = -8 = c/a = 40/(-5). Verified.

   Answer: Zeroes 4 and -2; hits the ground after 4 s; highest at t = 1 s, greatest height 45 m; sum 2 = -b/a, product -8 = c/a

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