Mathematics — Standard (041) — CBSE Class 10 board question

(i) Co-ordinates of Point A:

Point A is the vertex (highest point) of the parabola.

For $p(x) = -0.0025x^2 - 0.025x + 136$, vertex x-coordinate:

$x = -\dfrac{b}{2a} = -\dfrac{-0.025}{2(-0.0025)} = -\dfrac{-0.025}{-0.005} = -5$

$p(-5) = -0.0025(25) - 0.025(-5) + 136 = -0.0625 + 0.125 + 136 = 136.0625$

Co-ordinates of A = (−5, 136.0625)

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(ii) Span of the Arch:

The span is the distance between the two zeroes (roots) of $p(x)$.

Setting $p(x) = 0$: $-0.0025x^2 - 0.025x + 136 = 0$

Multiply by $-400$: $x^2 + 10x - 54400 = 0$

$x = \dfrac{-10 \pm \sqrt{100 + 217600}}{2} = \dfrac{-10 \pm \sqrt{217700}}{2} \approx \dfrac{-10 \pm 466.6}{2}$

$x_1 \approx 228.3,\quad x_2 \approx -238.3$

Span = $228.3 - (-238.3) \approx 466.6$ units

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(iii) Zeroes and Verification:

From the diagram, the arch meets the ground (x-axis) at two points — the zeroes are approximately x ≈ 228.3 and x ≈ −238.3.

Verification:

• Sum of zeroes $= 228.3 + (-238.3) = -10 = -\dfrac{b}{a} = -\dfrac{-0.025}{-0.0025} = -10$ ✓
• Product of zeroes $= 228.3 \times (-238.3) \approx -54,400 = \dfrac{c}{a} = \dfrac{136}{-0.0025} = -54,400$ ✓

Relationship is verified.

Source: Polynomials (Chapter 2), Zeroes and Geometrical Meaning / Relationship between zeroes and coefficients

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• (i): The vertex formula $x = -b/2a$ gives the x-coordinate of A; substitute back for y.
• (ii): Span = distance between the two roots (where the arch meets ground). Examiners expect the quadratic solved and difference computed.
• (iii): Examiners want you to read zeroes from diagram (where curve cuts x-axis), then verify using $\alpha+\beta = -b/a$ and $\alpha\beta = c/a$. Showing both relations earns full marks. In the OR option, simply substitute x = 100 and x = −100 and compare; they will not be equal since the parabola is not symmetric about x = 0 (axis of symmetry is x = −5).