Graph of Polynomial PYQs
Overview
Geometrical Meaning of Zeroes | Verified CBSE Board Questions. Practice Class 10 Maths Chapter 2 PYQs with step-by-step solutions on SJMaths.
Practice Previous Year Questions Topic-wise
Q1
2023
00:00
The graph of $y = p(x)$ is given. The number of zeroes of $p(x)$ from the graph is:
[Graph shows curve touching x-axis at exactly 1 point]
[Graph shows curve touching x-axis at exactly 1 point]
The number of zeroes of $p(x)$ is the number of points where the graph intersects or touches the X-axis.
Since the graph touches the X-axis at exactly one point, the number of zeroes is 1.
Ans: 1 zero (curve touches but does not cross x-axis)
Q2
2023
00:00
Assertion (A): The graph of $p(x) = (x-1)(x+1)(x-2)$ cuts the X-axis at 3 distinct points.
Reason (R): A polynomial of degree $n$ has at most $n$ zeroes.
Reason (R): A polynomial of degree $n$ has at most $n$ zeroes.
Assertion: The zeroes are $x=1, -1, 2$. These are 3 distinct points. Thus (A) is true.
Reason: It is a fundamental property that degree $n$ implies at most $n$ zeroes. Thus (R) is true.
The reason explains why it's possible for a cubic polynomial (degree 3) to have 3 points of intersection.
Ans: (A) Both A and R are true; R is the correct explanation
Q3
2020
00:00
What is the number of zeroes of $p(x)$ if its graph touches the x-axis at exactly one point and does not cross it?
The points where the graph touches or crosses the x-axis represent the zeroes of the polynomial.
Touching at one point implies exactly 1 zero.
Ans: 1
Q4
2026
00:00
PASSAGE: A parabolic arch of a railway bridge on the Chenab riverbed connects two hills. The arch is represented by $p(x) = -0.0025x^2 - 0.025x + 136$.
(i) Write the coordinates where the arch meets the y-axis. 1 Mark
(ii) What is the total span of the arch (distance PQ)? 1 Mark
(iii) (a) Write the zeroes of $p(x)$ and verify sum of zeroes using coefficients. 2 Marks
OR (b) Find $p(100)$ and $p(-100)$. Are they equal? Justify. 2 Marks
(i) Write the coordinates where the arch meets the y-axis. 1 Mark
(ii) What is the total span of the arch (distance PQ)? 1 Mark
(iii) (a) Write the zeroes of $p(x)$ and verify sum of zeroes using coefficients. 2 Marks
OR (b) Find $p(100)$ and $p(-100)$. Are they equal? Justify. 2 Marks
(i) Meets y-axis when $x=0$. $p(0) = 136$. Coordinates are $(0, 136)$.
(ii) Set $p(x)=0$: $-0.0025x^2 - 0.025x + 136 = 0 \Rightarrow x^2 + 10x - 54400 = 0$. Solving gives $x \approx 228.3, -238.3$. Span $\approx 466.6$ m.
(iii) (a) Zeroes $\alpha, \beta$ are roots of $p(x)=0$. Sum $\alpha+\beta = -b/a = -(-0.025)/(-0.0025) = -10$.
(iii) (b) $p(100) = -0.0025(10000) - 2.5 + 136 = 108.5$. $p(-100) = -25 + 2.5 + 136 = 113.5$. Not equal because of the linear term $-0.025x$.
Q5
2025
00:00
PASSAGE: Shibi decorated his door with garlands on Onam. Each garland forms a downward-opening parabola. The polynomial $p(x)$ represents one garland.
(i) Based on the downward-opening parabola, what is the sign of leading coefficient $a$? 1 Mark
(ii) Find a quadratic polynomial with sum = -1 and product = -2. 1 Mark
(iii) (a) For what value of $k$ is -1 a zero of $(k-2)x^2 - 2x - 5$? 2 Marks
OR (b) If $\alpha, \beta$ are zeroes of $x^2 - 7x + 12$, find $1/\alpha + 1/\beta$. 2 Marks
(i) Based on the downward-opening parabola, what is the sign of leading coefficient $a$? 1 Mark
(ii) Find a quadratic polynomial with sum = -1 and product = -2. 1 Mark
(iii) (a) For what value of $k$ is -1 a zero of $(k-2)x^2 - 2x - 5$? 2 Marks
OR (b) If $\alpha, \beta$ are zeroes of $x^2 - 7x + 12$, find $1/\alpha + 1/\beta$. 2 Marks
(i) Downward-opening parabola implies leading coefficient $a < 0$ (Negative).
(ii) $p(x) = x^2 - (\text{sum})x + (\text{product}) = x^2 - (-1)x + (-2) = x^2 + x - 2$.
(iii) (a) Substitute $x = -1$: $(k-2)(-1)^2 - 2(-1) - 5 = 0 \Rightarrow k - 2 + 2 - 5 = 0 \Rightarrow k = 5$.
(iii) (b) $1/\alpha + 1/\beta = (\alpha+\beta)/\alpha\beta$. From $x^2-7x+12$, sum=7, product=12. Ans: $7/12$.
Q6
2024
00:00
PASSAGE: A student draws a downward-opening parabola $p(x)$ that cuts the x-axis at (-1, 0) and (4, 0).
(i) What is the number of zeroes of $p(x)$? 1 Mark
(ii) Find the sum and product of the zeroes. 1 Mark
(iii) (a) Write one possible polynomial $p(x)$ consistent with the information. 2 Marks
OR (b) If $p(x) = -x^2 + kx + 4$, find the value of $k$. 2 Marks
(i) What is the number of zeroes of $p(x)$? 1 Mark
(ii) Find the sum and product of the zeroes. 1 Mark
(iii) (a) Write one possible polynomial $p(x)$ consistent with the information. 2 Marks
OR (b) If $p(x) = -x^2 + kx + 4$, find the value of $k$. 2 Marks
(i) Graph cuts X-axis at 2 points $\Rightarrow$ 2 zeroes.
(ii) Zeroes are -1 and 4. Sum = $-1 + 4 = 3$. Product = $(-1)(4) = -4$.
(iii) (a) $p(x) = -(x+1)(x-4) = -(x^2 - 3x - 4) = -x^2 + 3x + 4$.
(iii) (b) From (a), $k = 3$.
Q7
2023
00:00
PASSAGE: Radha, an aspiring landscape designer, is creating a pool design. Fountains throw water upward following the parabolic path: $p(x) = -x^2 + 5x - 4$, where $x$ is horizontal distance in metres.
(i) Find the zeroes of $p(x)$. 1 Mark
(ii) Find the value of $x$ at which water attains maximum height. 1 Mark
(iii) (a) If $h$ is the maximum height from water level, find $h$. 2 Marks
OR (b) At what point(s) on x-axis is height of water exactly 2 units? 2 Marks
(i) Find the zeroes of $p(x)$. 1 Mark
(ii) Find the value of $x$ at which water attains maximum height. 1 Mark
(iii) (a) If $h$ is the maximum height from water level, find $h$. 2 Marks
OR (b) At what point(s) on x-axis is height of water exactly 2 units? 2 Marks
(i) $-x^2+5x-4 = 0 \Rightarrow x^2-5x+4=0 \Rightarrow (x-1)(x-4)=0$. Zeroes are 1 and 4.
(ii) Max height occurs at vertex $x = -b/2a = -5/2(-1) = 2.5$ m.
(iii) (a) $h = p(2.5) = -(2.5)^2 + 5(2.5) - 4 = -6.25 + 12.5 - 4 = 2.25$ m.
(iii) (b) Set $p(x)=2 \Rightarrow -x^2+5x-4=2 \Rightarrow x^2-5x+6=0 \Rightarrow (x-2)(x-3)=0$. Ans: at $x=2$ and $x=3$.
Q8
2022
00:00
PASSAGE: A rainbow is an arch of 7 colours visible after rain. Each colour makes a parabola. The graph of $y = f(x)$ is shown representing a rainbow polynomial.
(i) From the graph, write the number of zeroes of the curve. 1 Mark
(ii) If the graph does not intersect x-axis but intersects y-axis at one point, how many zeroes? 1 Mark
(iii) (a) If $p(x) = x^2 + (a+1)x + b$ has zeroes 2 and -3, find $a$ and $b$. 2 Marks
OR (b) Find $p(x)$ at $x = 0$ and $x = -1$. Are they equal? 2 Marks
(i) From the graph, write the number of zeroes of the curve. 1 Mark
(ii) If the graph does not intersect x-axis but intersects y-axis at one point, how many zeroes? 1 Mark
(iii) (a) If $p(x) = x^2 + (a+1)x + b$ has zeroes 2 and -3, find $a$ and $b$. 2 Marks
OR (b) Find $p(x)$ at $x = 0$ and $x = -1$. Are they equal? 2 Marks
(i) The number of zeroes is the number of X-axis intersections.
(ii) Zeroes correspond only to X-axis intersections. If it doesn't intersect X-axis, number of zeroes = 0.
(iii) (a) Sum $= 2-3 = -1 = -(a+1) \Rightarrow a = 0$. Product $= (2)(-3) = -6 = b$. Ans: $a=0, b=-6$.
(iii) (b) $p(0) = b$. $p(-1) = 1 - (a+1) + b = -a + b$. If $a \neq 0$, they are not equal.
Q9
2021
00:00
PASSAGE: A ball is thrown upward from the ground. Its height $h$ (in metres) at time $t$ (in seconds) is $h(t) = -5t^2 + 20t$. The graph of $h(t)$ is a downward parabola.
(i) At what time(s) is the ball at ground level? Find zeroes of $h(t)$. 1 Mark
(ii) What is the maximum height reached by the ball? 1 Mark
(iii) (a) Verify sum of times when ball is at ground = 4, using zeroes-coefficient relationship. 2 Marks
OR (b) For what duration is the ball above 15 m? 2 Marks
(i) At what time(s) is the ball at ground level? Find zeroes of $h(t)$. 1 Mark
(ii) What is the maximum height reached by the ball? 1 Mark
(iii) (a) Verify sum of times when ball is at ground = 4, using zeroes-coefficient relationship. 2 Marks
OR (b) For what duration is the ball above 15 m? 2 Marks
(i) Ground level $h(t)=0 \Rightarrow -5t(t-4)=0 \Rightarrow t=0$ s (start) and $t=4$ s (end).
(ii) Max height at $t = -b/2a = -20/2(-5) = 2$ s. $h(2) = -5(4) + 20(2) = 20$ m.
(iii) (a) Sum of zeroes $= -b/a = -20/-5 = 4$. Verified.
(iii) (b) $-5t^2 + 20t > 15 \Rightarrow t^2 - 4t + 3 < 0 \Rightarrow (t-1)(t-3) < 0$. Ball is above 15m between $t=1$ and $t=3$. Duration = 2 seconds.
The graph of $y = p(x)$ is given. The number of zeroes of $p(x)$ from the graph is:
[Graph shows curve touching x-axis at exactly 1 point]
[Graph shows curve touching x-axis at exactly 1 point]
The number of zeroes of $p(x)$ is the number of points where the graph intersects or touches the X-axis.
Since the graph touches the X-axis at exactly one point, the number of zeroes is 1.
Ans: 1 zero (curve touches but does not cross x-axis)
Q2
2023
00:00
Assertion (A): The graph of $p(x) = (x-1)(x+1)(x-2)$ cuts the X-axis at 3 distinct points.
Reason (R): A polynomial of degree $n$ has at most $n$ zeroes.
Reason (R): A polynomial of degree $n$ has at most $n$ zeroes.
Assertion: The zeroes are $x=1, -1, 2$. These are 3 distinct points. Thus (A) is true.
Reason: It is a fundamental property that degree $n$ implies at most $n$ zeroes. Thus (R) is true.
The reason explains why it's possible for a cubic polynomial (degree 3) to have 3 points of intersection.
Ans: (A) Both A and R are true; R is the correct explanation
Q3
2020
00:00
What is the number of zeroes of $p(x)$ if its graph touches the x-axis at exactly one point and does not cross it?
The points where the graph touches or crosses the x-axis represent the zeroes of the polynomial.
Touching at one point implies exactly 1 zero.
Ans: 1
Q4
2026
00:00
PASSAGE: A parabolic arch of a railway bridge on the Chenab riverbed connects two hills. The arch is represented by $p(x) = -0.0025x^2 - 0.025x + 136$.
(i) Write the coordinates where the arch meets the y-axis. 1 Mark
(ii) What is the total span of the arch (distance PQ)? 1 Mark
(iii) (a) Write the zeroes of $p(x)$ and verify sum of zeroes using coefficients. 2 Marks
OR (b) Find $p(100)$ and $p(-100)$. Are they equal? Justify. 2 Marks
(i) Write the coordinates where the arch meets the y-axis. 1 Mark
(ii) What is the total span of the arch (distance PQ)? 1 Mark
(iii) (a) Write the zeroes of $p(x)$ and verify sum of zeroes using coefficients. 2 Marks
OR (b) Find $p(100)$ and $p(-100)$. Are they equal? Justify. 2 Marks
(i) Meets y-axis when $x=0$. $p(0) = 136$. Coordinates are $(0, 136)$.
(ii) Set $p(x)=0$: $-0.0025x^2 - 0.025x + 136 = 0 \Rightarrow x^2 + 10x - 54400 = 0$. Solving gives $x \approx 228.3, -238.3$. Span $\approx 466.6$ m.
(iii) (a) Zeroes $\alpha, \beta$ are roots of $p(x)=0$. Sum $\alpha+\beta = -b/a = -(-0.025)/(-0.0025) = -10$.
(iii) (b) $p(100) = -0.0025(10000) - 2.5 + 136 = 108.5$. $p(-100) = -25 + 2.5 + 136 = 113.5$. Not equal because of the linear term $-0.025x$.
Q5
2025
00:00
PASSAGE: Shibi decorated his door with garlands on Onam. Each garland forms a downward-opening parabola. The polynomial $p(x)$ represents one garland.
(i) Based on the downward-opening parabola, what is the sign of leading coefficient $a$? 1 Mark
(ii) Find a quadratic polynomial with sum = -1 and product = -2. 1 Mark
(iii) (a) For what value of $k$ is -1 a zero of $(k-2)x^2 - 2x - 5$? 2 Marks
OR (b) If $\alpha, \beta$ are zeroes of $x^2 - 7x + 12$, find $1/\alpha + 1/\beta$. 2 Marks
(i) Based on the downward-opening parabola, what is the sign of leading coefficient $a$? 1 Mark
(ii) Find a quadratic polynomial with sum = -1 and product = -2. 1 Mark
(iii) (a) For what value of $k$ is -1 a zero of $(k-2)x^2 - 2x - 5$? 2 Marks
OR (b) If $\alpha, \beta$ are zeroes of $x^2 - 7x + 12$, find $1/\alpha + 1/\beta$. 2 Marks
(i) Downward-opening parabola implies leading coefficient $a < 0$ (Negative).
(ii) $p(x) = x^2 - (\text{sum})x + (\text{product}) = x^2 - (-1)x + (-2) = x^2 + x - 2$.
(iii) (a) Substitute $x = -1$: $(k-2)(-1)^2 - 2(-1) - 5 = 0 \Rightarrow k - 2 + 2 - 5 = 0 \Rightarrow k = 5$.
(iii) (b) $1/\alpha + 1/\beta = (\alpha+\beta)/\alpha\beta$. From $x^2-7x+12$, sum=7, product=12. Ans: $7/12$.
Q6
2024
00:00
PASSAGE: A student draws a downward-opening parabola $p(x)$ that cuts the x-axis at (-1, 0) and (4, 0).
(i) What is the number of zeroes of $p(x)$? 1 Mark
(ii) Find the sum and product of the zeroes. 1 Mark
(iii) (a) Write one possible polynomial $p(x)$ consistent with the information. 2 Marks
OR (b) If $p(x) = -x^2 + kx + 4$, find the value of $k$. 2 Marks
(i) What is the number of zeroes of $p(x)$? 1 Mark
(ii) Find the sum and product of the zeroes. 1 Mark
(iii) (a) Write one possible polynomial $p(x)$ consistent with the information. 2 Marks
OR (b) If $p(x) = -x^2 + kx + 4$, find the value of $k$. 2 Marks
(i) Graph cuts X-axis at 2 points $\Rightarrow$ 2 zeroes.
(ii) Zeroes are -1 and 4. Sum = $-1 + 4 = 3$. Product = $(-1)(4) = -4$.
(iii) (a) $p(x) = -(x+1)(x-4) = -(x^2 - 3x - 4) = -x^2 + 3x + 4$.
(iii) (b) From (a), $k = 3$.
Q7
2023
00:00
PASSAGE: Radha, an aspiring landscape designer, is creating a pool design. Fountains throw water upward following the parabolic path: $p(x) = -x^2 + 5x - 4$, where $x$ is horizontal distance in metres.
(i) Find the zeroes of $p(x)$. 1 Mark
(ii) Find the value of $x$ at which water attains maximum height. 1 Mark
(iii) (a) If $h$ is the maximum height from water level, find $h$. 2 Marks
OR (b) At what point(s) on x-axis is height of water exactly 2 units? 2 Marks
(i) Find the zeroes of $p(x)$. 1 Mark
(ii) Find the value of $x$ at which water attains maximum height. 1 Mark
(iii) (a) If $h$ is the maximum height from water level, find $h$. 2 Marks
OR (b) At what point(s) on x-axis is height of water exactly 2 units? 2 Marks
(i) $-x^2+5x-4 = 0 \Rightarrow x^2-5x+4=0 \Rightarrow (x-1)(x-4)=0$. Zeroes are 1 and 4.
(ii) Max height occurs at vertex $x = -b/2a = -5/2(-1) = 2.5$ m.
(iii) (a) $h = p(2.5) = -(2.5)^2 + 5(2.5) - 4 = -6.25 + 12.5 - 4 = 2.25$ m.
(iii) (b) Set $p(x)=2 \Rightarrow -x^2+5x-4=2 \Rightarrow x^2-5x+6=0 \Rightarrow (x-2)(x-3)=0$. Ans: at $x=2$ and $x=3$.
Q8
2022
00:00
PASSAGE: A rainbow is an arch of 7 colours visible after rain. Each colour makes a parabola. The graph of $y = f(x)$ is shown representing a rainbow polynomial.
(i) From the graph, write the number of zeroes of the curve. 1 Mark
(ii) If the graph does not intersect x-axis but intersects y-axis at one point, how many zeroes? 1 Mark
(iii) (a) If $p(x) = x^2 + (a+1)x + b$ has zeroes 2 and -3, find $a$ and $b$. 2 Marks
OR (b) Find $p(x)$ at $x = 0$ and $x = -1$. Are they equal? 2 Marks
(i) From the graph, write the number of zeroes of the curve. 1 Mark
(ii) If the graph does not intersect x-axis but intersects y-axis at one point, how many zeroes? 1 Mark
(iii) (a) If $p(x) = x^2 + (a+1)x + b$ has zeroes 2 and -3, find $a$ and $b$. 2 Marks
OR (b) Find $p(x)$ at $x = 0$ and $x = -1$. Are they equal? 2 Marks
(i) The number of zeroes is the number of X-axis intersections.
(ii) Zeroes correspond only to X-axis intersections. If it doesn't intersect X-axis, number of zeroes = 0.
(iii) (a) Sum $= 2-3 = -1 = -(a+1) \Rightarrow a = 0$. Product $= (2)(-3) = -6 = b$. Ans: $a=0, b=-6$.
(iii) (b) $p(0) = b$. $p(-1) = 1 - (a+1) + b = -a + b$. If $a \neq 0$, they are not equal.
Q9
2021
00:00
PASSAGE: A ball is thrown upward from the ground. Its height $h$ (in metres) at time $t$ (in seconds) is $h(t) = -5t^2 + 20t$. The graph of $h(t)$ is a downward parabola.
(i) At what time(s) is the ball at ground level? Find zeroes of $h(t)$. 1 Mark
(ii) What is the maximum height reached by the ball? 1 Mark
(iii) (a) Verify sum of times when ball is at ground = 4, using zeroes-coefficient relationship. 2 Marks
OR (b) For what duration is the ball above 15 m? 2 Marks
(i) At what time(s) is the ball at ground level? Find zeroes of $h(t)$. 1 Mark
(ii) What is the maximum height reached by the ball? 1 Mark
(iii) (a) Verify sum of times when ball is at ground = 4, using zeroes-coefficient relationship. 2 Marks
OR (b) For what duration is the ball above 15 m? 2 Marks
(i) Ground level $h(t)=0 \Rightarrow -5t(t-4)=0 \Rightarrow t=0$ s (start) and $t=4$ s (end).
(ii) Max height at $t = -b/2a = -20/2(-5) = 2$ s. $h(2) = -5(4) + 20(2) = 20$ m.
(iii) (a) Sum of zeroes $= -b/a = -20/-5 = 4$. Verified.
(iii) (b) $-5t^2 + 20t > 15 \Rightarrow t^2 - 4t + 3 < 0 \Rightarrow (t-1)(t-3) < 0$. Ball is above 15m between $t=1$ and $t=3$. Duration = 2 seconds.