CBSE Class 10 · Maths
CBSE Class 10 Maths Polynomials Competency Based Questions
Help your child master CBSE Class 10 Maths Polynomials competency based questions — covering zeroes of polynomials, graphs, quadratic polynomial formation, and real-life parabola applications. Each question includes detailed step-by-step solutions prepared by Angle Belearn’s CBSE specialists to build deep conceptual understanding and board exam confidence.
CBSE Class 10 Maths Polynomials — Questions with Solutions
Multiple Choice Questions (MCQ)
The graph of a polynomial $p(x)$ cuts the $X$-axis at 3 points and touches it at 2 other points. The number of zeroes of $p(x)$ is
Solution
Option (D) is correct.
According to the property of polynomials, the number of zeroes $=$ number of points at which the graph intersects the $X$-axis. The graph intersects the $X$-axis at 5 different points (3 cuts + 2 touches). Therefore, number of zeroes $= 5$.
According to the property of polynomials, the number of zeroes $=$ number of points at which the graph intersects the $X$-axis. The graph intersects the $X$-axis at 5 different points (3 cuts + 2 touches). Therefore, number of zeroes $= 5$.
In the figure, the graph of a polynomial $p(x)$ is shown. The number of zeroes of $p(x)$ is
Solution
Option (C) is correct.
According to the property of polynomials, number of zeroes $=$ number of points at which the graph intersects the $x$-axis.
From the figure it is clear that the graph intersects the $X$-axis at three different points. Therefore, the polynomial has 3 zeroes.
According to the property of polynomials, number of zeroes $=$ number of points at which the graph intersects the $x$-axis.
From the figure it is clear that the graph intersects the $X$-axis at three different points. Therefore, the polynomial has 3 zeroes.
A quadratic polynomial, the product and sum of whose zeroes are 5 and 8 respectively, is
Solution
Option (A) is correct.
For any quadratic polynomial $ax^{2}+bx+c$:
$$\text{Sum of zeroes} = \frac{-b}{a} = 8 \implies b = -8k,\ a = k$$
$$\text{Product of zeroes} = \frac{c}{a} = 5 \implies c = 5k,\ a = k$$
Required polynomial $= kx^{2}-8kx+5k = k\left[x^{2}-8x+5\right]$
For any quadratic polynomial $ax^{2}+bx+c$:
$$\text{Sum of zeroes} = \frac{-b}{a} = 8 \implies b = -8k,\ a = k$$
$$\text{Product of zeroes} = \frac{c}{a} = 5 \implies c = 5k,\ a = k$$
Required polynomial $= kx^{2}-8kx+5k = k\left[x^{2}-8x+5\right]$
If $x-1$ is a factor of the polynomial $p(x)=x^{3}+ax^{2}+2b$ and $a+b=4$, then
Solution
Option (B) is correct.
Given $p(x) = x^{3}+ax^{2}+2b$ and $a+b=4$ …(i)
Since $x-1$ is a factor, $x=1$ is a zero: $p(1) = 0$
$$1 + a + 2b = 0 \implies a + 2b = -1 \quad \text{…(ii)}$$
Subtracting (i) from (ii): $b = -5$. Substituting back: $a = 9$.
Therefore $a = 9$ and $b = -5$.
Given $p(x) = x^{3}+ax^{2}+2b$ and $a+b=4$ …(i)
Since $x-1$ is a factor, $x=1$ is a zero: $p(1) = 0$
$$1 + a + 2b = 0 \implies a + 2b = -1 \quad \text{…(ii)}$$
Subtracting (i) from (ii): $b = -5$. Substituting back: $a = 9$.
Therefore $a = 9$ and $b = -5$.
If the zeroes of the quadratic polynomial $x^{2}+(a+1)x+b$ are 2 and $-3$, then
Solution
Option (D) is correct.
Given zeroes $\alpha = 2$ and $\beta = -3$.
Sum of zeroes $= 2+(-3) = -1$
Product of zeroes $= 2 \times (-3) = -6$
Quadratic polynomial: $x^{2}-(α+β)x+αβ = x^{2}+x-6$ …(ii)
Comparing with $x^{2}+(a+1)x+b$: $a+1 = 1 \implies a = 0$ and $b = -6$.
Given zeroes $\alpha = 2$ and $\beta = -3$.
Sum of zeroes $= 2+(-3) = -1$
Product of zeroes $= 2 \times (-3) = -6$
Quadratic polynomial: $x^{2}-(α+β)x+αβ = x^{2}+x-6$ …(ii)
Comparing with $x^{2}+(a+1)x+b$: $a+1 = 1 \implies a = 0$ and $b = -6$.
How many zero(es) does $(x-2)(x+3)$ have?
Solution
Option (C) is correct.
Setting $(x-2)(x+3) = 0$:
$$x – 2 = 0 \implies x = 2$$
$$x + 3 = 0 \implies x = -3$$
We get two values of $x$, i.e., $x = 2$ or $x = -3$. Hence, the polynomial has two zeroes.
Setting $(x-2)(x+3) = 0$:
$$x – 2 = 0 \implies x = 2$$
$$x + 3 = 0 \implies x = -3$$
We get two values of $x$, i.e., $x = 2$ or $x = -3$. Hence, the polynomial has two zeroes.
Which of these is the polynomial whose zeroes are $\dfrac{1}{3}$ and $\left(-\dfrac{3}{4}\right)$?
Solution
Option (A) is correct.
$$\text{Sum of zeroes} = \frac{1}{3}+\left(-\frac{3}{4}\right) = \frac{4-9}{12} = \frac{-5}{12}$$
$$\text{Product of zeroes} = \frac{1}{3} \times \left(-\frac{3}{4}\right) = -\frac{1}{4}$$
Required polynomial $= x^{2} – \left(\frac{-5}{12}\right)x+\left(-\frac{1}{4}\right) = \dfrac{12x^{2}+5x-3}{12}$
Hence $12x^{2}+5x-3$ is the required polynomial (with $\frac{1}{12}$ as a constant factor).
$$\text{Sum of zeroes} = \frac{1}{3}+\left(-\frac{3}{4}\right) = \frac{4-9}{12} = \frac{-5}{12}$$
$$\text{Product of zeroes} = \frac{1}{3} \times \left(-\frac{3}{4}\right) = -\frac{1}{4}$$
Required polynomial $= x^{2} – \left(\frac{-5}{12}\right)x+\left(-\frac{1}{4}\right) = \dfrac{12x^{2}+5x-3}{12}$
Hence $12x^{2}+5x-3$ is the required polynomial (with $\frac{1}{12}$ as a constant factor).
The number of quadratic polynomials having zeroes $-5$ and $-3$ is
Solution
Option (D) is correct.
Let zeroes $\alpha = -5$ and $\beta = -3$. The general form is:
$$k\left[x^{2}-(\alpha+\beta)x+\alpha\beta\right] = k\left[x^{2}+8x+15\right]$$
where $k$ is any non-zero real number. Since $k$ can take infinitely many values, more than 3 such polynomials exist.
Let zeroes $\alpha = -5$ and $\beta = -3$. The general form is:
$$k\left[x^{2}-(\alpha+\beta)x+\alpha\beta\right] = k\left[x^{2}+8x+15\right]$$
where $k$ is any non-zero real number. Since $k$ can take infinitely many values, more than 3 such polynomials exist.
$p$ and $q$ are the zeroes of the polynomial $4y^{2}-4y+1$. What is the value of $\dfrac{1}{p}+\dfrac{1}{q}+pq$?
Solution
Option (D) is correct.
For $4y^{2}-4y+1$: Sum of zeroes $p+q = -\left(\dfrac{-4}{4}\right) = 1$ and Product of zeroes $pq = \dfrac{1}{4}$
$$\frac{1}{p}+\frac{1}{q}+pq = \frac{p+q}{pq}+pq = \frac{1}{1/4}+\frac{1}{4} = 4+\frac{1}{4} = \frac{17}{4}$$
For $4y^{2}-4y+1$: Sum of zeroes $p+q = -\left(\dfrac{-4}{4}\right) = 1$ and Product of zeroes $pq = \dfrac{1}{4}$
$$\frac{1}{p}+\frac{1}{q}+pq = \frac{p+q}{pq}+pq = \frac{1}{1/4}+\frac{1}{4} = 4+\frac{1}{4} = \frac{17}{4}$$
If one zero of the quadratic polynomial $x^{2}+3x+k$ is 2, then the value of $k$ is
Solution
Option (B) is correct.
Let $p(x) = x^{2}+3x+k$. Since 2 is a zero, $p(2) = 0$:
$$(2)^{2}+3(2)+k = 0 \implies 4+6+k = 0 \implies k = -10$$
Let $p(x) = x^{2}+3x+k$. Since 2 is a zero, $p(2) = 0$:
$$(2)^{2}+3(2)+k = 0 \implies 4+6+k = 0 \implies k = -10$$
The zeroes of the polynomial $x^{2}-3x-m(m+3)$ are
Solution
Option (B) is correct.
Putting $x = -m$: $(-m)^{2}-3(-m)-m(m+3) = m^{2}+3m-m^{2}-3m = 0$ ✓
Putting $x = m+3$: $(m+3)^{2}-3(m+3)-m(m+3) = (m+3)[m+3-3-m] = 0$ ✓
Hence $-m$ and $m+3$ are the zeroes of the given polynomial.
Putting $x = -m$: $(-m)^{2}-3(-m)-m(m+3) = m^{2}+3m-m^{2}-3m = 0$ ✓
Putting $x = m+3$: $(m+3)^{2}-3(m+3)-m(m+3) = (m+3)[m+3-3-m] = 0$ ✓
Hence $-m$ and $m+3$ are the zeroes of the given polynomial.
The degree of polynomial having zeroes $-3$ and $4$ only is
Solution
Option (A) is correct.
By the definition of a polynomial, a polynomial of degree $n$ has at most $n$ zeroes. Since the polynomial has exactly two zeroes ($-3$ and $4$), the minimum degree required is 2.
By the definition of a polynomial, a polynomial of degree $n$ has at most $n$ zeroes. Since the polynomial has exactly two zeroes ($-3$ and $4$), the minimum degree required is 2.
If $\alpha$ and $\beta$ are the zeros of a polynomial $f(x)=px^{2}-2x+3p$ and $\alpha+\beta=\alpha\beta$, then $p$ is
Solution
Option (B) is correct.
For $f(x) = px^{2}-2x+3p$:
$$\alpha+\beta = \frac{2}{p} \qquad \text{and} \qquad \alpha\beta = \frac{3p}{p} = 3$$
Since $\alpha+\beta = \alpha\beta$:
$$\frac{2}{p} = 3 \implies p = \frac{2}{3}$$
For $f(x) = px^{2}-2x+3p$:
$$\alpha+\beta = \frac{2}{p} \qquad \text{and} \qquad \alpha\beta = \frac{3p}{p} = 3$$
Since $\alpha+\beta = \alpha\beta$:
$$\frac{2}{p} = 3 \implies p = \frac{2}{3}$$
Which of the following could be the graph of the polynomial $(x-1)^{2}(x+2)$?
- ✗(A)
- ✗(B)
- ✓(C)
- ✗(D)
Solution
Option (C) is correct.
Let $y = (x-1)^{2}(x+2)$.
At $X$-axis: $(x-1)^{2}(x+2)=0 \implies x = 1\ \text{(repeated)},\ x = -2$
At $Y$-axis ($x=0$): $y = (-1)^{2}(2) = 2$
The graph must touch the $X$-axis at $x=1$ (repeated root), cut it at $x=-2$, and cross the $Y$-axis at $y=2$. Option (C) satisfies all these conditions.
Let $y = (x-1)^{2}(x+2)$.
At $X$-axis: $(x-1)^{2}(x+2)=0 \implies x = 1\ \text{(repeated)},\ x = -2$
At $Y$-axis ($x=0$): $y = (-1)^{2}(2) = 2$
The graph must touch the $X$-axis at $x=1$ (repeated root), cut it at $x=-2$, and cross the $Y$-axis at $y=2$. Option (C) satisfies all these conditions.
The quadratic polynomial whose sum of zeroes is $3$ and product of zeroes is $-2$ is:
Solution
Option (D) is correct.
Required polynomial $= x^{2} – (\text{sum of zeroes})x + (\text{product of zeroes})$
$$= x^{2} – 3x + (-2) = x^{2}-3x-2$$
Required polynomial $= x^{2} – (\text{sum of zeroes})x + (\text{product of zeroes})$
$$= x^{2} – 3x + (-2) = x^{2}-3x-2$$
If $(x+1)$ is a factor of $2x^{3}+ax^{2}+2bx+1$, and $2a-3b=4$, find the values of $a$ and $b$.
Solution
Option (C) is correct.
By Factor Theorem, $f(-1) = 0$:
$$2(-1)^{3}+a(-1)^{2}+2b(-1)+1 = 0 \implies a-2b = 1 \quad \text{…(1)}$$
Also given: $2a-3b = 4$ …(2). From (1): $a = 1+2b$. Substituting in (2):
$$2(1+2b)-3b = 4 \implies b = 2,\ a = 5$$
By Factor Theorem, $f(-1) = 0$:
$$2(-1)^{3}+a(-1)^{2}+2b(-1)+1 = 0 \implies a-2b = 1 \quad \text{…(1)}$$
Also given: $2a-3b = 4$ …(2). From (1): $a = 1+2b$. Substituting in (2):
$$2(1+2b)-3b = 4 \implies b = 2,\ a = 5$$
The number of zeroes that the polynomial $f(x) = (x-2)^{2}+4$ can have is:
Solution
Option (C) is correct.
Simplify: $f(x) = (x-2)^{2}+4 = x^{2}-4x+8$
Discriminant: $D = (-4)^{2}-4(1)(8) = 16-32 = -16 < 0$
Since $D < 0$, the polynomial has no real zeroes.
Simplify: $f(x) = (x-2)^{2}+4 = x^{2}-4x+8$
Discriminant: $D = (-4)^{2}-4(1)(8) = 16-32 = -16 < 0$
Since $D < 0$, the polynomial has no real zeroes.
The zeroes of the polynomial $f(x) = 4x^{2}-12x+9$ are
Solution
Option (A) is correct.
Here $a=4,\ b=-12,\ c=9$. Discriminant: $D = (-12)^{2}-4(4)(9) = 144-144 = 0$
Since $D = 0$, the polynomial has two equal real zeroes:
$$x = \frac{-(-12)}{2 \times 4} = \frac{12}{8} = \frac{3}{2}$$
Here $a=4,\ b=-12,\ c=9$. Discriminant: $D = (-12)^{2}-4(4)(9) = 144-144 = 0$
Since $D = 0$, the polynomial has two equal real zeroes:
$$x = \frac{-(-12)}{2 \times 4} = \frac{12}{8} = \frac{3}{2}$$
If $p(x)$ is a polynomial of at least degree one and $p(k) = 0$, then $k$ is known as
Solution
Option (B) is correct.
If a polynomial $p(x)$ satisfies $p(k) = 0$, then the value $k$ for which the polynomial becomes zero is called the zero (or root) of the polynomial. Hence, $k$ is known as the zero of $p(x)$.
If a polynomial $p(x)$ satisfies $p(k) = 0$, then the value $k$ for which the polynomial becomes zero is called the zero (or root) of the polynomial. Hence, $k$ is known as the zero of $p(x)$.
If $p(x) = ax+b$, then the zero of $p(x)$ is
Solution
Option (D) is correct.
Set $p(x) = 0$: $ax + b = 0 \implies ax = -b \implies x = -\dfrac{b}{a}$
Set $p(x) = 0$: $ax + b = 0 \implies ax = -b \implies x = -\dfrac{b}{a}$
Case Study and Subjective Questions
The below pictures are a few natural examples of parabolic shape, represented by a quadratic polynomial. A parabolic arch is an arch in the shape of a parabola. In structures, their curve represents an efficient method of load distribution and can be found in bridges and in architecture in a variety of forms.
(i) In the standard form of quadratic polynomial $ax^{2}+bx+c$, $a$, $b$ and $c$ are:
(A) All are real numbers.
(B) All are rational numbers.
(C) ‘$a$’ is a non-zero real number and $b$ and $c$ are any real numbers.
(D) All are integers.
(ii) If the roots of the quadratic polynomial are equal, where discriminant $D = b^{2}-4ac$, then:
(A) $D>0$ (B) $D<0$ (C) $D \neq 0$ (D) $D=0$
(iii) If $\alpha$ and $\dfrac{1}{\alpha}$ are the zeroes of $2x^{2}-x+8k$, then $k$ is:
(A) 4 (B) $\dfrac{1}{4}$ (C) $-\dfrac{1}{4}$ (D) 2
(iv) The graph of $x^{2}+1=0$:
(A) Intersects $X$-axis at two distinct points.
(B) Touches $X$-axis at a point.
(C) Neither touches nor intersects $X$-axis.
(D) Either touches or intersects $X$-axis.
(v) If the sum of the roots is $-p$ and product of the roots is $-\dfrac{1}{p}$, then the quadratic polynomial is:
(A) $k\!\left(-px^{2}+\dfrac{x}{p}+1\right)$ (B) $k\!\left(px^{2}-\dfrac{x}{p}-1\right)$ (C) $k\!\left(x^{2}+px-\dfrac{1}{p}\right)$ (D) $k\!\left(x^{2}+px+\dfrac{1}{p}\right)$
(i) In the standard form of quadratic polynomial $ax^{2}+bx+c$, $a$, $b$ and $c$ are:
(A) All are real numbers.
(B) All are rational numbers.
(C) ‘$a$’ is a non-zero real number and $b$ and $c$ are any real numbers.
(D) All are integers.
(ii) If the roots of the quadratic polynomial are equal, where discriminant $D = b^{2}-4ac$, then:
(A) $D>0$ (B) $D<0$ (C) $D \neq 0$ (D) $D=0$
(iii) If $\alpha$ and $\dfrac{1}{\alpha}$ are the zeroes of $2x^{2}-x+8k$, then $k$ is:
(A) 4 (B) $\dfrac{1}{4}$ (C) $-\dfrac{1}{4}$ (D) 2
(iv) The graph of $x^{2}+1=0$:
(A) Intersects $X$-axis at two distinct points.
(B) Touches $X$-axis at a point.
(C) Neither touches nor intersects $X$-axis.
(D) Either touches or intersects $X$-axis.
(v) If the sum of the roots is $-p$ and product of the roots is $-\dfrac{1}{p}$, then the quadratic polynomial is:
(A) $k\!\left(-px^{2}+\dfrac{x}{p}+1\right)$ (B) $k\!\left(px^{2}-\dfrac{x}{p}-1\right)$ (C) $k\!\left(x^{2}+px-\dfrac{1}{p}\right)$ (D) $k\!\left(x^{2}+px+\dfrac{1}{p}\right)$
Answer
(i) Option (C) is correct.
In the standard form of quadratic polynomial, ‘$a$’ is a non-zero real number and $b$ and $c$ are any real numbers.
(ii) Option (D) is correct.
If the roots of the quadratic polynomial are equal, then the discriminant $D = b^{2}-4ac = 0$.
(iii) Option (B) is correct.
Product of zeroes $= \alpha \cdot \dfrac{1}{\alpha} = 1$ and $\dfrac{c}{a} = \dfrac{8k}{2} = 4k$.
So $4k = 1 \implies k = \dfrac{1}{4}$.
(iv) Option (C) is correct.
From the graph, $x^{2}+1 = 0$ has no real solution (since $x^{2} = -1$). The graph neither touches nor intersects the $X$-axis.
(v) Option (C) is correct.
$P(x) = k\!\left[x^{2}-(\text{sum})x+(\text{product})\right] = k\!\left[x^{2}-(-p)x+\left(-\dfrac{1}{p}\right)\right] = k\!\left(x^{2}+px-\dfrac{1}{p}\right)$
In the standard form of quadratic polynomial, ‘$a$’ is a non-zero real number and $b$ and $c$ are any real numbers.
(ii) Option (D) is correct.
If the roots of the quadratic polynomial are equal, then the discriminant $D = b^{2}-4ac = 0$.
(iii) Option (B) is correct.
Product of zeroes $= \alpha \cdot \dfrac{1}{\alpha} = 1$ and $\dfrac{c}{a} = \dfrac{8k}{2} = 4k$.
So $4k = 1 \implies k = \dfrac{1}{4}$.
(iv) Option (C) is correct.
From the graph, $x^{2}+1 = 0$ has no real solution (since $x^{2} = -1$). The graph neither touches nor intersects the $X$-axis.
(v) Option (C) is correct.
$P(x) = k\!\left[x^{2}-(\text{sum})x+(\text{product})\right] = k\!\left[x^{2}-(-p)x+\left(-\dfrac{1}{p}\right)\right] = k\!\left(x^{2}+px-\dfrac{1}{p}\right)$
Applications of Parabolas: Highway Overpasses/Underpasses
A highway underpass is parabolic in shape.
A parabola is the graph of $p(x) = ax^{2}+bx+c$. Parabolas are symmetric about a vertical line known as the Axis of Symmetry, which runs through the vertex (maximum or minimum point). 
(i) If the highway overpass is represented by $x^{2}-2x-8$, find its zeroes.
(ii) Find the product of zeroes of the polynomial in (i).
OR Find the number of zeroes that polynomial $f(x) = (x-2)^{2}+4$ can have.
(iii) Write the name of the graph which represents the above case.
A highway underpass is parabolic in shape.
A parabola is the graph of $p(x) = ax^{2}+bx+c$. Parabolas are symmetric about a vertical line known as the Axis of Symmetry, which runs through the vertex (maximum or minimum point). (i) If the highway overpass is represented by $x^{2}-2x-8$, find its zeroes.
(ii) Find the product of zeroes of the polynomial in (i).
OR Find the number of zeroes that polynomial $f(x) = (x-2)^{2}+4$ can have.
(iii) Write the name of the graph which represents the above case.
Answer
(i) Factorising $x^{2}-2x-8 = 0$:
$x^{2}-4x+2x-8 = 0 \implies x(x-4)+2(x-4) = 0 \implies (x-4)(x+2) = 0$
$$x = 4 \quad \text{or} \quad x = -2$$
(ii) For $x^{2}-2x-8$, comparing with $ax^{2}+bx+c$: $a=1,\ c=-8$.
$$\text{Product of zeroes} = \frac{c}{a} = \frac{-8}{1} = -8$$
OR: $f(x)=(x-2)^{2}+4 = x^{2}-4x+8$. Discriminant $D = 16-32 = -16 < 0$. No real zeroes exist — the polynomial has 0 zeroes.
(iii) The graph representing a quadratic polynomial is called a Parabola.
$x^{2}-4x+2x-8 = 0 \implies x(x-4)+2(x-4) = 0 \implies (x-4)(x+2) = 0$
$$x = 4 \quad \text{or} \quad x = -2$$
(ii) For $x^{2}-2x-8$, comparing with $ax^{2}+bx+c$: $a=1,\ c=-8$.
$$\text{Product of zeroes} = \frac{c}{a} = \frac{-8}{1} = -8$$
OR: $f(x)=(x-2)^{2}+4 = x^{2}-4x+8$. Discriminant $D = 16-32 = -16 < 0$. No real zeroes exist — the polynomial has 0 zeroes.
(iii) The graph representing a quadratic polynomial is called a Parabola.
Amit designs a flower vase using a graph of polynomial equations. The equation of the curve is given in the graph.
(i) Sara looks at the graphical model and observes: “The zero of the polynomial is at the origin.” Is she correct? If not, what are the coordinates of the zero of the polynomial?
(ii) The curve $m$ is a mirror image of $p(y)$ on the $y$-axis. Which polynomial represents curve $m$?
(iii) Sara changes the coefficient of $y^{3}$ in the polynomials for curves $l$ and $m$. How does this affect the shape of the flower pot?
OR Amit wants to decrease the minimum opening of the flower pot. Which term of the polynomials for curves $l$ and $m$ should he change?
(i) Sara looks at the graphical model and observes: “The zero of the polynomial is at the origin.” Is she correct? If not, what are the coordinates of the zero of the polynomial?
(ii) The curve $m$ is a mirror image of $p(y)$ on the $y$-axis. Which polynomial represents curve $m$?
(iii) Sara changes the coefficient of $y^{3}$ in the polynomials for curves $l$ and $m$. How does this affect the shape of the flower pot?
OR Amit wants to decrease the minimum opening of the flower pot. Which term of the polynomials for curves $l$ and $m$ should he change?
Answer
(i) Sara is incorrect. The zero of the polynomial $p(y)$ is not at the origin. The coordinates of the zero are $(0,\ -2)$.
(ii) Since curve $m$ is a mirror image of curve $l$ on the $y$-axis, all the signs are reversed. The polynomial representing curve $m$ is:
$$p(y) = 0.25y^{3}+0.1y^{2}+0.3y+1$$
(iii) When the coefficient of $y^{3}$ is changed, the curvature of the flower vase changes — the flower pot will have a different shape.
OR: To decrease the minimum opening of the flower pot, Amit should change the constant term (the term without $y$) of the polynomials for curves $l$ and $m$, which shifts the points where the curves intersect the $x$-axis.
(ii) Since curve $m$ is a mirror image of curve $l$ on the $y$-axis, all the signs are reversed. The polynomial representing curve $m$ is:
$$p(y) = 0.25y^{3}+0.1y^{2}+0.3y+1$$
(iii) When the coefficient of $y^{3}$ is changed, the curvature of the flower vase changes — the flower pot will have a different shape.
OR: To decrease the minimum opening of the flower pot, Amit should change the constant term (the term without $y$) of the polynomials for curves $l$ and $m$, which shifts the points where the curves intersect the $x$-axis.
Radha, an aspiring landscape designer, is tasked with creating a visually captivating pool design that incorporates a unique arrangement of fountains. The fountains are arranged so that when water is thrown upwards, it forms the shape of a parabola.
The height of each fountain rod above the water level is $10\ \text{cm}$. The equation of the downward-facing parabola representing the water fountain is: $$p(x) = -x^{2}+5x-4$$ (i) Find the zeroes of the polynomial $p(x)$.
(ii) Find the value of $x$ at which the water attains maximum height.
(iii)(a) If $h$ is the maximum height attained by the water stream from the water level, find the value of $h$.
OR
(iii)(b) At what point(s) on the $x$-axis is the height of water above the $x$-axis $2$ units?
The height of each fountain rod above the water level is $10\ \text{cm}$. The equation of the downward-facing parabola representing the water fountain is: $$p(x) = -x^{2}+5x-4$$ (i) Find the zeroes of the polynomial $p(x)$.
(ii) Find the value of $x$ at which the water attains maximum height.
(iii)(a) If $h$ is the maximum height attained by the water stream from the water level, find the value of $h$.
OR
(iii)(b) At what point(s) on the $x$-axis is the height of water above the $x$-axis $2$ units?
Answer
(i) Setting $p(x) = 0$: $-x^{2}+5x-4=0 \implies x^{2}-5x+4=0 \implies (x-1)(x-4)=0$
$$\text{Zeroes: } x = 1 \quad \text{and} \quad x = 4$$
(ii) The $x$-coordinate of the vertex (maximum) is: $$x = -\frac{b}{2a} = -\frac{5}{2(-1)} = \frac{5}{2}$$
(iii)(a) Substitute $x = \dfrac{5}{2}$ in $p(x)$: $$h = -\left(\frac{5}{2}\right)^{2}+5\left(\frac{5}{2}\right)-4 = -\frac{25}{4}+\frac{50}{4}-\frac{16}{4} = \frac{9}{4}\ \text{units}$$
OR (iii)(b) Set $p(x) = 2$: $-x^{2}+5x-4=2 \implies x^{2}-5x+6=0 \implies (x-2)(x-3)=0$
$$x = 2 \quad \text{and} \quad x = 3$$
$$\text{Zeroes: } x = 1 \quad \text{and} \quad x = 4$$
(ii) The $x$-coordinate of the vertex (maximum) is: $$x = -\frac{b}{2a} = -\frac{5}{2(-1)} = \frac{5}{2}$$
(iii)(a) Substitute $x = \dfrac{5}{2}$ in $p(x)$: $$h = -\left(\frac{5}{2}\right)^{2}+5\left(\frac{5}{2}\right)-4 = -\frac{25}{4}+\frac{50}{4}-\frac{16}{4} = \frac{9}{4}\ \text{units}$$
OR (iii)(b) Set $p(x) = 2$: $-x^{2}+5x-4=2 \implies x^{2}-5x+6=0 \implies (x-2)(x-3)=0$
$$x = 2 \quad \text{and} \quad x = 3$$
Mark the correct choice as:
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Assertion (A): The values of $k$ for which the quadratic polynomial $kx^{2}+x+k$ has equal zeroes are $\pm\dfrac{1}{2}$.
Reason (R): If all the three zeroes of a cubic polynomial $x^{3}+ax^{2}-bx+c$ are positive, then at least one of $a$, $b$ and $c$ is non-negative.
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Assertion (A): The values of $k$ for which the quadratic polynomial $kx^{2}+x+k$ has equal zeroes are $\pm\dfrac{1}{2}$.
Reason (R): If all the three zeroes of a cubic polynomial $x^{3}+ax^{2}-bx+c$ are positive, then at least one of $a$, $b$ and $c$ is non-negative.
Answer
Option (C) is correct — A is true but R is false.
Assertion: For $f(x) = kx^{2}+x+k$, equal roots require discriminant $= 0$:
$(1)^{2}-4(k)(k) = 0 \implies 1-4k^{2}=0 \implies k^{2}=\dfrac{1}{4} \implies k = \pm\dfrac{1}{2}$ ✓ Assertion is true.
Reason: All zeroes of a cubic polynomial are positive only when all constants $a$, $b$, and $c$ are negative (not non-negative). The Reason is false.
Assertion: For $f(x) = kx^{2}+x+k$, equal roots require discriminant $= 0$:
$(1)^{2}-4(k)(k) = 0 \implies 1-4k^{2}=0 \implies k^{2}=\dfrac{1}{4} \implies k = \pm\dfrac{1}{2}$ ✓ Assertion is true.
Reason: All zeroes of a cubic polynomial are positive only when all constants $a$, $b$, and $c$ are negative (not non-negative). The Reason is false.
Mark the correct choice as:
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Assertion (A): The graph of $y=p(x)$ shown below has 5 zeroes.
Reason (R): If the graph of a polynomial intersects the $x$-axis at exactly two points, it need not be a quadratic polynomial.
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Assertion (A): The graph of $y=p(x)$ shown below has 5 zeroes.
Reason (R): If the graph of a polynomial intersects the $x$-axis at exactly two points, it need not be a quadratic polynomial.
Answer
Option (B) is correct — Both A and R are true but R is NOT the correct explanation of A.
Assertion: The graph intersects the $x$-axis at 5 points, so the number of zeroes is 5. Assertion is true.
Reason: A polynomial of degree more than 2 can have two real zeroes with remaining zeroes imaginary, so its graph may intersect the $x$-axis at only two points. Reason is true.
However, the Reason does not explain why the Assertion holds — both are independently true.
Assertion: The graph intersects the $x$-axis at 5 points, so the number of zeroes is 5. Assertion is true.
Reason: A polynomial of degree more than 2 can have two real zeroes with remaining zeroes imaginary, so its graph may intersect the $x$-axis at only two points. Reason is true.
However, the Reason does not explain why the Assertion holds — both are independently true.
Mark the correct choice as:
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Assertion (A): If one of the zeroes of the quadratic polynomial $(k-1)x^{2}+kx+1$ is $-3$, then $k = \dfrac{4}{3}$.
Reason (R): If $-1$ is a zero of the polynomial $p(x) = kx^{2}-4x+k$, then the value of $k$ is $-2$.
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Assertion (A): If one of the zeroes of the quadratic polynomial $(k-1)x^{2}+kx+1$ is $-3$, then $k = \dfrac{4}{3}$.
Reason (R): If $-1$ is a zero of the polynomial $p(x) = kx^{2}-4x+k$, then the value of $k$ is $-2$.
Answer
Option (B) is correct — Both A and R are true but R is NOT the correct explanation of A.
Assertion: $p(-3)=0 \implies (k-1)(9)+k(-3)+1=0 \implies 9k-9-3k+1=0 \implies 6k=8 \implies k=\dfrac{4}{3}$. Assertion is true.
Reason: $p(-1)=0 \implies k(1)-4(-1)+k=0 \implies 2k+4=0 \implies k=-2$. Reason is true.
Both are true, but Reason is an independent result and does not explain Assertion.
Assertion: $p(-3)=0 \implies (k-1)(9)+k(-3)+1=0 \implies 9k-9-3k+1=0 \implies 6k=8 \implies k=\dfrac{4}{3}$. Assertion is true.
Reason: $p(-1)=0 \implies k(1)-4(-1)+k=0 \implies 2k+4=0 \implies k=-2$. Reason is true.
Both are true, but Reason is an independent result and does not explain Assertion.
Mark the correct choice as:
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Assertion (A): $x^{2}+7x+12$ has no real zeroes.
Reason (R): A quadratic polynomial can have at the most two zeroes.
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Assertion (A): $x^{2}+7x+12$ has no real zeroes.
Reason (R): A quadratic polynomial can have at the most two zeroes.
Answer
Option (D) is correct — A is false but R is true.
Assertion: $x^{2}+7x+12 = (x+3)(x+4)$. The zeroes are $x=-3$ and $x=-4$, which are real. So Assertion is false.
Reason: A quadratic polynomial has degree 2, so it can have at most two zeroes. Reason is true.
Assertion: $x^{2}+7x+12 = (x+3)(x+4)$. The zeroes are $x=-3$ and $x=-4$, which are real. So Assertion is false.
Reason: A quadratic polynomial has degree 2, so it can have at most two zeroes. Reason is true.
Mark the correct choice as:
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Assertion (A): If the sum of the zeroes of the quadratic polynomial $x^{2}-2kx+8$ is $2$, then the value of $k$ is $1$.
Reason (R): The sum of zeroes of a quadratic polynomial $ax^{2}+bx+c$ is $-\dfrac{b}{a}$.
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Assertion (A): If the sum of the zeroes of the quadratic polynomial $x^{2}-2kx+8$ is $2$, then the value of $k$ is $1$.
Reason (R): The sum of zeroes of a quadratic polynomial $ax^{2}+bx+c$ is $-\dfrac{b}{a}$.
Answer
Option (A) is correct — Both A and R are true and R is the correct explanation of A.
Assertion: For $x^{2}-2kx+8$, sum of zeroes $= -\dfrac{-2k}{1} = 2k = 2 \implies k = 1$. Assertion is true.
Reason: The sum of zeroes formula $-\dfrac{b}{a}$ is the standard algebraic result for any quadratic polynomial. Reason is true and correctly explains the Assertion.
Assertion: For $x^{2}-2kx+8$, sum of zeroes $= -\dfrac{-2k}{1} = 2k = 2 \implies k = 1$. Assertion is true.
Reason: The sum of zeroes formula $-\dfrac{b}{a}$ is the standard algebraic result for any quadratic polynomial. Reason is true and correctly explains the Assertion.
A ball is thrown in the air so that $t$ seconds after it is thrown, its height $h$ metre above its starting point is given by the polynomial:
$$h = 25t – 5t^{2}$$ 
(i) Write the zeroes of the given polynomial.
(ii) Find the maximum height achieved by the ball.
(iii)(a) After throwing upward, how much time did the ball take to reach the height of $30$ m?
OR
(iii)(b) Find the two different values of $t$ when the height of the ball was $20$ m.
(i) Write the zeroes of the given polynomial.
(ii) Find the maximum height achieved by the ball.
(iii)(a) After throwing upward, how much time did the ball take to reach the height of $30$ m?
OR
(iii)(b) Find the two different values of $t$ when the height of the ball was $20$ m.
Answer
(i) Setting $h=0$: $25t-5t^{2}=0 \implies 5t(5-t)=0$
$$\text{Zeroes: } t = 0 \quad \text{and} \quad t = 5$$
(ii) Maximum occurs at $t = -\dfrac{b}{2a} = -\dfrac{25}{2(-5)} = \dfrac{5}{2}$:
$$h = 25\left(\frac{5}{2}\right)-5\left(\frac{5}{2}\right)^{2} = \frac{125}{2}-\frac{125}{4} = \frac{125}{4}\ \text{m}$$
(iii)(a) Setting $h=30$: $25t-5t^{2}=30 \implies t^{2}-5t+6=0 \implies (t-2)(t-3)=0$
$t=2\ \text{s}$ or $t=3\ \text{s}$. The ball first reaches $30$ m at $\mathbf{t = 2\ \text{seconds}}$.
OR (iii)(b) Setting $h=20$: $25t-5t^{2}=20 \implies t^{2}-5t+4=0 \implies (t-1)(t-4)=0$
$$t = 1\ \text{s} \quad \text{and} \quad t = 4\ \text{s}$$
$$\text{Zeroes: } t = 0 \quad \text{and} \quad t = 5$$
(ii) Maximum occurs at $t = -\dfrac{b}{2a} = -\dfrac{25}{2(-5)} = \dfrac{5}{2}$:
$$h = 25\left(\frac{5}{2}\right)-5\left(\frac{5}{2}\right)^{2} = \frac{125}{2}-\frac{125}{4} = \frac{125}{4}\ \text{m}$$
(iii)(a) Setting $h=30$: $25t-5t^{2}=30 \implies t^{2}-5t+6=0 \implies (t-2)(t-3)=0$
$t=2\ \text{s}$ or $t=3\ \text{s}$. The ball first reaches $30$ m at $\mathbf{t = 2\ \text{seconds}}$.
OR (iii)(b) Setting $h=20$: $25t-5t^{2}=20 \implies t^{2}-5t+4=0 \implies (t-1)(t-4)=0$
$$t = 1\ \text{s} \quad \text{and} \quad t = 4\ \text{s}$$
In a pool at an aquarium, a dolphin jumps out of the water travelling at $20\ \text{cm/s}$. Its height above the water level after $t$ seconds is given by:
$$h = 20t – 16t^{2}$$ 
(i) Find the zeroes of the polynomial $p(t)$.
(ii) Which type of graph represents $p(t)$?
(iii) What would be the value of $h$ at $t = \dfrac{3}{2}$? Interpret the result.
(i) Find the zeroes of the polynomial $p(t)$.
(ii) Which type of graph represents $p(t)$?
(iii) What would be the value of $h$ at $t = \dfrac{3}{2}$? Interpret the result.
Answer
(i) Setting $p(t)=0$: $20t-16t^{2}=0 \implies 4t(5-4t)=0$
$$\text{Zeroes: } t = 0 \quad \text{and} \quad t = \frac{5}{4}$$
(ii) $p(t) = -16t^{2}+20t$ is a quadratic polynomial with leading coefficient $a=-16 < 0$. The graph is a downward-opening parabola. Option (a) is correct.
(iii) Substitute $t = \dfrac{3}{2}$: $$h = 20\left(\frac{3}{2}\right)-16\left(\frac{3}{2}\right)^{2} = 30-16\left(\frac{9}{4}\right) = 30-36 = -6$$ The value $h = -6$ is negative. This means at $t = \dfrac{3}{2}$ seconds, the dolphin has already re-entered the water and is $6$ units below the water surface.
$$\text{Zeroes: } t = 0 \quad \text{and} \quad t = \frac{5}{4}$$
(ii) $p(t) = -16t^{2}+20t$ is a quadratic polynomial with leading coefficient $a=-16 < 0$. The graph is a downward-opening parabola. Option (a) is correct.
(iii) Substitute $t = \dfrac{3}{2}$: $$h = 20\left(\frac{3}{2}\right)-16\left(\frac{3}{2}\right)^{2} = 30-16\left(\frac{9}{4}\right) = 30-36 = -6$$ The value $h = -6$ is negative. This means at $t = \dfrac{3}{2}$ seconds, the dolphin has already re-entered the water and is $6$ units below the water surface.
A rainbow is an arch of colours visible in the sky after rain. Each colour of the rainbow makes a parabola. Any quadratic polynomial $p(x) = ax^{2}+bx+c\ (a \neq 0)$ represents a parabola on the graph paper.
(i) The graph of a rainbow $y = f(x)$ is shown in the figure. Write the number of zeroes of the curve.
(ii) If the graph of a rainbow does not intersect the $x$-axis but intersects the $y$-axis at one point, then how many zeroes will it have?
(iii) If a rainbow is represented by the quadratic polynomial $p(x) = x^{2}+(a+1)x+b$ whose zeroes are $2$ and $-3$, find the values of $a$ and $b$.
(i) The graph of a rainbow $y = f(x)$ is shown in the figure. Write the number of zeroes of the curve.
(ii) If the graph of a rainbow does not intersect the $x$-axis but intersects the $y$-axis at one point, then how many zeroes will it have?
(iii) If a rainbow is represented by the quadratic polynomial $p(x) = x^{2}+(a+1)x+b$ whose zeroes are $2$ and $-3$, find the values of $a$ and $b$.
Answer
(i) From the given figure, the rainbow-shaped curve intersects the $x$-axis at two points. Number of zeroes $= \mathbf{2}$.
(ii) Zeroes are the $x$-coordinates where the graph cuts the $x$-axis. If the graph does not intersect the $x$-axis at all, then $f(x) \neq 0$ for all real $x$. Intersection with the $y$-axis does not give a zero. Therefore, the polynomial has $\mathbf{0}$ zeroes.
(iii) Given zeroes: $\alpha = 2,\ \beta = -3$.
Sum of zeroes $= 2+(-3) = -1 = -(a+1) \implies a+1=1 \implies \mathbf{a=0}$
Product of zeroes $= 2\times(-3) = -6 = b \implies \mathbf{b=-6}$
(ii) Zeroes are the $x$-coordinates where the graph cuts the $x$-axis. If the graph does not intersect the $x$-axis at all, then $f(x) \neq 0$ for all real $x$. Intersection with the $y$-axis does not give a zero. Therefore, the polynomial has $\mathbf{0}$ zeroes.
(iii) Given zeroes: $\alpha = 2,\ \beta = -3$.
Sum of zeroes $= 2+(-3) = -1 = -(a+1) \implies a+1=1 \implies \mathbf{a=0}$
Product of zeroes $= 2\times(-3) = -6 = b \implies \mathbf{b=-6}$
Shibi decorated the door of his house with garlands on the occasion of Onam. Each garland forms the shape of a parabola.
Based on the above information, answer the following questions:
(i) Suppose the quadratic polynomial for the given curve is $ax^{2}+bx+c$. What will be the sign of $a$?
(ii) Find a quadratic polynomial with the sum and product of its zeroes as $-1$ and $-2$ respectively.
(iii) If $\alpha$ and $\beta$ are the zeroes of the polynomial $f(x) = x^{2}-7x+12$, find the value of $\dfrac{1}{\alpha}+\dfrac{1}{\beta}$.
Based on the above information, answer the following questions:
(i) Suppose the quadratic polynomial for the given curve is $ax^{2}+bx+c$. What will be the sign of $a$?
(ii) Find a quadratic polynomial with the sum and product of its zeroes as $-1$ and $-2$ respectively.
(iii) If $\alpha$ and $\beta$ are the zeroes of the polynomial $f(x) = x^{2}-7x+12$, find the value of $\dfrac{1}{\alpha}+\dfrac{1}{\beta}$.
Answer
(i) A garland hanging from a door forms a downward-opening parabola. For a quadratic polynomial, $a < 0$ when the parabola opens downward. Therefore $\mathbf{a < 0}$.
(ii) Required polynomial $= x^{2}-Sx+P = x^{2}-(-1)x+(-2)$
$$= x^{2}+x-2$$
(iii) For $f(x) = x^{2}-7x+12$: $\alpha+\beta = 7$ and $\alpha\beta = 12$.
$$\frac{1}{\alpha}+\frac{1}{\beta} = \frac{\alpha+\beta}{\alpha\beta} = \frac{7}{12}$$
(ii) Required polynomial $= x^{2}-Sx+P = x^{2}-(-1)x+(-2)$
$$= x^{2}+x-2$$
(iii) For $f(x) = x^{2}-7x+12$: $\alpha+\beta = 7$ and $\alpha\beta = 12$.
$$\frac{1}{\alpha}+\frac{1}{\beta} = \frac{\alpha+\beta}{\alpha\beta} = \frac{7}{12}$$
Mark the correct choice as:
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Assertion (A): $P(x) = 4x^{3}-x^{2}+5x^{4}+3x-2$ is a polynomial of degree $3$.
Reason (R): The highest power of $x$ in the polynomial $P(x)$ is the degree of the polynomial.
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Assertion (A): $P(x) = 4x^{3}-x^{2}+5x^{4}+3x-2$ is a polynomial of degree $3$.
Reason (R): The highest power of $x$ in the polynomial $P(x)$ is the degree of the polynomial.
Answer
Option (D) is correct — A is false but R is true.
Assertion: Rewriting in descending order: $P(x) = 5x^{4}+4x^{3}-x^{2}+3x-2$. The highest power is $4$, so the degree is $4$, not $3$. Assertion is false.
Reason: The degree of a polynomial is defined as the highest power of the variable with a non-zero coefficient. Reason is true.
Assertion: Rewriting in descending order: $P(x) = 5x^{4}+4x^{3}-x^{2}+3x-2$. The highest power is $4$, so the degree is $4$, not $3$. Assertion is false.
Reason: The degree of a polynomial is defined as the highest power of the variable with a non-zero coefficient. Reason is true.
Mark the correct choice as:
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Assertion (A): $x^{3}+x$ has only one real zero.
Reason (R): A polynomial of $n^{\text{th}}$ degree must have $n$ real zeroes.
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Assertion (A): $x^{3}+x$ has only one real zero.
Reason (R): A polynomial of $n^{\text{th}}$ degree must have $n$ real zeroes.
Answer
Option (C) is correct — A is true but R is false.
Assertion: $x^{3}+x = x(x^{2}+1)=0 \implies x=0$ or $x^{2}=-1$ (imaginary). Only $x=0$ is real. Assertion is true.
Reason: A polynomial of degree $n$ has at most $n$ zeroes — they need not all be real. The claim that it must have $n$ real zeroes is false.
Assertion: $x^{3}+x = x(x^{2}+1)=0 \implies x=0$ or $x^{2}=-1$ (imaginary). Only $x=0$ is real. Assertion is true.
Reason: A polynomial of degree $n$ has at most $n$ zeroes — they need not all be real. The claim that it must have $n$ real zeroes is false.
Mark the correct choice as:
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Assertion (A): If one zero of the polynomial $p(x) = (k^{2}+4)x^{2}+13x+4k$ is the reciprocal of the other, then $k = 2$.
Reason (R): If $(x-a)$ is a factor of $p(x)$, then $p(a) = 0$, i.e., $a$ is a zero of $p(x)$.
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Assertion (A): If one zero of the polynomial $p(x) = (k^{2}+4)x^{2}+13x+4k$ is the reciprocal of the other, then $k = 2$.
Reason (R): If $(x-a)$ is a factor of $p(x)$, then $p(a) = 0$, i.e., $a$ is a zero of $p(x)$.
Answer
Option (B) is correct — Both A and R are true but R is NOT the correct explanation of A.
Assertion: If one zero is the reciprocal of the other, product of zeroes $= 1$:
$$\frac{c}{a} = \frac{4k}{k^{2}+4} = 1 \implies 4k = k^{2}+4 \implies (k-2)^{2}=0 \implies k=2$$ Assertion is true.
Reason: The Factor Theorem is a correct and true statement. However, it was not applied to find $k$ in this problem. So Reason is true but does not explain the Assertion.
Assertion: If one zero is the reciprocal of the other, product of zeroes $= 1$:
$$\frac{c}{a} = \frac{4k}{k^{2}+4} = 1 \implies 4k = k^{2}+4 \implies (k-2)^{2}=0 \implies k=2$$ Assertion is true.
Reason: The Factor Theorem is a correct and true statement. However, it was not applied to find $k$ in this problem. So Reason is true but does not explain the Assertion.
Mark the correct choice as:
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Assertion (A): $x^{2}+4x+5$ has two zeroes.
Reason (R): A quadratic polynomial can have at the most two zeroes.
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Assertion (A): $x^{2}+4x+5$ has two zeroes.
Reason (R): A quadratic polynomial can have at the most two zeroes.
Answer
Option (D) is correct — A is false but R is true.
Assertion: Discriminant $D = 4^{2}-4(1)(5) = 16-20 = -4 < 0$. Since $D < 0$, the polynomial has no real zeroes. Assertion is false.
Reason: A quadratic polynomial (degree 2) can have at most two zeroes. Reason is true.
Assertion: Discriminant $D = 4^{2}-4(1)(5) = 16-20 = -4 < 0$. Since $D < 0$, the polynomial has no real zeroes. Assertion is false.
Reason: A quadratic polynomial (degree 2) can have at most two zeroes. Reason is true.
Mark the correct choice as:
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Assertion (A): The graph $y = f(x)$ shown in the figure has 3 zeroes.
Reason (R): The number of zeroes of a polynomial $f(x)$ is equal to the number of points at which the graph of $y = f(x)$ cuts or touches the $x$-axis.
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Assertion (A): The graph $y = f(x)$ shown in the figure has 3 zeroes.
Reason (R): The number of zeroes of a polynomial $f(x)$ is equal to the number of points at which the graph of $y = f(x)$ cuts or touches the $x$-axis.
Answer
Option (A) is correct — Both A and R are true and R is the correct explanation of A.
Assertion: From the given graph, the curve intersects the $x$-axis at three distinct points, so the polynomial has 3 zeroes. Assertion is true.
Reason: By definition, a zero of a polynomial is a value of $x$ for which $f(x)=0$, which graphically occurs exactly where $y=f(x)$ cuts or touches the $x$-axis. Reason is true and correctly explains the Assertion.
Assertion: From the given graph, the curve intersects the $x$-axis at three distinct points, so the polynomial has 3 zeroes. Assertion is true.
Reason: By definition, a zero of a polynomial is a value of $x$ for which $f(x)=0$, which graphically occurs exactly where $y=f(x)$ cuts or touches the $x$-axis. Reason is true and correctly explains the Assertion.
Frequently Asked Questions
What is the Polynomials chapter about in CBSE Class 10 Maths? ⌄
The Polynomials chapter in CBSE Class 10 Maths introduces your child to the definition, types, and properties of polynomials. Key concepts include understanding the zeroes of a polynomial and their geometric meaning from graphs, the relationship between zeroes and coefficients of quadratic and cubic polynomials, and forming polynomials when their zeroes are given. These are foundational skills that are tested across MCQs, short-answer, and case-study questions in board exams.
How many marks does Polynomials carry in the CBSE Class 10 board exam? ⌄
Polynomials typically carries 6–8 marks in the CBSE Class 10 Maths board exam. Questions appear as 1-mark MCQs, 2–3 mark short-answer questions, and as part of 4–5 mark case study questions. Practising all question formats ensures your child is fully prepared for any combination that may appear on the paper.
What are the most important topics in Class 10 Polynomials? ⌄
The most important topics are: (1) Zeroes of a polynomial and reading them from graphs — including the difference between cutting and touching the $x$-axis, (2) Relationship between zeroes and coefficients — sum $= -b/a$ and product $= c/a$ for a quadratic polynomial, (3) Forming a quadratic polynomial when sum and product of zeroes are given, and (4) Assertion-Reason questions based on properties of zeroes, discriminant, and degree of polynomials.
What are common mistakes students make in Class 10 Polynomials? ⌄
The most frequent mistakes are: confusing the sum of zeroes formula ($-b/a$) with the product formula ($c/a$); incorrectly counting zeroes from a graph by missing points where the curve only touches the $x$-axis; forgetting to include the arbitrary constant $k$ when writing the general form of a polynomial for given zeroes; and incorrectly claiming a polynomial “must” have $n$ real zeroes for degree $n$ (it can have at most $n$ real zeroes).
How does Angle Belearn help students master CBSE Class 10 Polynomials? ⌄
Angle Belearn provides a comprehensive bank of competency based questions that mirror the latest CBSE exam pattern — covering MCQs, case studies, and assertion-reason questions with full step-by-step solutions. Our CBSE specialists verify every explanation so your child understands not just the answer but the reasoning behind it. With consistent practice on Angle Belearn, students build the confidence and speed needed to score full marks in Polynomials on exam day.
