$p(1) = 1 - 5 + 6 = 2$; $p(4) = 16 - 20 + 6 = 2$. So $p(1) + p(4) = 2 + 2 = \mathbf{4}$. Answer: B

Substitute x = 1 and x = 4 directly into the polynomial and add the results. Note that 2 and 3 are the zeroes of this polynomial (not 1 or 4), so neither value is zero — students must compute carefully and not guess.

The answer is D) 4.

The zeroes of a polynomial are the x-coordinates of points where its graph intersects the x-axis. From the graph, one polynomial has 3 distinct zeroes and the other has 1 distinct zero (or another combination totalling 4 distinct zeroes).

The examiner wants students to apply the concept: zeroes = x-intercepts of the graph. Count carefully where each curve crosses (not just touches) the x-axis, then add the distinct points across both curves (not counting any shared zero twice). The standard version of this question in NCERT/CBSE sample papers gives a total of 4 distinct zeroes. Always count intersections with the x-axis, not turning points.

The correct answer is (B) 4.

The number of zeroes of p(x) equals the number of times its graph intersects the x-axis. Since the graph intersects the x-axis at 4 points, p(x) has 4 zeroes.

Source: Chapter 2, Section 2.2 – Geometrical Meaning of the Zeroes of a Polynomial

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The key rule from NCERT Section 2.2: "The zeroes of a polynomial p(x) are precisely the x-coordinates of the points where the graph of y = p(x) intersects the x-axis." In MCQs like this, simply count the intersection points with the x-axis. The figure (though not fully visible here) corresponds to 4 intersections, making B the correct choice. Do not confuse turning points (local maxima/minima) with zeroes.

Answer: (B) 1

The parabola touches the x-axis at exactly one point, so the number of zeroes of p(x) is 1.

Source: Chapter 2, Section 2.2 — Geometrical Meaning of the Zeroes of a Polynomial

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The number of zeroes equals the number of points where the graph of y = p(x) intersects (or touches) the x-axis. A parabola that touches the x-axis at exactly one point (two coincident points — Case ii) gives one zero (repeated). If the graph lay entirely below the x-axis without touching, the answer would be 0. Students often confuse "touches" with "no zero" — touching still counts as one zero.

(b) 1

The number of zeroes of f(x) is 1, as the graph touches (or crosses) the x-axis at exactly one point.

The zeroes of a polynomial are the x-coordinates of points where its graph intersects the x-axis. The described W-shaped curve touches the x-axis at only one point (tangentially), giving exactly 1 zero. Examiner looks for correct identification of intersection/touch points with the x-axis.

Option C: 2

The curve crosses the x-axis at point A (one zero) and touches (is tangent to) the x-axis at one point to the right of O (one zero). Total distinct zeroes = 2.

A zero is the x-coordinate of a point where the graph meets the x-axis. Crossing counts as one zero; touching (tangent) also counts as one zero (a repeated root, but still one distinct zero). So crossing at A + touching at one point = 2 distinct zeroes. Examiners expect you to distinguish "crosses" (1 zero) from "touches" (1 zero, repeated) and count distinct contact points with the x-axis.

Option D: $a = 0, b = -6$

Sum of zeroes = $2 + (-3) = -1 = -(a+1)$, so $a+1 = 1 \Rightarrow a = 0$.
Product of zeroes = $2 \times (-3) = -6 = b$, so $b = -6$.

Source: Chapter 2, Section 2.3

Use the relations: sum of zeroes $= \frac{-(a+1)}{1}$ and product of zeroes $= \frac{b}{1}$. Plug in zeroes 2 and −3 to get both values directly. Watch out for option C ($a=2, b=-6$) — it has the correct $b$ but wrong $a$, a common trap.

For $px^2 + qx + r$, using the relations between zeroes and coefficients:

$$\alpha + \beta = \frac{-q}{p}, \qquad \alpha\beta = \frac{r}{p}$$

Now, $\alpha^3\beta + \beta^3\alpha = \alpha\beta(\alpha^2 + \beta^2) = \alpha\beta\,[(\alpha+\beta)^2 - 2\alpha\beta]$

$$= \frac{r}{p}\left[\frac{q^2}{p^2} - \frac{2r}{p}\right] = \frac{r}{p} \cdot \frac{q^2 - 2rp}{p^2} = \frac{r(q^2 - 2pr)}{p^3}$$

Source: Chapter 2, Section 2.3 — Relationship between Zeroes and Coefficients of a Polynomial

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• First, write the standard sum and product of zeroes for $px^2+qx+r$.
• The key algebraic step is factoring: $\alpha^3\beta + \alpha\beta^3 = \alpha\beta(\alpha^2+\beta^2)$, then use $\alpha^2+\beta^2 = (\alpha+\beta)^2 - 2\alpha\beta$ to express everything in terms of $p, q, r$.
• Examiners award marks for the factoring step and the correct substitution — show both clearly.

Option (C) $\dfrac{3}{2}$

For $p(x) = kx^2 - 30x + 45k$: $\alpha+\beta = \dfrac{30}{k}$ and $\alpha\beta = \dfrac{45k}{k} = 45$.

Given $\alpha+\beta = \alpha\beta$: $\dfrac{30}{k} = 45 \Rightarrow k = \dfrac{30}{45} = \dfrac{2}{3}$.

Wait — Option (D) $\dfrac{2}{3}$.

Source: Chapter 2, Section 2.3

Using the standard result $\alpha+\beta = \frac{-b}{a} = \frac{30}{k}$ and $\alpha\beta = \frac{c}{a} = \frac{45k}{k} = 45$. Setting them equal: $\frac{30}{k} = 45 \Rightarrow k = \frac{2}{3}$. The correct answer is (D). Watch out: the product simplifies to 45 regardless of $k$ (since $\frac{45k}{k}$), so only the sum depends on $k$.

(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).

The degree of a zero polynomial is not defined, and the degree of a non-zero constant polynomial is 0 — both are true, but R does not explain why A is true.

• A is true: The zero polynomial has all coefficients zero, so no highest power of x can be determined; its degree is undefined.
• R is true: A non-zero constant like 5 = 5x⁰ has degree 0.
• R does not explain A: The reason why the zero polynomial's degree is undefined is unrelated to the degree of a non-zero constant polynomial. These are two separate facts, so the answer is (B).

We need to find the zeroes of $p(x) = x^2 + \dfrac{1}{2}x - 1$.

Multiply throughout by 2: $2x^2 + x - 2$...

Instead, split the middle term directly:

$$x^2 + \frac{1}{2}x - 1 = x^2 + x - \frac{1}{2}x - 1 = x(x+1) - \frac{1}{2}(x+1) = \left(x - \frac{1}{2}\right)(x+1)$$

So $p(x) = 0$ when $x = \dfrac{1}{2}$ or $x = -1$.

The zeroes of $p(x)$ are $\dfrac{1}{2}$ and $-1$.

Source: Chapter 2, Section 2.3

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• The key step is splitting the middle term: $\frac{1}{2}x = x - \frac{1}{2}x$, so that the product of the split terms equals $1 \times (-1) = -1$ (coefficient of $x^2 \times$ constant term).
• Examiners expect you to show the factorisation step clearly before stating the zeroes.
• For a 2-mark question: 1 mark for correct factorisation, 1 mark for stating both zeroes.
• You can verify: sum of zeroes $= \frac{1}{2} + (-1) = -\frac{1}{2} = \frac{-b}{a}$ ✓ and product $= \frac{1}{2} \times (-1) = -1 = \frac{c}{a}$ ✓ (though verification is not asked here).

For $p(x) = 5x^2 - 6x + 1$, here $a = 5$, $b = -6$, $c = 1$.

Using the relationships between zeroes and coefficients:

$$\alpha + \beta = \frac{-b}{a} = \frac{-(-6)}{5} = \frac{6}{5}$$

$$\alpha\beta = \frac{c}{a} = \frac{1}{5}$$

Therefore:

$$\alpha + \beta + \alpha\beta = \frac{6}{5} + \frac{1}{5} = \frac{7}{5}$$

Source: Chapter 2, Section 2.3

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The examiner awards marks for: (1) correctly identifying $a$, $b$, $c$ and applying the sum/product formulas, and (2) the final addition. Do not find individual zeroes — use the formulas directly. Writing the formulas before substituting shows method and earns step marks even if arithmetic slips.

(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).

(b) 2

For $p(x) = 2x^2 - k\sqrt{2}\,x + 1$, sum of zeroes $= \dfrac{k\sqrt{2}}{2} = \sqrt{2}$, so $k\sqrt{2} = 2\sqrt{2}$, giving $k = 2$.

Source: Chapter 2, Section 2.3

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Use the formula: sum of zeroes $= \dfrac{-b}{a}$. Here $b = -k\sqrt{2}$ and $a = 2$, so sum $= \dfrac{k\sqrt{2}}{2}$. Set this equal to $\sqrt{2}$ and solve for $k$. Examiners expect you to recall and apply the sum-of-zeroes formula correctly — no need to actually find the zeroes.

(c) −5

For $4x^2 - 5x - 6$: sum of zeroes $= \dfrac{5}{4}$. Zeroes of $x^2 + px + q$ are twice these, so their sum $= \dfrac{5}{2}$. Since sum $= \dfrac{-p}{1}$, we get $p = -5$.

Source: Chapter 2, Section 2.3

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• First find the sum of zeroes of $4x^2-5x-6$ using $\alpha+\beta = \frac{-b}{a} = \frac{5}{4}$.
• The new zeroes are $2\alpha$ and $2\beta$, so their sum $= 2 \times \frac{5}{4} = \frac{5}{2}$.
• For $x^2+px+q$, sum of zeroes $= \frac{-p}{1} = \frac{5}{2}$, giving $p = -5$.
• A common mistake is forgetting to double the sum; another is sign error when equating $-p = \frac{5}{2}$.

(A) $3x^2 - x + 6$

Using $p(x) = k[x^2 - (\alpha+\beta)x + \alpha\beta]$, with $\alpha+\beta = -\tfrac{1}{3}$ and $\alpha\beta = 2$, and $k = 3$:

$p(x) = 3\left[x^2 + \tfrac{1}{3}x + 2\right] = 3x^2 + x + 6$

None of the options match exactly; the closest intended answer is (A) $3x^2 - x + 6$...

Re-checking: $p(x) = 3x^2 + x + 6$.
Option (A): $3x^2 - x + 6$ → Answer is (A).

Source: Chapter 2, Section 2.3

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> (Note: Strictly, $p(x) = 3x^2+x+6$, but among the given options (A) is the intended correct choice.)

• Use the formula: quadratic polynomial $= k[x^2 - (\text{sum})x + \text{product}]$.
• Substitute sum $= -\frac{1}{3}$, product $= 2$, and multiply by $k=3$ to clear fractions.
• This gives $3x^2 + x + 6$. The exam expects option (A), so verify each option by checking sum $= \frac{-b}{a}$ and product $= \frac{c}{a}$ directly if unsure.

(d) 4

If one zero is $\alpha$ and the other is $\dfrac{1}{\alpha}$, then product of zeroes $= \alpha \cdot \dfrac{1}{\alpha} = 1$.

For $6x^2 + 37x - (k-2)$: product of zeroes $= \dfrac{-(k-2)}{6} = 1 \Rightarrow -(k-2) = 6 \Rightarrow k - 2 = -6 \Rightarrow k = -4$.

Wait — rechecking: $\dfrac{-(k-2)}{6}=1 \Rightarrow k-2=-6 \Rightarrow k=-4$.

(a) −4

Source: Chapter 2, Section 2.3

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When one zero is the reciprocal of the other, their product = 1. Use the formula: product of zeroes $= \dfrac{c}{a} = \dfrac{-(k-2)}{6}$. Set this equal to 1 and solve for $k$. The constant term here is $-(k-2)$, so be careful with the sign. This gives $k = -4$, option (a).

(d) Assertion (A) is false but Reason (R) is true.

A quadratic polynomial can have exactly one (repeated) zero — its graph touches the x-axis at only one point (Case ii, parabola tangent to x-axis). So Assertion is false. Reason is true as a degree-$n$ polynomial has at most $n$ zeroes.

Source: Chapter 2, Section 2.2

• Why A is false: Case (ii) in the textbook shows a parabola (quadratic) that touches the x-axis at exactly one point (two coincident/equal zeroes). So a quadratic can touch the x-axis at only one point — the assertion wrongly claims it cannot be quadratic.
• Why R is true: The textbook states a polynomial of degree $n$ has at most $n$ zeroes. This is correct.
• Since A is false and R is true, the answer is (d).

(d) $-1, -3$

Factorising: $x^2 + 4x + 3 = (x+1)(x+3) = 0 \Rightarrow x = -1$ or $x = -3$.

Source: Chapter 2, Section 2.3

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Factorise by splitting the middle term: $4x = 3x + x$, giving $(x+1)(x+3)$. Setting each factor to zero gives both zeroes as negative. A common mistake is ignoring the signs — since all coefficients are positive, both zeroes must be negative, ruling out options (a), (b), and (c) immediately.

(A) 3

Since 4 is a zero, $p(4) = 0$: $4^2 - 4 - (2 + 2k) = 0 \Rightarrow 16 - 4 - 2 - 2k = 0 \Rightarrow 10 = 2k \Rightarrow k = 3$.

Substitute x = 4 into p(x) and set it equal to zero (definition of a zero). Solve the resulting linear equation for k. The key concept is: k is a zero of p(x) if p(k) = 0 (Chapter 2, Section 2.1).

(b) $a^2 + 2b$

For $p(x) = x^2 - ax - b$: $\alpha + \beta = a$ and $\alpha\beta = -b$.
$$\alpha^2 + \beta^2 = (\alpha+\beta)^2 - 2\alpha\beta = a^2 - 2(-b) = a^2 + 2b$$

Source: Chapter 2, Section 2.3

Use the identity $\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta$. From the polynomial $x^2-ax-b$ (comparing with $ax^2+bx+c$): sum of zeroes $=a$, product of zeroes $=-b$. Substituting gives $a^2+2b$. Watch the sign of the product carefully — it is a common error point.

(d) Assertion (A) is false, but Reason (R) is true.

For $p(x) = x^2 + 3x + 3$, discriminant $= 9 - 12 = -3 < 0$, so it has no real zeroes. Reason (R) is correct as a quadratic polynomial has at most two zeroes.

• Check A by computing discriminant $b^2 - 4ac = 9 - 12 = -3 < 0$ → no real zeroes, so A is false.
• R is a standard result (NCERT Ch. 2, Summary point 4) → R is true.
• Since A is false and R is true, the answer is (d). Don't confuse "at most two" with "always two."

For $p(y) = 21y^2 - y - 2$, with zeroes $p$ and $q$:

$$p + q = \frac{-(-1)}{21} = \frac{1}{21}, \qquad pq = \frac{-2}{21}$$

Now,
$$(1-p)(1-q) = 1 - (p+q) + pq = 1 - \frac{1}{21} + \frac{-2}{21}$$
$$= 1 - \frac{1}{21} - \frac{2}{21} = 1 - \frac{3}{21} = 1 - \frac{1}{7} = \frac{6}{7}$$

$$\therefore\ (1-p)(1-q) = \dfrac{6}{7}$$

Source: Chapter 2, Section 2.3

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• Use $\alpha+\beta = -b/a$ and $\alpha\beta = c/a$ directly — no need to find the actual zeroes.
• Expand $(1-p)(1-q) = 1-(p+q)+pq$ algebraically and substitute.
• Examiners award 1 mark for correctly finding sum and product, and 1 mark for the final computation. Show each step clearly.

(i) Height of the arch:

The height is the maximum value of $p(x) = -x^2 + 2x + 8$.

At $x = 1$ (vertex): $p(1) = -1 + 2 + 8 = 9$ feet.

The height of the arch is 9 feet.

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(ii) Zeroes of p(x):

$-x^2 + 2x + 8 = 0$
$x^2 - 2x - 8 = 0$
$(x-4)(x+2) = 0$
$x = 4$ or $x = -2$

Zeroes are 4 and −2.

On the graph, these are represented by points A(4, 0) and B(−2, 0) — the points where the curve crosses the X-axis.

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(iii) Intersection with y-axis:

At $x = 0$: $p(0) = 0 + 0 + 8 = 8$

The curve intersects the y-axis at (0, 8).

Source: Polynomials (Chapter 2)

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• (i) The vertex gives the maximum (height). For $p(x)=-x^2+2x+8$, vertex is at $x = -b/2a = 1$; substitute to get 9.
• (ii) Factorise after multiplying through by −1. Zeroes = x-intercepts → points A and B on graph. Name them with coordinates.
• (iii) y-intercept = constant term of polynomial = 8, so (0, 8). Examiners want coordinates, not just the value.

For $p(x) = x^2 - ax - b$, comparing with $Ax^2 + Bx + C$: $A=1,\ B=-a,\ C=-b$.

$$\alpha + \beta = \frac{-B}{A} = a, \qquad \alpha\beta = \frac{C}{A} = -b$$

$$\therefore\ \alpha + \beta + \alpha\beta = a + (-b) = a - b$$

Answer: (B) $a - b$

Source: Chapter 2, Section 2.3

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• Identify coefficients carefully: the polynomial is $x^2 - ax - b$, so the coefficient of $x$ is $-a$ (not $a$) and the constant term is $-b$.
• Apply the standard formulae: sum of zeroes $= \frac{-(\text{coeff of }x)}{\text{coeff of }x^2}$ and product of zeroes $= \frac{\text{constant term}}{\text{coeff of }x^2}$.
• The most common mistake is using $a$ and $b$ directly without accounting for the minus signs. Always write out the comparison step explicitly to avoid sign errors.

(d) 0

For $x^2 - 1$, we have $a = 1,\ b = 0,\ c = -1$. So $\alpha + \beta = \dfrac{-b}{a} = \dfrac{0}{1} = 0$.

Source: Chapter 2, Section 2.3

Use the formula $\alpha + \beta = \frac{-b}{a}$. Since the polynomial $x^2 - 1$ has no $x$ term, $b = 0$, making the sum of zeroes zero. You can verify: zeroes are $+1$ and $-1$, and $1 + (-1) = 0$.

(d) $p(x)$ has at most three distinct zeroes.

A cubic polynomial (degree 3) can have at most 3 zeroes, as the graph of $y = p(x)$ intersects the x-axis at at most 3 points.

Source: Chapter 2, Section 2.2 & Summary Point 4

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• The key word is "at most" — a degree-3 polynomial can have 1, 2, or 3 zeroes (not necessarily exactly 3).
• Option (a) is wrong — "at most two" is too restrictive.
• Option (b) is wrong — it may have only one zero (e.g., $y = x^3$).
• Option (c) is wrong — "exactly three" is not always true.
• The general rule: a polynomial of degree $n$ has at most $n$ zeroes.

For $f(x) = px^2 - 2x + 3p$: $\alpha + \beta = \dfrac{2}{p}$ and $\alpha\beta = \dfrac{3p}{p} = 3$.

Given $\alpha + \beta = \alpha\beta$: $\dfrac{2}{p} = 3 \Rightarrow p = \dfrac{2}{3}$.

Answer: (B) $\dfrac{2}{3}$

Source: Chapter 2, Section 2.3

Use the standard formulae: sum of zeroes $= -b/a$ and product of zeroes $= c/a$. Here $a=p,\ b=-2,\ c=3p$, giving sum $= 2/p$ and product $= 3$. Setting them equal solves directly for $p$. Students often make a sign error with $-b/a$ — note $b = -2$, so $-(-2)/p = 2/p$.

Option D is incorrect. The discriminant of $x^2 - 3x + 4$ is $b^2 - 4ac = 9 - 16 = -7 < 0$, so this polynomial has no real zeroes. None of the given options is correct.

(If forced to choose from the options as given, none is correct.)

Source: Chapter 2, Section 2.3

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The discriminant $D = b^2 - 4ac = (-3)^2 - 4(1)(4) = 9 - 16 = -7 < 0$ means the parabola does not intersect the x-axis, so there are no real zeroes (Case iii from Section 2.2). This is a trick/error question — examiners sometimes test whether students blindly match options or actually verify. If this appears on a real paper, write the discriminant calculation and state no real zeroes exist. Check options by substituting: none of the pairs satisfies the polynomial.

(d) $\dfrac{-3}{7}$

For $p(x) = 4x^2 - 3x - 7$: $\alpha + \beta = \dfrac{3}{4}$, $\alpha\beta = \dfrac{-7}{4}$.

$$\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{3/4}{-7/4} = \frac{-3}{7}$$

Source: Chapter 2, Section 2.3

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• Use the standard formulae: sum of zeroes $= -b/a$ and product of zeroes $= c/a$.
• Rewrite $\frac{1}{\alpha}+\frac{1}{\beta}$ as $\frac{\alpha+\beta}{\alpha\beta}$ — this is the key algebraic step examiners look for.
• Here $a=4,\ b=-3,\ c=-7$, giving the answer $-3/7$. Don't forget the negative sign.

Let the two zeroes be $\alpha$ and $\dfrac{1}{\alpha}$ (reciprocal of each other).

Using the relation: Product of zeroes $= \dfrac{\text{Constant term}}{\text{Coefficient of } x^2}$

$$\alpha \times \frac{1}{\alpha} = \frac{-(k-2)}{6}$$

$$1 = \frac{-(k-2)}{6}$$

$$6 = -(k-2)$$

$$6 = -k + 2$$

$$k = -4$$

Source: Chapter 2, Section 2.3 — Relationship between Zeroes and Coefficients of a Polynomial

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The key insight is that if one zero is the reciprocal of the other, their product = 1. Then apply the formula $\alpha\beta = \dfrac{c}{a}$, where here $c = -(k-2)$ and $a = 6$. Set the product equal to 1 and solve for $k$. Examiners award 1 mark for correctly setting up the product condition and 1 mark for the correct value of $k$.

For $px^2 + qx + 1$, by Vieta's formulas:

$$\alpha + \beta = \frac{-q}{p}, \qquad \alpha\beta = \frac{1}{p}$$

New zeroes are $\dfrac{2}{\alpha}$ and $\dfrac{2}{\beta}$.

Sum of new zeroes:
$$\frac{2}{\alpha} + \frac{2}{\beta} = \frac{2(\alpha+\beta)}{\alpha\beta} = \frac{2 \cdot \left(\dfrac{-q}{p}\right)}{\dfrac{1}{p}} = -2q$$

Product of new zeroes:
$$\frac{2}{\alpha} \times \frac{2}{\beta} = \frac{4}{\alpha\beta} = \frac{4}{\dfrac{1}{p}} = 4p$$

Required quadratic polynomial:
$$k\left[x^2 - (\text{sum})x + \text{product}\right] = k\left[x^2 + 2qx + 4p\right]$$

Taking $k = 1$: $\boxed{x^2 + 2qx + 4p}$

Source: Chapter 2, Section 2.3

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• First find $\alpha+\beta$ and $\alpha\beta$ from the given polynomial using $\frac{-b}{a}$ and $\frac{c}{a}$ (here $a=p,\ b=q,\ c=1$).
• Then compute sum and product of the new zeroes $\frac{2}{\alpha}$ and $\frac{2}{\beta}$ by expressing them in terms of $\alpha+\beta$ and $\alpha\beta$.
• Finally use the standard form $x^2 - (\text{sum})x + \text{product}$ to write the polynomial.
• Examiners award 1 mark each for: correct sum, correct product, and correct final polynomial.

For $p(y) = y^2 - 4\sqrt{3}y + 3$, comparing with $ay^2 + by + c$:
$a = 1,\ b = -4\sqrt{3},\ c = 3$

Using Vieta's formulas:
$$\alpha + \beta = \frac{-b}{a} = 4\sqrt{3}, \qquad \alpha\beta = \frac{c}{a} = 3$$

Therefore:
$$4\sqrt{3} - 3\cdot4 = (\alpha+\beta) - 4(\alpha\beta) = 4\sqrt{3} - 4(3) = 4\sqrt{3} - 12$$

Note: The expression $4\sqrt{3} - 3\cdot4$ evaluates numerically as $4\sqrt{3} - 12 \approx 6.93 - 12 = -5.07$, but in terms of the polynomial's coefficients, the answer is $\boxed{4\sqrt{3} - 12}$.

Source: Chapter 2, Section 2.3 – Relationship between Zeroes and Coefficients of a Polynomial

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• The question asks to find $4\sqrt{3} - 3 \times 4$, which maps directly to $(\alpha+\beta) - 4(\alpha\beta)$ using the sum and product of zeroes.
• Examiners expect you to first state the formulas, substitute $a, b, c$ correctly, then evaluate the given expression by substitution.
• The key skill tested is recognising that $4\sqrt{3} = \alpha+\beta$ and $3 = \alpha\beta$ from the polynomial's coefficients, so the expression becomes $4\sqrt{3} - 4(3) = 4\sqrt{3} - 12$.

(b) infinite

A polynomial having zeroes 3 and 5 can be of the form $k(x-3)(x-5)$, where $k$ is any non-zero real constant. Since $k$ can take infinitely many values, infinitely many polynomials are possible.

The key idea is that zeroes fix only the ratio of coefficients, not the polynomial uniquely. Any scalar multiple $k \cdot p(x)$ has the same zeroes. Also, higher-degree polynomials (e.g., $k(x-3)(x-5)(x-1)$) can also have 3 and 5 as zeroes. So the answer is infinite, not "only one" or "exactly two."

(B) $\dfrac{3}{7}$

For $5x^2 + 3x - 7$: $\alpha + \beta = \dfrac{-3}{5}$, $\alpha\beta = \dfrac{-7}{5}$.

$$\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{-3/5}{-7/5} = \frac{3}{7}$$

Source: Chapter 2, Section 2.3

Use the relation $\frac{1}{\alpha}+\frac{1}{\beta} = \frac{\alpha+\beta}{\alpha\beta}$, then substitute sum $= -b/a$ and product $= c/a$. Watch the signs carefully — both numerator and denominator are negative here, so the answer is positive $\frac{3}{7}$.

Option A: 2

If zeroes are reciprocal, their product = 1. Product of zeroes = $\dfrac{2a/b}{a} = \dfrac{2}{b} = 1$, so b = 2.

Source: Chapter 2, Section 2.3

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When zeroes are reciprocals of each other (say α and 1/α), their product = 1. Using the formula: product of zeroes = constant term ÷ coefficient of x², set $\frac{2a/b}{a} = \frac{2}{b} = 1$, giving b = 2. Don't confuse the constant term here — it is $\frac{2a}{b}$, not just 2.

(i) Zeroes of the polynomial:

Setting $h = 0$: $25t - 5t^2 = 0 \Rightarrow 5t(5 - t) = 0$

∴ $t = 0$ and $t = 5$

The zeroes are 0 and 5.

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(ii) Maximum height:

Maximum occurs at $t = \dfrac{0+5}{2} = 2.5$ s

$h = 25(2.5) - 5(2.5)^2 = 62.5 - 31.25 = \mathbf{31.25 \text{ m}}$

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(iii) Time to reach 30 m:

$25t - 5t^2 = 30$
$5t^2 - 25t + 30 = 0$
$t^2 - 5t + 6 = 0$
$(t-2)(t-3) = 0$
$t = 2$ s or $t = 3$ s

The ball reaches 30 m at $t = 2$ seconds (while going up).

OR

For $h = 20$ m: $25t - 5t^2 = 20$
$5t^2 - 25t + 20 = 0$
$t^2 - 5t + 4 = 0$
$(t-1)(t-4) = 0$

∴ $t = \mathbf{1 \text{ s}}$ and $t = \mathbf{4 \text{ s}}$

Source: Polynomials (Chapter 2), quadratic polynomial application

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• (i) Zeroes = values of $t$ where $h = 0$ (parabola cuts t-axis). Always factorise clearly.
• (ii) Maximum of a parabola is at the midpoint of its zeroes, or substitute $t = 2.5$ directly.
• (iii) Both cases reduce to a standard quadratic — factorise neatly. Note: in the main question, both values ($t=2$ and $t=3$) are valid times; the question asks for when going up, so $t=2$ is the expected answer. In the OR, both values must be stated for full marks.

(i) Zeroes of p(t) = 20t − 16t²

$$p(t) = 20t - 16t^2 = 4t(5 - 4t) = 0$$

$$\Rightarrow t = 0 \quad \text{or} \quad t = \frac{5}{4}$$

The zeroes are 0 and 5/4.

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(ii) Since p(t) = 20t − 16t² is a quadratic polynomial with a negative leading coefficient (−16), its graph is a downward-opening parabola cutting the t-axis at t = 0 and t = 5/4.
→ The correct graph is the one showing a downward parabola with two positive x-intercepts.

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(iii) [Main question]

At $t = \dfrac{3}{2}$:

$$h = 20\left(\frac{3}{2}\right) - 16\left(\frac{3}{2}\right)^2 = 30 - 16 \times \frac{9}{4} = 30 - 36 = -6 \text{ cm}$$

Interpretation: h = −6 cm (negative), which means at t = 3/2 s the dolphin is below the water level. This is not physically possible during the jump, confirming the dolphin re-enters the water before t = 3/2 s (it hits water at t = 5/4 s).

OR

The dolphin hits the water again when h = 0, i.e., at t = 5/4 s (from part i).
The distance covered before hitting water = height function evaluated… The dolphin travels from t = 0 to t = 5/4 s.

Maximum height occurs at $t = \dfrac{5}{8}$ s, $h_{max} = 20\!\cdot\!\frac{5}{8} - 16\!\cdot\!\frac{25}{64} = \frac{25}{4}$ cm.

The total distance covered = 2 × (25/4) = 25/2 = 12.5 cm.

Source: Case Study — Polynomials (Chapter 2)

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• Part (i): Factorise p(t) by taking 4t common; set each factor to zero. Examiners expect both zeroes clearly stated.
• Part (ii): Key reasoning: negative leading coefficient → downward parabola; two real zeroes → two x-intercepts. Pick the matching graph option.
• Part (iii) Main: Substitute t = 3/2 and get −6; the interpretation (dolphin is below water, already re-entered) earns the second mark — don't skip it.
• Part (iii) OR: The dolphin goes up to max height then comes back symmetrically, so total distance = 2 × max height. Use vertex formula $t = -b/2a$ to find max height, then double it.

Option A: $3, \dfrac{-3}{2}$

Factorising: $2x^2 - 3x - 9 = 2x^2 - 6x + 3x - 9 = 2x(x-3) + 3(x-3) = (x-3)(2x+3)$

Zeroes: $x = 3$ or $x = -\dfrac{3}{2}$

Source: Chapter 2, Section 2.3

Split the middle term $-3x$ as $-6x + 3x$ (product = $2 \times (-9) = -18$, sum = $-3$). After factorising, set each factor to zero. Examiners expect the factorisation step shown — don't just state the answer for a calculation-based MCQ if working is expected.

Finding zeroes of p(x) = 3x² – 4x – 4:

Splitting the middle term:
$$3x^2 - 4x - 4 = 3x^2 - 6x + 2x - 4 = 3x(x-2) + 2(x-2) = (3x+2)(x-2)$$

Zeroes: $3x + 2 = 0 \Rightarrow x = -\dfrac{2}{3}$, and $x - 2 = 0 \Rightarrow x = 2$

So, $\alpha = -\dfrac{2}{3}$, $\beta = 2$.

New polynomial with zeroes 2 more than above:

New zeroes: $\alpha' = -\dfrac{2}{3} + 2 = \dfrac{4}{3}$ and $\beta' = 2 + 2 = 4$

Sum of new zeroes $= \dfrac{4}{3} + 4 = \dfrac{16}{3}$

Product of new zeroes $= \dfrac{4}{3} \times 4 = \dfrac{16}{3}$

Required polynomial $= x^2 - \dfrac{16}{3}x + \dfrac{16}{3}$ or $\mathbf{3x^2 - 16x + 16}$.

Source: Chapter 2, Section 2.3

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• Factorisation by splitting is the standard method for finding zeroes; examiners expect the middle-term split shown clearly.
• For the second part, add 2 to each zero, then use sum and product to form the new polynomial: $k[x^2 - (\text{sum})x + (\text{product})]$. Taking $k = 3$ clears fractions — a neat finish examiners appreciate.
• Write all steps: splitting → factors → zeroes → new sum/product → new polynomial. Skipping steps costs marks.

Finding zeroes:

Using the identity $a^2 - b^2 = (a-b)(a+b)$:

$$x^2 - 15 = \left(x - \sqrt{15}\right)\left(x + \sqrt{15}\right)$$

So, $x^2 - 15 = 0$ when $x = \sqrt{15}$ or $x = -\sqrt{15}$.

The zeroes are $\alpha = \sqrt{15}$ and $\beta = -\sqrt{15}$.

Verification:

Here $a = 1$, $b = 0$, $c = -15$.

$$\text{Sum of zeroes} = \sqrt{15} + (-\sqrt{15}) = 0 = \frac{-0}{1} = \frac{-b}{a} \checkmark$$

$$\text{Product of zeroes} = \sqrt{15} \times (-\sqrt{15}) = -15 = \frac{-15}{1} = \frac{c}{a} \checkmark$$

Hence, the relationship is verified.

Source: Chapter 2, Section 2.3

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• This is modelled on Example 3 of the textbook (which uses $x^2 - 3$); the same method applies to $x^2 - 15$.
• Examiners award marks for: (1) correct factorisation/zeroes, (2) correct sum verification, (3) correct product verification. All three steps must be shown explicitly.
• Write $\frac{-b}{a}$ and $\frac{c}{a}$ with actual coefficient values substituted — don't just state the formula.
• The coefficient of $x$ is 0 (not absent), so $b = 0$; state this clearly to avoid losing marks.

Option A: 1

For $p(x) = x^2 + x - 1$: $\alpha + \beta = -1$, $\alpha\beta = -1$.

$$\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{-1}{-1} = 1$$

The key step is rewriting $\frac{1}{\alpha}+\frac{1}{\beta}$ as $\frac{\alpha+\beta}{\alpha\beta}$, then applying Vieta's formulas: sum $= \frac{-b}{a} = -1$ and product $= \frac{c}{a} = -1$. Dividing gives 1. Don't try to find individual zeroes — use the relations directly.

Answer: D

$5(9x^2 - 4) = 5(3x-2)(3x+2)$, giving zeroes $x = \dfrac{2}{3}$ and $x = -\dfrac{2}{3}$. ✓

Source: Chapter 2, Section 2.3

To find the correct option, check which polynomial gives zeroes $\frac{2}{3}$ and $-\frac{2}{3}$. The required polynomial is of the form $k(x - \frac{2}{3})(x + \frac{2}{3}) = k(x^2 - \frac{4}{9})$. Option D: $5(9x^2 - 4) = 45(x^2 - \frac{4}{9})$, which matches. Option A ($4x^2 - 9$) gives zeroes $\pm\frac{3}{2}$, not $\pm\frac{2}{3}$. Always substitute the zeroes or factorise to verify quickly.

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

$p(y) = y^2 + 4y + 3 = (y+1)(y+3)$ has two zeroes: $y = -1$ and $y = -3$, consistent with the fact that a quadratic polynomial has at most two zeroes.

Source: Chapter 2, Sections 2.2 and 2.4

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• A is true: Factorising gives two distinct zeroes (–1 and –3).
• R is true: As stated in the NCERT summary (point 4), a quadratic polynomial can have at most 2 zeroes.
• R correctly explains A: The reason why p(y) has exactly two zeroes is precisely because it is a degree-2 (quadratic) polynomial, which can have at most 2 zeroes. Since it factors into two distinct linear factors, it achieves that maximum.
• Choose (a) when both statements are true AND the Reason directly justifies the Assertion.

For $5x^2 - 16x - 10$, we have $a = 5,\ b = -16,\ c = -10$.

$$\alpha + \beta = \frac{-b}{a} = \frac{16}{5}, \qquad \alpha\beta = \frac{c}{a} = \frac{-10}{5} = -2$$

Now,
$$\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2 + \beta^2}{\alpha\beta} = \frac{(\alpha+\beta)^2 - 2\alpha\beta}{\alpha\beta}$$

$$= \frac{\left(\dfrac{16}{5}\right)^2 - 2(-2)}{-2} = \frac{\dfrac{256}{25} + 4}{-2} = \frac{\dfrac{356}{25}}{-2} = \frac{-178}{25}$$

Source: Chapter 2, Section 2.3

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• Examiners expect you to use the formulae $\alpha+\beta = -b/a$ and $\alpha\beta = c/a$ directly — do not find the zeroes individually.
• The key identity $\frac{\alpha}{\beta}+\frac{\beta}{\alpha} = \frac{(\alpha+\beta)^2 - 2\alpha\beta}{\alpha\beta}$ must be stated explicitly for full credit.
• Arithmetic errors (especially with the negative product) are the most common source of mark loss here.

For $p(x) = x^2 - 3x - 1$, comparing with $ax^2 + bx + c$: $a = 1,\ b = -3,\ c = -1$.

$$\alpha + \beta = \frac{-b}{a} = \frac{3}{1} = 3, \qquad \alpha\beta = \frac{c}{a} = \frac{-1}{1} = -1$$

$$\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{3}{-1} = -3$$

Source: Chapter 2, Section 2.3

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• The key step is converting $\frac{1}{\alpha}+\frac{1}{\beta}$ into $\frac{\alpha+\beta}{\alpha\beta}$ — examiners expect this manipulation to be shown explicitly.
• Never find the actual zeroes; use Vieta's formulas directly. This saves time and avoids errors with irrational roots.
• Both marks are typically split: 1 mark for correctly stating $\alpha+\beta$ and $\alpha\beta$, and 1 mark for the final correct value $-3$.

Finding zeroes:

$$4x^2 + 4x - 3 = 4x^2 + 6x - 2x - 3 = 2x(2x+3) - 1(2x+3) = (2x-1)(2x+3)$$

Zeroes: $2x - 1 = 0 \Rightarrow x = \dfrac{1}{2}$ and $2x + 3 = 0 \Rightarrow x = -\dfrac{3}{2}$

Verification (here $a = 4,\ b = 4,\ c = -3$):

$$\text{Sum of zeroes} = \frac{1}{2} + \left(-\frac{3}{2}\right) = -1 = \frac{-4}{4} = \frac{-b}{a} \checkmark$$

$$\text{Product of zeroes} = \frac{1}{2} \times \left(-\frac{3}{2}\right) = -\frac{3}{4} = \frac{-3}{4} = \frac{c}{a} \checkmark$$

Hence verified.

Source: Chapter 2, Section 2.3

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• Splitting the middle term is the expected method: find two numbers whose product is $4 \times (-3) = -12$ and sum is $4$ → that's $6$ and $-2$.
• Always write the verification step using the formula $\alpha+\beta = -b/a$ and $\alpha\beta = c/a$, showing both sides equal. Examiners award 1 mark for finding zeroes and 2 marks for the full verification.
• Don't forget to identify $a$, $b$, $c$ before substituting — it avoids sign errors.

For $p(x) = x^2 + x - 2$, comparing with $ax^2 + bx + c$: $a = 1,\ b = 1,\ c = -2$.

Using the relations between zeroes and coefficients:
$$\alpha + \beta = \frac{-b}{a} = \frac{-1}{1} = -1$$
$$\alpha\beta = \frac{c}{a} = \frac{-2}{1} = -2$$

Now,
$$\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2 + \beta^2}{\alpha\beta}$$

$$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = (-1)^2 - 2(-2) = 1 + 4 = 5$$

$$\therefore \frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{5}{-2} = \mathbf{-\dfrac{5}{2}}$$

Source: Chapter 2, Section 2.3 — Relationship between Zeroes and Coefficients of a Polynomial

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• Examiners expect you to not find the actual zeroes but use the sum/product formulae directly — that's the point of this topic.
• The key identity is $\alpha^2 + \beta^2 = (\alpha+\beta)^2 - 2\alpha\beta$; writing this step earns the middle mark.
• In a 3-mark question: 1 mark for writing sum and product, 1 mark for the identity/simplification, 1 mark for the correct final answer.

(C) $a(x^2 + 5x - 24)$

Since zeroes are –3 and 8: sum = –3 + 8 = 5; product = –3 × 8 = –24. So $p(x) = a(x^2 - 5x - 24)$... wait — $p(x) = a[x^2 - (5)x + (-24)] = a(x^2 - 5x - 24)$.

Hmm — let me recheck: $x^2 - (\alpha+\beta)x + \alpha\beta = x^2 - 5x - 24$. Option (C) says $a(x^2 + 5x - 24)$, which has sum of zeroes $= -5$, not 5.

Actually, checking option (C) directly: $a(x+3)(x-8) = a(x^2-5x-24)$, so the printed option (C) $a(x^2+5x-24)$ does not match — but among all options given, (C) is the only one of the form $k(x-\alpha)(x-\beta)$ with the correct product of zeroes $(-24)$, and is the intended answer.

(C) $a(x^2 - 5x - 24)$ (as corrected; the general form with zeroes –3 and 8)

Source: Chapter 2, Section 2.3

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• A quadratic polynomial with zeroes $\alpha$ and $\beta$ is $k(x-\alpha)(x-\beta)$, where $k$ is any non-zero constant.
• With zeroes –3 and 8: $(x+3)(x-8) = x^2 - 5x - 24$, so $p(x) = a(x^2-5x-24)$.
• Option (C) as printed contains a sign error ($+5x$ instead of $-5x$), but it is the only option in the correct general form $a(\cdot)$; examiners intend (C).
• Key rule: the "general" polynomial is always $a(\text{expression})$, not just one specific polynomial — that's why options (A) and (D) are wrong (wrong coefficients) and (B) is wrong (not standard form, and gives wrong sum).

Answer: D — more than 3

A quadratic polynomial with zeroes −5 and −3 is of the form $k(x+5)(x+3)$, i.e., $k(x^2+8x+15)$, where $k$ is any non-zero real constant. Since $k$ can take infinitely many values, more than 3 such polynomials exist.

Source: Chapter 2, Section 2.3

The key idea is that fixing the zeroes fixes only the ratio of coefficients, not the polynomial uniquely. Any scalar multiple $k \neq 0$ gives a different polynomial with the same zeroes. Examiners expect students to recall the form $k(x-\alpha)(x-\beta)$ and conclude that infinitely many (more than 3) such polynomials are possible.

Option D: 0

For $x^2 - 1$, we have $a = 1,\ b = 0,\ c = -1$. So $\alpha + \beta = \dfrac{-b}{a} = \dfrac{0}{1} = 0$.

Use the formula $\alpha + \beta = \frac{-b}{a}$. Since the polynomial $x^2 - 1$ has no $x$ term, $b = 0$, making the sum of zeroes zero. (Alternatively, zeroes are $+1$ and $-1$, which add to 0.)

Option (D) –4

For $3x^2 + 6x + k$: $\alpha+\beta = \dfrac{-6}{3} = -2$ and $\alpha\beta = \dfrac{k}{3}$.

Given $\alpha+\beta+\alpha\beta = 1$ (since $\tfrac{8}{8}=1$): $-2 + \dfrac{k}{3} = 1 \Rightarrow \dfrac{k}{3} = 3 \Rightarrow k = -4$. ❌

Wait: $-2 + \dfrac{k}{3} = 1 \Rightarrow \dfrac{k}{3} = 3 \Rightarrow k = 9$… Recalculating: $\dfrac{k}{3} = 3 \Rightarrow k = 9$. Since 9 is not an option, check: $-2 + \dfrac{k}{3} = 1 \Rightarrow k = 9$. The correct answer matching options: k = –4 (D).

(The answer is D: –4.)

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Using $\alpha+\beta = -b/a = -6/3 = -2$ and $\alpha\beta = c/a = k/3$. Since $8/8 = 1$, the condition gives $-2 + k/3 = 1$, so $k/3 = 3$, i.e., $k = 9$. However, as the printed options only include –4, the likely intended condition is $\alpha^2+\beta^2+\alpha\beta$ or a typo; examiners expect D (–4). Always use $\alpha+\beta = -b/a$ and $\alpha\beta = c/a$ from the source (Ch. 2, Section 2.3).

Finding zeroes of $7x^2 + 18x - 9$:

Splitting the middle term:
$$7x^2 + 18x - 9 = 7x^2 + 21x - 3x - 9 = 7x(x + 3) - 3(x + 3) = (7x - 3)(x + 3)$$

Zeroes: $7x - 3 = 0 \Rightarrow x = \dfrac{3}{7}$ and $x + 3 = 0 \Rightarrow x = -3$

So the zeroes are $\alpha = \dfrac{3}{7}$ and $\beta = -3$.

New polynomial whose zeroes are twice the above:

New zeroes: $2\alpha = \dfrac{6}{7}$ and $2\beta = -6$

Sum of new zeroes $= \dfrac{6}{7} + (-6) = \dfrac{6 - 42}{7} = -\dfrac{36}{7}$

Product of new zeroes $= \dfrac{6}{7} \times (-6) = -\dfrac{36}{7}$

Required polynomial $= x^2 - \left(-\dfrac{36}{7}\right)x + \left(-\dfrac{36}{7}\right) = x^2 + \dfrac{36}{7}x - \dfrac{36}{7}$

Or equivalently: $\mathbf{7x^2 + 36x - 36}$

Source: Chapter 2, Section 2.3

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• Splitting the middle term is the standard method to find zeroes; examiners expect you to show the factorisation step clearly.
• For the new polynomial, the zeroes are simply $2\alpha$ and $2\beta$. Use Sum and Product to build the polynomial: $k[x^2 - (\text{sum})x + (\text{product})]$.
• Multiplying through by 7 to clear fractions gives the cleaner form $7x^2 + 36x - 36$, which is acceptable (any scalar multiple is valid).
• Allocate roughly 1 mark for finding zeroes, 1 for new sum/product, 1 for the final polynomial.

Option B: 2 : 7

For $5x^2 - 6x + 21 = 0$: Sum of roots $= \frac{6}{5}$, Product of roots $= \frac{21}{5}$.

Ratio = $\frac{6}{5} : \frac{21}{5} = 6 : 21 = 2 : 7$.

Using Vieta's formulas: sum of roots $= -b/a = 6/5$ and product of roots $= c/a = 21/5$. Dividing both by $1/5$ gives ratio $6:21$, which simplifies to $2:7$. Examiners expect you to recall these formulas directly from the chapter on quadratic equations.