SEE Model Paper 1

When a polynomial p(x) is divided by (x − a), the remainder is p(a).

$$\begin{aligned} V(2) &= 2^3 + 4 \times 2^2 + 2 - 6 \\ &= 8 + 16 + 2 - 6 \\ &= 20 \end{aligned}$$
The remainder is 20: the box made with x = 2 dm holds 20 dm³ (20 litres).

$$\begin{aligned} g \circ f(x) &= g(f(x)) = g(0.9x) \\ &= 1.13 \times 0.9x \\ &= 1.017x \end{aligned}$$
g ∘ f(x) = 1.017x: the customer pays 1.017 times the marked price.

$$g \circ f(2000) = 1.017 \times 2000 = 2034$$
Rs 2,034.

VAT first, then the discount:
$$\begin{aligned} f \circ g(x) &= f(1.13x) \\ &= 0.9 \times 1.13x \\ &= 1.017x = g \circ f(x) \end{aligned}$$
He is right: both orders multiply the marked price by 0.9 × 1.13 = 1.017, and multiplication can be done in either order.

The inverse of h(x) = 1.017x turns the price paid back into the marked price: $h^{-1}(x) = \frac{x}{1.017}$.
$$h^{-1}(4068) = \frac{4068}{1.017} = 4000$$
The marked price was Rs 4,000.

The feasible region is the region of the graph whose points satisfy all the constraints (inequalities) together: the common solution region.

Boundary lines: 2x + y = 10 through (5, 0) and (0, 10); x + 2y = 8 through (8, 0) and (0, 4). The origin satisfies both inequalities (0 ≤ 10, 0 ≤ 8), so the region is on the origin's side of each line, in the first quadrant.

The two lines meet where 2x + y = 10 and x + 2y = 8: doubling the second, 2x + 4y = 16; subtracting the first, 3y = 6, so y = 2 and x = 4.

Corner points: O(0, 0), P(5, 0), Q(4, 2), R(0, 4).

Profit P = 1500x + 2000y at the corners:

4 chairs and 2 tables, for a profit of Rs 10,000.

$$\sin 3A = 3 \sin A - 4 \sin^3 A$$

$$\begin{aligned} \sin 3A &= 3 \times \frac{1}{3} - 4 \times \left(\frac{1}{3}\right)^3 \\ &= 1 - \frac{4}{27} = \frac{23}{27} \end{aligned}$$
$\sin 3A = \frac{23}{27}$.

AB is the building (12 m), CD the tower (h m) and BD = AE = x m the distance between them. AE is horizontal at the roof's height; the angle $EAC = 60^\circ$ (elevation of the top) and the angle $EAD = 30^\circ$ (depression of the foot), so ED = AB = 12 m.

In right-angled △AED, ED = 12 m and the angle $EAD = 30^\circ$:
$$\begin{aligned} \tan 30^\circ &= \frac{ED}{AE} \\ \frac{1}{\sqrt{3}} &= \frac{12}{x} \\ x &= 12\sqrt{3} \text{ m} \end{aligned}$$
In right-angled △AEC, the angle $EAC = 60^\circ$:
$$\begin{aligned} \tan 60^\circ &= \frac{CE}{AE} \\ CE &= 12\sqrt{3} \times \sqrt{3} = 36 \text{ m} \end{aligned}$$
$$h = CE + ED = 36 + 12 = 48 \text{ m}$$
The tower is 48 m high (and about 20.78 m from the building).

A conditional identity is a trigonometric relation that is true only when a given condition holds (such as $A + B + C = 180^\circ$), not for all angles.

The angles add up to 180∘°, so the identity should hold.
$$\begin{aligned} \text{L.H.S.} &= \cos 90^\circ + \cos 60^\circ + \cos 30^\circ \\ &= 0 + \frac{1}{2} + \frac{\sqrt{3}}{2} \\ &= \frac{1 + \sqrt{3}}{2} \end{aligned}$$
For the R.H.S., the half angles are 45∘°, 30∘° and 15∘°, and
$$\begin{aligned} \sin 15^\circ &= \sin(45^\circ - 30^\circ) \\ &= \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}} \cdot \frac{1}{2} \\ &= \frac{\sqrt{3} - 1}{2\sqrt{2}} \end{aligned}$$
$$\begin{aligned} \text{R.H.S.} &= 1 + 4 \sin 45^\circ \sin 30^\circ \sin 15^\circ \\ &= 1 + 4 \times \frac{1}{\sqrt{2}} \times \frac{1}{2} \times \frac{\sqrt{3} - 1}{2\sqrt{2}} \\ &= 1 + \frac{\sqrt{3} - 1}{2} \\ &= \frac{1 + \sqrt{3}}{2} \end{aligned}$$
L.H.S. = R.H.S., verified.

In a triangle each angle is between 0∘° and 180∘°, so each half angle $\frac{A}{2}$, $\frac{B}{2}$, $\frac{C}{2}$ is between 0∘° and 90∘°, where the sine is positive. So $4 \sin\frac{A}{2} \sin\frac{B}{2} \sin\frac{C}{2} > 0$, and the right-hand side is more than 1. Hence cos A +  cos B +  cos C > 1 (Part b's triangle gives about 1.37).

$$\tan\theta = \pm\frac{m_1 - m_2}{1 + m_1 m_2}$$
(the + sign gives the acute angle when the value is positive).

Slope of the road: $m_1 = \frac{7 - 1}{4 - 1} = 2$. The canal, x − 3y + 6 = 0, is $y = \frac{1}{3}x + 2$, so $m_2 = \frac{1}{3}$.
$$\tan\theta = \frac{2 - \frac{1}{3}}{1 + 2 \times \frac{1}{3}} = \frac{\frac{5}{3}}{\frac{5}{3}} = 1$$
$\theta = 45^\circ$.

$$(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$$

The center is the midpoint of the diameter AB:
$$\left(\frac{-1 + 5}{2}, \frac{2 + 10}{2}\right) = (2, 6)$$
The radius is the distance from the center to B:
$$\begin{aligned} r &= \sqrt{(5 - 2)^2 + (10 - 6)^2} \\ &= \sqrt{9 + 16} = 5 \end{aligned}$$
Center (2, 6), radius 5 m.

$$\begin{aligned} (x - 2)^2 + (y - 6)^2 &= 25 \\ x^2 - 4x + 4 + y^2 - 12y + 36 &= 25 \end{aligned}$$
$x^2 + y^2 - 4x - 12y + 15 = 0$. (The diameter form gives the same: (x + 1)(x − 5) + (y − 2)(y − 10) = 0.)

At P(6, 3): 36 + 9 − 24 − 36 + 15 = 0. The point satisfies the equation, so the lamp post is on the edge (it is exactly 5 m from the center).

R first: P(x, y) → P′(−y, x).
$$\begin{aligned} A(1, 1) &\to A'(-1, 1) \\ B(3, 1) &\to B'(-1, 3) \\ C(2, 3) &\to C'(-3, 2) \end{aligned}$$
Then T: P′(x, y) → P′′(x + 2, y − 1).
$$\begin{aligned} A'(-1, 1) &\to A''(1, 0) \\ B'(-1, 3) &\to B''(1, 2) \\ C'(-3, 2) &\to C''(-1, 1) \end{aligned}$$
A′′(1, 0), B′′(1, 2), C′′(−1, 1).

Plot A(1, 1), B(3, 1), C(2, 3) and A′′(1, 0), B′′(1, 2), C′′(−1, 1), and join each set of three.

$$\cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}|\,|\vec{b}|}$$

AB = (5 − 1, 4 − 2) = (4, 2), AC = (3 − 1, 8 − 2) = (2, 6).
$$\begin{aligned} \cos A &= \frac{4 \times 2 + 2 \times 6}{\sqrt{16 + 4}\,\sqrt{4 + 36}} \\ &= \frac{20}{\sqrt{20}\,\sqrt{40}} = \frac{20}{20\sqrt{2}} \\ &= \frac{1}{\sqrt{2}} \end{aligned}$$
The angle at A is 45∘°.

BA = (−4, −2), BC = (−2, 4):
$$\begin{aligned} \overrightarrow{BA} \cdot \overrightarrow{BC} &= (-4)(-2) + (-2)(4) \\ &= 8 - 8 = 0 \end{aligned}$$
The scalar product of two non-zero vectors is 0 only when they are perpendicular, so he is right: the angle at B is 90∘° (and with 45∘° at A, the angle at C is 45∘° too).

$$\text{Coefficient of Q.D.} = \frac{Q_3 - Q_1}{Q_3 + Q_1}$$

$$\frac{180 - 120}{180 + 120} = \frac{60}{300} = 0.2$$
0.2. (The quartile deviation itself is $\frac{180 - 120}{2} = 30$ cm.)

In the first field. Its coefficient, 0.2, is smaller than 0.25: the middle half of its plants are spread less, relative to their size, so its heights are more uniform. (The coefficient, not the quartile deviation, is compared because the two fields' plants may differ in average height.)

1. $\lim_{x \to a} f(x)$ exists (the left-hand and right-hand limits are equal);
2. f(a) is defined;
3. $\lim_{x \to a} f(x) = f(a)$.

For t ≠ 2:
$$\begin{aligned} v(t) &= \frac{5(t^2 - 4)}{t - 2} \\ &= \frac{5(t - 2)(t + 2)}{t - 2} \\ &= 5(t + 2) \end{aligned}$$
As t → 2, v(t) → 5 × 4 = 20. (From a table: v(1.99) = 19.95, v(2.01) = 20.05.) $\lim_{t \to 2} v(t) = 20$.

$$v(2) = \frac{5 \times 4 - 20}{2 - 2} = \frac{0}{0}$$
No: $\frac{0}{0}$ is an indeterminate form, so v(2) is not defined.

Not continuous: the limit exists (20) but v(2) is not defined. Defining v(2) = 20 makes it continuous. The value 20 m/s is the stone's speed at the instant t = 2 s, which the average speeds over shorter and shorter times close in on.

$$\begin{aligned} & \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 0 & 2 & 2 & 0 \\ 0 & 0 & 2 & 2 \end{pmatrix} \\ &= \begin{pmatrix} 0 & 2 & 4 & 2 \\ 0 & 0 & 2 & 2 \end{pmatrix} \end{aligned}$$
A′(0, 0), B′(2, 0), C′(4, 2), D′(2, 2): a parallelogram.

Through B′(2, 0) and C′(4, 2):
$$y - 0 = \frac{2 - 0}{4 - 2}(x - 2) \Rightarrow y = x - 2$$
x − y − 2 = 0.

|M| = 1 × 1 − 1 × 0 = 1 ≠ 0, so M is non-singular and has an inverse, which undoes it:
$$M^{-1} = \frac{1}{1}\begin{pmatrix} 1 & -1 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & -1 \\ 0 & 1 \end{pmatrix}$$
Yes, with $M^{-1} = \begin{pmatrix} 1 & -1 \\ 0 & 1 \end{pmatrix}$.

$$\begin{aligned} M^2 &= \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix} \\ 2M &= \begin{pmatrix} 2 & 2 \\ 0 & 2 \end{pmatrix} \end{aligned}$$
$M^2 \ne 2M$. For C(2, 2): $M^2$ gives (2 + 4, 2) = (6, 2), but 2M gives (4 + 4, 4) = (8, 4). No, I don't agree: slanting twice keeps the motif's height, while 2M also doubles every length (an enlargement as well).

In a rhombus ABCD, let AB = a and AD = b. All sides are equal, so |a| = |b|. Then
$$\begin{aligned} \overrightarrow{AC} &= \overrightarrow{AB} + \overrightarrow{BC} = \vec{a} + \vec{b} \\ \overrightarrow{DB} &= \overrightarrow{AB} - \overrightarrow{AD} = \vec{a} - \vec{b} \end{aligned}$$
$$\begin{aligned} \overrightarrow{AC} \cdot \overrightarrow{DB} &= (\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b}) \\ &= |\vec{a}|^2 - |\vec{b}|^2 = 0 \end{aligned}$$
The scalar product is 0, so $AC \perp BD$. Proved.

No. The proof used |a| = |b|, which holds only when two neighboring sides are equal, as in a rhombus. In a parallelogram with sides 4 and 2 (a rectangle, say), $|\vec{a}|^2 - |\vec{b}|^2 = 16 - 4 = 12 \ne 0$, so its diagonals are not perpendicular.

No. A circle through A(0, 4) and C(0, −4) has its center on the perpendicular bisector of AC, the X-axis; one through B(3, 0) and D(−3, 0) has its center on the perpendicular bisector of BD, the Y-axis. So the center would have to be the origin, but OA = 4 and OB = 3 are not equal. No circle passes through all four corners.

$$h(t) = t^2 - 8t + 16 + 4 = (t - 4)^2 + 4$$
The vertex of the parabola is (4, 4), and it opens upwards (a = 1 > 0), so the lowest height is 4 m, after 4 seconds; then the kite climbs again.

$$\begin{aligned} \frac{\sin 2\theta}{1 + \cos 2\theta} &= \frac{2\sin\theta\cos\theta}{1 + 2\cos^2\theta - 1} \\ &= \frac{2\sin\theta\cos\theta}{2\cos^2\theta} \\ &= \frac{\sin\theta}{\cos\theta} = \tan\theta \end{aligned}$$
Proved.

$$\begin{aligned} \sin 2\theta &= \frac{2\tan\theta}{1 + \tan^2\theta} = \frac{\frac{2}{12}}{1 + \frac{1}{144}} \\ &= \frac{1}{6} \times \frac{144}{145} = \frac{24}{145} \end{aligned}$$
$$\begin{aligned} \cos 2\theta &= \frac{1 - \tan^2\theta}{1 + \tan^2\theta} \\ &= \frac{\frac{143}{144}}{\frac{145}{144}} = \frac{143}{145} \end{aligned}$$
$$\begin{aligned} \frac{\sin 2\theta}{1 + \cos 2\theta} &= \frac{\frac{24}{145}}{\frac{288}{145}} = \frac{24}{288} \\ &= \frac{1}{12}, \quad \text{so } h = \frac{x}{12} \end{aligned}$$

With $f(x) = \frac{x}{12}$:
$$g \circ f(x) = g\left(\frac{x}{12}\right) = \frac{x}{12} + 90$$
At 6 m, x = 600 cm: g ∘ f(600) = 50 + 90 = 140. The handrail is 140 cm (1.4 m) above the ground there.

$$\begin{aligned} \tan^2\theta &= \frac{1 - \cos 2\theta}{1 + \cos 2\theta} = \frac{\frac{2}{17}}{\frac{32}{17}} = \frac{1}{16} \\ \tan\theta &= \frac{1}{4} \end{aligned}$$
$\frac{1}{4} > \frac{1}{12}$: it rises 3 cm for every 12 cm. No, it is too steep.

$$\begin{aligned} \bar{X} &= \frac{\Sigma fm}{N} = \frac{100}{20} = 5 \text{ GB} \\ \sigma &= \sqrt{\frac{\Sigma fd^2}{N}} = \sqrt{\frac{80}{20}} = 2 \text{ GB} \end{aligned}$$
$$\text{C.V.} = \frac{\sigma}{\bar{X}} \times 100\% = \frac{2}{5} \times 100\% = 40\%$$
Standard deviation 2 GB; coefficient of variation 40%.

From the left, c(x) = 30x → 30 × 4 = 120; from the right, c(x) = 20x + 50 → 80 + 50 = 130. (A table: c(3.99) = 119.70, c(4.01) = 130.20.) Left-hand limit Rs 120, right-hand limit Rs 130.

No. The left-hand and right-hand limits differ (120 ≠ 130), so c(x) has no limit at x = 4 and is discontinuous there. The bill jumps by about Rs 10 the moment a student passes 4 GB: a student using 4.01 GB pays about Rs 130, not Rs 120. The company's claim is wrong.