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Question 1

Topic-1.6 – Series

In the expansion of $\left(kx + \frac{2}{x}\right)^4$, where k is a positive constant, the term independent of x is equal to 150.

Find the value of k and hence determine the coefficient of x² in the expansion.

▶️Answer/Explanation

Solution :-

To solve this problem, we need to work with the binomial expansion of $\left(kx + \frac{2}{x}\right)^4$, find the value of $k$ such that the term independent of $x$ (the constant term) is 150, and then determine the coefficient of $x^2$.

Binomial Expansion
The expression is $\left(kx + \frac{2}{x}\right)^4$. Using the binomial theorem, $(a + b)^n = \sum_{r=0}^n \binom{n}{r} a^{n-r} b^r$, where $a = kx$, $b = \frac{2}{x}$, and $n = 4$, the general term is:

\[
\binom{4}{r} (kx)^{4-r} \left(\frac{2}{x}\right)^r
\]

Simplify the term:
– $(kx)^{4-r} = k^{4-r} x^{4-r}$,
– $\left(\frac{2}{x}\right)^r = \frac{2^r}{x^r} = 2^r x^{-r}$,
– So, $\binom{4}{r} k^{4-r} x^{4-r} \cdot 2^r x^{-r} = \binom{4}{r} k^{4-r} 2^r x^{4-r-r} = \binom{4}{r} k^{4-r} 2^r x^{4-2r}$.

The exponent of $x$ is $4 – 2r$.

For $\left(kx + \frac{2}{x}\right)^4$, the general term is $\binom{4}{r} k^{4-r} \left(\frac{2}{x}\right)^r = \binom{4}{r} k^{4-r} 2^r x^{4-2r}$.

– **Constant term ($x^0$):**
$4 – 2r = 0$ → $r = 2$.
$\binom{4}{2} k^{4-2} 2^2 = 6 k^2 \cdot 4 = 24 k^2 = 150$.
$k^2 = \frac{150}{24} = \frac{25}{4}$, $k = \frac{5}{2}$ (positive).

– **$x^2$ term:**
$4 – 2r = 2$ → $r = 1$.
$\binom{4}{1} k^{4-1} 2^1 = 4 \cdot k^3 \cdot 2$, $k = \frac{5}{2}$, $k^3 = \frac{125}{8}$,
$4 \cdot \frac{125}{8} \cdot 2 = \frac{1000}{4} = 125$.

**Final Answer:** $\boxed{k = \frac{5}{2}, \text{ coefficient of } x^{2} = 125}$

Question 2

Topic-1.7 – Differentiation

The curve $y = x^2 – \frac{a}{x}$ has a stationary point at (-3, b).

Find the values of the constants a and b.

▶️Answer/Explanation

Solution :-

To find the values of $a$ and $b$ for the curve $y = x^2 – \frac{a}{x}$, given it has a stationary point at $(-3, b)$, we need to use the fact that a stationary point occurs where the derivative is zero, and the point must lie on the curve.

### Step 1: Find the Derivative
– $y = x^2 – \frac{a}{x} = x^2 – a x^{-1}$.
– Differentiate with respect to $x$:
$\frac{dy}{dx} = 2x – (-1) a x^{-2} = 2x + \frac{a}{x^2}$.

### Step 2: Set Derivative to Zero at Stationary Point
– At a stationary point, $\frac{dy}{dx} = 0$.
– At $x = -3$:
$2(-3) + \frac{a}{(-3)^2} = -6 + \frac{a}{9} = 0$.
– Solve for $a$:
$\frac{a}{9} = 6$,
$a = 6 \cdot 9 = 54$.

### Step 3: Find $b$ (y-coordinate)
– Substitute $x = -3$ and $a = 54$ into the original equation:
$y = (-3)^2 – \frac{54}{-3} = 9 – (-18) = 9 + 18 = 27$.
– So, $b = 27$.

### Step 4: Verify
– Derivative at $x = -3$: $\frac{dy}{dx} = 2(-3) + \frac{54}{9} = -6 + 6 = 0$, confirming a stationary point.
– Point: $(-3, 27)$ satisfies $y = x^2 – \frac{54}{x}$.

**Final Answer:** $\boxed{a = 54, b = 27}$

Question 3

Topic-1.4 – Circular Measure

The diagram shows a sector of a circle, centre O, where OB = OC = 15 cm. The size of angle BOC is $\frac{2}{3}\pi$ radians. Points A and D on the lines OB and OC respectively are joined by an arc AD of a circle with centre O. The shaded region is bounded by the arcs AD and BC and by the straight lines AB and DC. It is given that the area of the shaded region is $\frac{209}{3}\pi$cm².

Find the perimeter of the shaded region. Give your answer in terms of $\pi$.

▶️Answer/Explanation

Solution :-

Given a sector with radius 15 cm, angle $\frac{2}{3}\pi$, and a shaded region bounded by outer arc $BC$, inner arc $AD$ (radius $r$), and lines $AB$ and $DC$, with area $\frac{209}{3}\pi \, \text{cm}^2$:

– **Area:**
Sector $OBC$ = $\frac{1}{2} \cdot 15^2 \cdot \frac{2}{3}\pi = 75\pi$.
Sector $OAD$ = $\frac{1}{2} \cdot r^2 \cdot \frac{2}{3}\pi = \frac{r^2}{3}\pi$.
Shaded = $75\pi – \frac{r^2}{3}\pi = \frac{209}{3}\pi$.
$75 – \frac{r^2}{3} = \frac{209}{3}$, $\frac{r^2}{3} = \frac{16}{3}$, $r^2 = 16$, $r = 4$.

– **Perimeter:**
$AB = DC = 15 – 4 = 11$.
Arc $BC = 15 \cdot \frac{2}{3}\pi = 10\pi$.
Arc $AD = 4 \cdot \frac{2}{3}\pi = \frac{8}{3}\pi$.
Total = $11 + 10\pi + 11 + \frac{8}{3}\pi = 22 + \frac{38}{3}\pi$.

**Final Answer:** $\boxed{22 + \frac{38}{3}\pi}$

Question 4

Topic-1.3 – Coordinate Geometry

Show that the curve with equation $x^2 – 3xy – 40 = 0$ and the line with equation $3x + y + k = 0$ meet for all values of the constant k.

▶️Answer/Explanation

Solution :-

To show the curve $x^2 – 3xy – 40 = 0$ and line $3x + y + k = 0$ meet for all $k$:

– From the line: $y = -3x – k$.
– Substitute into the curve: $x^2 – 3x(-3x – k) – 40 = 0$,
– $x^2 + 9x^2 + 3kx – 40 = 10x^2 + 3kx – 40 = 0$.
– Quadratic discriminant: $\Delta = (3k)^2 – 4 \cdot 10 \cdot (-40) = 9k^2 + 160$.
– Since $9k^2 + 160 > 0$ for all $k$, there are always real roots for $x$.

Thus, the curve and line intersect for all $k$.

Question 5

(a) Topic-1.7 – Differentiation

(b) Topic-1.7 – Differentiation

The equation of a curve is such that $\frac{dy}{dx} = 4x – 3\sqrt{x} + 1$.

(a) Find the x-coordinate of the point on the curve at which the gradient is $\frac{11}{2}$.

(b) Given that the curve passes through the point (4, 11), find the equation of the curve.

▶️Answer/Explanation

Solution :-

Let’s tackle this problem step-by-step, starting with the given derivative $\frac{dy}{dx} = 4x – 3\sqrt{x} + 1$. We need to solve two parts: (a) find the $x$-coordinate where the gradient is $\frac{11}{2}$, and (b) find the curve’s equation given it passes through $(4, 11)$.

### Part (a): Find the $x$-coordinate where the gradient is $\frac{11}{2}$
The gradient is the derivative, so set $\frac{dy}{dx} = \frac{11}{2}$:
$4x – 3\sqrt{x} + 1 = \frac{11}{2}$.

Move all terms to one side:
$4x – 3\sqrt{x} + 1 – \frac{11}{2} = 0$.

Simplify:
$4x – 3\sqrt{x} + 1 – \frac{11}{2} = 4x – 3\sqrt{x} + \frac{2}{2} – \frac{11}{2} = 4x – 3\sqrt{x} – \frac{9}{2} = 0$.

Multiply through by 2 to clear the fraction:
$8x – 6\sqrt{x} – 9 = 0$.

Let $u = \sqrt{x}$ (so $x = u^2$, $u \geq 0$):
$8u^2 – 6u – 9 = 0$.

Solve this quadratic using the discriminant:
$a = 8$, $b = -6$, $c = -9$,
$\Delta = b^2 – 4ac = (-6)^2 – 4 \cdot 8 \cdot (-9) = 36 + 288 = 324$,
$u = \frac{-b \pm \sqrt{\Delta}}{2a} = \frac{6 \pm \sqrt{324}}{16} = \frac{6 \pm 18}{16}$.

$u = \frac{6 + 18}{16} = \frac{24}{16} = \frac{3}{2}$,
$u = \frac{6 – 18}{16} = \frac{-12}{16} = -\frac{3}{4}$ (discard, since $u \geq 0$).

So, $u = \frac{3}{2}$, thus $x = u^2 = \left(\frac{3}{2}\right)^2 = \frac{9}{4}$.

**Check:**
At $x = \frac{9}{4}$, $\sqrt{x} = \frac{3}{2}$,
$\frac{dy}{dx} = 4 \cdot \frac{9}{4} – 3 \cdot \frac{3}{2} + 1 = 9 – \frac{9}{2} + 1 = 9 – 4.5 + 1 = 5.5 = \frac{11}{2}$.

Correct!

**Answer for (a):** $x = \frac{9}{4}$.

—

### Part (b): Find the Equation of the Curve
Given $\frac{dy}{dx} = 4x – 3\sqrt{x} + 1$ and the curve passes through $(4, 11)$, integrate to find $y$:
$y = \int (4x – 3\sqrt{x} + 1) \, dx$,
$= \int 4x \, dx – \int 3\sqrt{x} \, dx + \int 1 \, dx$,
$= 4 \cdot \frac{x^2}{2} – 3 \cdot \frac{x^{3/2}}{3/2} + x + C$,
$= 2x^2 – 3 \cdot \frac{2}{3} x^{3/2} + x + C$,
$= 2x^2 – 2x^{3/2} + x + C$.

Use the point $(4, 11)$ to find $C$:
$y = 11$ when $x = 4$,
$11 = 2(4^2) – 2(4^{3/2}) + 4 + C$,
$4^2 = 16$, $4^{3/2} = (4^{1/2})^3 = 2^3 = 8$,
$11 = 2 \cdot 16 – 2 \cdot 8 + 4 + C = 32 – 16 + 4 + C = 20 + C$,
$C = 11 – 20 = -9$.

So, the equation is:
$y = 2x^2 – 2x^{3/2} + x – 9$.

**Check:**
Derivative: $\frac{d}{dx} (2x^2 – 2x^{3/2} + x – 9) = 4x – 2 \cdot \frac{3}{2} x^{1/2} + 1 = 4x – 3\sqrt{x} + 1$, matches.
At $x = 4$: $2 \cdot 16 – 2 \cdot 8 + 4 – 9 = 32 – 16 + 4 – 9 = 11$, matches.

**Final Answers:**
(a) $x = \frac{9}{4}$,
(b) $y = 2x^2 – 2x^{3/2} + x – 9$.

\[
\boxed{\text{(a) } \dfrac{9}{4}, \text{ (b) } y = 2x^{2} – 2x^{\frac{3}{2}} + x – 9}
\]

Question 6

(a) Topic-1.3 – Coordinate Geometry
(b) Topic-1.3 – Coordinate Geometry

Circles $C_{1}$ and $C_{2}$ have equations

$$x^{2}+y^{2}+6x-10y+18=0 \quad \text{and} \quad (x-9)^{2}+(y+4)^{2}-64=0$$

respectively.

(a)Find the distance between the centres of the circles.

P and Q are points on $C_{1}$ and $C_{2}$ respectively. The distance between P and Q is denoted by d.

(b)Find the greatest and least possible values of d.

▶️Answer/Explanation

Solution :-

Step 1: Find the centers and radii of the circles

We begin by rewriting each equation in standard form.

Circle $C_1$

$x^2 + y^2 + 6x – 10y + 18 = 0$

To find the center and radius, we complete the square.

Completing the square for
$x$
:
$x^2 + 6x = (x+3)^2 – 9$
Completing the square for
$y$
:
$y^2 – 10y = (y-5)^2 – 25$

Rewriting the equation:

$(x+3)^2 – 9 + (y-5)^2 – 25 + 18 = 0$ $(x+3)^2 + (y-5)^2 – 16 = 0$ $(x+3)^2 + (y-5)^2 = 16$

Thus,

C_1

has center

(-3,5)

and radius

r_1 = \sqrt{16} = 4

Circle $C_2$

$(x-9)^2 + (y+4)^2 – 64 = 0$ $(x-9)^2 + (y+4)^2 = 64$

Thus,

C_2

has center

(9,-4)

and radius

r_2 = \sqrt{64} = 8

Step 2: Distance Between the Centers

Using the distance formula:

$d = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2}$

where

(x_1, y_1) = (-3,5)

and

(x_2, y_2) = (9,-4)

$d = \sqrt{(9 – (-3))^2 + (-4 – 5)^2}$ $d = \sqrt{(9+3)^2 + (-4-5)^2}$ $d = \sqrt{12^2 + (-9)^2}$ $d = \sqrt{144 + 81} = \sqrt{225} = 15$

Thus, the distance between the centers is 15.

Step 3: Greatest and Least Values of $d$

The distance

d

between points

P

and

Q

varies between the sum and difference of the radii.

Greatest distance occurs when
$P$
and
$Q$
are on opposite sides along the line connecting the centers:
$d_{\max} = d + r_1 + r_2 = 15 + 4 + 8 = 27$
Least distance occurs when
$P$
and
$Q$
are on the same side along the line connecting the centers:
$d_{\min} = d – (r_1 + r_2) = 15 – (4 + 8) = 3$

Final Answers:

(a) Distance between centers:

15

(b) Greatest possible value of

d

27

Least possible value of

d

3

Question 7

(a) Topic-1.7 – Differentiation
(b) Topic-1.8 – Integration

The diagram shows part of the curve with equation $y = \frac{12}{\sqrt[3]{{2x + 1}}}$. The point A on the curve has coordinates $\left(\frac{7}{2}, 6\right)$.

(a) Find the equation of the tangent to the curve at A. Give your answer in the form $y = mx + c$.

(b) Find the area of the region bounded by the curve and the lines $x = 0$, $x = \frac{7}{2}$, and $y = 0$.

▶️Answer/Explanation

Solution :-

Let’s solve the problem step by step.

(a) Finding the Equation of the Tangent to the Curve at A

The equation of the curve is given by:

$y = \frac{12}{\sqrt[3]{2x + 1}}$

Step 1: Differentiate $y$ with respect to $x$

Rewrite the function using exponent notation:

$y = 12 (2x + 1)^{-\frac{1}{3}}$

Differentiate using the chain rule:

$\frac{dy}{dx} = 12 \times \left(-\frac{1}{3}\right) (2x+1)^{-\frac{4}{3}} \times 2$ $\frac{dy}{dx} = -\frac{24}{3} (2x+1)^{-\frac{4}{3}}$ $\frac{dy}{dx} = -8 (2x+1)^{-\frac{4}{3}}$

Step 2: Evaluate $\frac{dy}{dx}$ at $A \left(\frac{7}{2}, 6\right)$

Substituting $x = \frac{7}{2}$ :

$2x+1 = 2\left(\frac{7}{2}\right) + 1 = 8$ $\frac{dy}{dx} = -8 (8)^{-\frac{4}{3}}$

We calculate $8^{-\frac{4}{3}}$ :

$8^{\frac{1}{3}} = 2, \quad \Rightarrow 8^{-\frac{4}{3}} = \frac{1}{2^4} = \frac{1}{16}$ $\frac{dy}{dx} = -8 \times \frac{1}{16} = -\frac{8}{16} = -\frac{1}{2}$

Thus, the gradient of the tangent line is $-\frac{1}{2}$ .

Step 3: Find the Equation of the Tangent Line

Using the point-slope form:

$y – y_1 = m(x – x_1)$

Substituting $A \left(\frac{7}{2}, 6\right)$ and $m = -\frac{1}{2}$ :

$y – 6 = -\frac{1}{2} \left(x – \frac{7}{2} \right)$ $y – 6 = -\frac{1}{2}x + \frac{7}{4}$ $y = -\frac{1}{2}x + \frac{7}{4} + 6$

Since $6 = \frac{24}{4}$ ,

$y = -\frac{1}{2}x + \frac{7}{4} + \frac{24}{4}$ $y = -\frac{1}{2}x + \frac{31}{4}$

Thus, the equation of the tangent line is:

$\boxed{y = -\frac{1}{2}x + \frac{31}{4}}$

(b) Finding the Area Under the Curve

The required area is given by:

$\text{Area} = \int_{0}^{\frac{7}{2}} \frac{12}{\sqrt[3]{2x+1}} \, dx$

Step 1: Substituting $u = 2x + 1$

Let $u = 2x + 1$ , so $du/dx = 2$ or $dx = \frac{du}{2}$ .

When $x = 0$ , then $u = 1$ .
When $x = \frac{7}{2}$ , then $u = 8$ .

Rewriting the integral:

$\int_{1}^{8} \frac{12}{\sqrt[3]{u}} \cdot \frac{du}{2}$ $= \int_{1}^{8} \frac{12}{2} u^{-\frac{1}{3}} \, du$ $= \int_{1}^{8} 6u^{-\frac{1}{3}} \, du$

Step 2: Integrating

$\int u^{-\frac{1}{3}} \, du = \frac{u^{\frac{2}{3}}}{\frac{2}{3}} = \frac{3}{2} u^{\frac{2}{3}}$ $\int_{1}^{8} 6u^{-\frac{1}{3}} \, du = 6 \times \frac{3}{2} u^{\frac{2}{3}} \Big|_1^8$ $= 9 \left( u^{\frac{2}{3}} \Big|_1^8 \right)$ $= 9 \left( 8^{\frac{2}{3}} – 1^{\frac{2}{3}} \right)$

Since $8^{\frac{2}{3}} = (8^{\frac{1}{3}})^2 = 2^2 = 4$ ,

$= 9 \left( 4 – 1 \right)$ $= 9 \times 3 = 27$

Thus, the area is:

$\boxed{27}$

Question 8

(a) Topic-1.5 – Trigonometry
(b) Topic-1.5 – Trigonometry

(a) It is given that β is an angle between 90° and 180° such that sin β = a.

Express tan²β – 3 sin β cos β in terms of a.

(b) Solve the equation $\sin^2 \theta + 2 \cos^2 \theta = 4 \sin \theta + 3$ for $0^\circ < \theta < 360^\circ$.

▶️Answer/Explanation

Solution :-

(a) Express $\tan^2\beta – 3 \sin\beta \cos\beta$ in terms of $a$

We are given $\sin\beta = a$ and know that:

$\cos\beta = -\sqrt{1 – a^2} \quad \text{(since $ \beta $ is in the second quadrant)}$ $\tan^2\beta = \frac{a^2}{1 – a^2}$ $-3\sin\beta\cos\beta = -3a(-\sqrt{1 – a^2}) = 3a\sqrt{1 – a^2}$

Thus,

$\tan^2\beta – 3\sin\beta\cos\beta = \frac{a^2}{1 – a^2} – 3a\sqrt{1 – a^2}$ $\boxed{\frac{a^2}{1 – a^2} – 3a\sqrt{1 – a^2}}$

(b) Solve $\sin^2\theta + 2\cos^2\theta = 4\sin\theta + 3$ for $0^\circ < \theta < 360^\circ$

Step 1: Express in Terms of $\sin\theta$

Using $\cos^2\theta = 1 – \sin^2\theta$ :

$\sin^2\theta + 2(1 – \sin^2\theta) = 4\sin\theta + 3$ $\sin^2\theta + 2 – 2\sin^2\theta = 4\sin\theta + 3$ $-\sin^2\theta + 2 = 4\sin\theta + 3$ $-\sin^2\theta – 4\sin\theta – 1 = 0$

Multiply by -1:

$\sin^2\theta + 4\sin\theta + 1 = 0$

Step 2: Solve the Quadratic Equation

Using the quadratic formula:

$\sin\theta = \frac{-4 \pm \sqrt{16 – 4}}{2}$ $= \frac{-4 \pm \sqrt{12}}{2} = \frac{-4 \pm 2\sqrt{3}}{2}$ $= -2 \pm \sqrt{3}$

Approximating:

$\sin\theta = -2 + \sqrt{3} \approx -0.268$

(Only this root is valid, as $-2 – \sqrt{3}$ is out of range for sine.)

Step 3: Find $\theta$

$\theta_r = \sin^{-1}(0.268) \approx 15.6^\circ$

Since $\sin\theta$ is negative, solutions are in Quadrants III and IV:

$\theta = 180^\circ + 15.6^\circ = 195.6^\circ$ $\theta = 360^\circ – 15.6^\circ = 344.4^\circ$

Final Answer:

$\boxed{195.6^\circ, \quad 344.4^\circ}$

Question 9

(a) Topic-1.7 – Differentiation
(b) Topic-1.7 – Differentiation

The equation of a curve is $y = 4 + 5x + 6x^2 – 3x^3$.

(a) Find the set of values of x for which y decreases as x increases.

(b) It is given that $y = 9x + k$ is a tangent to the curve.

Find the value of the constant k.

▶️Answer/Explanation

Solution :-

(a) Find the set of values of

x

for which

y

decreases as

x

increases

We differentiate the given function:

$\frac{dy}{dx} = 5 + 12x – 9x^2$

Set

$\frac{d y}{d x} < 0to find decreasing intervals:$

$5 + 12x – 9x^2 < 0$

Rearrange:

$9x^2 – 12x – 5 > 0$

Solve

$9x^2 – 12x – 5 = 0$

using the quadratic formula:

$x = \frac{12 \pm \sqrt{144 + 180}}{18} = \frac{12 \pm 18}{18}$

$x = \frac{5}{3}, \quad x = -\frac{1}{3}$

Since the parabola opens upwards, the inequality holds for:

$x < -\frac{1}{3} \quad \text{or} \quad x > \frac{5}{3}$

Final Answer:

$\boxed{x < -\frac{1}{3} \quad \text{or} \quad x > \frac{5}{3}}$

(b) Find the value of

k

such that

y = 9x + k

is a tangent

Differentiate the curve and set it equal to 9:

$5 + 12x – 9x^2 = 9$

$9x^2 – 12x + 4 = 0$

Solve using the quadratic formula:

$x = \frac{12 \pm \sqrt{144 – 144}}{18} = \frac{12}{18} = \frac{2}{3}$

Substituting

$x = \frac{2}{3}$

into the curve equation:

$y = 4 + 5\left(\frac{2}{3}\right) + 6\left(\frac{4}{9}\right) – 3\left(\frac{8}{27}\right)$

$= 4 + \frac{10}{3} + \frac{24}{9} – \frac{24}{27}$

$= \frac{82}{9}$

Since

$y = 9x + k$

x = \frac{2}{3}

$\frac{82}{9} = 9\left(\frac{2}{3}\right) + k$

$\frac{82}{9} = 6 + k$

$k = \frac{82}{9} – 6 = \frac{28}{9}$

Final Answer:

\boxed{\frac{28}{9}}

Question 10

(a) Topic-1.6 – Series
(b) Topic-1.6 – Series

An arithmetic progression has first term 5 and common difference d, where $d > 0$. The second, fifth
and eleventh terms of the arithmetic progression, in that order, are the first three terms of a geometric
progression.

(a) Find the value of d.

(b) The sum of the first 77 terms of the arithmetic progression is denoted by S_{77}. The sum of the first
10 terms of the geometric progression is denoted by G_{10}.

Find the value of $S_{77} – G_{10}$.

▶️Answer/Explanation

Solution :-

### Part (a): Find $d$

The arithmetic progression (AP) has first term 5 and common difference $d > 0$. Its terms are:

– $a_2 = 5 + d$
– $a_5 = 5 + 4d$
– $a_{11} = 5 + 10d$

These are the first three terms of a geometric progression (GP), so the ratio between consecutive terms is equal:

\[ \frac{5 + 4d}{5 + d} = \frac{5 + 10d}{5 + 4d} \]

Cross-multiply:

\[ (5 + 4d)^2 = (5 + d)(5 + 10d) \]

Expand:

\[ 25 + 40d + 16d^2 = 25 + 55d + 10d^2 \]

Simplify by subtracting 25 and rearranging:

\[ 16d^2 + 40d = 10d^2 + 55d \]

\[ 6d^2 – 15d = 0 \]

\[ 3d (2d – 5) = 0 \]

Since $d > 0$, $d = \frac{5}{2}$.

**Verification**: $a_2 = \frac{15}{2}$, $a_5 = 15$, $a_{11} = 30$. Ratios: $\frac{15}{\frac{15}{2}} = 2$, $\frac{30}{15} = 2$. Confirmed.

\[ d = \frac{5}{2} \]

### Part (b): Find $S_{77} – G_{10}$

#### $S_{77}$: AP sum

AP sum: $S_n = \frac{n}{2} (a_1 + a_n)$

– $a_1 = 5$
– $a_{77} = 5 + 76 \cdot \frac{5}{2} = 5 + 190 = 195$
– $S_{77} = \frac{77}{2} (5 + 195) = \frac{77}{2} \cdot 200 = 77 \cdot 100 = 7700$

#### $G_{10}$: GP sum

GP: first term $a_2 = \frac{15}{2}$, ratio $r = 2$, $n = 10$.

\[ G_n = b_1 \frac{r^n – 1}{r – 1} \]

\[ G_{10} = \frac{15}{2} \cdot \frac{2^{10} – 1}{2 – 1} = \frac{15}{2} (1024 – 1) = \frac{15}{2} \cdot 1023 = \frac{15345}{2} \]

#### Difference:

\[ S_{77} – G_{10} = 7700 – \frac{15345}{2} = \frac{15400}{2} – \frac{15345}{2} = \frac{55}{2} \]

### Final Answers

(a) $d = \frac{5}{2}$

(b) $S_{77} – G_{10} = \frac{55}{2}$ (or 27.5)

Question 11

(a) Topic-1.2 – Functions
(b) Topic-1.2 – Functions
(c) Topic-1.2 – Functions
(d) Topic-1.2 – Functions

The function f is defined by $f(x) = 3 + 6x – 2x^2$ for $x \in \mathbb{R}$.

(a) Express $f(x)$ in the form $a – b(x – c)^2$, where $a$, $b$, and $c$ are constants, and state the range
of f.

(b) The graph of $y = f(x)$ is transformed to the graph of $y = h(x)$ by a reflection in one of the axes
followed by a translation. It is given that the graph of $y = h(x)$ has a minimum point at the origin.

Give details of the reflection and translation involved.

(c) Sketch the graph of y = g(x) and explain why g is a one-one function. You are not required to find
the coordinates of any intersections with the axes.

(d) Sketch the graph of $y=g^{-1}(x)$ on your diagram in (c), and find an expression for g⁻¹(x).
You should label the two graphs in your diagram appropriately and show any relevant mirror line.

▶️Answer/Explanation

Solution :-

Let’s simplify the solution for this problem step by step, focusing on clarity and brevity.

### (a) Express $f(x) = 3 + 6x – 2x^2$ in the form $a – b(x – c)^2$ and state the range

Rewrite $f(x)$ by completing the square. Start with the quadratic:

\[ f(x) = -2x^2 + 6x + 3 \]

Factor out $-2$ from the $x$ terms:

\[ f(x) = -2(x^2 – 3x) + 3 \]

Complete the square inside:

\[ x^2 – 3x = (x – \frac{3}{2})^2 – \frac{9}{4} \]

So:

\[ f(x) = -2\left[(x – \frac{3}{2})^2 – \frac{9}{4}\right] + 3 \]

\[ = -2(x – \frac{3}{2})^2 + 2 \cdot \frac{9}{4} + 3 \]

\[ = -2(x – \frac{3}{2})^2 + \frac{18}{4} + \frac{12}{4} \]

\[ = -2(x – \frac{3}{2})^2 + \frac{30}{4} \]

\[ = -2(x – \frac{3}{2})^2 + \frac{15}{2} \]

Thus, $a = \frac{15}{2}$, $b = 2$, $c = \frac{3}{2}$.

**Range**: Since $-2(x – \frac{3}{2})^2 \leq 0$ (negative because $b = 2 > 0$), the maximum value occurs when $x = \frac{3}{2}$:

\[ f\left(\frac{3}{2}\right) = \frac{15}{2} \]

As $x \to \pm \infty$, $f(x) \to -\infty$. So, the range is:

\[ f(x) \leq \frac{15}{2} \]

### (b) Reflection and translation to get $h(x)$ with minimum at the origin

$f(x) = \frac{15}{2} – 2(x – \frac{3}{2})^2$ is a downward parabola with vertex (maximum) at $(\frac{3}{2}, \frac{15}{2})$. We need $h(x)$ to have a minimum at $(0, 0)$, implying an upward parabola.

1. **Reflection**: Reflect $y = f(x)$ in the $x$-axis to flip it upward:

\[ y = -f(x) = -\left[3 + 6x – 2x^2\right] = 2x^2 – 6x – 3 \]

Vertex becomes $(\frac{3}{2}, -\frac{15}{2})$, now a minimum.

2. **Translation**: Shift the vertex from $(\frac{3}{2}, -\frac{15}{2})$ to $(0, 0)$:
– Move left by $\frac{3}{2}$: replace $x$ with $x + \frac{3}{2}$
– Move up by $\frac{15}{2}$: add $\frac{15}{2}$

\[ h(x) = 2\left(x + \frac{3}{2}\right)^2 – 6\left(x + \frac{3}{2}\right) – 3 + \frac{15}{2} \]

Simplify:

\[ = 2\left(x^2 + 3x + \frac{9}{4}\right) – 6x – 9 – 3 + \frac{15}{2} \]

\[ = 2x^2 + 6x + \frac{18}{4} – 6x – 12 + \frac{15}{2} = 2x^2 + \frac{9}{2} – 12 + \frac{15}{2} = 2x^2 + \frac{9 – 24 + 15}{2} = 2x^2 \]

So, $h(x) = 2x^2$, minimum at $(0, 0)$.

**Details**: Reflect in the $x$-axis, then translate left by $\frac{3}{2}$ and up by $\frac{15}{2}$.

### (c) Sketch $y = g(x)$ and explain why it’s one-to-one

The problem references $g(x)$, but it’s undefined. Assuming $g(x) = f(x)$ (common in such contexts unless $h(x)$ is intended), use $f(x) = 3 + 6x – 2x^2$:

– **Sketch**: Downward parabola, vertex at $(\frac{3}{2}, \frac{15}{2})$, $x$-intercepts via $2x^2 – 6x – 3 = 0$ (solve if needed, but not required).
– **One-to-one?**: $f(x)$ is not one-to-one (parabola fails horizontal line test). If $g(x) = h(x) = 2x^2$, it’s also not one-to-one. Assuming intent is a mislabel and we need a related one-to-one function, consider a restriction later. For now, assume $g(x) = f(x)$ and note it’s not one-to-one unless restricted.

### (d) Sketch $y = g^{-1}(x)$ and find its expression

Assuming $g(x) = f(x)$ needs inversion with restriction for one-to-one:

Restrict $f(x)$ to $x \geq \frac{3}{2}$ (right side, decreasing):

\[ y = 3 + 6x – 2x^2 \]

Solve for $x$:

\[ 2x^2 – 6x + (3 – y) = 0 \]

Discriminant: $\Delta = (-6)^2 – 4 \cdot 2 \cdot (3 – y) = 36 – 24 + 8y = 12 + 8y$

\[ x = \frac{6 \pm \sqrt{12 + 8y}}{4} \]

Take $+$ for $x \geq \frac{3}{2}$:

\[ x = \frac{6 + \sqrt{12 + 8y}}{4} = \frac{6 + 2\sqrt{3 + 2y}}{4} = \frac{3 + \sqrt{3 + 2y}}{2} \]

\[ g^{-1}(x) = \frac{3 + \sqrt{3 + 2x}}{2}, \quad x \leq \frac{15}{2} \]

**Sketch**: $y = g(x)$ (right branch), $y = g^{-1}(x)$ reflected over $y = x$.

### Final Answers

(a) $f(x) = \frac{15}{2} – 2\left(x – \frac{3}{2}\right)^2$, range $f(x) \leq \frac{15}{2}$

(b) Reflect in $x$-axis, translate left $\frac{3}{2}$, up $\frac{15}{2}$

(d) $g^{-1}(x) = \frac{3 + \sqrt{3 + 2x}}{2}$, sketch with $y = x$ mirror line

Question 6

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