**Question**

**Question**

A piece of wire of length 24cm is bent to form the perimeter of a sector of a circle of radius r cm.

(i)Show that the area of the sector ,\(Acm^{2}\) is given by \(A=12r-r^{2}\)

(ii)Express A in the form \(a-\left ( r-b \right )^{2}\),where a and b are constants.

(iii)Given that r can vary,state the greatest value of A and find the corresponding angle of the sector.

**▶️Answer/Explanation**

(i) To find the area of the sector, we first need to find the angle subtended by the sector at the center of the circle.

The perimeter of the sector is equal to the length of the wire, which is 24 cm. The perimeter of the sector consists of the arc length and the two radii.

Let’s denote the angle subtended by the sector at the center of the circle as θ. The arc length can be calculated using the formula \(s=r\Theta\) , where s is the arc length and r is the radius.

The perimeter of the sector is given by s+2r=24.

Substituting the value of s, we have\( r\Theta +2r=24\)

We can rearrange this equation to express θ in terms of r:\(\Theta =\frac{24-2r}{r}\)

Now, to find the area of the sector, we use the formula A\(=\frac{1}{2}r^{2}\Theta\)

Substituting the value of \(\Theta \),we get \(A=\frac{1}{2}r^{2}.\frac{24-2r}{r}\)

Simplifying, we have \(A=12r-r^{2}\)

Hence, the area of the sector is given by \(A=12r-r^{2}\)

(ii) To express A in the form \(a-(r-b)^{2}\),we can expand\( A=12r-r^{2}\)

\(A=12r-r^{2}=-(r^{2}-12r)=-(r^{2}-2.6.r+6^{2}-6^{2})=-(r-6)^{2}+6^{2}\)

Therefore,\( A=-(r-6)^{2}+36\)

Hence, A can be expressed in the desired form as\( a-(r-b)^{2}\) where a = 36 and b = 6.

(iii) The greatest value of A occurs when the term \(-(r-6)^{2}\) is zero, which means r−6=0. So the greatest value of A is obtained when r = 6.

Substituting r = 6 into the expression for A, we have:

\(A=-(6-6)^{2}+36=36\)

Therefore, the greatest value of A is 36.

To find the corresponding angle of the sector, we substitute r=6 into the equation for \(\Theta :\)

\(\Theta =\frac{24-2r}{r}=\frac{24-2.6}{6}=2\)

The corresponding angle of the sector is 2 radians.

**Question**

**Question**

(i)Express in the form \(4x^{2}-12x\) in the form \(\left ( 2x+a \right )^{2}+b\)

(ii)Hence ,or otherwise,find the set of values of x satisfying \(4x^{2}-12x> 7\).

**▶️Answer/Explanation**

(i)To express \((2x-2)^{2}-9 \)in the form\( (2x+a)^{2}+b\) , we can expand and simplify:

\(\left ( 2x-2 \right )^{2}-9=\left ( 4x^{2}-8x+4 \right )-9=4x^{2}-8x-5\)

So,\(\left ( 2x-2 \right )^{2}-9 \)can be expressed as \(4x^{2}-8x-5 \)in the desired form

(ii)To solve the inequality 2x-3>4 , we can isolate x by adding 3 to both sides:

2x>4+3

2x>7

\(x>\frac{7}{2}\)

This means x must be greater than\( \frac{7}{2}\)

Now, let’s solve the inequality 2x-3<-4.. Adding 3 to both sides, we get:

2x<-4+3

2x<-1

\(x<-\frac{1}{2}\)

This means x must be less than\( -\frac{1}{2}\)

Combining these results, we have:

\(x>\frac{7}{2} or x<-\frac{1}{2}\)

Therefore, the values of x satisfying the inequality are\( x>\frac{7}{2}\) or \(x<-\frac{1}{2}\)

**Question**

**Question**

The function \( f:x\rightarrow x^{2}-4x+k\) is defined for the domain \(x\geq p\) ,where k and p are constants.**(i)Express \(f\left ( x \right )\) in the form ,\(\left ( x+a \right )^{2}+b+k\) where a and b are constants.**

(ii)State the range of f in terms of k.

(iii)State the smallest value of p for which f is one-one.

(iv)For the value of p found in part (iii),find an expression for \(f^{-1}\left ( x \right )\) and state the domain of \(f^{-1}\) giving your answer in terms of k.

**▶️Answer/Explanation**

(i) To express \(f(x)=x^{2}-4x+k\) in the form\( (x+a)^{2}+b+k \), we can complete the square.

\(f(x)=x^{2}-4x+k\)

\(=(x^{2}-4x+4)-4+k\)

\(=(x-2)^{2}-4+k\)

\(=(x-2)^{2}+(k-4)\)

Therefore, f(x) can be expressed as \((x-2)^{2}+(k-4)\) in the desired form, where a=−2 and b=k−4.

(ii) The range of f can be determined by analyzing the graph of \(f(x)=(x-2)^{2}+k-4.\)Since \((x-2)^{2}\) is always non-negative, the minimum value of f(x) occurs when \((x-2)^{2}=0 \)which happens when x=2.Therefore, the minimum value of f is\( f(2)=(2-2)^{2}+(k-4)=k-4.\)

Hence, the range of f in terms of k is \([k-4,\infty )\)

(iii) For f to be one-one (injective), each value of x in the domain must correspond to a unique value of f(x). In other words, different values of x should not produce the same output for f(x). Since\( f(x)=(x-2)^{2}+(k-4)\), we need to ensure that the quadratic term \((x-2)^{2}\)is always positive for\( x\geq p.\)

The smallest value of p occurs when\( (x-2)^{2}=0 \), which means x=2 . Therefore, the smallest value of

p for which f is one-one is p=2.

(iv) To find the inverse function\( f^{-1}(x),\) we need to solve the equation \(y=(x-2)^{2}+(k-4)\) for x in terms of y. Rearranging the equation:

\(y=(x-2)^{2}+(k-4)\)

\(y-(k-4)=(x-2)^{2}\)

\(\sqrt{y-(k-4)}=x-2\)

\(x=\sqrt{y-(k-4)}+2\)

Therefore, the expression for \(f^{-1}(x) \)is\( f^{-1}(x)=\sqrt{y-(k-4)}+2\)

The domain of \(f^{-1}\) corresponds to the range of f. From part (ii), we know that the range of f is \([k-4,+\infty )\). Thus, the domain of\( [k-4,+\infty )\), given in terms of k.

**Question**

**Question**

(i) Express \(x^{2}+6x+2\) in the form \(\left ( x+a \right )^{2}+b\),where a and b are constants.

(ii)Hence,or otherwise, find the set of values of x for which \(x^{2}+6x+2> 9\).

**▶️Answer/Explanation**

(i) To express the quadratic equation \(x^{2}+6x+2 \)in the form\( (x+a)^{2}+b , \)we need to complete the square.

To complete the square, we need to add the square of half the coefficient of the x term. In this case, half of 6 is 3, so we add \(3^{2}=9 \) to the equation

\(x^{2}+6x+2 = x^{2}+6x+9+2-9 \)

\(= (x^{2}+6x+9) + (2-9) \)

\(= (x+3)^{2} – 7.\)

Therefore,\(x^{2}+6x+2\) can be expressed in the form\( \left ( x+3 \right )^{2}-7 \), where a=3 and b=−7.

(ii) Now, we can use the expression \(\left ( x+3 \right )^{2}-7 \)to find the set of values of x for which \(x^{2}+6+2> 9\)

\((x+3)^{2}-7>9\)

Adding 7 to both sides, we get:

\((x+3)^{2}>16\)

Taking the square root of both sides, we have:

\(\left | x+3 \right |>4\)

This inequality represents two separate conditions:

x+3>4 and x+3<-4

Solving the first condition, we have:x>1

Solving the second condition, we have:x<-7

Therefore, the set of values of x for which \(x^{2}+6x+2>9\) is x<-7,x>1.

**Question **

**Question**

The position vectors of points A and B relative to an origin O are a and b respectively. The position vectors of points C and D relative to O are 3a and 2b respectively. It is given that

\(a=\begin{pmatrix}2\\ 1\\ 2\end{pmatrix}\) and \(b=\begin{pmatrix}4\\ 0\\ 6\end{pmatrix}\)(i) Find the unit vector in the direction of \(\vec{CD}\)

(ii) The point E is the mid-point of CD. Find angle EOD.

**▶️Answer/Explanation**

(i)To find the unit vector in the direction of \(\vec{CD}\),we need to calculate the vector \(\vec{CD}\) first.

The position vector of point C relative to O is 3A , and the position vector of point D relative to O is 2b. Therefore,\(\vec{CD}=2b-3a\)

Substituting the given values of a and b, we have:

\(\vec{CD}=2\begin{pmatrix}4\\ 0\\ 6\end{pmatrix}-3\begin{pmatrix}2\\ 1\\ 2\end{pmatrix}\)

Expanding the calculation, we get:

\(\vec{CD}=\begin{pmatrix}8\\ 0\\ 12\end{pmatrix}-\begin{pmatrix}6\\ 3\\ 6\end{pmatrix}\)

\(\vec{CD}=\begin{pmatrix}2\\ -3\\ 6\end{pmatrix}\)

To find the unit vector in the direction of \(\vec{CD}\),we divide \(\vec{CD} \)by its magnitude

\(\left \| \vec{CD} \right \|=\sqrt{2^{2}+(-3)^{2}+6^{2}}=\sqrt{49}=7\)

Therefore, the unit vector\( \vec{CD}\) in the direction of \(\vec{CD}\) is :

\(\hat{CD}=\frac{1}{7}\begin{pmatrix}2\\ -3\\ 6\end{pmatrix}\)

\(=\begin{pmatrix}\frac{2}{7}\\ -\frac{3}{7}\\ \frac{6}{7}\end{pmatrix}\)

(ii) The point E is the midpoint of CD. To find the position vector of point E, we can use the midpoint formula:

\(\vec{OE}=\frac{1}{2}\left ( \vec{OC}+\vec{OD} \right )\).The position vector of point C relative to O is

3a, and the position vector of point D relative to O is 2b. Therefore:

\(\vec{OC}=3a \) and \(\vec{OD}=2b\)

Substituting the given values of a and b, we have:

\(\vec{OC}=3\begin{pmatrix}2\\ 1\\ 2\end{pmatrix}=\begin{pmatrix}6\\ 3\\ 6\end{pmatrix}\)

\(\vec{OD}=2\begin{pmatrix}4\\ 0\\ 6\end{pmatrix}=\begin{pmatrix}8\\ 0\\ 12\end{pmatrix}\)

Using the midpoint formula:

\(\vec{OE}=\frac{1}{2}\left ( \vec{OC}+\vec{OD} \right )\)

\(=\frac{1}{2}\left ( \begin{pmatrix}6\\ 3\\6 \end{pmatrix} +\begin{pmatrix}8\\ 0\\ 12\end{pmatrix}\right )\)

\(=\frac{1}{2}\begin{pmatrix}14\\ 3\\ 18\end{pmatrix}=\begin{pmatrix}7\\ \frac{3}{2}\\ 9\end{pmatrix}\)

To find the angle EOD, we can use the dot product of vectors. The dot product between two vectors is given by:

\(\vec{a}.\vec{b}=\left | \vec{a} \right |\left | \vec{b} \right |\cos \Theta\)

where \(\vec{a} \) and \(\vec{b}\) are vectors, \(\left | \vec{a} \right | and \left | \vec{b} \right | \)are their magnitudes, and \(\Theta\) is the angle between them.

\(\vec{EO}=\begin{pmatrix}7\\ \frac{3}{2}\\ 9\end{pmatrix}\) (position vector of point E relative to O)

\(\vec{OD}=2b=2\begin{pmatrix}4\\ 0\\ 6\end{pmatrix}=\begin{pmatrix}8\\ 0\\ 12\end{pmatrix}\) (position vector of point D relative to O)

To find the angle between these vectors, we can use the dot product formula:

\(\vec{EO}.\vec{OD}=\left | \vec{EO} \right |.\left | \vec{OD} \right |\cos \Theta\)

Let’s calculate the dot product and magnitudes:

\(\vec{EO}.\vec{OD}=\begin{pmatrix}7\\ \frac{3}{2}\\ 9 \end{pmatrix}.\begin{pmatrix}8\\ 0\\ 12\end{pmatrix}\)

\(=(7)(8)+(\frac{3}{2})(0)+(9)(12)=56+0+108 =164\)

\(\left | \vec{EO}\right |=\sqrt{7^{2}+\left ( \frac{3}{2} \right )^{2}+9^{2}}=\sqrt{49+\frac{9}{4}+81}=\sqrt{\frac{289}{4}}=\frac{17}{2}\)

\(\left | \vec{OD} \right |=\sqrt{8^{2}+0^{2}+12^{2}}=\sqrt{64+0+144}=\sqrt{208}=4\sqrt{13}\)

Substituting the values into the formula:

\(\vec{EO}.\vec{OD}=\left | \vec{EO} \right |\left |\vec{OD} \right |\cos \Theta\)

\(164=\frac{17}{2}.4\sqrt{13}.\cos \Theta\)

\(\cos \Theta =\frac{164}{\frac{17}{2}.4\sqrt{13}}\)

Now, we can find the value of the angle \Theta using the inverse cosine function:

\(\Theta =\cos ^{-1}\left ( \frac{328}{17\sqrt{13}} \right )\)

\(\Theta \approx 8.6^{\circ}\)

Therefore, the angle EOD is approximately \(8.6^{\circ}.\)

**Question**

**Question**

(i) Express \( -x^{2}+6x-5\) in the form \(a(x+b)^{2}+c\), where a, b and c are constants.

The function \(f:x\rightarrow -x^{2}+6x-5\) defined for \(x\geq m\) ,where m is constant.

(ii) State the smallest value of m for which f is one-one.

(iii) For the case where m = 5, find an expression for \(f^{-1}(x)\) and state the domain of \(f^{-1}\).

**▶️Answer/Explanation**

(i) To express the quadratic function\( -x^{2}+6x-5\) in the form \(a(x+b)^{2}+c\) we need to complete the square.

First, let’s rewrite the quadratic function as follows:

\(-x^{2}+6x-5=-(x^{2}-6x+5)\)

Now, we want to complete the square inside the parentheses:

\(x^{2}-6x+5=(x^{2}-6x+9)-9+5=(x-3)^{2}-4\)

Substituting this back into the original expression:

\(-x^{2}+6x-5=-\left [ (x-3)^{2}-4 \right ]=-1\left ( x-3 \right )^{2}+4\)

So, the quadratic function can be expressed in the form \(a(x+b)^{2}+c\) as

\(-x^{2}+6x-5=-1(x-3)^{2}+4\)

Therefore, we have a=-1,b=3 and c=4.

(ii) For a quadratic function to be one-to-one (i.e., injective), it should pass the horizontal line test, which means each horizontal line intersects the graph at most once.

In this case, the quadratic function is f:x\rightarrow \(-x^{2}+6x-5\) for \(x\geq m\) , where m is a constant .

For the function to be one-to-one, the vertex of the parabola (which occurs at\( x=-\frac{b}{2a})\) should lie on or to the right of the line x=m.

Using the coefficients from the quadratic function, we have:

\(x=\frac{-6}{2(-1)}=3\)

Since the vertex is at x=3, we need m\leq 3 for the function to be one-to-one.

The smallest value of m that satisfies this condition is m=3.

(iii)\((x-3)^{2}=4-y\)

Correct order of operations

\(f^{-1}(x)=3+\sqrt{4-x}\)

Domain is \( x\leq 0\)

**Question**

**(i) Express \(x^{2}-2x-15\) in the form \((x+a)^{2}+b\).**

The function f is defined for p ≤ x ≤ q, where p and q are positive constants, by f : x → \(x^{2}\) − 2x − 15.

The range of f is given by c ≤ fx ≤ d, where c and d are constants.

(ii) State the smallest possible value of c.

For the case where c = 9 and d = 65,

(iii) find p and q,

(iv) find an expression for \(f^{-1}(x)\).

**▶️Answer/Explanation**

(i) To express \(x^{2}-2x-15\) in the form \((x+a)^{2}+b\), we complete the square. Let’s go through the steps:

\(x^{2}-2x-15 = (x^{2}-2x+1) – 16 = (x-1)^{2} – 16\)

Therefore, \(x^{2}-2x-15\) can be expressed as \((x-1)^{2} – 16\) in the desired form.

(ii) To determine the smallest possible value of \(c\), we need to find the minimum value of \(f(x)\) over its given domain. Since the coefficient of \(x^2\) is positive, the parabola opens upwards, and the minimum value occurs at the vertex.

The x-coordinate of the vertex can be found using the formula \(x = -\frac{b}{2a}\), where \(a\) is the coefficient of \(x^2\) and \(b\) is the coefficient of \(x\). In this case, \(a = 1\) and \(b = -2\). Therefore, the x-coordinate of the vertex is:

\(x = -\frac{-2}{2(1)} = 1\)

Substituting \(x = 1\) into the function \(f(x)\):

\(f(1) = 1^2 – 2(1) – 15 = 1 – 2 – 15 = -16\)

Thus, the minimum value of the function \(f\) is \(-16\). Therefore, the smallest possible value of \(c\) is \(-16\).

(iii) For the case where c=9 and d=65, we know that the range of is given by f \(c\leq f(x)\leq d\).Substituting the values, we have:

\(9\leq (x-1)^{2}-16\leq 65\)

\(25\leq (x-1)^{2}\leq 81\)

\(5\leq x-1\leq 9\)

\(6\leq x\leq 10\)

Therefore, the values of p and q are 6 and 10 respectively,

(iv) To find an expression for \(f^{-1}(x)\), we swap the roles of \(x\) and \(y\) and solve for \(y\) in terms of \(x\).

Starting from the original function \(f(x) = x^{2} – 2x – 15\):

\(x = y^{2} – 2y – 15\)

Rearranging the equation:

\(y^{2} – 2y = x + 15\)

Completing the square on the left side:

\(y^{2} – 2y + 1 = x + 15 + 1\)

\((y – 1)^{2} = x + 16\)

Taking the square root of both sides:

\(y – 1 = \pm \sqrt{x + 16}\)

Solving for \(y\):

\(y = 1 \pm \sqrt{x + 16}\)

So, the expression for \(f^{-1}=1 \pm \sqrt{x + 16}\)

**Question**

**Question**

A curve for which \( \frac{\mathrm{d} y}{\mathrm{d} x}=7-x^{2}-6x \) pases tghrough the p[oint (3,-10)

(i)Find the equation of the curve.

(ii)Express \(7-x^{2}-6x\) in the form \(a-(x+b)^{2}\),where a and b are constants.

(iii)Find the set of values of x for which the gradient of the curve is positive.

**▶️Answer/Explanation**

(i) To find the equation of the curve, we integrate the given expression for \(\frac{{dy}}{{dx}}\):

\(\frac{{dy}}{{dx}} = 7 – x^{2} – 6x\)

Integrating both sides with respect to \(x\):

\(\int \frac{{dy}}{{dx}} \,dx = \int (7 – x^{2} – 6x) \,dx\)

Integrating each term separately:

\(y = \int 7 \,dx – \int x^{2} \,dx – \int 6x \,dx\)

Simplifying the integrals:

\(y = 7x – \frac{{x^{3}}}{3} – 3x^{2} + C\)

where \(C\) is the constant of integration.

To determine the value of \(C\), we use the fact that the curve passes through the point (3, -10). Substituting \(x = 3\) and \(y = -10\) into the equation:

\(-10 = 7(3) – \frac{{3^{3}}}{3} – 3(3)^{2} + C\)

\(-10 = 21 – 9 – 27 + C\)

\(-10 = -15 + C\)

\(C = 5\)

Therefore, the equation of the curve is:

\(y = 7x – \frac{{x^{3}}}{3} – 3x^{2} + 5\)

(ii) To express \(7 – x^{2} – 6x\) in the form \(a – (x + b)^{2}\), we complete the square. Let’s go through the steps:

\[7 – x^{2} – 6x = -(x^{2} + 6x) + 7 = -(x^{2} + 6x + 9) + 7 + 9 = -(x + 3)^{2} + 16\]

Therefore, \(7 – x^{2} – 6x\) can be expressed as \(- (x + 3)^{2} + 16\) in the desired form.

(iii) To find the set of values of \(x\) for which the gradient of the curve is positive, we examine the expression \(7 – x^{2} – 6x\) and determine when it is greater than zero.

\(7 – x^{2} – 6x > 0\)

Rearranging the inequality:

\(-x^{2} – 6x + 7 > 0\)

To solve this inequality, we can find the roots of the quadratic expression \(f(x) = -x^{2} – 6x + 7\) by setting \(f(x)\) equal to zero:

\(-x^{2} – 6x + 7 = 0\)

Factoring the quadratic equation:

\(-(x – 1)(x + 7) = 0\)

Setting each factor equal to zero:

\(x – 1 = 0 \quad \text{or} \quad x + 7 = 0\)

Solving for \(x\):

\(x = 1 \quad \text{or} \quad x = -7\)

These are the values of \(x\) where the gradient of the curve is zero. To determine the set of values where the gradient is positive, we need to consider the intervals between these points.

Considering the intervals \(x < -7\), \(-7 < x < 1\), and \(x > 1\)

**Question **

The diagram shows part of the curve\( y = \frac{3}{\sqrt({1+4x})} \) and a point P (2, 1) lying on the curve. The normal to the curve at P intersects the x-axis at Q.

**(i) Show that the x-coordinate of Q is \(\frac{16}{9}\)**

(ii) Find, showing all necessary working, the area of the shaded region.

**Answer/Explanation**

**Question**

**Question**

The function f is defined by \(f(x)=-2x^{2}+12x-3\) for \(\varepsilon R\)

(i) Express \(-2x^{2}+12x-3 \)in the from \(2(x+a)^{a}+b,\)where a and b are constants.

(ii) State the greatest value of f(x).

The function g is defined by g(x) = 2x + 5 .

(iii) Find the values of x for which gf(x )+1=0.

**▶️Answer/Explanation**

(i) To express the function \(-2x^2 + 12x – 3\) in the form \(2(x + a)^2 + b\), we need to complete the square. Let’s go through the steps:

\(-2x^2 + 12x – 3 = -2(x^2 – 6x) – 3 = -2(x^2 – 6x + 9 – 9) – 3\)

Expanding and simplifying:

\(-2(x^2 – 6x + 9) + 18 – 3 = -2(x – 3)^2 + 15\)

Therefore, \(-2x^2 + 12x – 3\) can be expressed as \(2(x – 3)^2 + 15\) in the desired form, where \(a = 3\) and \(b = 15\).

(ii) To find the greatest value of \(f(x)\), we observe that the quadratic term \(-2x^2\) has a negative coefficient. This means the graph of \(f(x)\) opens downward and the vertex represents the maximum point. The x-coordinate of the vertex can be found using the formula \(-\frac{b}{2a}\) for a quadratic in the form \(ax^2 + bx + c\).

For \(f(x) = -2x^2 + 12x – 3\), the coefficient of the quadratic term is \(a = -2\) and the coefficient of the linear term is \(b = 12\). Using the formula, we can find:

\(x_{\text{vertex}} = -\frac{12}{2(-2)} = -\frac{12}{-4} = 3\)

Substituting this x-coordinate into the function, we find:

\(f(3) = -2(3)^2 + 12(3) – 3 = -18 + 36 – 3 = 15\)

Therefore, the greatest value of \(f(x)\) is 15.

(iii) To find the values of \(x\) for which \(g(f(x)) + 1\), we need to substitute \(f(x)\) into \(g(x)\) and solve for \(x\):

\(g(f(x)) + 1 = 2f(x) + 5 + 1 = 2(-2x^2 + 12x – 3) + 6 = -4x^2 + 24x – 1+1\)

Now, we set this expression equal to \(x\) and solve for \(x\):

\(-4x^2 + 24x = 0\)

\(4x^{2}-24x=0\)

4x(x-6)=0

x=6 and x=0

### Question

(i) Express \(x^{2}-4x+7\) in the form\((x+a)^{2}+b.\)

The function f is defined by \(f(x) = x^{2} − 4x + 7 \) for x < k, where k is a constant.

(ii) State the largest value of k for which f is a decreasing function.

The value of k is now given to be 1.

(iii) Find an expression for \(f^{-1}(x ) \)and state the domain of \( f^{-1}(x) \)

(iv) The function g is defined by \(g(x )= \frac{2}{x-1} \)for x > 1. Find an expression for gof(x) and state the range of gof(x).

**▶️Answer/Explanation**

(i) To express the quadratic function \(x^2 – 4x + 7\) in the form \((x + a)^2 + b\), we need to complete the square. Let’s go through the steps:

\(x^2 – 4x + 7 = (x^2 – 4x + 4) + 7 – 4 = (x – 2)^2 + 3\)

Therefore, \(x^2 – 4x + 7\) can be expressed as \((x – 2)^2 + 3\) in the desired form, where \(a = -2\) and \(b = 3\).

(ii) To determine the largest value of \(k\) for which \(f(x)\) is a decreasing function, we can analyze the coefficient of the linear term (-4x) in the quadratic function. Since the coefficient is negative, the quadratic function is decreasing when \(x\) is less than the x-coordinate of the vertex.

The x-coordinate of the vertex can be found using the formula \(-\frac{b}{2a}\) for a quadratic in the form \(ax^2 + bx + c\). For \(f(x) = x^2 – 4x + 7\), the coefficient of the linear term is \(b = -4\) and the coefficient of the quadratic term is \(a = 1\).

\(x_{\text{vertex}} = -\frac{-4}{2(1)} = -\frac{-4}{2} = 2\)

Therefore, the largest value of \(k\) for which \(f(x)\) is a decreasing function is \(k = 2\).

(iii) To find an expression for \(f^{-1}(x)\), we need to interchange the roles of \(x\) and \(f(x)\) and solve for \(x\). Let \(y = f(x)\):

\(y = x^2 – 4x + 7\)

Swap \(x\) and \(y\):

\(x = y^2 – 4y + 7\)

Rearrange the equation:

\(y^2 – 4y + 7 – x = 0\)

Now, solve for \(y\) using the quadratic formula:

\(y = \frac{-(-4) \pm \sqrt{(-4)^2 – 4(1)(7 – x)}}{2(1)}\)

Simplifying:

\(y = \frac{4 \pm \sqrt{16 – 28 + 4x}}{2}\)

\(y = \frac{4 \pm \sqrt{-12 + 4x}}{2}\)

\(y = 2 \pm \sqrt{x – 3}\)

Therefore, the expression for \(f^{-1}(x)\) is \(f^{-1}(x) = 2 \pm \sqrt{x – 3}\). The domain of \(f^{-1}(x)\) is the set of all real numbers \(x\) such that \(x – 3 \geq 0\), which simplifies to \(x \geq 3\).

(iv) To find an expression for \(g \circ f(x)\), we substitute \(f(x)\) into \(g(x)\):

\(g \circ f(x) = g(f(x)) = \frac{2}{f(x) – 1}\)

Using the expression for \(f(x) = x^2 – 4x + 7\), we substitute it into the above equation:

\(g \circ f(x) = \frac{2}{(x^2 – 4x + 7) – 1} = \frac{2}{x^2 – 4x + 6}\)

Therefore, the expression for \(g \circ f(x)\) is \(\frac{2}{x^2 – 4x + 6}\).

To find the range, we need to consider the values that the expression \(x^{2}−4x+6 \) can take.

Range of gf(x) is \(\left ( 0.\frac{2}{3} \right )\)

**Question**

**Question**

By using a suitable substitute, solve the equation

\((2x-3)^2-\frac{4}{(2x-3)^2}-3=0\).

**▶️Answer/Explanation**

We have the equation \((2x-3)^2 – \frac{4}{(2x-3)^2} – 3 = 0\).

Let’s make a substitution to simplify the equation. Let’s substitute \(u = (2x-3)\). This means \(u\) represents \(2x-3\).

Using this substitution, the equation becomes:

\(u^2 – \frac{4}{u^2} – 3 = 0\).

Multiplying through by \(u^2\) to eliminate the fraction, we get:

\(u^4 – 3u^2 – 4 = 0\).

Now let’s introduce another variable. Let’s substitute \(v = u^2\). This means \(v\) represents \(u^2\).

Using this substitution, the equation becomes:

\(v^2 – 3v – 4 = 0\).

We can now solve this quadratic equation for \(v\).

Using the quadratic formula, \(v = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}\), with \(a = 1\), \(b = -3\), and \(c = -4\), we get:

\(v = \frac{3 \pm \sqrt{(-3)^2 – 4(1)(-4)}}{2(1)}\).

Simplifying further, we have:

\(v = \frac{3 \pm \sqrt{9 + 16}}{2}\).

\(v = \frac{3 \pm \sqrt{25}}{2}\).

\(v = \frac{3 \pm 5}{2}\).

This gives us two possible values for \(v\):

1. \(v = \frac{3 + 5}{2} = 4\).

2. \(v = \frac{3 – 5}{2} = -1\).

Now, let’s substitute back to \(u = \sqrt{v}\) to solve for \(u\):

For \(v = 4\), we have \(u = \sqrt{4} = 2\) or \(u = -\sqrt{4} = -2\).

For \(v = -1\), taking the square root of a negative number would result in imaginary solutions, which we can disregard since we are looking for real solutions.

Finally, substituting back to \(x = \frac{u + 3}{2}\), we get:

For \(u = 2\), \(x = \frac{2 + 3}{2} = \frac{5}{2}\).

For \(u = -2\), \(x = \frac{-2 + 3}{2} = \frac{1}{2}\).

Therefore, the solutions to the equation are \(x = \frac{5}{2}\) and \(x = \frac{1}{2}\).

**Question**

**Question**

(a) Express −3×2 + 12x + 2 in the form −3 (x-a)2 + b, where a and b are constants.

The one-one function f is defined by f : x → −\(3x^2 + 12x + 2\) for x ≤ k.

(b) State the largest possible value of the constant k.

It is now given that k = −1.

(c) State the range of f.

(d) Find an expression for f −1 (x).

The result of translating the graph of y = f (x) by \(\binom{-3}{1}\) is the graph of y = g (x).

(e) Express g(x) in the form \(px^2+ qx + r\), where p, q and r are constants.

**▶️Answer/Explanation**

(a) To express \(-3x^2 + 12x + 2\) in the form \(-3(x-a)^2 + b\), we need to complete the square. Let’s go through the steps:

\(-3x^2 + 12x + 2\)

First, factor out -3 from the quadratic terms:

\(-3(x^2 – 4x) + 2\)

Next, take half of the coefficient of \(x\) (-4) and square it to get 4:

\(-3(x^2 – 4x + 4 – 4) + 2\)

Now, rewrite the expression inside the parentheses as a perfect square:

\(-3((x – 2)^2 – 4) + 2\)

Distribute -3:

\(-3(x – 2)^2 + 12 + 2\)

Simplify:

\(-3(x – 2)^2 + 14\)

So, the expression \(-3x^2 + 12x + 2\) can be expressed in the form \(-3(x-a)^2 + b\) as \(-3(x – 2)^2 + 14\), where \(a = 2\) and \(b = 14\).

(b) To determine the largest possible value of the constant \(k\), we need to find the vertex of the parabola represented by the quadratic equation \(-3x^2 + 12x + 2\). The x-coordinate of the vertex can be found using the formula \(x = -\frac{b}{2a}\), where \(a\) and \(b\) are the coefficients of the quadratic equation. In this case, \(a = -3\) and \(b = 12\). Plugging these values into the formula, we have:

\(x = -\frac{12}{2(-3)} = 2\)

Therefore, the largest possible value of \(k\) is 2.

(c) To determine the range of \(f\), we need to analyze the graph of the function. The coefficient of \(x^2\) is negative (-3), indicating that the parabola opens downwards. Therefore, the range of \(f\) will be all real numbers less than or equal to the maximum value of the parabola. The maximum value occurs at the vertex, which can be found using the formula \(x = -\frac{b}{2a}\). Plugging in the values, we have:

\(x = -\frac{12}{2(-3)} = 2\)

Substituting \(x = 2\) back into the equation for \(f\), we get:

\(f(2) = -3(2 – 2)^2 + 14 = 14\)

Therefore, the range of \(f\) is \(\{y \in \mathbb{R} : y \leq 14\}\).

(d) To find an expression for \(f^{-1}(x)\), we need to solve the equation \(y = -3(x – 2)^2 + 14\) for \(x\). Let’s rearrange the equation:

\(y – 14 = -3(x – 2)^2\)

Divide both sides by -3:

\(-\frac{y – 14}{3} = (x – 2)^2\)

Take the square root of both sides:

\(\sqrt{-\frac{y – 14}{3}} = x – 2\)

Add 2 to both sides:

\(\sqrt{-\frac{y – 14}{3}} + 2 = x\)

So, the expression for \(f^{-1}(x)\) is:

\(f^{-1}(x) = \sqrt{-\frac{x – 14}{3}} + 2\)

The domain of \(f^{-1}\) will be the range of \(f\), which is \(\{x \in \mathbb{R} : x \leq 14\}\).

(e) To express the translated function \(g(x)\) in the form \(px^2 + qx + r\), where \(p\), \(q\), and \(r\) are constants, we need to apply the translation \(\binom{-3}{1}\) to the function \(f(x) = -3x^2 + 12x + 2\).

The translation \(\binom{-3}{1}\) corresponds to shifting the function \(f(x)\) three units to the left and one unit upward. To achieve this, we replace \(x\) with \(x + 3\) and \(f(x)\) with \(g(x)\).

\(g(x) = -3(x + 3)^2 + 12(x + 3) + 2\)

Simplifying the equation, we have:

\(g(x) = -3(x^2 + 6x + 9) + 12x + 36 + 2\)

\(g(x) = -3x^2 – 18x – 27 + 12x + 38\)

\(g(x) = -3x^2 – 6x + 11\)

So, the expression for \(g(x)\) in the form \(px^2 + qx + r\) is \(-3x^2 – 6x + 11\), where \(p = -3\), \(q = -6\), and \(r = 11\).

*Question*

(a)Express \(2x^{2} − 8x + 14 \)in the form \(2[(x – a)^2 + b] \)

The functions f and g are defined by

\(f(x) = x^{2} \) for x ∈ R >,

\(g(x) = 2x^{2} − 8x + 14 \) for x ∈ R>.

(b) Describe fully a sequence of transformations that maps the graph of y = f(x) onto the graph of y= g(x), making clear the order in which the transformations are applied.

**▶️Answer/Explanation**

(a) To express \(2x^2 – 8x + 14\) in the form \(2[(x – a)^2 + b]\), we need to complete the square. Let’s begin:

\(2x^2 – 8x + 14 = 2(x^2 – 4x) + 14 = 2(x^2 – 4x + 4 – 4) + 14 = 2((x – 2)^2 – 4] + 14\)

Expanding further:

\(2((x – 2)^2 – 4)+ 14 = 2(x – 2)^2 – 8 + 14 = 2(x – 2)^2 + 6\)

So, the expression \(2x^2 – 8x + 14\) can be written in the form \(2((x – 2)^2 + 3)\), where \(a = 2\) and \(b = 3\).

(b) To map the graph of \(y = f(x)\) onto the graph of \(y = g(x)\), we need to apply a sequence of transformations. Here is a description of the transformations in the order they are applied:

1. Reflection: Reflect the graph of \(y = f(x)\) across the x-axis to obtain the graph of \(y = -f(x)\).

2. Vertical Stretch: Stretch the graph vertically by a factor of 2 to obtain the graph of \(y = -2f(x)\).

3. Vertical Translation: Translate the graph 6 units upward to obtain the graph of \(y = -2f(x) + 6\).

4. Horizontal Translation: Translate the graph 2 units to the right to obtain the graph of \(y = -2f(x – 2) + 6\).

Each transformation is applied in the given order, resulting in the desired mapping from the graph of \(y = f(x)\) to the graph of \(y = g(x)\).