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 $$4x^{2}-12x> 7$$

(i)To express $$4x^{2}-12x$$ in the form $$\left(2x+a\right)^{2}+b$$, we need to complete the square.
We start with $$4x^{2}-12x$$. First, we factor out the common factor of 4:
$$4x^{2}-12x = 4(x^{2}-3x)$$
Next, we take half of the coefficient of the $$x$$ term, which is $$-3/2$$, square it, and add it inside the parentheses:
$$4\left [ \left ( x-\frac{3}{2} \right )^{2}-\left ( \frac{3}{2} \right )^{2} \right ]$$
$$4\left [ \left ( x-\frac{3}{2} \right )^{2}-\frac{9}{4} \right ]$$
$$4 \left ( x-\frac{3}{2} \right )^{2}-4\frac{9}{4}$$
$$\left [ 2\left ( x-\frac{3}{2} \right ) \right ]^{2}-9$$
$$\left [ 2\left ( x-\frac{3}{2} \right ) \right ]^{2}-9$$
$$\left ( 2x-3 \right )^{2}-9$$
So, $$4x^{2}-12x$$ can be expressed in the form $$\left ( 2x-3 \right )^{2}-9$$
(ii) To solve the inequality $$4x^{2}-12x > 7$$, we can first rewrite it as $$4x^{2}-12x-7 > 0$$.
$$x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$$
In this case, $$a = 4$$, $$b = -12$$, and $$c = -7$$.
$$x = \frac{-(-12) \pm \sqrt{(-12)^{2}-4(4)(-7)}}{2(4)}$$
$$x = \frac{12 \pm \sqrt{144+112}}{8}$$
$$x = \frac{12 \pm \sqrt{256}}{8}$$
$$x = \frac{12 \pm 16}{8}$$
$$x_1 = \frac{12+16}{8} = \frac{28}{8} = \frac{7}{2}$$
$$x_2 = \frac{12-16}{8} = \frac{-4}{8} = -\frac{1}{2}$$
Therefore, the set of values of $$x$$ satisfying $$4x^{2}-12x > 7$$ is $$x > \frac{7}{2}$$ or $$x < -\frac{1}{2}$$.

### 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 $$f^{-1}$$.

(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 $$x^{2}+6x+2$$ 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 $$x^{2}+6x+2> 9$$.

(i) To express $$x^{2}+6x+2$$ in the form $$(x+a)^{2}+b$$, we complete the square.
We start with $$x^{2}+6x+2$$. To determine the value of $$a$$, we take half of the coefficient of the $$x$$ term, which is 6, and square it:
$$\left(\frac{6}{2}\right)^{2} = 9$$
$$x^{2}+6x+9+2-9 = (x+3)^{2}+(-7)$$
Therefore, $$x^{2}+6x+2$$ can be expressed as $$(x+3)^{2}-7$$.
(ii) We have the inequality $$x^{2}+6x+2>9$$. Let’s use the expression we derived in part (i) to simplify the inequality:
$$(x+3)^{2}-7>9$$
$$(x+3)^{2}>16$$
Taking the square root of both sides (considering both positive and negative roots):
$$x+3>4$$ or $$x+3<-4$$
For $$x+3>4$$: $$x>1$$
For $$x+3<-4$$: $$x<-7$$
Therefore, the set of values of $$x$$ that satisfies $$x^{2}+6x+2>9$$ is $$x<-7$$ or $$x>1$$.

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 $$f^{-1}(x)$$.

(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

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.

(i) To find the equation of the curve, we need to integrate the given derivative with respect to $$x$$. Integrating $$\frac{\mathrm{d} y}{\mathrm{d} x}=7-x^{2}-6x$$ will give us the original function $$y$$.
Integrating $$7-x^{2}-6x$$ with respect to $$x$$, we get:
$$y = \int (7-x^{2}-6x) \, dx = 7x – \frac{x^{3}}{3} – 3x^{2} + C$$
where $$C$$ is the constant of integration.
(ii) To express $$7-x^{2}-6x$$ in the form $$a-(x+b)^{2}$$, we complete the square. Let’s start by rearranging the terms:
$$7-x^{2}-6x = -x^{2}-6x+7$$
Now, we want to find the value of $$b$$ such that $$-x^{2}-6x+7$$ can be written as $$-(x+b)^{2}$$. We take half of the coefficient of the $$x$$ term, which is $$-6$$, and square it:
$$\left(\frac{-6}{2}\right)^{2} = 9$$
$$-x^{2}-6x+7 = -(x+3)^{2} + 9 + 7 = – (x+3)^{2} + 16$$
Therefore, $$7-x^{2}-6x$$ can be expressed as $$- (x+3)^{2} + 16$$.
(iii) The gradient of the curve is given by $$\frac{\mathrm{d} y}{\mathrm{d} x}$$, which is $$7-x^{2}-6x$$. To find the set of values of $$x$$ for which the gradient is positive, we need to solve the inequality $$7-x^{2}-6x > 0$$.
$$16-\left ( x+3 \right )^{2}>0$$
$$16-\left ( x^{2}+9+6x \right )>0$$
$$16-x^{2}-9-6x>0$$
$$-x^{2}-6x+7>0$$
$$x^{2}+6x-7<0$$
$$(x-1)(x+7)<0$$
$$-7<x<1$$
Therefore, the set of values of $$x$$ for which the gradient of the curve is positive is $$-7<x<1.$$

### Question

The graph of y = f(x) is transformed to the graph of $$y = 1 + f(\frac{1}{2}x)$$.
Describe fully the two single transformations which have been combined to give the resulting transformation.

The given transformation $$y = 1 + f\left(\frac{1}{2}x\right)$$ involves two single transformations:
Horizontal compression by a factor of 2: This transformation is represented by the expression $$\frac{1}{2}x$$. It indicates that the x-values of the original function $$f(x)$$ have been halved, resulting in a horizontal compression. This means that the graph is narrower compared to the original graph. Points on the graph are shifted closer to the y-axis.
Vertical shift upward by 1 unit: This transformation is represented by the constant term $$1$$ added to the function $$f\left(\frac{1}{2}x\right)$$. It indicates that the entire graph has been shifted vertically by 1 unit in the positive y-direction. The effect of this transformation is that all the points on the graph are moved upward by 1 unit.
In summary, the original graph of $$y = f(x)$$ is first horizontally compressed by a factor of 2, making it narrower, and then shifted vertically upward by 1 uni

### 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.

(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 $$x=0$$

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