# CIE A level -Pure Mathematics 1 : Topic : 1.1 Quadratics: completing the square for a quadratic polynomial : Exam Questions Paper 1

### 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.

(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, $$r\Theta +2r=24$$
Rearranging  this equation :$$\Theta =\frac{24-2r}{r}$$
Now,  A$$=\frac{1}{2}r^{2}\Theta$$
$$A=\frac{1}{2}r^{2}.\frac{24-2r}{r}$$
$$A=12r-r^{2}$$
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.
$$A=-(6-6)^{2}+36=36$$
Therefore, the greatest value of A is 36.
$$\Theta =\frac{24-2r}{r}=\frac{24-2.6}{6}=2$$
The corresponding angle of the sector is 2 radians.

### 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 $$(2x-2)^{2}-9$$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

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.

(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.

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