CIE A level -Pure Mathematics 1 : Topic : 1.2 Function: function, domain, range : Exam Questions Paper 1

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 $$f(x)$$ in the form $$(x + a)^2 + b + k$$, we expand 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 $$(x + a)^2 + b + k$$ with $$a = -2$$ and $$b = k – 4$$.
(ii) The range of $$f$$ represents all possible values of $$f(x)$$ as $$x$$ varies within its domain. Since $$f(x) = (x – 2)^2 + (k – 4)$$, we can observe that the square term $$(x – 2)^2$$ is always non-negative.
Thus, the smallest value $$f(x)$$ can attain is when $$(x – 2)^2 = 0$$, which occurs when $$x = 2$$. In this case, $$f(x) = (2 – 2)^2 + (k – 4) = k – 4$$. Hence, the smallest value in the range is $$k – 4$$.
Therefore, the range of $$f$$ is given by $$f(x) \geq k – 4$$ for all $$x$$ in the domain $$x \geq p$$.
(iii) For $$f$$ to be one-to-one (injective), each distinct value in the domain must correspond to a distinct value in the range. This means that no two different $$x$$ values can produce the same $$f(x)$$ value.
To ensure this, we need to set the smallest value of the range $$k – 4$$ to be greater than any other value in the range. In other words, $$k – 4$$ must be the minimum value of $$f(x)$$ for all $$x$$ in the domain $$x \geq p$$.
Therefore, the smallest value of $$p$$ for which $$f$$ is one-to-one is $$p = 2$$.
(iv) To find the inverse function $$f^{-1}(x)$$, we swap the roles of $$x$$ and $$f(x)$$ in the equation $$f(x) = (x – 2)^2 + (k – 4)$$ and solve for $$x$$:
$$x = (f^{-1}(x) – 2)^2 + (k – 4)$$
Let’s solve for $$f^{-1}(x)$$:
$$x – (k – 4) = (f^{-1}(x) – 2)^2$$
$$\sqrt{x – (k – 4)} = f^{-1}(x) – 2$$
$$f^{-1}(x) = \sqrt{x – (k – 4)} + 2$$
The domain of $$f^{-1}(x)$$ is determined by the range of $$f(x)$$. From part (ii), we know that $$f(x) \geq k – 4$$. Therefore, for $$f^{-1}(x)$$ to be defined, we need $$x – (k – 4) \geq k – 4$$, which simplifies to $$x \geq 2k – 4$$.
Thus, the domain of $$f^{-1}(x)$$ is $$x \geq 2k – 4$$.

Question

The functions f and g are defined by

$$f(x)=\frac{4}{x}-2$$ for$$x> 0$$

$$g(x)=\frac{4}{5x+2}$$ for $$x\geq 0$$

(i)Find and simplify an expression for fg(x) and state the range of fg.

(ii)Find an expression for $$g^{-1}(x)$$ and find the domain of $$g^{-1}$$.

(i) To find the expression for $$f \circ g$$ or $$fg(x)$$, we substitute the definition of $$g(x)$$ into $$f(x)$$:
$$fg(x) = f(g(x)) = f\left(\frac{4}{5x+2}\right) = \frac{4}{\frac{4}{5x+2}} – 2 = 5x.$$
Range of f(x) is greater than zero.
So the simplified expression for $$fg(x)$$ is $$5x$$. The range of $$fg$$ is all real numbers since there are no restrictions on $$x$$ in the given expressions.
(ii) To find the expression for $$g^{-1}(x)$$ and its domain, we follow these steps:
Replace $$g(x)$$ with $$y$$:
$$y=\frac{4}{5x+2}.$$
$$5x+2 = \frac{4}{y}.$$
$$5x = \frac{4}{y} – 2.$$
$$x = \frac{4-2y}{5y}.$$
Therefore, the expression for $$g^{-1}(x)$$ is:
$$g^{-1}(x) = \frac{4-2x}{5x}.$$
The domain of $$g^{-1}$$ consists of 5x>0 \cup 4-2x\neq \geq 0 \Rightarrow x\leq 2.
Theefore,domain of $$g^{-1}$$ is 0<x\leq 2.

Question

$$f:x\rightarrow \frac{1}{x^{2}-9}$$

$$g:x\rightarrow 2x-3$$

(i)Find and simplify an expression for g(g(x)).

(ii)Find an expression for  $$f^{-1}\left ( x \right )$$ and state the domain of $$f^{-1}$$

(iii)Solve the equation $$fg(x)=\frac{1}{7}$$

(i) Expression for $$g(g(x))$$:
$$g(x) = 2x – 3$$
$$gg(x) = g(2x – 3) = 2(2x – 3) – 3 = 4x – 9.$$
(ii) Expression for $$f^{-1}(x)$$ and its domain:
To find the inverse of a function $$f(x)$$, we need to swap the roles of $$x$$ and $$y$$ and solve for $$x$$. So, for $$f(x) = \frac{1}{x^2 – 9}$$:
Swap $$x$$ and $$y$$: $$x = \frac{1}{y^2 – 9}$$
Solve for $$y$$: $$y^2 – 9 = \frac{1}{x}$$
Take the square root of both sides: $$y = \pm \sqrt{\frac{1}{x} + 9}$$
The domain of $$f^{-1}(x)$$ is determined by the original function $$f(x)$$.
In this case, the denominator of $$f(x)$$ cannot be zero, so $$x^2 – 9 \neq 0$$, which means $$x \neq \pm 3$$. Additionally, since we’re taking the square root of $$\frac{1}{x} + 9$$, we need $$\frac{1}{x} + 9 \geq 0$$, which implies $$x > 0$$. Therefore, the domain of $$f^{-1}(x)$$ is $$x > 0$$ and $$x \neq 3$$.
(iii)EITHER:
$$\frac{1}{\left ( 2x-3 \right )^{2}-9}=\frac{1}{7}$$
$$\left ( 2x-3 \right )^{2}=16$$ or $$4x^{2}-12x-7=0$$
$$x=\frac{7}{2}$$ or $$-\frac{1}{2}$$\
$$x=\frac{7}{2}$$

OR:

$$g(x)=f^{-1}\left ( \frac{1}{7} \right )$$
$$g(x)=4$$
$$2x-3=4$$
$$x=\frac{7}{2}$$

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