Question
The graph of \( y=f(\left| x\right|)\) for \(-6\leq x\leq 6\) is shown in the following diagram.
(a) On the following axes, sketch the graph of \(y=\left| f(\left| x\right|)\right|\) for \(-6\leq x\leq 6\).
It is given that f is an odd function.
(b) On the following axes, sketch the graph of \(y=f(x)\) for \(-6\leq x\leq 6\).
It is also given that \(\int_{0}^{4}f(\left| x\right|)dx=1.6\).
(c) Write down the value of
(i) \(\int_{-4}^{0}f(x)dx\);
(ii) \(\int_{-4}^{4}(f(\left| x\right|)+f(x))dx\).
▶️Answer/Explanation
Detailed Solution
(a) To find the graph for \(y=\left| f(\left| x\right|)\right|\) , the part below the -axis is taken as mirror image above the X-axis (reflection of all negative sections in x-axis), therefore the graph is as follows
(a) 
(b) To find the graph for \(y=f(x)\) for \(-6\leq x\leq 6\) , since the mod is removed from x and it is given that y=f(x) is an odd function, therefore the graph will be same for x≥0 whereas for x<0 it will be symmetric about the origin that is just same behavior as the graph of y = \(x^{3}\) as shown below
(b) 
(c) (i) From the graph in part (b), we can conclude that the area \(\int_{-4}^{0}f(x)dx\) = \(-\int_{0}^{4}f(\left| x\right|)dx\) = \(-1.6\)
(ii) Since f(x) is an odd function therefore, \(\int_{-4}^{4} f(x))dx\) = 0, and \(\int_{-4}^{4}(f(\left| x\right|) = 2 \int_{0}^{4}(f(\left| x\right|) \)=\(2\times (-1.6)\) = \(3.2\)
Question
Consider the function f(x) = \(2^{x} – \frac{1}{2^{x}},\) x ∈ R.
(a) Show that f is an odd function.
The function g is given by g(x) = \(\frac{x-1}{x^{2}-2x-3},\) where x ∈ R, x ≠ -1 , x ≠ 3.
(b) Solve the inequality f (x) ≥ g(x).
▶️Answer/Explanation
Ans:
(a) attempt to replace x with −x
f(-x) = \(2^{-x}-\frac{1}{2^{-x}}\)
EITHER
\(=\frac{1}{2^{x}}-2^{x} = -f(x)\)
OR
\(=-\left (2^{x} \frac{1}{2^{x}} \right )\left ( =-f(x) \right )\)
Note: Award M1A0 for a graphical approach including evidence that either the graph is invariant after rotation by 180°about the origin or the graph is invariant after a reflection in the
y -axis and then in the x -axis (or vice versa).
so f is an odd function
(b) attempt to find at least one intersection point
x = -1.26686….., x = 0.177935…..,x = 3.06167
x = -1.27, x = 0.178, x = 3.06
-1.27 ≤ x < – 1,
0.178 ≤ x < 3,
x ≥ 3.06
Question
A function f is defined by f ( x ) = \(x\sqrt{1-x^{2}} where -1\leq x\leq 1.\)
The graph of y = f (x) is shown below.
(a) Show that f is an odd function.
The range of f is a ≤ y ≤ b , where a, b ∈ R.
(b) Find the value of a and the value of b.
▶️Answer/Explanation
Ans:
(a) attempts to replace x with –x
\(f(-x) = -x\sqrt{1-(-x)^{2}}\)
\(= -x\sqrt{1-(-x)^{2}} \left ( =-f(x) \right )\)
Note: Award M1A1 for an attempt to calculate both f (-x ) and – f (-x) independently, showing that they are equal.
Note: Award M1A0 for a graphical approach including evidence that either the graph is invariant after rotation by 180° about the origin or the graph is invariant after a reflection in the y-axis and then in the x-axis (or vice versa).
so f is an odd function
(b) attempts both product rule and chain rule differentiation to find f¢(x)
Note: Award M1 for an attempt to evaluate f(x) at least at one of their f¢(x) = 0 roots.
\(a = -\frac{1}{2} and b = \frac{1}{2}\)
Note: Award A1 for \(-\frac{1}{2}\leq y\leq \frac{1}{2}.\)
Question
A function f is defined by f ( x ) = \(x\sqrt{1-x^{2}} where -1\leq x\leq 1.\)
The graph of y = f (x) is shown below.
(a) Show that f is an odd function.
The range of f is a ≤ y ≤ b , where a, b ∈ R.
(b) Find the value of a and the value of b.
▶️Answer/Explanation
Ans:
(a) attempts to replace x with –x
\(f(-x) = -x\sqrt{1-(-x)^{2}}\)
\(= -x\sqrt{1-(-x)^{2}} \left ( =-f(x) \right )\)
Note: Award M1A1 for an attempt to calculate both f (-x ) and – f (-x) independently, showing that they are equal.
Note: Award M1A0 for a graphical approach including evidence that either the graph is invariant after rotation by 180° about the origin or the graph is invariant after a reflection in the y-axis and then in the x-axis (or vice versa).
so f is an odd function
(b) attempts both product rule and chain rule differentiation to find f¢(x)
Note: Award M1 for an attempt to evaluate f(x) at least at one of their f¢(x) = 0 roots.
\(a = -\frac{1}{2} and b = \frac{1}{2}\)
Note: Award A1 for \(-\frac{1}{2}\leq y\leq \frac{1}{2}.\)