Question

The equation of a curve is \(y=x^{3}+px^{2}\) ,where p is a positive constant .

(i)Show that the origin is a stationary point on the curve and find the coordinates of the other stationary point in terms of p.

(ii)Find the nature of each of the stationary points.

Another curve has equation \(y=x^{3}+px^{2}+px\)

(iii)Find the set of values of p for which this curve has n o stationary points.

Answer/Explanation

(i)\(\frac{\mathrm{d} y}{\mathrm{d} x}=3x^{2}+2px\)

Sets to 0\(\rightarrow x=0\) or \(-\frac{2p}{3}\)

→(0,0) or\((\frac{2p}{3},\frac{4p^{3}}{27})\)

(ii)\(\frac{\mathrm{d} ^{2}y}{\mathrm{d} x^{2}}=6x+2p\)

At (0,0)→2p  positive Minimum

At\((\frac{2p}{3},\frac{4p^{3}}{27})\)→-2p -ve Maximum

(iii)\(y=x^{3}+px^{2}+px\rightarrow 3x^{2}+2px+p=0\)

Uses \(b^{2}-4ac\)

\(\rightarrow 4p^{2}-12p< 0\)

\(\rightarrow 0< p< 3\)

Question

The equation of a curve is \(y=x^{3}+px^{2}\) ,where p is a positive constant .

(i)Show that the origin is a stationary point on the curve and find the coordinates of the other stationary point in terms of p.

(ii)Find the nature of each of the stationary points.

Another curve has equation \(y=x^{3}+px^{2}+px\)

(iii)Find the set of values of p for which this curve has n o stationary points.

Answer/Explanation

(i)\(\frac{\mathrm{d} y}{\mathrm{d} x}=3x^{2}+2px\)

Sets to 0\(\rightarrow x=0\) or \(-\frac{2p}{3}\)

→(0,0) or\((\frac{2p}{3},\frac{4p^{3}}{27})\)

(ii)\(\frac{\mathrm{d} ^{2}y}{\mathrm{d} x^{2}}=6x+2p\)

At (0,0)→2p  positive Minimum

At\((\frac{2p}{3},\frac{4p^{3}}{27})\)→-2p -ve Maximum

(iii)\(y=x^{3}+px^{2}+px\rightarrow 3x^{2}+2px+p=0\)

Uses \(b^{2}-4ac\)

\(\rightarrow 4p^{2}-12p< 0\)

\(\rightarrow 0< p< 3\)

Question

The diagram shows that the function f defined  for \(-1\leq x\leq 4\) ,where

(i)State the range of f.

(ii)Copy the diagram and on your copy sketch the graph of \(y=f^{-1}\left ( x \right )\)

(iii)Obtain expressions to define the function \(f^{-1}\),giving also the set of values for which each exp[ression is valid.

Answer/Explanation

(i) \(-5\leq f(x)\leq 4\) For f(x) allow x or y;

(ii) \(f^{-1}\left ( x \right )\) approximately correct (independent of f)

Closed region between (1,1)and (4,4) line reaches x-axis.

(iii)LINE: \(f^{-1}\left ( x \right )=\frac{1}{3}\left ( x+2 \right )\)

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

CURVE:\( 5-y=\frac{4}{x}\) OR \(x=5-\frac{4}{y}\)

\(f^{-1}\left ( x \right )=5-\frac{4}{x}\)

for \(1< x\leq 4\)

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

(ii)\(f\left ( x \right )> k-4\) or \(\left [ k-4,\infty \right ]\) or \(\left ( k-4,\infty \right )\)

(ii)smallest value of p=2

(iv) \(x-2=\left ( \pm \right )\sqrt{y+4-k}\)

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

\(f^{-1}\left ( x \right )=2+\sqrt{x+4-k}\)

Domain is \(x> k-4\) or \(\left [ k-4,\infty \right ]\) 

or \(\left ( k-4,\infty \right )\)

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}\0.

Answer/Explanation

 (i)\(fg(x)=5x\)

Range of fg is\( y\geq \)

(ii)\(y=\frac{4}{\left ( 5x+2 \right )}\Rightarrow x=\frac{4-2y}{5y}\)

\(g^{-1}\left ( x \right )=\frac{\left ( 4-2x \right )}{5x}\)

0,2 with no incorrect inequality

\(0< x\leq 2\)

Question

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

\(g:x\rightarrow 2x-3\)

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

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

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

Answer/Explanation

(i)\(gg\left ( x \right )=g\left ( 2x-3 \right )=2\left ( 2x-3 \right )-3=4x-9\)

(ii)\(y=\frac{1}{\left ( x^{2} -9\right )}\rightarrow x^{2}=\frac{1}{y}+9\)

\(f^{-1}\left ( c \right )=\sqrt{\frac{1}{x}+9}\)

Attempt solution of \(\sqrt{\frac{1}{x}+9}> 3\) or attempt to find range of f \(\left ( y> 0 \right )\)

Domain is \(x> 0\) 

(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:

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

\(g(x)=4\)

\(2x-3=4\)

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

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

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

(ii)\(\vec{OE}=\begin{pmatrix}6\\ 3\\ 6\end{pmatrix}+\begin{pmatrix}1\\-1\tfrac{1}{2}\\ 3\end{pmatrix}=\begin{pmatrix}7\\ 1\tfrac{1}{2}\\ 9\end{pmatrix}\)

\(\vec{OE}.\vec{OD}=56+0+108=164\)

\(\left | \vec{OE} \right |=\sqrt{132.25}=11.5\)

\(\left | \vec{OD} \right |=\sqrt{208}\)

\(164=\sqrt{132.25}\times \sqrt{208}\times \cos \Theta\)

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

 

 

 

Question

 The function f is defined by \(f:x\rightarrow x^{2}+1\) for \(x\geq \)

(i) Define in a similar way the inverse function \(f^{-1}\).

(ii) Solve the equation \(f(x)=\frac{185}{16}\).

Answer/Explanation

(i)\(x=\pm \sqrt{y-1}\)

\(f^{-1}:x\rightarrow \sqrt{x-1}\) for \(x> 1\)

(ii) \(f(x)=(x^{2}+1)^{2}+1\)

\(x^{2}+1=\pm \frac{13}{4}\)

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

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)\((x-1)^{2}-16\)

(ii)-16

(iii)\(9\leq (x-1)^{2}-16\leq 65\) OR \(x^{2}-2x-15=9\rightarrow 6,-4\)

\(25\leq (x-1)^{2}\leq 81 \) \(x^{2}-2x-15=65\rightarrow 10,-8\)

\(5\leq x-1\leq 9 \)             p=6

\(6\leq x\leq 10 \)        q=10

(iv)\(x=\left ( y-1 \right )^{2}-16\)      (interchange x/y)

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

f^{-1}\left ( x \right )=1+\sqrt{x+16}

Question

Find the set of values of k for which the line y = 2x − k meets the curve \(y=x^{2}+kx-2\) at two distinct points.

Answer/Explanation

\(x^{2}+x(k-2)+(k-2)=0\)

\((k-2)^{2}-4(k-2)> 0\)

\((k-2)(k-6)> 0\)

k<2 or k>6

Allow \(\left ( -\infty ,2 \right )\cup \left ( 6,\infty \right )\)

 

Question

The function f is defined by \(f:x\rightarrow \frac{2}{3-2x}\) for \(x\epsilon R\) ,\(x\neq \frac{3}{2}\)

(i) Find an expression for \(f^{-1}\left ( x \right )\).

The function g is defined by \(g:x\rightarrow 4x+a\)  for \(x\epsilon R\) ,where a is ac constant.

(ii)Find the value of a for which gf(-1)=3

(iii) Find the possible values of a given that the equation\( f^{-1}\left ( x \right )=g^{-1}\left ( x \right )\) has two equal roots.

Answer/Explanation

\(f:x\rightarrow \frac{2}{3-2x}\) \(g:x\rightarrow 4x+a\),

(i)\(y=\frac{2}{3-2x}\rightarrow y\left ( 3-2x \right )=2\rightarrow 3-2x=\frac{2}{y}\)

\(\rightarrow 2x=3-\frac{2}{y}\rightarrow f^{-1}\left ( x \right )=\frac{3}{2}-\frac{1}{x}\)

(ii)\(gf(-1)=3f(-1)=\frac{2}{5}\)

\(\frac{8}{5}+a=3\rightarrow a=\frac{7}{5}\)

(iii)\(g^{-1}\left ( x \right )=\frac{x-a}{4}=f^{-1}\left ( x \right )\)

\(\rightarrow x^{2}-x\left ( a+6 \right )+4(=0)\)

Solving \left ( a+6 \right )^{2}=16 or \(a^{2}+12a+20(=0)\)

\(\rightarrow a=-2\) or -10

Question.

The function f is defined by \(f: x\rightarrow 4 sin x – 1\) for \(\frac{-\Pi}{2}\leq x\leq\frac{\Pi}{2}\).
(i) State the range of f.
(ii) Find the coordinates of the points at which the curve y = f(x) intersects the coordinate axes.
(iii) Sketch the graph of y = f(x).
(iv) Obtain an expression for \(f^{-1}(x)\) , stating both the domain and range of \(f^{-1}\).

Answer/Explanation

(i) \(f:x\rightarrow 4\sin x-1\)  for \(-\frac{\pi }{2}\leq x\leq \frac{\pi }{2}\)

Range \(-5\leq f(x)\leq 3\)

(ii) 4s-1=0→\(s=\frac{1}{4}\rightarrow x=0.253\)

\(x=0\rightarrow y=-1\)

(iv)range-\(\frac{1}{2}\pi \leq f^{-1}\left ( x \right )\leq\frac{1}{2}\pi \)

domain \(-5\leq x\leq 3\)

Inverse \(f^{-1}(x)=\sin ^{-1}\left ( \frac{x+1}{4} \right )\)

Question.

The functions f and g are defined for x ≥ 0 by

                                                                                          

(i) Show that \(gf(x) = 6x^2 + 11\) and obtain an unsimplified expression for \(fg(x)|).

(ii) Find an expression for \((fg)^{−1}x\) and determine the domain of \((fg)^{−1}\).

(iii) Solve the equation \(gf(2x) = fg(x)\).

Answer/Explanation

(i) \(gf\left ( x \right )=3\left ( 2x^{2} +3\right )+2=6x^{2}+11\) 

\(fg\left ( x \right )=2\left ( 3x +2\right )^{2}+3\0 Allow \(18x^{2}+24x+11\)

(ii)\(y=2\left ( 3x+2 \right )^{2}+3 \Rightarrow 3x+2=\left ( \pm \right )\sqrt{\left ( \frac{y-3}{2} \right )}oe\)

\(\Rightarrow x=\left ( \pm \right )\frac{1}{3}\sqrt{\frac{\left ( y-3 \right )}{2}}-\frac{2}{3}oe\)

\(\Rightarrow \left ( fg \right )^{-1}\left ( x \right )=\frac{1}{3}\sqrt{\frac{\left ( y-3 \right )}{2}}-\frac{2}{3}oe\)

Solve their \(\left ( fg \right )^{-1}\left ( x \right )\geq 0\) or attempt range of fg

Domain is \(x\geq 11\)

(iii) \(6\left ( 2x \right )^{2}+11=2\left ( 3x+2 \right )^{2}+3\)

\(6x^{2}-24x=0 oe\)

x=0,4

Question

The function f is such that \(f(x) =a^2 x^2-ax=3b\) for \(x\leq\frac{1}{2a}\), where a and b are constants.

(i) For the case where \(f(-2) = 4a^2 – b + 8\)  and \(f(-3) = 7a^2 – b + 14\), find the possible values of a
and b.

(ii) For the case where a = 1 and b = -1, find an expression for \(f^-{1}(x)\) and give the domain of \(f^{-1}\)

Answer/Explanation

Ans:(i) \(2a+4b=8\)
\(2a^2+3a+4b+14\)
\(2a^2+3a+(8-2a)=14\rightarrow (a+2)(2a-3)=0\)
\(a=-2\) or \(\frac{3}{2}\)
\(b=3\) or \(\frac{5}{4}\)

(ii)

Question.

Functions f and g are defined for \(x\in R\) by

where a and b are constants.
(a) Given that \(gg(2) = 10\) and \(f^{−1}(2) = 14\), find the values of a and b.

(b) Using these values of a and b, find an expression for \(gf(x)\) in the form \(cx + d\), where c and d are constants.

Answer/Explanation

(a) \(3(3x + b) +b = 9x + 4b → 10 = 18 + 4b\)

         \( b =−2\)

          Either \(f (14) = 2\) or \(f^{-1}(x) = 2(x + a)\) etc. 

           \(a = 5\)

 (b) \(gf(x)=3\left ( \frac{1}{2}x-5 \right )-2\)

           \(gf(x)=\frac{3}{2}x-17\)

Question

The function f is defined by f(x) = 2 − 3 cos x for 0 ≤ x ≤ 20.

(i) State the range of f.

(ii) Sketch the graph of y=f(x).

The function g is defined by g(x)= 2 − 3 cos x for 0 ≤ x ≤ p, where p is a constant.

(iii) State the largest value of p for which g has an inverse.

(iv) For this value of p, find an expression for \(g^{-1}(x)\)

Answer/Explanation

(i)
−1 ⩽ f(x) ⩽ 5 or [–1, 5] (may use y or f instead of f(x))

(iii) (greatest value of p =)π

9(iv)
\(x = 2− 3cosx → cosx =1⁄3(2-x)\) 

\(g^{-1}=cos^{-1}\frac{2-x}{3}\)