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

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 

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)
\(x = 2− 3cosx → cosx =1⁄3(2-x)\) 

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

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)
\(x = 2− 3cosx → cosx =1⁄3(2-x)\) 

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

Queastion

(i) Express \(x^{2}-4x+7\) in the form\((x+a)^{2}+b.\)

The function f is defined by \(f(x) = x^{2} − 4x + 7 \) for x < k, where k is a constant.  

(ii) State the largest value of k for which f is a decreasing function.

The value of k is now given to be 1.

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

(iv) The function g is defined by \(g(x )= \frac{2}{x-1} \)for x > 1. Find an expression for gof(x) and state the range of gof(x).

.

Answer/Explanation

(i)

\(\left [ (x-2^{2}) \right ]+\left [ 3 \right ]\)

(ii)

Largest k is 2 Accept \(k\leq 2\)

(iii) 

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

\(\Rightarrow F^{-1}(x)=2-\sqrt{x-3}forx>4\)

Question

          

     The diagram shows a curve with equation \(y=4x^{\frac{1}{2}}-2x\ for \ x\geqslant 0,\) and a straight line with equation
      y = 3 − x. The curve crosses the x-axis at A (4, 0) and crosses the straight line at B and C.
       (a) Find, by calculation, the x-coordinates of B and C.                                                                                             [4]

       (b) Show that B is a stationary point on the curve.                                                                                                    [2]

       (c) Find the area of the shaded region.                                                                                                                         [6]

Answer/Explanation

Ans

12 (a) \(y=4x^{\frac{1}{2}}-2x=3-x\rightarrow x-4x^{\frac{1}{2}}+3(=0)\)

            \(\left ( x^{\frac{1}{2}}-1 \right )\left ( x^{\frac{1}{2}}-3 \right )(=0)\ or \ (u-1)(u-3)(=0)\)

             \(x^{\frac{1}{2}}=1,3\)

              x =1, 9 
Alternative method for question 12(a)

            \(\left ( 4x^{\frac{1}{2}} \right )^{2}=(3+x)^{2}\)

             \(16x=9+6x+x^{2}\rightarrow x^{2}-10x+9(=0)\)

              (x – 1 )( x – 9 ) ( =0 )

              x = 1, 9

12(b) \(\frac{dy}{dx}2x^{1/2}-2\)

           \(\frac{dy}{dx}2x^{1/2}-2=0\) when x =1 hence B is a stationary point 

12 (c) Area of correct triangle \(\frac{1}{2}(9-3)\times 6\)

           \(\int (4x^{\frac{1}{2}}-2x)(dx)=\left [ \frac{4x^{\frac{3}{2}}}{\frac{3}{2}}-x^{2} \right ]\)

            \((72-81)-\left ( \frac{64}{3}-16 \right )\)

             \(-14\frac{1}{3}\)

              \(Shaded \ region=18-14\frac{1}{3}=3\frac{2}{3}\)

Question.

(a) Express −3x2 + 12x + 2 in the form −3 (x-a)2 + b, where a and b are constants.

The one-one function f is defined by f : x → −3x2 + 12x + 2 for x ≤ k.

(b) State the largest possible value of the constant k.

It is now given that k = −1.

(c) State the range of f.

(d) Find an expression for f −1 (x).

The result of translating the graph of y = f (x) by \(\binom{-3}{1}\)  is the graph of y = g (x).

(e) Express g(x) in the form px2 + qx + r, where p, q and r are constants.

Answer/Explanation

(a)  \(\left \{ -3\left ( x – 2 \right )^{2} \right \}\)    {+14}

(b)   [k =] 2

(c) [Range is y] ⩽ –13

(d) y = -3 (x-2)2 + 14 leading to (x-2)2 =  \(\frac{14-y}{3}\)

x = 2 (±) \(\sqrt{\frac{14-y}{3}}\)

[f-1 (x)] = 2 – \(\sqrt{\frac{14-y}{3}}\)

(e)  \(\left [ g(x) =\right ] \left \{ -3\left ( x+3-2 \right )^{2} \right \} + \left \{ 14+1 \right \}\)

g(x) = -3x2 – 6x + 12

Question

Functions f and g are defined as follows:
f:x  →x2 + 2x + 3 for x ≤ -1,
g:x →2x + 1 for x ≥ -1.

(a) Express f(x) in the form (x+a)2 + b and state the range of f.

(b) Find the expression for f-1(x).
(c) Solve the equation gf(x)=13

Answer/Explanation

Ans:

(a) \([f(x)=](x+1)^2+2\)
Range [of f is (y)] ≥ 2
(b) \(y=(x+1)^2+2\) leading to \(x=[±]=-\sqrt{y-2}-1\)
\(f^{-1}=-\sqrt{x-2}-1\)
(c) \(2(x^2+2x+3)+1=13\)
\(2x^2+4x-6[=0]\) leading to (2)(x-1)(x+3)[=0]
x=-3 only

Question

  Functions f and g are defined as follows:
             \(f(x)=(x-2)^{2}-4 \ for \ x\geqslant 2\)

              \(g(x)=ax+2 \ for \ x\epsilon \mathbb{R}\)
     where a is a constant.
      (a) State the range of f.                                                                                                                    [1]
      (b) Find f −1 (x).                                                                                                                                 [2]
      (c) Given that a = \(-\frac{5}{3}\), solve the equation f(x) = g(x)                                         [3]

      (d) Given instead that ggf −1 (12) = 62, find the possible values of a.                                   [5]

Answer/Explanation

Ans

9 (a) Range of f is f(x) ⩾−4

9 (b) \(y=(x-2)^{2}-4\Rightarrow (x-2)^{2}=y+4\Rightarrow x-2=+\sqrt{y+4}\ or \ \pm \sqrt{y+4}\)

           \((3x+2)(x-3)[=0] \ or \ \frac{7\pm \sqrt{7^{2}-4(3)(-6)}}{6}\ OE\ \)

            x = 3 only 

9(d) f-1 (12) = 6

         g(f–1 (12)) = 6a + 2

        g(g(f–1 (12))) =a(6a + 2) + 2 = 62 

      6a2 + 2a – 60 [= 0] 

      \(a=-\frac{10}{3}\ or \ 3\)

Alternative method for Question 9(d)

     \(g(f^{-1}(x))=a\left ( \sqrt{x+4+2} \right )+2 \ or \ gg(x)=a(ax+2)+2\)

     \(g(g(f^{-1}(12)))=a(6a+2)+2=62\)

      6a2 + 2a – 60 [= 0] 

      \(a=-\frac{10}{3}\ or \ 3\)

Question.

The diagram shows the graph of \(y = f(x)\), where \(f(x) =3/2cos 2x +1/2\) for 0 ≤ x ≤ π.

(a) State the range of f.

A function g is such that \(g(x) = f(x) + k\) , where k is a positive constant. The x-axis is a tangent to the curve \(y = g(x)\).

(b) State the value of k and hence describe fully the transformation that maps the curve \(y = f(x)\) onto  \(y = g(x)\).

(c) State the equation of the curve which is the reflection of \(y = f(x)\) in the x-axis. Give your answer in the form \(y = a cos 2x + b\), where a and b are constants.

Answer/Explanation

(a) −1 ⩽ f(x) ⩽ 2

(b) k = 1 
          Translation by 1 unit upwards parallel to the y-axis

(c)  \(y=-\frac{3}{2}\cos 2x-\frac{1}{2}\)