Question

A curve has parametric equations.

x = t + ln (t + 1),          \(y=3te^{2t}\)

(i) Find the equation of the tangent to the curve at the origin.                        

(ii) Find the coordinates of the stationary point, giving each coordinate correct to 2 decimal places.

Answer/Explanation

                                                                                                                                                                                                                                         

5(i) Use product rule to differentiate y obtaining \(k_{1}e^{2t}+k_{2}te^{2t}\)

Obtain correct \(3e^{2t}+6te^{2t}\)

State derivative of x is \(1+\frac{1}{t+1}\)

Use\( \frac{dy}{dx}=\frac{dy}{dt}/\frac{dx}{dt}\)

with t =0 to find gradient

Obtain\( y=\frac{3}{2}x\) or equivalent

5(ii) Equate\(  \frac{dy}{dx} or \frac{dy}{dt}\) to zero and solve for t 

Obtain\( t=-\frac{1}{2}\)

Obtain x =−1.19

Obtain y = −0.55

Question

A curve is such that \(\frac{\mathrm{d} y}{\mathrm{d} x}=x^{\frac{1}{2}}-x^{-\frac{1}{2}}\) .The curve passes through the point \(\left ( 4,\frac{2}{3} \right )\)

(i)Find the equation of the curve .

(ii)Find \(\frac{\mathrm{d} ^{2}y}{\mathrm{d} x^{2}}\)

(iii)Find the coordinates of the stationary point and determine its nature.

Answer/Explanation

(i)\(y=\frac{2}{3}x^{\frac{3}{2}}-2x^{\frac{1}{2}}+c\)

\(\frac{2}{3}=\frac{16}{3}-4+c\)

\(c=-\frac{2}{3}\)

(ii)\(\frac{1}{2}x^{-\frac{1}{2}}+\frac{1}{2}^{-\frac{3}{2}}\)

(iii)\(x^{\frac{1}{2}}-x^{-\frac{1}{2}}=0\)\(\rightarrow \frac{x-1}{\sqrt{x}}=0\)

x=1

When x=1,\(y=\frac{2}{3}-2-\frac{2}{3}=-2\)

When x=1 ,\(\frac{\mathrm{d} ^{2}y}{\mathrm{d} x^{2}}=1> 0\) Hence minimum

Question

(i)Express \(2x^{2}-12x+13\) in the form \(a\left ( x+b \right )^{2}+c\) where a,b and c are constants.

(ii)The function f is defined by \(f\left ( x \right )=2x^{2}-12x+13 \) for \( x\geq k\) where k is a constant.It is given that f is a one-one function.State the smallest possible value of k.

The value of k is now given to be 7.

(iii) Find the range of f.

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

Answer/Explanation

(i) \(2\left ( x-3 \right )^{2}-5\) or a=2,b=-3,c=-5

(ii) 3

(iii)\(\left ( y \right )\geq 27\)

(iv)\(2\left ( x-3 \right )^{2}=\left ( y+5 \right )\)

\(x-3=\left ( \pm \right )\sqrt{\frac{1}{2}\left ( y+5 \right )}\)

\(x=3+\left ( \pm \right )\sqrt{\frac{1}{2}\left ( y+5 \right )}\)

\(\left ( f^{-1}\left ( x \right ) \right )=3+\sqrt{\frac{1}{2}\left ( x+5 \right )}\) for \(x\geq 27\)

Question 

The diagram shows the line y=1 and the part of the curve \(y=\frac{2}{\sqrt{\left ( x+1 \right )}}\).

(i)Show that the equation \(y=\frac{2}{\sqrt{\left ( x+1 \right )}}\) can be written in the form \(x=\frac{4}{y^{2}}-1\)

(ii)Find \(\int \left ( \frac{4}{y^{2}}-1 \right )dy\).Hence find the area of the shaded region.

(ii)The shaded region is rotated through about the y-axis.Find the exact value of the volume of revolution  obtained.

Answer/Explanation

(i) \(x=\frac{4}{y^{2}}-1\)

(ii)\(\int \left ( \frac{4}{y^{2}}-1 \right )dy=\left [- \frac{4}{y} -y\right ]\)

\(^{2}_{1}\left [ -\frac{4}{2}-2-\left ( -4-1 \right ) \right ]\)

(iii)\(\pi \int x^{2}dy=\left ( \pi \right )\int \left ( \frac{16}{y^{4}}-\frac{8}{y^{2}} +y\right )\)

\(\left ( \pi \right )\left [ -\frac{16}{3y^{3}}+\frac{8}{y} +y\right ]\)

\(\left ( \pi \right )\left [\left ( -\frac{16}{24}+4+2 \right )-\left ( -\frac{-16}{3}+8+1\right )\right ]\)

\(\frac{5\pi }{3}\)

Question

A curve has equation \(y=f(x)\) and it is given that \(f{}'(x)=3x^{\frac{1}{2}}-2x^{-\frac{1}{2}}\),The point A is the onl;y point in the curve at which gradient is -1

(i)Find the x-coordinate of A.

(ii)Given that curve also passes through the point (4,10),find the coordinate of A, giving your answer as a fraction.

Answer/Explanation

(i) \(3z-\frac{2}{z}=-1\Rightarrow 3z^{2}+z-2=0\0

\(x^{\frac{1}{2}}(or z)\) \(=\frac{2}{3}\) or -1

\(x=\frac{4}{9}\) only

(ii)\(f(x)=\frac{3x^{\frac{3}{2}}}{\frac{3}{2}}-\frac{2x^{\frac{1}{2}}}{\frac{1}{2}}\)(+c)

Sub  x=4,y=10  \(10=16-8+c\Rightarrow c=2\)

When \(x=\frac{4}{9}\),\(y=2\left ( \frac{4}{9} \right )^{\frac{3}{2}}-4\left ( \frac{4}{9} \right )^{\frac{1}{2}}+2\)

\(-\frac{2}{27}\)

 

Question

A curve is such that \(\frac{\mathrm{d} y}{\mathrm{d} x}=-\frac{8}{x^{3}}-1\) and the point (2,4)lies on the curve.Find the equation of the curve.

Answer/Explanation

\(y=\frac{4}{x^{2}}-x\)

Sub(2,4)\(\rightarrow c=5\)

Question

A curve has equation \(y=f(x)\) . It is given that \(f{}’\left ( x \right )=\frac{1}{\sqrt{x+6}}+\frac{6}{x^{2}}\) and that \(f(3)=1\) . Find f(x).

Answer/Explanation

Attempt integration

\(f(x)=2(x+6)^{\frac{1}{2}}-\frac{6}{x}\)

\(2(3)-\frac{6}{3}+c=1\)

c=-3

 

Question

The function f is defined for x > 0 and is such that \( f{}’\left ( x \right )=2x-\frac{2}{x^{2}}\). The curve y = f(x) passes through the point P(2, 6.)
(i) Find the equation of the normal to the curve at P.
(ii) Find the equation of the curve.
(iii) Find the x-coordinate of the stationary point and state with a reason whether this point is a maximum or a minimum.

Answer/Explanation

(i)\(f{}'(2)=4-\frac{1}{2}=\frac{7}{2}\rightarrow \)gradient of  normal =\(-\frac{2}{7}\)

\(y-6=-\frac{2}{7}(x-2)\)

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

6=4+1+c⇒c=1

(iii)\(2x-\frac{2}{x^{2}}=0\)⇒\(2x^{3}-2=0\)

x=1

\(f{}”(x)=2+\frac{4}{x^{3}}\) or any valid method

\(f{}”(1)=6\) >0 hence minimum\)

Question

 A curve for which \( \frac{\mathrm{d} y}{\mathrm{d} x}=7-x^{2}-6x \)   pases tghrough the p[oint (3,-10)

(i)Find the equation of the curve.

(ii)Express \(7-x^{2}-6x\) in the form \(a-(x+b)^{2}\),where a and b are constants.

(iii)Find the set of values of x for which the gradient of the curve is positive.

Answer/Explanation

(i)\(y=7x-\frac{x^{3}}{3}-\frac{6x^{2}}{2}+c\)

Uses (3,-10) →c=5

(ii)\(7-x^{2}-6x=16-(x+3)^{2}\)

(iii)\(16-\left ( x+3 \right )^{2}> 0\rightarrow \left ( x+3 \right )^{2}< 16\),and solve

End-points x=1 or -7

\(\rightarrow -7< x< 1\)

Question.

A curve is such that \(\frac{\partial y}{\partial x}==2-8(3x+4)^\frac{1}{2}\)
(i) A point P moves along the curve in such a way that the x-coordinate is increasing at a constant rate of 0.3 units per second.
Find the rate of change of the y-coordinate as P crosses the y-axis.

The curve intersects the y-axis where y =\frac{4}{3}
(ii) Find the equation of the curve.

Answer/Explanation

(i) \(\frac{\mathrm{d} y}{\mathrm{d} x}=2-8\left ( 3x+4 \right )^{-\frac{1}{2}}\)

\(\left ( x=0,\rightarrow \frac{\mathrm{d} y}{\mathrm{d} x} =-2\right )\)

\(\frac{\mathrm{d} y}{\mathrm{d} t}=\frac{\mathrm{d} y}{\mathrm{d} x}\times \frac{\mathrm{d} x}{\mathrm{d} t}\rightarrow -0.6\)

(ii) \(y=\left \{ 2x \right \}\left \{ -\frac{8\sqrt{3x+4}}{\frac{1}{2}}\div 3 \right \}\left ( +c \right )\)

\(x=0,y=\frac{4}{3}\rightarrow c=12\)

Question.

                                                             

The diagram shows the curve \(y = f(x).\) defined for x > 0. The curve has a minimum point at A and

crosses the x-axis at B and C. It is given that \(\frac{\partial y}{\partial x}=2x-\frac{2}{x^3}\)  and that the curve passes through the point \((4,\frac{189}{6})\)

(i) Find the x-coordinate of A

(ii) Find f(x).

(iii) Find the x-coordinates of B and C.

(iv) Find, showing all necessary working, the area of the shaded region.

Answer/Explanation

(i) \(\frac{2x-2}{x^{3}}=0\)

\(x^{4}=1\Rightarrow x=1\) at A cao

(ii)\(f\left ( x \right )=\frac{x^{2}+1}{x^{2}}\left ( +c \right )\) cao

\(\frac{189}{16}=\frac{16+1}{16+c}\)

\(c=-\frac{17}{4}\)

(iii) \(\frac{x^{2}+1}{x^{2}-\frac{17}{4}}=0\Rightarrow 4x^{4}-17x^{2}+4\left ( =0 \right )\)

\(\left ( 4x^{2} -1\right )\left ( x^{2}-4 \right )\left ( =0 \right )\)

\(x=\frac{1}{2},2\)

(iv)\( \int \left (\frac{ x^{2}+x^{-2}-17 }{4}\right )dx=\frac{x^{3}}{3}-\frac{1}{x}-\frac{17x}{4}\)

\(\left ( \frac{8}{3}-\frac{1}{2}-\frac{17}{2} \right )-\left ( \frac{1}{24}-2-\frac{17}{8} \right )\)

\(Area=\frac{9}{4}\)

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)
\(x = 2− 3cosx → cosx =1⁄3(2-x)\) 

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

Question 

A curve with equation y = f(x) passes through the points ( 0, 2) and (3, −1 ) It is given that \({f}'(x)=kx^{2}-2x\)  , where k is a constant. Find the value of k.

Answer/Explanation

\(y=\frac{1}{3}kx^{3}-x^{2}+(c)\) Sub (0, 2) Sub (3,‒1)→ \(−1= 9k-9 + their  c\)

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

Question

The equation of a curve is such that \(\frac{dy}{dx}=\frac{1}{(x-3)^{2}}+x\). . It is given that the curve passes through the
     point (2, 7).
     Find the equation of the curve.                                                                                                                                               [4]

Answer/Explanation

Ans

2  \((y=)\left [ -(x-3)^{-1} \right ]\left [ +\frac{1}{2}x^{2} \right ](+c)\)

     7 = 1 + 2 + c 

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

Question


The diagram shows the curve with equation \(y=9(x^{-\frac{1}{2}}-4x^{-\frac{3}{2}})\). The curve crosses the x-axis at the point A.
(a) Find the x-coordinate of A.
(b) Find the equation of the tangent to the curve at A.
(c) Find the x-coordinate of the maximum point of the curve.
(d) Find the area of the region bounded by the curve, the x-axis and the line x=9.

Answer/Explanation

Ans:

(a) \(9(x^{-\frac{1}{2}}-4x^{-\frac{3}{2}})=0\) leading to \(9x^{-\frac{3}{2}(x-4)=0\)
x=4 only

(b) \(\frac{dy}{dx}=9(-\frac{1}{2}x^{-\frac{3}{2}}+6x^{-\frac{5}{2}})\)
At x = 4 gradient \(=9(-\frac{1}{16}+\frac{6}{32})=\frac{9}{8}\)
Equation is \(y=\frac{9}{8}(x-4)\)

(c) \(9x^{-\frac{5}{2}}(-\frac{1}{2}x+6)=0\)
x=12

(d) \(\int 9(x^{-\frac{1}{2}}-4x^{-\frac{3}{2}})dx=9(\frac{x^{\frac{1}{2}}}{\frac{1}{2}}-\frac{4x^{-\frac{1}{2}}}{-\frac{1}{2}})\)
\(9[(6+\frac{8}{3})-(4+4)]\)
6

Question

  The equation of a curve is such that  \(\frac{dx}{dy}=\frac{3}{x^{4}}+32x^{3}\). 

    It is given that the curve passes through the point  \(\left ( \frac{1}{2},4 \right )\)

    Find the equation of the curve.

Answer/Explanation

Ans

1   \([y=]-\frac{1}{x^{3}}+8x^{4}[+c]\)

      \(4=-8+\frac{1}{2}+c\)

       \(y=-\frac{1}{x^{3}}+8x^{4}+\frac{23}{2}\)