### 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.

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.

(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 $$f^{-1}$$.

(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.

(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.

(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.

$$y=\frac{4}{x^{2}}-x$$

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).

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.

(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.

(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.

(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.

(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 $$g^{-1}(x)$$

(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 $$g^{-1}(x)$$

(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.

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

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.

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.

1   $$[y=]-\frac{1}{x^{3}}+8x^{4}[+c]$$
$$4=-8+\frac{1}{2}+c$$
$$y=-\frac{1}{x^{3}}+8x^{4}+\frac{23}{2}$$