### Question

A curve is defined by the parametric equations
x = 3t − 2 sin t,            y = 5t + 4 cos t,

where $$0\leqslant t\leqslant 2\pi$$. At each of the points P and Q on the curve, the gradient of the curve is $$\frac{5}{2}$$.

(a) Show that the values of t at P and Q satisfy the equation 10 cos t − 8 sin t = 5.                                                                                        [3]

(b) Express 10 cost − 8 sin t in the form $$R\cos (t+\alpha ), where \ R> 0> and \ 0< \alpha < \frac{1}{2}\pi$$. Give the exact
value of R and the value of $$\alpha$$  correct to 3 significant figures.                                                                                                      [3]

(c) Hence find the values of t at the points P and Q.                                                                                                                                              [4]

Ans

(a) Carry out division at least as far as  3x2+ kx

Obtain quotient 3x2 – 4 – 4

Confirm remainder is 9 AG

(b) Integrate to obtain at least  k1 x3 and k2 ln(3x + 2) terms

Obtain x3 – 2x2 – 4x + 3 ln(3x + 2)
(FT from quotient in part (a))

Apply limits correctly
Apply appropriate logarithm properties correctly
Obtain 125 ln64

(c) State or imply $$9x^{3}-6x^{2}-20x-8=(3x+2)(3x^{2}-4x-4)$$

(FT from quotient in part (a))
Attempt to solve cubic eqn to find positive value of x (or of 3ey )
Use logarithms to solve equation of form 3ey = k where k > 0 M1
Obtain $$\frac{1}{3}ln2$$ or exact equivalent

### Question

The diagram shows part of the curve with equation
$$y=4sin^2x+8sinx+3$$.
where x is measured in radians. The curve crosses the x-axis at the point A and the shaded region is bounded by the curve and the lines x=0 and y=0.
(a) Find the exact x-coordinate of A.
(b) Find the exact gradient of the curve at A.
(c) Find the exact area of the shaded region.

Ans:

(a) Solve equation y = 0 to find the value of x
Obtain $$\frac{7}{6}\pi$$
(b) Attempt first derivative using chain rule
Obtain $$\frac{dy}{dx}=8sinxcosx+8cos x$$
Substitute value from part (a) to find gradient $$-2\sqrt{3}$$
(c) Express integrand in the form $$k_1+k_2 cos 2x + k_3 sin x$$
Obtain correct 5 – 2cos 2x + 8sin x
Integrate to obtain 5x – sin2x – 8cos x
Apply limits 0 and their value from part (a) correctly
Obtain $$\frac{35}{6}\pi+\frac{7}{2}\sqrt{3}+8$$ or exact equivalent

### Question

A curve has equation
$$3x^2-y^2-4 In(2y+3)=26$$.
Find the equation of the tangent to the curve at the point (3, -1).

Ans:

Differentiate $$-y^2$$ to obtain $$-2y\frac{dy}{dx}$$
Differentiate -4In(2y+3) to obtain $$\frac{-8}{2y+3}\frac{dy}{dx}$$
Attempt differentiation of all terms
Substitute x = 3, y =-1 to find numerical value of $$\frac{dy}{dx}$$
Obtain $$\frac{dy}{dx}=3$$
Obtain equation y = 3x – 10

### Question

The diagram shows part of the curve with equation $$y=\frac{5x}{4x^3+1}$$. The shaded region is bounded by the curve and the lines x=1, x=3 and y=0.

(a) Find $$\frac{dy}{dx}$$ and hence find the the x-coordinate of the maximum point.
(b) Use the trapezium rule with two intervals to find an approximation to the area of the shaded region. Give your answer correct to 2 significant figures.
(c) State, with a reason, whether your answer to part (b) is an over-estimate or under-estimate of the exact area of the shaded region.

Ans:

1.  Differentiate using quotient rule (or product rule)
Obtain $$\frac{5(4x^3+1)-60x^3}{(4x^3+1)^2}$$
Equate first derivative to zero and attempt solution
Obtain $$x=\frac{1}{2}$$
2. Use y values $$\frac{5}{5}, \frac{10}{33}, \frac{15}{109}$$ or decimal equavalents
Use correct formula, or equivalent, with h=1
Obtain $$\frac{1}{2}(1+\frac{20}{30}+\frac{15}{109})$$ or equivalent and hence 0.87
3. State over-estimate with reference to top of each trapezium above curve

### Question

The parametric equations of a curve are
$$x=e^{2t}cos4t$$,         y=3sin2t.
Find the gradient of the curve at the point for which t=0.

Ans:

Attempt use of product rule to find $$\frac{dx}{dt}$$
Obtain $$2e^{2t}cos4t-4e^{2t}sin 4t$$
Obtain $$\frac{dy}{dt}=6cos2t$$
Divide to find $$\frac{dy}{dx}$$ with t = 0 substituted
Obtain 3

### Question

A curve has equation $$y=f(x) \ where \ f(x)=\frac{4x^{3}+8x-4}{2x-1}$$

(a) Find an expression for$$\frac{dy}{dx}$$ and hence find the coordinates of each of the stationary points of the
curve $$y=f(x)$$                                                                                                                                                                               [5]

(b) Divide $$4x^{3}+8x-4 \ by \ (2x-1), \ and \ hence \ find \ \int f(x)dx$$                                                                           [5]

Ans

8(a) Differentiate using the quotient rule (or product rule)
Obtain $$\frac{(2x-1)(12x^{3}+8)-2(4x^{3}+8x-4)}{(2x-1)^{2}}$$

Equate first derivative to zero and attempt solution
Obtain (0, 4)
Obtain $$\left ( \frac{3}{4},\frac{59}{8} \right )$$

8 (b) Carry out division to obtain quotient of form 2x2 + kx  +  m

Obtain correct quotient $$2x^{2}+x+\frac{9}{2}$$

Obtain remainder $$\frac{1}{2}$$

Integrate to obtain at least $$k_{1}x^{3}\ and \ k_{2} \ ln(2x-1) \ terms$$

Obtain $$\frac{2}{3}x^{3}+\frac{1}{2}x^{2}+\frac{9}{2}x+\frac{1}{4}ln(2x-1)$$ as final answer

### Question

A curve is defined by the parametric equations
x = 3t − 2 sin t,                    y = 5t + 4 cos t,
where 0 ≤ t ≤ 2π. At each of the points P and Q on the curve, the gradient of the curve is $$\frac{5}{2}$$

(a) Show that the values of t at P and Q satisfy the equation 10 cost − 8 sin t = 5.                                                                                       [3]

(b) Express 10 cost − 8 sin t in the form $$R \cos(t+\alpha ), \ where \ R> 0\ and \ 0< \alpha \frac{1}{2}\pi$$. Give the exact
value of R and the value of $$\alpha$$ correct to 3 significant figures.                                                                                                    [3]

(c) Hence find the values of t at the points P and Q.                                                                                                                                            [4]

Ans

7 (a) Obtain $$\frac{dx}{dt}=3-2\cos t \ and \ \frac{dy}{dt}=5-4\sin t$$

Equate expression for $$\frac{dy}{dx}\ to \ \frac{5}{2}$$

Obtain $$10\cos t-8\sin t=5$$

7 (b) State $$R=\sqrt{164}$$ or exact equivalent

Use appropriate trigonometry to find $$\alpha$$

Obtain 0.675 with no errors seen

7 (c) Carry out correct method to find one value of t
Obtain 0.495
Carry out correct method to find second value of t
Obtain 4.44

### Question

The diagram shows the curve with equation  $$y=\frac{3x+2}{ln x}$$. The curve has a minimum point M.

(a) Find an expression for $$\frac{dy}{dx}$$  and show that the x-coordinate of M satisfies the equation $$x=\frac{3x+2}{3 lnx}$$

(b) Use the equation in part (a) to show by calculation that the x-coordinate of M lies between
3 and 4.                                                                                                                                                                                                                [2]

(c) Use an iterative formula, based on the equation in part (a), to find the x-coordinate of M correct
to 5 significant figures. Give the result of each iteration to 7 significant figures.                                                                                [3]

Ans

5 (a) Use quotient rule (or equivalent) to find first derivative

Obtain $$\frac{dy}{dx}=\frac{3ln x-\frac{1}{x}(3x+2)}{(lnx)^{2}}$$

Equate first derivative to zero and confirm $$x=\frac{3x+2}{3lnx}$$

5 (b) Consider $$x-\frac{3x+2}{3lnx}$$ or equivalent for values 3 and 4

Obtain  −0.33… and 0.63… or equivalents and justify conclusion

5 (c) Use iteration process correctly at least once

Show sufficient iterations to 7 s.f. to justify answer or show sign
change in the interval [3.32225, 3.32235]

### Question

A curve has parametric equations
x = ln (2t + 6) − ln t,           y = t ln t.
(a) Find the value of t at the point P on the curve for which x = ln 4.                                           [3]

(b) Find the exact gradient of the curve at P.                                                                                      [5]

Ans

4 (a) Equate x to ln 4 and use relevant logarithm property

Obtain equation with no logarithm present, $$\frac{2t+6}{t}=4$$

Obtain t = 3

4 (b) Obtain $$\frac{dx}{dt}=\frac{2}{2t+6}-\frac{1}{t}$$

Use product rule to find  $$\frac{dy}{dt}$$

Obtain   $$ln t+t\times \times \frac{1}{t}$$

Divide to obtain  $$\frac{dy}{dx}\ using \ their\ \frac{dy}{dt}\ and \ \frac{dx}{dt}\ correctly$$

Obtain 6(ln3 1)

### Question

.

The diagram shows the curve with parametric equations

$$x = ln (2t + 3), y = \frac{2t-3}{2t+3}$$

The curve crosses the y-axis at the point A and the x-axis at the point B

(a)Show that $$\frac{dy}{dx} = \frac{6}{2t + 3}$$.

(b)Find the gradient of the curve at A.

(c) Find the gradient of the curve at B.(a)Obtain $$\frac{dy}{dt} = \frac{6}{2t + 3}$$.

Use quotient rule, or equivalent, to find $$\frac{dy}{dt}$$

Obtain $$\frac{dy}{dt}$$ = $$\frac{2(2t+3)-2(2t-3)}{(2t+3)^{2}}$$

Divide to confirm $$\frac{dy}{dx}$$ =  $$\frac{6}{2t+3}$$

(b)Attempt to find value of t corresponding to x=0

Obtain t = −1 and hence gradient is 6

(c)Attempt to find value of t corresponding to y=0

Obtain t = $$\frac{3}{2}$$ and hence gradient is 1

### Question

The curve with equation y = xe2x + 5e−x has a minimum point M.

(a)Show that the x-coordinate of M satisfies the equation x =  $$\frac{1}{3}$$ ln 5 − $$\frac{1}{3}$$ ln (1 + 2x).

(b)Use an iterative formula, based on the equation in part (a), to find the x-coordinate of M correct to 3 significant figures. Use an initial value of 0.35 and give the result of each iteration to 5 significant figures.

(a)Attempt use of product rule to differentiate xe2x

Obtain  e2x + 2xe2x – 5e-x

Equate first derivative to zero and multiply by ex to obtain an equation involving e3x

Obtain  e3x(1 + 2x) = 5 or equivalent

Confirm given result  x = $$\frac{1}{3}$$ ln5 – $$\frac{1}{3}$$ ln (1 + 2x)  with sufficient detail.

(b)Use iteration process correctly at least once

Show sufficient iterations to 5sf to justify answer or show sign change in interval [0.3565, 0.3575]

### Question

A curve has equation e2xy – ey = 100.

(a)Show that $$\frac{dy}{dx}=\frac{2e^{2x}y}{e^{y-e^{2x}}}.$$

(b)Show that the curve has no stationary points.

It is required to find the x-coordinate of P, the point on the curve at which the tangent is parallel to the y-axis.

(c)Show that the x-coordinate of P satisfies the equation.

x = ln 10 $$-\frac{1}{2}In (2x-1)$$State or imply ey – e2x = 0 and hence y = 2x

(d)Use an iterative formula, based on the equation in part (c), to find the x-coordinate of P correct to 3 significant figures. Use an initial value of 2 and give the result of each iteration to 5 significant figures.

(a)Use product rule to differentiate e2x y

Obtain $$2e^{2x}y+e^{2x}\frac{dy}{dx}$$

Obtain $$2e^{2x}y+e^{2x}\frac{dy}{dx}- e^{y}\frac{dy}{dx} = 0$$ and rearrange to confirm given result.

(b)Consider e2xy = 0 and either state e2x  ≠ 0 or substitute y = 0 in equation of curve

Complete argument with e2x ≠ 0 or  e2x> 0 and substitution to show y cannot be zero.

(c)Substitute for y in equation of curve and attempt rearrangement as far as e2x = …..

Use relevant logarithm properties

Confirm equation $$x = In 10 – \frac{1}{2}In (2x-1)$$

(d)Use iteration process correctly at least once

Show sufficient iterations to 5 sf to justify answer or show sign change in interval [1.815, 1.825]

### Question

(a)Given that y = $$tan^{2}x,$$ show that $$\frac{dy}{dx} = 2 tan x + 2tan^{3}x.$$

(b)Find the exact value of $$\int_{\frac{1}{4}\pi }^{\frac{1}{3}\pi } \left ( tan x + tan^{2}x + tan^{3}x \right )dx$$ .

(a)Differentiate to obtain 2 tan x sec2 x

Use sec2 x = 1 + tan2 x  to confirm 2 tan x + 2 tan3x

(b)Attempt to use part (a) result to integrate tan x +  tan 3 x

Obtain $$\frac{1}{2}tan^{2}x$$

Use relevant identity to integrate  tan2x

Obtain sec2 x−1 and hence tan x−x

Use limits correctly for integrand of form k1 tan2 x + k2 tan x + k3x

Obtain $$\sqrt{3} – \frac{1}{12}\pi$$

### Question

(a)Show that sin 2θ cot θ − cos 2θ ≡ 1.

(bHence find the exact value of $$sin\frac{1}{6}\pi cot\frac{1}{12}\pi$$.

(c) Find the smallest positive value of 1 (in radians) satisfying the equation

sin 2θ  cotθ − 3 cos 2θ = 1.

(a)Use at least two of sin 2θ =  2sin θ cos  θ, cos 2θ =  cos2θ – sin2θ,  cotθ = $$\frac{cos\theta }{sin\theta }$$

Express LHS in terms of sin θ and cos θ only and attempt valid simplification

Obtain cos2 θ  + sin2 θ  or equivalent and hence 1

(b)Substitute θ = $$\frac{1}{12}\pi$$ and show or imply $$sin\frac{1}{6}\pi cot\frac{1}{12}\pi = 1 + cos\frac{1}{6}\pi$$

Obtain 1+ $$\frac{1}{2}\sqrt{3}$$  or exact equivalent

(c)Use the identity from part (a) to obtain − 2cos 2θ  = 0 or equivalent

Obtain $$\theta = \frac{1}{4}\pi$$

### Question

The variables x and y satisfy the equation y = 3 2a a x , where a is a constant. The graph of ln y against x is a straight line with gradient 0.239.

(a) Find the value of a correct to 3 significant figures.

(b)Hence find the value of x when y = 36. Give your answer correct to 3 significant figures.

Ans:

(a)State or imply equation is  ln y = ln 3 2a  + x In a

Equate gradient of line involving a to 0.239

Obtain ln 0.239 a = and hence a =1.27

(b)Substitute y = 36 in ln y =…  equation and solve for x

Obtain 3.32

### Question

The equation of a curve is $$x^{2}-4xy-2y^{2}=1$$

(i)Find an expression for $$\frac{dy}{dx}$$ and show that the gradient of the curve at the point(−1, 2) is-$$\frac{5}{2}$$

(ii) Show that the curve has no stationary points.

(iii) Find the x-coordinate of each of the points on the curve at which the tangent is parallel to the y-axis

.

7(i) Obtain$$-4y-4x\frac{dy}{dx}$$
from use of the product rule Differentiate $$-2y^{2}$$to obtain$$-4y\frac{dy}{dx}$$

Obtain 2 , 0 x = with no  extra terms

Rearrange to obtain expression for$$\frac{dy}{dx}$$ and substitute x = −1, y = 2

Obtain$$\frac{dy}{dx}=\frac{2x-4y}{4x+4y}$$ OE and hence -$$\frac{5}{2}$$

7(ii) Equate numerator of derivative to zero to produce equation in x and y Substitute into equation of curve to produce equation in x or y

Obtain $$-6y^{2}=1 or -\frac{3}{2}x^{2}=1$$

7(iii) Use denominator of derivative equated to zero with equation of curve to produce equation in x

Obtain $$3x^{2}=1 and hence x=\pm \frac{1}{\sqrt{3}}$$

### Question

A curve has equation y =$$\frac{3+2Inx}{1+Inx}$$.Find the exact gradient of the curve at the point for which y = 4.

Use quotient rule (or product rule) to find first derivative Obtain$$-\frac{1}{x(1+Inx)^{2}}$$ or (unsimplified) equivalent  Use y=4 to obtain In$$x=-\frac{1}{2}$$ or exact equivalent for x

Substitute for x in their first derivative

Obtain$$-4e^{\frac{1}{2}}$$ or exact equivalent

### Question

The diagram shows the curve with equation y=$$\frac{8+x^{3}}{2-5x}$$.The maximum point is denoted by M.

(i) Find an expression for $$\frac{dy}{dx}$$ and determine the gradient of the curve at the point where the curve crosses the x-axis.

(ii) Show that the x-coordinate of the point M satisfies the equation x =$$\sqrt{(0.6x+4x^{-1})}.$$

(iii) Use an iterative formula, based on the equation in part (ii), to find the x-coordinate of M correct to 3 significant figures. Give the result of each iteration to 5 significant figures.

<p(i) Use quotient rule (or product rule) to differentiate .

Obtain $$\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{3x^{2}\left ( 2-5x \right )-\left ( -5 \right )\left ( 8+x^{3} \right )}{\left ( 2-5x \right )^{2}}$$

State or imply curve crosses  x-axiz when x=-2.

Substitute -2 to obtain 1

(ii)Equate numerator of first derivative to zero and rearrange as far as $$kx^{3}$$=….or equivalent

Confirm given result $$x=\sqrt{0.6x+4x^{-1}}$$

(iii) Use iterative process correctly atleast once

Show sufficient iterations to 5sf to justify answer or show a sign change in the interval $$\left [ 1.805,1.815 \right ]$$

### Question

Find the equation of the normal to the curve$$x^{2}In y+2x+5y$$=11

at the point(3, 1).

Obtain $$2x\ln y+x^{2}\times \frac{1}{y}\times \frac{\mathrm{d} y}{\mathrm{d} x}$$

Obtain …..$$+2+5\frac{\mathrm{d} y}{\mathrm{d} x}=0$$

Substitute x=3 and y=1 to find value of their $$\frac{\mathrm{d} y}{\mathrm{d} x}$$

Obtain $$\frac{\mathrm{d} y}{\mathrm{d} x}=-\frac{2}{14}$$

Attempt equation of line through(3,1) with gradient of normal.

Obtain y=7x-20 or equivalent Unsimplified

### Question

The diagram shows the curve with equation y = sin 2x + 3 cos 2x for 0 ≤ x ≤ 0. At the points P and Q on the curve, the gradient of the curve is 3.

(i) Find an expression for $$\frac{dy}{dx}$$

(ii) By first expressing $$\frac{dy}{dx}$$  in the form R cos (2x + \alpha), where $$R>0 and 0<\alpha <\frac{1}{2}\pi$$  R , find the x-coordinates of P and Q, giving your answers correct to 4 significant figures.

7(i) State expression of form $$k_{1}cos2x+k_{2}sin2x$$

7(ii) State$$R=\sqrt{40} or 6.324…$$

Use appropriate trigonometry to find α

Obtain 1.249…

Equate their $$Rcos(2x+\alpha ) to 3 and find cos^{-1}(3\div R)$$

Carry out correct process to find one value of α

Obtain 1.979

Carry out correct process to find second value of α within the range

Obtain 3.055

### 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

The equation of a curve is $$x^{2} − 4xy − 2y^{2} = 1$$.

(i) Find an expression for $$\frac{\mathrm{d}y }{\mathrm{d} x}$$  and show that the gradient of the curve at the point $$(−1, 2)$$ is $$– \frac{5}{2}$$.

(ii) Show that the curve has no stationary points.

(iii) Find the x-coordinate of each of the points on the curve at which the tangent is parallel to the y-axis.

(i) State $$R= \sqrt{29}$$ or $$5.385…$$

Use appropriate trigonometry to find  $$\alpha$$

Obtain $$0.3805$$ with no errors seen

(ii)State that equation is $$5cos\Theta -2sin\Theta =4$$

Evaluate  $$cos ^{-1} (k/R) – \alpha$$   to find one value of θ

Obtain $$0.353$$

Carry out correct method to find second value
Obtain 5.17 and no extra solutions in the range.

(iii) State integrand as $$\frac{1}{29}sec^{2}(\frac{x}{2}+0.385)$$
Integrate to obtain from $$ktan(\frac{x}{2}+their\alpha )$$
Obtain $$\frac{2}{29}tan(\frac{x}{2}+0.385)+c$$

### Question

The parametric equations of a curve are

x = 2t − sin 2t, y = 5t + cos 2t, for$$0 ≤ t \frac{1}{2}\pi$$ .

At the point P on the curve, the gradient of the curve is 2.

(i) Show that the value of the parameter at P satisfies the equation 2 sin 2t − 4 cos 2t = 1.

(ii) By first expressing 2 sin 2t − 4 cos 2t in the form$$R sin2t − \Theta$$, where$$R > 0 and 0 < ! <\Theta$$

(i) Obtain expression for $$\frac{dy}{dx}$$ with numerator quadratic, denominator linear

Obtain$$\frac{3t^{2}-6t}{2t+4}$$

Identify t = 3 at P
Obtain $$\frac{9}{10}$$or equivalent

(ii) Equate first derivative to zero and obtain non-zero value of t
Obtain t = 2
Substitute to obtain (12,- 4)

(iii) Equate expression for gradient to m and rearrange to confirm$$3t^{2}-(2m+6)-4m$$=0

Attempt solution of quadratic inequality or equation resulting from
discriminant

Obtain critical values -$$\sqrt{72}-9,m\geqslant \sqrt{72-9}$$

Conclude
$$m\leqslant \sqrt{72-9},m\geqslant \sqrt{72}-9$$

### Question

The equation of a curve is $$y=\frac{e^{2x}}{4x+1}$$ and the point P on the curve has y-coordinate 10.

(i) Show that the x coordinate of P satisfies the equation$$x=\frac{1}{2}In(40x+10).$$

(ii) Use the iterative formula $$x_{n+1}=\frac{1}{2}In(40x_{n}+10)with x_{1}= 2.3$$ to find the x-coordinate of P correct
to 4 significant figures. Give the result of each iteration to 6 significant figures.       [3]

(iii) Find the gradient of the curve at P, giving the answer correct to 3 significant figures.

5(i) Attempt rearrangement of$$\frac{e^{2x}}{4x+1}=10 to x=…..$$involving ln

Confirm$$x=\frac{1}{2} In(40x+10)$$

5(ii) Use iteration process correctly at least once Obtain final answer 2.316 A1
Show sufficient iterations to 6 sf to justify answer or show a sign change in the interval [2.3155, 2.3165]

5(iii) Use quotient rule (or product rule) to find derivative Obtain $$\frac{2e^{2x}(4x+1)-4e^{2x}}{(4x+1)^{2}}$$or equivalent Substitute answer from part (ii) (or more accurate value) into attempt at first derivative  Obtain 16.1

### Question

Find the gradient of the curve $$x^{2}siny+cos3y=4$$
at the point$$(2,\frac{1}{2}\pi ).$$

Use product rule for derivative of$$x^{2}sin y$$

Obtain$$2xsiny +x^{2}cosy\frac{dy}{dx}$$

Obtain$$-3sin3y\frac{dy}{dx}$$ as derivative of cos 3y

Obtain $$2xsiny+x^{2}cosy\frac{dy}{dx}-3sin3y\frac{dy}{dx}$$=0

Substitute$$x=2 y=\frac{1}{2}\pi$$to find value of $$\frac{dy}{dx}$$

Obtain$$-\frac{4}{3}$$

### Question

The equation of a curve is y =$$\frac{3x^{2}}{x^{2}+4}$$ At the point on the curve with positive x-coordinate p, the gradient of the curve is -$$\frac{1}{2}$$

(i) Show that p =$$\sqrt{\left ( \frac{48p-16}{p^{2}+8} \right )}$$

(ii) Show by calculation that 2 < p < 3.

(iii) Use an iterative formula based on the equation in part (i) to find the value of p correct to 4 significant figures. Give the result of each iteration to 6 significant figures

.

(i) Use quotient rule or equivalent

Obtain  or equivalent

Equate first derivative to and remove algebraic denominators dep on

Obtain  or equivalent

Confirm given result

(ii) Consider sign of at 2 and 3 or equivalent

Complete argument correctly with appropriate calculations

(iii) Carry out iteration process correctly at least once  Obtain final answer 2.728  Show sufficient iterations to justify accuracy to 4 sf or show sign change
in interval (2.7275, 2.7285)

### Question

The equation of a curve is y =$$\frac{3x^{2}}{x^{2}+4}$$ At the point on the curve with positive x-coordinate p, the gradient of the curve is -$$\frac{1}{2}$$

(i) Show that p =$$\sqrt{\left ( \frac{48p-16}{p^{2}+8} \right )}$$

(ii) Show by calculation that 2 < p < 3.

(iii) Use an iterative formula based on the equation in part (i) to find the value of p correct to 4 significant figures. Give the result of each iteration to 6 significant figures

.

(i) Use quotient rule or equivalent

Obtain  or equivalent

Equate first derivative to and remove algebraic denominators dep on

Obtain  or equivalent

Confirm given result

(ii) Consider sign of at 2 and 3 or equivalent

Complete argument correctly with appropriate calculations

(iii) Carry out iteration process correctly at least once  Obtain final answer 2.728  Show sufficient iterations to justify accuracy to 4 sf or show sign change
in interval (2.7275, 2.7285)

Question

A curve is defined by the parametric equations

$$x=2tan\Theta$$ , $$y=3sin2\Theta$$

for $$0 ≤ \Theta <\frac{1}{2}\pi$$

(i) Show that$$\frac{dy}{dx}=6cos^{4}\Theta -3cos^{2}\Theta$$

(ii) Find the coordinates of the stationary point.

(iii) Find the gradient of the curve at the point $$(2\sqrt{3},\frac{3}{2}\sqrt{3})$$

(i) Obtain$$\frac{dx}{d\Theta }=2sec^{2}\Theta and \frac{dy}{d\Theta }=6cos2\Theta$$

Use$$cos2\Theta =2cos^{2}\Theta -1$$ or equivalent

Form expression for $$\frac{dy}{dx}$$ in terms of $$cos\Theta$$

Confirm$$6cos^{4}\Theta -3cos^{2}\Theta$$with no errors seen

(ii) Equate first derivative to zero and obtain at least $$cos\Theta =\pm \frac{1}{\sqrt{2}}$$

Obtain$$\Theta =\frac{1}{4}\pi$$  or equivalent

Obtain (2, 3)

(iii) State or imply $$\Theta =\frac{1}{3}\pi$$ or equivalent

Obtain$$-\frac{3}{8}$$ or equivalent only

### Question

Find the gradient of the curve$$y=3e^{4x}-61 In(2x+3)$$ at the point for which x = 0.

Obtain first derivative of form $$k_{1}e^{4x}+\frac{k_{2}}{2x+3}$$

Obtain correct $$12e^{4x}+\frac{12}{2x+3}$$

Obtain 8

### Question

The diagram shows the part of the curve y = 3ex sin 2x for $$0\leqslant x\leqslant \frac{1}{2}\pi$$ , and the stationary point M.

(i) Find the equation of the tangent to the curve at the origin.[4]

(ii) Find the coordinates of M, giving each coordinate correct to 3 decimal places.[4]

Ans:

6 (i) Use product rule to obtain expression of form $$k_{1}e^{-x}sin2x+k_{2}e^{-x}cos2x$$
Obtain correct  $$-3e^{-x}sin2x+6e^{-x}cos2x$$
Substitute x = 0 in first derivative to obtain equation of form y = mx
Obtain y = 6x or equivalent with no errors in solution

(ii) Equate first derivative to zero and obtain tan 2x = k
Carry out correct process to find value of x
Obtain x = 0.554
Obtain y =1.543

### Question

The equation of a curve is 2x3 + y3 = 24.

(i) Express $$\frac{dy}{dx}$$ in terms of x and y, and show that the gradient of the curve is never positive.[4]

(ii) Find the coordinates of the two points on the curve at which the gradient is −2.[5]

Ans:

7 (i) State $$3y^{2}\frac{dy}{dx}$$ as derivative of y3

Equate derivative of left-hand side to zero and solve for $$\frac{dy}{dx}$$

Obtain $$\frac{dy}{dx}=-\frac{6x^{2}}{3y^{2}}$$ or equivalent

Observe x2 and y2 never negative and conclude appropriately

(ii) Equate first derivative to −2 and rearrange to y2 = x2 or equivalent

Substitute in original equation to obtain at least one equation in x3 or y3

Obtain 3x3 = 24 or x3 = 24 or 3y3 = 24 or – y3 = 24

Obtain (2, 2)

Obtain $$\left ( \sqrt[3]{24},- \sqrt[3]{24}\right )\left ( \sqrt[3]{24},- \sqrt[3]{24}\right )$$  − or (2.88, 2.88) − and no others

### Question

The parametric equations of a curve are

x = 6 sin2t, y = 2 sin 2t + 3 cos 2t,

for 0 ≤ t < π. The curve crosses the x-axis at points B and D and the stationary points are A and C, as shown in the diagram.

(i) Show that $$\frac{dy}{dx}=\frac{2}{3}$$ cot 2t − 1. [5]

(ii) Find the values of t at A and C, giving each answer correct to 3 decimal places. [3]

(iii) Find the value of the gradient of the curve at B. [3]

Ans:

7 (i) Obtain 12 sin t cos t or equivalent for $$\frac{dy}{dt}$$
Obtain 4 cos 2t − 6sin 2t or equivalent for $$\frac{dy}{dt}$$
Obtain expression for $$\frac{dy}{dx}$$ in terms of t
Use 2sin t cos t = sin 2t
Confirm given answer $$\frac{dy}{dx}=\frac{2}{3}$$ cot 2t − 1 with no errors seen

(ii) State or imply tan 2t $$=\frac{2}{3}$$
Obtain t = 0.294
Obtain t = 1.865

(iii) Attempt solution of 2sin 2t + 3cos 2t = 0 at least as far as tan 2t = …
Obtain tan 2t $$=-\frac{3}{2}$$ or equivalent
Substitute to obtain $$-\frac{13}{9}$$

Question

A curve has equation

$$y=\frac{3x+1}{x-5}.$$

Find the coordinates of the points on the curve at which the gradient is −4. [5]

Ans:

2 Use quotient rule or, after adjustment, product rule
Obtain $$\frac{3x-15-3x-1}{\left ( x-5 \right )^{2}}$$ or equivalent
Equate first derivative to –4 and solve for x
Obtain x-coordinates 3 and 7 or one correct pair of coordinates
Obtain y-coordinates –5 and 11 respectively or other correct pair of coordinates

Question

The parametric equations of a curve are $$x=\cos 2\Theta -\cos \Theta ,y=4\sin ^{2}\Theta$$

,for $$0\leq \Theta \leq \pi$$.
(i) Show that $$\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{8\cos \Theta }{1-4\cos \Theta }$$.

(ii) Find the coordinates of the point on the curve at which the gradient is −4.

(i) State $$\frac{\mathrm{d} x}{\mathrm{d} \Theta }=-2\sin 2\Theta +\sin \Theta$$ or $$\frac{\mathrm{d} y}{\mathrm{d} \Theta }=8\sin 2\Theta \cos \Theta$$

Use $$\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{\mathrm{d} y}{\mathrm{d} \Theta }\div \frac{\mathrm{d} x}{\mathrm{d} \Theta }$$.

Use sin 2θ = 2sinθ cosθ
(ii) Equate derivative to −4 and solve for cos θ M1
Obtain cos θ = 1⁄2
Obtain x = −1
Obtain y = 3

Question

The parametric equations of a curve are

$$x=e^{2t},y=4te^{^{t}}$$.
(i) Show that $$\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{2(t+1)}{e^{t}}$$
(ii) Find the equation of the normal to the curve at the point where t = 0.

(i) Use product rule to differentiate y M1
Obtain correct derivative in any form A1
Use $$\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{\mathrm{d} y}{\mathrm{d} t}\div \frac{\mathrm{d} x}{\mathrm{d} t}$$

(ii) Substitute t = 0 in $$\frac{\mathrm{d} y}{\mathrm{d} x}$$ and both parametric equations

Obtain $$\frac{\mathrm{d} y}{\mathrm{d} x}=2$$ and coordinates (1, 0) .

Form equation of the normal at their point, using negative reciprocal of their $$\frac{\mathrm{d} y}{\mathrm{d} x}$$
State correct equation of normal $$y=-\frac{1}{2}x+\frac{1}{2}$$ or equivalent

Question

(i) By differentiating $$\frac{1}{\cos \Theta }$$, show that if $$y=\sec \Theta$$ then $$\frac{\mathrm{d} y}{\mathrm{d} \Theta }=\tan \Theta \sec \Theta$$.

(ii)Hence show that  $$\frac{\mathrm{d} ^{2}y}{\mathrm{d} \Theta ^{2}}=a\sec ^{3}\Theta +b\sec \Theta$$ giving the values of a and b.

(iii)Find the exact value of $$\int _{0}^{\frac{1}{4}\pi }(1+\tan ^{2}\Theta -3\sec \Theta \tan \Theta )d\Theta$$.

(i)Differentiate using chain or quotient rule

Obtain derivative in any correct form.

(ii)Differentiate using product rule

State derivative of $$\tan \Theta =\sec ^{2}\Theta$$

Use trig identity $$1+\tan ^{2}\Theta =\sec ^{2}\Theta$$  correctly

Obtain $$2\sec ^{3}\Theta -\sec \Theta$$

(iii)Use $$\tan ^{2}\Theta =\sec ^{2}\Theta -1$$ to integrate $$\tan ^{2}\Theta$$

Obtain $$3\sec \Theta$$  from integration of $$3\sec \Theta \tan \Theta$$

Obtain $$\tan \Theta -3\sec \Theta$$

Attempt to substitute limits ,using exact values

Obtain answer $$4-3\sqrt{2}$$

Question

The parametric equation of a curve are $$x=\ln (1-2t)$$,$$y=\frac{2}{t}$$ ,for $$t< 0$$.

(i)Show that $$\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{1-2t}{r^{2}}$$.

(ii) Find the exact coordinates of the only point on the curve at which the gradient is 3.

(i) State $$\frac{\mathrm{d} x}{\mathrm{d} t}=\frac{-2}{1-2t}$$ or $$\frac{\mathrm{d} y}{\mathrm{d} t}=-2t^{-2}$$
Use $$\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{\mathrm{d} y}{\mathrm{d} t}\div \frac{\mathrm{d} x}{\mathrm{d} t}$$
(ii) Equate derivative to 3 and solve for t
State or imply that t = –1
Obtain coordinates (ln 3, –2)

Question

A curve has parametric equations $$x=\frac{1}{(2t+1)^{2}},y=\sqrt{(t+2)}$$.
The point P on the curve has parameter p and it is given that the gradient of the curve at P is −1.
(i) Show that $$p=(p+2)^{\frac{1}{6}}-\frac{1}{2}$$.
(ii) Use an iterative process based on the equation in part (i) to find the value of p correct to 3 decimalplaces. Use a starting value of 0.7 and show the result of each iteration to 5 decimal places

(i) Obtain derivative of form$$k(2t+1)^{-3}$$.
Obtain  $$-4(2t+1)^{-3}$$ or equivalent as derivative of x .
Obtain $$\frac{1}{2}(t+2)^{-\frac{1}{2}}$$ or equivalent as derivative of y

Equate attempt at $$\frac{\mathrm{d} y}{\mathrm{d} x}$$ to -1
Obtain $$(2p+1)^{3}=8(p+2)^{\frac{1}{2}}$$ or equivalent

Confirm given answer $$p=(p+2)^{\frac{1}{6}}-\frac{1}{2}$$

(ii) Use iteration process correctly at least once
Show sufficient iterations to 5 decimal places to justify answer or show a sign change in the interval (0.6775, 0.6785)
[0.7 → 0.68003 → 0.67857 → 0.67847 → 0.6784

Question

The diagram shows the curve$$y=4e^{\frac{1}{2}x}-6x+3$$ and its minimum point M.

(i) Show that the x-coordinate of M can be written in the form ln a, where the value of a is to be stated.
(ii) Find the exact value of the area of the region enclosed by the curve and the lines x = 0, x = 2 and y = 0.

(i) Differentiate to obtain expression of form  $$ke^{\frac{1}{2}x}+m$$

Obtain correct $$2e^{\frac{1}{2}x}-6$$
Equate attempt at first derivative to zero and attempt solution
Obtain $$\frac{1}{2}x=\ln 3$$ or equivalent
Conclude $$x=\ln 9$$ or a=9

(ii) Integrate to obtain expression of form $$ae^{\frac{1}{2}x}+bx^{2}+cx$$

Obtain correct $$8e^{\frac{1}{2}x}-3x^{2}+3x$$

Substitute correct limits and attempt simplification
Obtain 8e – 14

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