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]

Answer/Explanation

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.

Answer/Explanation

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).

Answer/Explanation

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.

Answer/Explanation

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.

Answer/Explanation

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]

Answer/Explanation

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]

Answer/Explanation

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]

Answer/Explanation

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

          Obtain final answer 3.3223

         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]

Answer/Explanation

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

Answer/Explanation

(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.

Answer/Explanation

(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

Obtain final answer 0.357

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.

Answer/Explanation

(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

Obtain final answer 1.82

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 \(\int_{\frac{1}{4}\pi }^{\frac{1}{3}\pi } \left ( tan x + tan^{2}x + tan^{3}x \right )dx\) .

Answer/Explanation

(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. 

Answer/Explanation

(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.

Answer/Explanation

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

Answer/Explanation

.

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.

Answer/Explanation

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.

Answer/Explanation

<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 

Obtain final answer 1.81

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).

Answer/Explanation

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. 

Answer/Explanation

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.

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

     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.

Answer/Explanation

(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\( R > 0 and 0 < ! <\Theta\)

Answer/Explanation

(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.                               

Answer/Explanation

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\( (2,\frac{1}{2}\pi ).\)

Answer/Explanation

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

Answer/Explanation

.

(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

Answer/Explanation

.

(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 \((2\sqrt{3},\frac{3}{2}\sqrt{3})\)

Answer/Explanation

(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.

Answer/Explanation

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]

Answer/Explanation

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]

Answer/Explanation

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]

Answer/Explanation

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]

Answer/Explanation

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.

Answer/Explanation

(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 
Obtain given answer correctly 
(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.

Answer/Explanation

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

Obtain given answer correctly.
(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 \(\int _{0}^{\frac{1}{4}\pi }(1+\tan ^{2}\Theta -3\sec \Theta \tan \Theta )d\Theta \).

Answer/Explanation

(i)Differentiate using chain or quotient rule

Obtain derivative in any correct form.

Obtain given answer correctly 

(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.

Answer/Explanation

(i) State \(\frac{\mathrm{d} x}{\mathrm{d} t}=\frac{-2}{1-2t}\) or \(\frac{\mathrm{d} y}{\mathrm{d} t}=-2t^{-2}\)
Use \(\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{\mathrm{d} y}{\mathrm{d} t}\div \frac{\mathrm{d} x}{\mathrm{d} t}\)
Obtain given answer correctly
(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

Answer/Explanation

(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 
Obtain final answer 0.678 
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.

Answer/Explanation

(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

Scroll to Top