Tangent and Normal Lines Problems Solutions.Pdf

New test - November 03, 2015 [79 marks]

Let f(x) = e2x cosx , −1 ≤ x ≤ 2 .

′ 2x 1a. Show that f (x) = e (2cosx − sin x) . [3 marks]

Markscheme correctly finding the derivative of e2x , i.e. 2e2x A1 correctly finding the derivative of cosx , i.e. −sin x A1 evidence of using the product rule, seen anywhere M1

e.g. f′ (x) = 2e2x cosx − e2x sin x f′ (x) = 2e2x (2cosx − sin x) AG N0 [3 marks]

Examiners report A good number of candidates demonstrated the ability to apply the product and chain rules to obtain the given derivative.

1b. Let the line L be the normal to the curve of f at x = 0 . [5 marks] Find the equation of L .

Markscheme evidence of finding f(0) = 1 , seen anywhere A1 attempt to find the gradient of f (M1)

e.g. substituting x = 0 into f′ (x) value of the gradient of f A1

e.g. f′ (0) = 2 , equation of tangent is y = 2x + 1 gradient of normal 1 (A1) = − 2 y 1 x y 1 x A1 N3 − 1 = − 2 ( = − 2 + 1) [5 marks]

Examiners report Where candidates recognized that the gradient of the tangent is the derivative, many went on to correctly find the equation of the normal.

1c. The graph of f and the line L intersect at the point (0, 1) and at a second point P. [6 marks] (i) Find the x-coordinate of P. (ii) Find the area of the region enclosed by the graph of f and the line L . Markscheme (i) evidence of equating correct functions M1 e.g. 2x x 1 x , sketch showing intersection of graphs e cos = − 2 + 1 x = 1.56 A1 N1 (ii) evidence of approach involving subtraction of integrals/areas (M1)

e.g. ∫ [f(x) − g(x)]dx , ∫ f(x)dx − area under trapezium fully correct integral expression A2

e.g. 1.56 2x x 1 x x , 1.56 2x x x ∫0 [e cos − (− 2 + 1)]d ∫0 e cos d − 0.951… area = 3.28 A1 N2 [6 marks]

Examiners report Few candidates showed the setup of the equation in part (c) before writing their answer from the GDC. Although a good number of candidates correctly expressed the integral to find the area between the curves, surprisingly few found a correct answer. Although this is a GDC paper, some candidates attempted to integrate this function analytically.

Let f(x) = ex(1 − x2 ) .

′ x 2 [3 marks] 2a. Show that f (x) = e (1 − 2x − x ) .

Markscheme evidence of using the product rule M1

f′ (x) = ex(1 − x2 ) + ex(−2x) A1A1 Note: Award A1 for ex(1 − x2 ) , A1 for ex(−2x) .

f′ (x) = ex(1 − 2x − x2 ) AG N0 [3 marks]

Examiners report Many candidates clearly applied the product rule to correctly show the given derivative. Some candidates missed the multiplicative nature of the function and attempted to apply a chain rule instead.

y = f(x) Part of the graph of y = f(x), for −6 ≤ x ≤ 2 , is shown below. The x-coordinates of the local minimum and maximum points are r and s respectively.

2b. Write down the equation of the horizontal asymptote. [1 mark]

Markscheme y = 0 A1 N1 [1 mark]

Examiners report For part (b), the equation of the horizontal asymptote was commonly written as x = 0 .

2c. Write down the value of r and of s. [4 marks]

Markscheme at the local maximum or minimum point

f′ (x) = 0 (ex(1 − 2x − x2 ) = 0) (M1) ⇒ 1 − 2x − x2 = 0 (M1) r = −2.41 s = 0.414 A1A1 N2N2 [4 marks]

Examiners report Although part (c) was a “write down” question where no working is required, a good number of candidates used an algebraic method of solving for r and s which sometimes returned incorrect answers. Those who used their GDC usually found correct values, although not always to three significant figures.

2d. Let L be the normal to the curve of f at P(0, 1) . Show that L has equation x + y = 1 . [4 marks] Markscheme f′ (0) = 1 A1 gradient of the normal = −1 A1 evidence of substituting into an equation for a straight line (M1) correct substitution A1

e.g. y − 1 = −1(x − 0) , y − 1 = −x , y = −x + 1 x + y = 1 AG N0 [4 marks]

Examiners report In part (d), many candidates showed some skill showing the equation of a normal, although some tried to work with the gradient of the tangent.

2e. Let R be the region enclosed by the curve y = f(x) and the line L. [5 marks] (i) Find an expression for the area of R. (ii) Calculate the area of R.

Markscheme (i) intersection points at x = 0 and x = 1 (may be seen as the limits) (A1) approach involving subtraction and integrals (M1) fully correct expression A2 N4

e.g. 1 x 2 , 1 1 ∫0 (e (1 − x ) − (1 − x))dx ∫0 f(x)dx − ∫0 (1 − x)dx (ii) area R = 0.5 A1 N1 [5 marks]

Examiners report Surprisingly few candidates set up a completely correct expression for the area between curves that considered both integration and the correct subtraction of functions. Using limits of −6 and 2 was a common error, as was integrating on f(x) alone. Where candidates did write a correct expression, many attempted to perform analytic techniques to calculate the area instead of using their GDC.

Let f(x) = e6x .

′ [1 mark] 3a. Write down f (x) .

Markscheme f′ (x) = 6e6x A1 N1 [1 mark]

P(0, b) 3b. The tangent to the graph of f at the point P(0, b) has gradient m . [4 marks] (i) Show that m = 6 . (ii) Find b .

Markscheme (i) evidence of valid approach (M1)

e.g. f′ (0) , 6e6×0 correct manipulation A1

e.g. 6e0 , 6 × 1 m = 6 AG N0 (ii) evidence of finding f(0) (M1) e.g. y = e6(0) b = 1 A1 N2 [4 marks]

Examiners report On the whole, candidates handled this question quite well with most candidates correctly applying the chain rule to an exponential function and successfully finding the equation of the tangent line. Some candidates lost a mark in (b)(i) for not showing sufficient working leading to the given answer.

[1 mark] 3c. Hence, write down the equation of this tangent.

Markscheme y = 6x + 1 A1 N1 [1 mark]

Examiners report On the whole, candidates handled this question quite well.

Let f x kx4 . The point k lies on the curve of f . At P, the normal to the curve is parallel to y 1 x . Find the value [6 marks] 4. ( ) = P(1, ) = − 8 of k.

Markscheme gradient of tangent = 8 (seen anywhere) (A1) f′ (x) = 4kx3 (seen anywhere) A1 recognizing the gradient of the tangent is the derivative (M1) setting the derivative equal to 8 (A1)

e.g. 4kx3 = 8 , kx3 = 2 substituting x = 1 (seen anywhere) (M1) k = 2 A1 N4 [6 marks] Examiners report Candidates‟ success with this question was mixed. Those who understood the relationship between the derivative and the gradient of the normal line were not bothered by the lack of structure in the question, solving clearly with only a few steps, earning full marks. Those who were unclear often either gained a few marks for finding the derivative and substituting x = 1 , or no marks for working that did not employ the derivative. Misunderstandings included simply finding the equation of the tangent or normal line, setting the derivative equal to the gradient of the normal, and equating the function with the normal or tangent line equation. Among the candidates who demonstrated greater understanding, more used the gradient of the normal (the equation 1 k 1 ) than the − 4 = − 8 gradient of the tangent (4k = 8 ) ; this led to more algebraic errors in obtaining the final answer of k = 2 . A number of unsuccessful candidates wrote down a lot of irrelevant mathematics with no plan in mind and earned no marks.

2 5. Consider the curve with equation f(x) = px + qx , where p and q are constants. The point A(1, 3) lies on the curve. The [7 marks] tangent to the curve at A has gradient 8. Find the value of p and of q .

Markscheme substituting x = 1 , y = 3 into f(x) (M1) 3 = p + q A1 finding derivative (M1)

f′ (x) = 2px + q A1 correct substitution, 2p + q = 8 A1 p = 5 , q = −2 A1A1 N2N2 [7 marks]

Examiners report A good number of candidates were able to obtain an equation by substituting the point 1, 3 into the function’s equation. Not as many knew how to find the other equation by using the derivative. Some candidates thought they needed to find the equation of the tangent line rather than recognising that the information about the tangent provided the gradient of the function at the point. While they were usually able to find this equation correctly, it was irrelevant to the question asked.

x [6 marks] 6. Let f(x) = e cosx . Find the gradient of the normal to the curve of f at x = π .

Markscheme evidence of choosing the product rule (M1)

f′ (x) = ex × (−sin x) + cosx × ex (= ex cosx − ex sin x) A1A1 substituting π (M1) e.g. f′ (π) = eπ cosπ − eπ sin π , eπ(−1 − 0) , −eπ taking negative reciprocal (M1)

e.g. − 1 f ′(π) gradient is 1 A1 N3 eπ [6 marks]

Examiners report Candidates familiar with the product rule easily found the correct derivative function. Many substituted π to find the tangent gradient, but surprisingly few candidates correctly considered that the gradient of the normal is the negative reciprocal of this answer.

y = ln(3x − 1) Consider the curve y = ln(3x − 1) . Let P be the point on the curve where x = 2 .

7a. Write down the gradient of the curve at P. [2 marks]

Markscheme gradient is 0.6 A2 N2 [2 marks]

Examiners report Although the command term "write down" was used in part (a), many candidates still opted for an analytic method for finding the derivative value. Although this value was often incorrect, many candidates knew how to find the equation of the normal and earned follow through marks in part (b).

7b. The normal to the curve at P cuts the x-axis at R. Find the coordinates of R. [5 marks]

Markscheme at R, y = 0 (seen anywhere) A1 at x = 2 , y = ln5 (= 1.609…) (A1) gradient of normal = −1.6666… (A1) evidence of finding correct equation of normal A1

e.g. y 5 x , y x c = ln5 = − 3 ( − 2) = −1.67 + x = 2.97 (accept 2.96) A1 coordinates of R are (2.97,0) N3 [5 marks]

Let f x x 1 x2 x , for x π . ( ) = sin + 2 − 2 0 ≤ ≤

′ [3 marks] 8a. Find f (x) .

Markscheme f′ (x) = cosx + x − 2 A1A1A1 N3 Note: Award A1 for each term. [3 marks]

Examiners report In part (a), most candidates were able to correctly find the derivative of the function.

g g(0) = 5 x = 2 g g g(0) = 5 . The line g

8b. Find g(4) . [3 marks]

Markscheme recognizing g(0) = 5 gives the point (0, 5) (R1) recognize symmetry (M1) eg vertex, sketch

g(4) = 5 A1 N3 [3 marks]

Examiners report In part (b), many candidates did not understand the significance of the axis of symmetry and the known point (0, 5), and so were unable to find g(4) using symmetry. A few used more complicated manipulations of the function, but many algebraic errors were seen.

The function g can be expressed in the form g(x) = a(x − h)2 + 3 .

[4 marks] 8c. (i) Write down the value of h . (ii) Find the value of a .

Markscheme (i) h = 2 A1 N1

(ii) substituting into g(x) = a(x − 2)2 + 3 (not the vertex) (M1) eg 5 = a(0 − 2)2 + 3 , 5 = a(4 − 2)2 + 3 working towards solution (A1)

eg 5 = 4a + 3 , 4a = 2 a 1 A1 N2 = 2

[4 marks] Examiners report In part (c), a large number of candidates were able to simply write down the correct value of h, as intended by the command term in this question. A few candidates wrote down the incorrect negative value. Most candidates attempted to substitute the x and y values of the known point correctly into the function, but again many arithmetic and algebraic errors kept them from finding the correct value for a.

8d. Find the value of x for which the tangent to the graph of f is parallel to the tangent to the graph of g . [6 marks]

Markscheme g x 1 x 2 1 x2 x ( ) = 2 ( − 2) + 3 = 2 − 2 + 5 correct derivative of g A1A1 eg 1 x , x 2 × 2 ( − 2) − 2 evidence of equating both derivatives (M1)

eg f′ = g′ correct equation (A1)

eg cosx + x − 2 = x − 2 working towards a solution (A1)

eg cosx = 0 , combining like terms x π A1 N0 = 2 Note: Do not award final A1 if additional values are given. [6 marks]

Examiners report Part (d) required the candidates to find the derivative of g, and to equate that to their answer from part (a). Although many candidates were able to simplify their equation to cosx = 0, many did not know how to solve for x at this point. Candidates who had made errors in parts (a) and/or (c) were still able to earn follow-through marks in part (d).

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