Pure Math 1

What is the equation of the tangent to the curve y=x^(3/2)-3x-4x^(1/2)+4 at the point (4, 0)?

• y=2(x-8)

• y=2(x+8)

• y=2(x-4)

• y=2x+4

A function f is defined by f:x→x^3 −x^2 −8x+5 for x<a. It is given that f is an increasing function.

Find the largest possible value of the constant a.

• a=-4/3

• a=-3/4

• a=4/3

• a=3/4

(a) A geometric progression has first term 3a and common ratio r. A second geometric progression has first term a and common ratio −2r. The two progressions have the same sum to infinity. Find the value of r.

(b) The first two terms of an arithmetic progression are 15 and 19 respectively. The first two terms of a second arithmetic progression are 420 and 415 respectively. The two progressions have the same sum of the first n terms. Find the value of n.

• r=-2/7
  n=91

• r=-4/7
  n=91

• r=-2/7
  n=87

• r=-4/7
  n=87

Machines in a factory make cardboard cones of base radius r cm and vertical height h cm. The volume,

V cm^3, of such a cone is given by V = (π r^2 h)/3. The machines produce cones for which h + r = 18. 3

(i) Find V.
(ii) Given that r can vary, find the non-zero value of r for which V has a stationary value and find if it is maximum or minimum.
(iii) Find the maximum volume of a cone that can be made by these machines.

• (i) V = (6 − r) π r^2
  (ii) Min
  (iii) 859
  

• (i) V = (6 − r) π r
  (ii) Max
  (iii)859

• (i) V = (6 − r) π r^2
  (ii) Max
  (iii) 905

• (i) V = (6 − r) π r
  (ii) Min
  (iii) 905

The diagram shows an isosceles triangle ABC in which AC = 16 cm and AB = BC = 10 cm. The circular arcs BE and BD have centres at A and C respectively, where D and E lie on AC.
(i)  Find angle BAC, correct to 4 decimal places.
(ii)  Find the area of the shaded region.

• (i) BAC = 0.6435 radians
  (ii) A = 15.8

• (i) BAC = 0.6435 radians
  (ii) A = 16.4

• (i) BAC = 0.6425 radians
  (ii) A = 16.4

• (i) BAC = 0.6425 radians
  (ii) A = 15.8

(a) The diagram shows part of the graph of y = a + b sin x. Find the values of the constants a and b.
(b) (i) Simplify the equation
(sin Θ + 2 cos Θ) (1 + sin Θ − cos Θ)
= sin Θ (1 + cos Θ)

(ii) Hence solve the equation
(sin Θ + 2 cos Θ) (1 + sin Θ − cos Θ)
= sin Θ (1 + cos Θ)

for −180° ≤ Θ ≤ 180°

• (a) a = -2, b = 3

  (b)
  (i) 3 (cosΘ)^2 − 2 cosΘ − 1 = 0
  (ii) Θ = 0, ± 109.5

• (a) a = 2, b = -3

  (b)
  (i) 3 (cosΘ)^2 − 2 cosΘ − 1 = 0
  (ii) Θ = 0, ± 109.5

• (a) a = -2, b = 3

  (b)
  (i) 3(cosΘ)^2+2cosΘ − 1 = 0
  (ii) Θ = 0, ± 70.5

• (a) a = 2, b = -3

  (b)
  (i) 3(cosΘ)^2+2cosΘ − 1 = 0
  (ii) Θ = 0, ± 70.5

(a) Relative to an origin O, the position vectors of two points P and Q are p and q respectively. The point R is such that PQR is a straight line with Q the mid-point of PR. Find the position vector of R in terms of p and q, simplifying your answer.
(b) The vector 6i + aj + bk has magnitude 21 and is perpendicular to 3i + 2j + 2k. Find the possible

values of a and b, showing all necessary working.

• (a) 2p-q
  (b) a=9 or−18
  b=−18 or 9

• (a) 2q‒p
  (b) a=9 or−18
  b=−18 or 9

• (a) 2q‒p
  (b) a=-9 or18
  b=18 or -9

• (a) 2p‒q
  (b) a=-9 or18
  b=18 or -9

The diagram shows part of the curve y = 1/2 (x^4 − 1)
, defined for x ≥ 0.

1. (i)  Find the area of the shaded region.
   (ii)  Find the volume obtained when the shaded region is rotated through 360° about the x-axis.
   (iii) Find, showing all necessary working, the volume obtained when the shaded region is rotated through 360° about the y-axis.

• (i) 2/5
  (ii) 0.559

  (iii) 1.05

• (i) 3/5
  (ii) 0.559

  (iii) 0.95

• (i) 2/5
  (ii) 0.441

  (iii) 0.95

• (i) 3/5
  (ii) 0.441

  (iii) 1.05