Sue Pemberton, Julianne Hughes and, Julian Gilbey Solutions for Chapter: Cross-Topic Review Exercise 1, Exercise 1: CROSS-TOPIC REVIEW EXERCISE 1

Author:Sue Pemberton, Julianne Hughes & Julian Gilbey

Sue Pemberton Mathematics Solutions for Exercise - Sue Pemberton, Julianne Hughes and, Julian Gilbey Solutions for Chapter: Cross-Topic Review Exercise 1, Exercise 1: CROSS-TOPIC REVIEW EXERCISE 1

Attempt the free practice questions on Chapter 14: Cross-Topic Review Exercise 1, Exercise 1: CROSS-TOPIC REVIEW EXERCISE 1 with hints and solutions to strengthen your understanding. Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book solutions are prepared by Experienced Embibe Experts.

Questions from Sue Pemberton, Julianne Hughes and, Julian Gilbey Solutions for Chapter: Cross-Topic Review Exercise 1, Exercise 1: CROSS-TOPIC REVIEW EXERCISE 1 with Hints & Solutions

HARD
AS and A Level
IMPORTANT

Hence, using logarithms, solve the equation 3×2y+4=3×2y-11, giving the answer correct to 3 significant figures.

HARD
AS and A Level
IMPORTANT

Solve the equation 25x-1=35x, giving your answer correct to 3 significant figures.

HARD
AS and A Level
IMPORTANT

solve the equation 5 sin 2θ+2 cos 2θ=4, giving all solutions in the interval 0°θ360°,

MEDIUM
AS and A Level
IMPORTANT

Determine the least value of 1(10 sin 2θ+4 cos 2θ)2 as θ varies.

HARD
AS and A Level
IMPORTANT

The polynomial p(x) is defined by p(x)=ax3+3x2+bx+12, where a and b are constants. It is given that (x+3) is a factor of p(x). It is also given that the remainder is 18 when p(x) is divided by (x+2).

Find the values of a and b.

HARD
AS and A Level
IMPORTANT

The polynomial p(x) is defined by p(x)=ax3+3x2+bx+12, where a and b are constants. It is given that (x+3) is a factor of p(x). It is also given that the remainder is 18 when p(x) is divided by (x+2). When a=2 and b=-5 have these values, show that the equation p(x)=0 has exactly one real root,

HARD
AS and A Level
IMPORTANT

The polynomial p(x) is defined by p(x)=ax3+3x2+bx+12, where a and b are constants. It is given that (x+3) is a factor of p(x). It is also given that the remainder is 18 when p(x) is divided by (x+2). When a=2 and b=-5 have these values, solve the equation p(sec y)=0 for -180°<y<180°