Solve x x - 1 ⩾ 3 .

Given that, x x - 1 ⩾ 3 , x - 1 ≠ 0 ⇒ x x - 1 - 3 ≥ 0 , x ≠ 1 ⇒ x - 3 x + 3 x - 1 ≥ 0 ⇒ 3 - 2 x x - 1 ≥ 0 if, m n ≥ 0 , then m n ≥ 0 ⇒ 3 - 2 x x - 1 ≥ 0 ⇒ ( 2 x - 3 ) ( x - 1 ) ≤ 0 ⇒ x - 3 2 x - 1 ≤ 0 If ( x - a ) ( x - b ) ≤ 0 , then a ≤ x ≤ b , where a < b . Here, a = 1 and b = 3 2 Therefore, solution of given inequality is 1 < x ⩽ 3 2 , since x ≠ 1

Solve x ( x - 1 ) x + 1 > x .

Given that, x ( x - 1 ) x + 1 > x , x + 1 ≠ 0 ⇒ x ( x - 1 ) x + 1 - x > 0 , x ≠ - 1 ⇒ x 2 - x - x 2 - x x + 1 > 0 ⇒ - 2 x x + 1 > 0 If, m n > 0 , then m n > 0 ⇒ - 2 x x + 1 > 0 ⇒ ( 2 x ) ( x + 1 ) < 0 If ( x - a ) ( x - b ) < 0 , then a < x < b , where a < b . Here, a = - 1 and b = 0 Therefore, solution of given inequality is - 1 < x < 0

Solve x 2 - 9 x - 1 ⩾ 4 .

Given that, x 2 - 9 x - 1 ⩾ 4 , x - 1 ≠ 0 ⇒ x 2 - 9 x - 1 - 4 ≥ 0 , x ≠ 1 ⇒ x 2 - 9 - 4 x + 4 x - 1 ≥ 0 ⇒ x 2 - 4 x - 5 x - 1 ≥ 0 if, m n ≥ 0 , then m n ≥ 0 ⇒ x 2 - 4 x - 5 x - 1 ≥ 0 ⇒ ( x - 5 ) x + 1 ( x - 1 ) ≥ 0 Using, wavy curve method. Zeroes of above inequality are - 1 , 1 & 5 . x > 5 ⇒ ( x - 5 ) ( x + 1 ) ( x - 1 ) > 0 Therefore, starting from RHS, we get We need to consider the range where ( x - 5 ) ( x + 1 ) ( x - 1 ) ≥ 0 . Therefore, solution of given inequality is - 1 ⩽ x < 1 or x ≥ 5 , since x ≠ 1

Solve x 2 - 2 x - 15 x - 2 ⩾ 0 .

Given that, x 2 - 2 x - 15 x - 2 ⩾ 0 , x - 2 ≠ 0 ⇒ x 2 - 2 x - 15 x - 2 ⩾ 0 , x ≠ 2 ⇒ x 2 - 5 x + 3 x - 15 x - 2 ≥ 0 ⇒ x x - 5 + 3 x - 5 x - 2 ≥ 0 ⇒ x + 3 ( x - 5 ) x - 2 ≥ 0 if, m n ≥ 0 , then m n ≥ 0 ⇒ ( x + 3 ) x - 2 ( x - 5 ) ≥ 0 Using, wavy curve method. Zeroes of above inequality are - 3 , 2 & 5 . x > 5 ⇒ ( x + 3 ) ( x - 2 ) ( x - 5 ) > 0 Therefore, starting from RHS, we get We need to consider the range where ( x - 5 ) ( x + 1 ) ( x - 1 ) ≥ 0 . Therefore, solution of given inequality is - 5 ⩽ x < - 2 or x ≥ 5 , since x ≠ 2 .

Solve x 2 + 4 x - 5 x 2 - 4 ⩽ 0 .

Given that, x 2 + 4 x - 5 x 2 - 4 ⩽ 0 , x 2 - 4 ≠ 0 ⇒ x 2 + 4 x - 5 x 2 - 4 ⩽ 0 , x ≠ 2 , - 2 ⇒ x 2 + 5 x - x - 5 x 2 - 4 ≤ 0 ⇒ x x + 5 - 1 x + 5 x - 2 x + 2 ≤ 0 ⇒ x + 5 ( x - 1 ) x - 2 x + 2 ≤ 0 if, m n ≤ 0 , then m n ≤ 0 ⇒ x + 5 ( x + 2 ) x - 1 ( x - 2 ) ≤ 0 Using, wavy curve method. Zeroes of above inequality are - 5 , - 2 , 1 & 2 . x < - 5 ⇒ ( x + 5 ) ( x + 2 ) ( x - 1 ) ( x - 2 ) > 0 Therefore, starting from LHS, we get We need to consider the range where ( x + 5 ) ( x + 2 ) ( x - 1 ) ( x - 2 ) ≤ 0 . Therefore, solution of given inequality is - 5 ⩽ x < - 2 or 1 ⩽ x < 2 , since x ≠ 2 , - 2 .

Solve x - 3 x + 4 ⩾ x + 2 x - 5 .

Given that, x - 3 x + 4 ⩾ x + 2 x - 5 , x + 4 ≠ 0 & x - 5 ≠ 0 ⇒ x - 3 x + 4 - x + 2 x - 5 ⩾ 0 , x ≠ - 4 , 5 ⇒ x - 3 x - 5 - x + 2 x + 4 x + 4 x - 5 ≥ 0 ⇒ x 2 - 8 x + 15 - x 2 + 6 x + 8 ( x + 4 ) ( x - 5 ) ≥ 0 ⇒ - 14 x + 7 ( x + 4 ) ( x - 5 ) ≥ 0 if, m n ≥ 0 , then m n ≥ 0 ⇒ - 7 ( 2 x - 1 ) ( x + 4 ) ( x - 5 ) ≥ 0 ⇒ ( 2 x - 1 ) ( x + 4 ) ( x - 5 ) ≤ 0 Using, wavy curve method. Zeroes of above inequality are - 4 , 1 2 & 5 . x > 5 ⇒ ( 2 x - 1 ) ( x + 4 ) ( x - 5 ) > 0 Therefore, starting from RHS, we get We need to consider the range where ( 2 x - 1 ) ( x + 4 ) ( x - 5 ) ≤ 0 . Therefore, solution of given inequality is x < - 4 or 1 2 ⩽ x < 5 , since x ≠ - 4 , 5 .

E M B I B E

Cambridge International AS & A Level Mathematics : Pure Mathematics 1 Course Book>Chapter 1 - Quadratics>EXERCISE 1G

Sue Pemberton Solutions for Chapter: Quadratics, Exercise 7: EXERCISE 1G

Author:Sue Pemberton

Sue Pemberton Mathematics Solutions for Exercise - Sue Pemberton Solutions for Chapter: Quadratics, Exercise 7: EXERCISE 1G

Attempt the practice questions on Chapter 1: Quadratics, Exercise 7: EXERCISE 1G with hints and solutions to strengthen your understanding. Cambridge International AS & A Level Mathematics : Pure Mathematics 1 Course Book solutions are prepared by Experienced Embibe Experts.