Advanced mathematics 1 by C. W. Celia, A. T. F. Nice, K. F. Elliott

x-1 30 Advanced mathematics 1 (k) 4x-1 < -2 (I) 3 -4x > 2 (m) 4-5x > -1 (n) 5-6x<7 (o) 1-3x > -6. (a) when x > 1 (b) when x < 1. State the complete set of values of x for which 4x+8 3 - - 1 >. x- S Find the set of values of x for which (a) 3 +4x x <3 (b) 3-4x 2-x <2 (c) 3x +4 x+3 2x-4 (d) X -1 1-4x x+2 >2 (e) - - > 4 <3 2-3x (f) 1-2x < S.

X+3 x-! ) -ve +ve +ve -ve -ve +ve +ve -ve +ve 2(x+3)(x-t) >0 when x < -3 and when x > t, 2x 2 +5x-3 > 0 when x < -3 and when x > t, and the required set is the union of the sets {x e IR: x < -3} and {x e IR: x > t}. Alternatively, Fig. 3 shows a sketch of the graph of y = 2x 2 + 5x- 3. From the graph it is clear that y is positive when x < - 3 and when x > t. Hence 2x 2 + 5x - 3 > 0 when x < - 3 and when x > t. X Fig. 3 Example 2 Find the set of values of x for which 1 + 2x - 3x 2 < 0. The roots of the equation 1 + 2x - 3x 2 = 0 are -!

A+ d)+ a = (a+ I)+ (a+ /)+(a+ /) + ... n(a +I). n[2a + (n -1)d]. e. n(n + 1)- n n giving r L r = 1 Example n L Evaluate r r(r + 1). = 1 &2 Advanced mathematics 1 2 = i-n(n + 1)(2n + 1). The series can be written as 1(2)+2(3)+3(4)+ ... +n(n+ 1). Let the sum of the series be sn. n(n + 1) 1 = 6 n(n+ 1)[(2n+ 1)+3] 1 = 3 n(n+ 1) (n+2). 3 1 Evaluate n (a) L n (2r-1) L (e) n (b) L n (3r+2) (f) I