Page 29 - mathsvol1ch1to3ans

29

3
x − 1
13.
x + x + 1
2
2
3
x − 1 (x − 1)(x + x + 1)
Since = ,
2
2
x + x + 1 x + x + 1
We have = x − 1

Exercise - 2.15

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Determine the region in the plane determined by the inequalities:
(1) x ≤ 3y, x ≥ y.
(2) y ≥ 2x, −2x + 3y ≤ 6.
(3) 3x + 5y ≥ 45, x ≥ 0, y ≥ 0.
(4) 2x + 3y ≤ 35, y ≥ 2, x ≥ 5.
(5) 2x + 3y ≤ 6, x + 4y ≤ 4, x ≥ 0, y ≥ 0.
(6) x − 2y ≥ 0, 2x − y ≤ −2, x ≥ 0, y ≥ 0.
(7) 2x + y ≥ 8, x + 2y ≥ 8, x + y ≤ 6.

Exercise - 2.16

1. Simplify:
2 2

3
(a) (125) 3 = (5 ) 3 = 5 2 = 25
−3 −3

4
(b) (16) 4 , = (2 ) 4 = 2 (−3) = 1
8
−2 −2

3
(c) (−1000) 3 , = (−(10) ) 3 = −10 (−2) = 1
100
!
1
3
(d) (3) (−6) , = (3) (−2) = 1
9
−2

(27) 3 −1
(e) . = (27) 3 = (3) (−1) = 1
−1 3

(27) 3
−1
3
2. Evaluate (256) −1/2 4 .
Solution:
3
−1 −1/2 4 −1
3 −1 3
3
(256) −1/2 4 = 2 (8) = 2 (−4) 4 = 2 = 8
) = 9/2, then find the value of (x
3. If (x 1/2 + x −1/2 2 1/2 − x −1/2 ) for x > 1.
p
−1/2
1/2
Solution: We know that (x − x ) = (x 1/2 + x −1/2 2
) − 4.
q q
Hence we have 9 − 4 = 1 = 1
2 4 2
2n 2 −n
3 9 3
4. Simplify and hence find the value of n: = 27.
3 3n
Solution: