Page 113 - Elementary Algebra Exercise Book I

ELEMENTARY ALGEBRA EXERCISE BOOK I inequAlities

3.75 Given f(x) = ax + bx , and 1 ≤ f(−1) ≤ 3, 2 ≤ f(1) ≤ 4, find the range of
2
f(−3).

Solution: f(−1) = a − b, f(1) = a + b, f(−3) =9a − 3b . Let f(−3) = mf(−1) + nf(1)

where m, n are parameters ready to be determined. 9a − 3b = m(a − b)+ n(a + b) =(m + n)a − (m − n)b

9a − 3b = m(a − b)+ n(a + b) =(m + n)a − (m − n)b . Comparing the coefficients to obtain m + n =9 ⇒ m =6,n =3. Thus
m − n =3
f(−3) = 6f(−1) + 3f(1) . Since 1 ≤ f(−1) ≤ 3, 2 ≤ f(1) ≤ 4 , we have 12 ≤ 6f(−1) + 3f(1) ≤ 30

12 ≤ 6f(−1) + 3f(1) ≤ 30, then 12 ≤ f(−3) ≤ 30. Therefore the range of f(−3) is [12, 30].

π
3.76 Given 0 <b< 1, 0 <a < , and x = (sin α) log sin α ,y = (cos α) log cos α ,z = (sin α) log cos α
b
b
b
4
x = (sin α) log sin α ,y = (cos α) log cos α ,z = (sin α) log cos α , determine the order of x,y,z .
b
b
b
π
Solution: 0 <b< 1, thus f(x) = log x is a decreasing function. 0 <a < , thus
b
4
0 < sin α< cos α< 10 < sin α< cos α< 1. Therefore, log sin α> log cos α> 0, then

b
b
(sin α) log sin α < (sin α) log cos α ,
b
b
i.e. x< z . And (sin α) log cos α < (cos α) log cos α , i.e. z <y . Hence,
b
b
we obtain the order x <z <y .
3.77 Consider a triangle with side lengths a, b, c , and its area is 1/4, the radius of
√ √ √
1
its circumcircle is 1. If s = a + b + c, t = 1 + 1 + . Compare s and t .
a b c
Solution: Let C be the angle whose opposite side length is c , and the radius of circumcircle
1
R =1, then c =2R sin C = 2 sin C . In addition, ab sin C = . Therefore abc =1. Then
1
2 4
√
√ √
1 1
1
1
1
1
1
1
1 1
1
1
1 1

t = 1 a + + = ( + )+ ( + )+ ( + ) ≥ 11 ab + 11 bc + 11 ca = 11 c+ a+ b √ √

√ √
√ √
= bb
√
c+ a+ a+
c+
1 11
2 c 1
1 11
11
2 b 1
2 a 1
11
11
11
1 11
11
t = =
c
c
b
+ + + = ( + )+ ( + )+ ( + )
b
+
a
2
√ √ t aa √ bb cc = ( + )+ ( + )+ ( + ) ≥ ≥ ab ab + + bc bc + + ca ca = = abc √ √ abc abc = =
cc
2 aa
bb
2 cc
aa
2
2
2 bb
√ √
√
a + √ b + c = s
√ √
a
a + + b + + c = ss . The equal sign can only be obtained if a = b = c = R =1, which is
b
c =
impossible. Hence, s <t .
3 1 1 1
3.78 a, b, c are positive numbers and a + b + c ≤ 3, show ≤ a+1 + b+1 + c+1 < 3.
2
1
1
1
1
1
1
Proof: Since a, b, c > 0, we have a+1 < 1, b+1 < 1, c+1 < 1 , then a+1 + b+1 + c+1 < 3.
Mean Inequality implies
Download free eBooks at bookboon.com
113