Page 73 - Elementary Algebra Exercise Book I

ELEMENTARY ALGEBRA EXERCISE BOOK I equAtions

2.72 Solve the functional equation f(x)+ f( x−1 ) = 1+ x (x =0,x =1) (i).
x
Solution: Replace x with x−1 in (i): f( x−1 )+ f( −1 ) = 2x−1 (ii). Replace x with −1 in
x x x−1 x x−1
2
2
3
3
(i): f( −1 )+ f(x) = x−2 (iii). (i)+(iii)-(ii) ⇒ f(x)= x −x −1 = x −x −1 , which is the only
x−1 x−1 2x(x−1) 2x −2x
2
solution of the original functional equation (i).
√
√ 9 3
2.73 Solve the equation ( 3) tan 2x − =0.
3 tan 2x
√
Solution: Let ( 3) tan 2x = y ( y> 0), then the equation becomes y> 0
√ √ √ √
3
9 3 √ y −9 3 √ 3 √ 5/6 √ 5/6
tan 2x
⇒ ( 3)
⇒
=0 ⇒ y − 9 3= 0 ⇒ y = 3
=3
3
3
9 3
− 2 =0 ⇒ y −9 3 =0 ⇒ y − 9 3= 0 ⇒ y = 3 5/6 ⇒ ( 3) tan 2x =3 5/6
y− 2 =0 ⇒ y 2 2 ⇒
tan 2x 5 y ⇒ tan 2x = 5 y ⇒ 2x = kπ + arctan 5 5
5
5
tan
5
1
3
2 = 2x 6= ⇒ tan 2x 3= ⇒ 2x = kπ + arctan (k ∈ N ) ⇒ x = kπ + arctan (k ∈ N ).
2 6 3 3 2 2 3
kπ 1 5
Hence, the solution set of the original equation is {x|x = + arctan ,k ∈ N} .
2 2 3
2.74 Solve the system of equations
lg |x + y| =1,
1
lg y − lg |x| = .
log 100
4

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