Fsc Part 2 Mathematics (Complete Solutions)

Q9
Determine whether the given function f is even or odd.

Determine whether the given function ƒ is even or odd.

(i)     ƒ (x) = x3 + x
(ii)     ƒ (x) = (x + 2)2
(iii)     ƒ (x) = x  X2 + 5 
(iv)     ƒ (x) =
x - 1
x + 1
, x ≠ -1
(v)     ƒ (x) = x 2/3 + 6
(vi)     ƒ (x) =
x3 - x
x2 + 1

Solution

(i)     ƒ (x) = x3 + x
Consider     ƒ (-x) = (-x)3 + (-x)
ƒ (-x) = -x3 - x
ƒ (-x) = -(x3 + x)
ƒ (-x) = -ƒ (x)
Hence ƒ (x) is an odd function.

(ii)     ƒ (x) = (x + 2)2
Consider     ƒ (-x) = (-x + 2)2
ƒ (-x) = (2 - x)2
Whish cannot be compared with ƒ (x)
Hence ƒ (x) is neither even nor odd function.

(iii)     ƒ (x) = x  X2 + 5 
Consider     ƒ (-x) = (-x)  (-x)2 + 5 
Consider     ƒ (-x) = -x  x2 + 5 
ƒ (-x) = -ƒ (x)
Hence ƒ (x) is an odd function.

(iv)     ƒ (x) =
x - 1
x + 1
, x ≠ -1
Consider     ƒ (-x) =
-x - 1
-x + 1
ƒ (-x) =
-(x + 1)
-(x - 1)
ƒ (-x) =
x + 1
x - 1
Whish cannot be compared with ƒ (x)
Hence ƒ (x) is neither even nor odd function.

(v)     ƒ(x) = x 2/3 + 6
Consider     ƒ (-x) = (-x) 2/3 + 6
ƒ (-x) = (-x 2) 1/3 + 6
ƒ (-x) = (x 2) 1/3 + 6
ƒ (-x) = x 2/3 + 6
ƒ (-x) = ƒ (x)
Hence ƒ (x) is an even function.

(vi)     ƒ(x) =
x3 - x
x2 + 1
Consider     ƒ (-x) =
(-x)3 - (-x)
(-x)2 + 1
ƒ (-x) =
-x3 + x
x2 + 1
ƒ (-x) =
-(x3 - x)
x2 + 1
ƒ (-x) = -ƒ (x)
Hence ƒ (x) is an odd function.

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