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.