Fsc Part 2 Mathematics (Complete Solutions)

# Q1 Given that

(a)     f(x) = x2 - x (b)     f(x) =
 √ x + 4

 Find (i)     f(-2) (ii)     f(0) (iii)     f(x-1) (iv)     f(x2 + 4)

Solution

(a) f(x) = x2 - x

(i) f(-2) = ?
since f(x) = x2 - x, So
f(-2) = (-2)2 - (-2)
f(-2) = 4 + 2
f(-2) = 6

(ii) f(0) = ?
since f(x) = x2 - x, So
f(0) = (0)2 - 0
f(0) = 0 - 0
f(0) = 0

(iii) f(x - 1) = ?
since f(x) = x2 - x, So
f(x - 1) = (x - 1)2 - (x - 1)
f(x - 1) = x2 - 2x + 1 - x + 1
f(x - 1) = x2 - 3x + 2

(iv) f(x2 + 4) = ?
since f(x) = x2 - x, So
f(x2 + 4) = (x2 + 4)2 - (x2 + 4)
f(x2 + 4) = x4 + 2(x2)(4) + 42 - x2 - 4
f(x2 + 4) = x4 + 8x2 + 16 - x2 - 4
f(x2 + 4) = x4 + 7x2 + 12

(b) f(x) =
 √ x + 4

(i) f(-2) = ?
since f(x) =
 √ x + 4
, So
f(-2) =
 √ -2 + 4
f(-2) =
 √ 2

(ii) f(0) = ?
since f(x) =
 √ x + 4
, So
f(0) =
 √ 0 + 4
f(0) =
 √ 4
f(0) = 2

(iii) f(x - 1) = ?
since f(x) =
 √ x + 4
, So
f(x - 1) =
 √ (x - 1) + 4
f(x - 1) =
 √ x - 1 + 4
f(x - 1) =
 √ x + 3

(iv) f(x2 + 4) = ?
since f(x) =
 √ x + 4
, So
f(x2+ 4) =
 √ (x2 + 4) + 4
f(x2+ 4) =
 √ x2 + 4 + 4
f(x2+ 4) =
 √ x2 + 8