Fsc Part 2 Mathematics (Complete Solutions)

# Q2 Find and Simplify

Find
 f(a + h) - f(a) h
and simplify where,

 (i) f(x) = 6x - 9 (ii) f(x) = sin x (iii) f(x) = x3 + 2x2 - 1 (iv) f(x) = cos x

Solutions

i) f(x) = 6x - 9
We have to find
 f(a + h) - f(a) h
= ?      ---------------------------------- (i)
f(a + h) = ?
as f(x) = 6x - 9

f(a + h) = 6(a + h) - 9
f(a + h) = 6a + 6h - 9

f(a) = ?
as f(x) = 6x - 9

f(a) = 6(a) - 9
f(a) = 6a - 9

Put the value of f(a + h) and f(a) in the equation (i)

=
 (6a + 6h - 9) - (6a - 9) h
=
 6a + 6h - 9 - 6a - 9 h
=
 6h h
 = 6

ii) f(x) = sin x
We have to find
 f(a + h) - f(a) h
= ?      ---------------------------------- (i)
f(a + h) = ?
as f(x) = sin x

f(a + h) = sin(a + h)

f(a) = ?
as f(x) = sin x

f(a) = sin a

Put the value of f(a + h) and f(a) in the equation (i)

=
 sin(a + h) - sin a h
=
 1 h
[ sin(a + h) - sin a ]
=
 1 h
[
2 cos (
 a + h + a 2
) sin (
 a + h - a 2
)
] ∴ [sin x - sin y = 2 sin( (x - y)/2 ) cos( (x + y)/2 )]
=
 2 h
[
cos (
 2a + h 2
) sin (
 h 2
)
]
=
 2 h
[
cos (
 2a 2
+
 h 2
) sin (
 h 2
)
]
=
 2 h
cos ( a +
 h 2
) sin (
 h 2
)

iii) f(x) = x3 + 2x2 - 1
We have to find
 f(a + h) - f(a) h
= ?      ---------------------------------- (i)
f(a + h) = ?
as f(x) = x3 + 2x2 - 1

f(a + h) = (a + h)3 + 2(a + h)2 - 1

f(a) = ?
as f(x) = x3 + 2x2 - 1

f(a) = a3 + 2a2 - 1

Put the value of f(a + h) and f(a) in the equation (i)

=
 [(a + h)3 + 2(a + h)2 - 1] - (a3 + 2a2 - 1) h
=
 [(a + h)3 + 2(a + h)2 - 1] - a3 - 2a2 + 1 h
=
 [a3 + h3 + 3ah(a + h) + 2(a2 + h2 + 2ah) - 1] - a3 - 2a2 + 1 h
=
 a3 + h3 + 3a2h + 3ah2 + 2a2 + 2h2 + 4ah - 1 - a3 - 2a2 + 1 h
=
 h3 + 3a2h + 3ah2 + 2h2 + 4ah h
=
 h h
[ h2 + 3a2 + 3ah + 2h + 4a ]
 = h2 + 3a2 + 3ah + 2h + 4a
 = h2 + 3ah + 2h + 3a2 + 4a
 = h2 + (3a + 2)h + 3a2 + 4a

iv) f(x) = cos x
We have to find
 f(a + h) - f(a) h
= ?      ---------------------------------- (i)
f(a + h) = ?
as f(x) = cos x

f(a + h) = cos(a + h)

f(a) = ?
as f(x) = cos x

f(a) = cos a

Put the value of f(a + h) and f(a) in the equation (i)

=
 cos(a + h) - cos a h
=
 1 h
[ cos(a + h) - cos a ]
=
 1 h
[
-2 sin (
 a + h + a 2
) sin (
 a + h - a 2
)
] ∴ [cos x - cos y = -2 sin( (x - y)/2 ) sin( (x + y)/2 )]
=
 -2 h
[
sin (
 2a + h 2
) sin (
 h 2
)
]
=
 -2 h
[
sin (
 2a 2
+
 h 2
) sin (
 h 2
)
]
=
 -2 h
sin ( a +
 h 2
) sin (
 h 2
)