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
)

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