Test 10 - Sol

Pure Mathematics – Calculus

p.1

King’s College 2007 – 2008 F.6 Pure Mathematics Revision Test 10 Suggested Solution 1.

1− t2 2dt x Let t = tan . Then, cos x = and dx = . 2 2 1+ t 1+ t2 dx

2dt

∫ 2 + cos x = ∫ 3 + t

2.

(a)

2

=

⎛ tan −1 ⎜⎜ 3 ⎝

2

lim(tan 3 x + cos 4 x ) = e 1 x

x →0

lim

x →0

⎛ ⎛ x⎞⎞ ⎜ tan⎜ ⎟ ⎟ 2 t ⎞ ⎝2⎠⎟+C −1 ⎜ ⎟⎟ + C = tan ⎜ 3⎠ 3 3 ⎟ ⎜ ⎟ ⎝ ⎠

ln ( tan 3 x + cos 4 x ) x

=e

lim l

x →0

3 sec 2 x − 4 sin 4 x tan 3 x + cos 4 x

3− 0

= e 0+1 = e 3

(b) Note that 2004 + x + x 2004 + x − x sin 2 2 2004 + x + x 1002 = −2 sin sin for all x > 0 2 2004 + x + x 1002 for all x > 0. So, 0 ≤ cos 2004 + x − cos x ≤ 2 sin 2004 + x + x 1002 Since 2 sin → 0 as x → ∞ , 2004 + x + x cos 2004 + x − cos x = −2 sin

(

)

we have lim cos 2004 + x − cos x = 0 (by Sandwich Theorem). x →∞

Let f ( x ) = cos x for all x > 0.

Then, f ' ( x ) =

− sin x 2 x

. By Mean Value Theorem, for some ξ ∈ ( x,2004 + x ) , we have

cos 2004 + x − cos x = f (2004 + x ) − f ( x ) = 2004 f ' (ξ ) = −

Note that

1

ξ

Therefore, −

1002 sin ξ

→ 0 as ξ → ∞ and sin ξ ≤ 1 for all ξ > 0 . 1002 sin ξ

ξ

→ 0 as ξ → ∞ .

Further note that as x → ∞ , ξ → ∞ .

(

)

Thus, we have lim cos 2004 + x − cos x = 0 . x →∞

ξ

.

Pure Mathematics – Calculus

3.

∫ sec

3

p.2

θdθ = ∫ sec θd tan θ = sec θ tan θ − ∫ tan 2 θ sec θdθ = sec θ tan θ − ∫ sec 3 θdθ + ∫ sec θdθ

So we have

∫ sec 4.

3

θdθ =

sec θ tan θ 1 sec θ tan θ 1 + ∫ sec θdθ = + ln sec θ + tan θ + C , where C is a constant. 2 2 2 2

Note that − 1 ≤ sin

1 ≤ 1 for all x ≠ 0 . x

So we have − sin x ≤ sin x sin

1 ≤ sin x for all x ≠ 0 . x

Since lim(− sin x ) = 0 = lim sin x , we have lim sin x sin x →0

Note that sin

x →0

x →0

(a) (i)

1 = 0 (by Sandwich Theorem). x

1 ≤ 1 for all x ≠ 0 and lim sin x = 0 . x →0 x

Thus, we have lim sin x sin

5.

x →0

1 =0 x

h(a) = f(a) – f(a) – 0 = 0 h(b) = f(b) – f(a) – (f(b) – f(a)) = 0

(ii) By Mean Value Theorem, there exists β ∈ (a, b ) such that h' (β ) = Now, we have h' (β ) = f ' (β ) − Since g ' (β ) ≠ 0 , thus we have

f (b ) − f (a ) g ' (β ) = 0 . g (b ) − g (a )

f ' (β ) f (b ) − f (a ) = . g ' (β ) g (b ) − g (a )

h(b ) − h(a ) = 0. b−a

(b) Note that F and G are differentiable in R, G (x ) ≠ G (c ) and G ' (t ) ≠ 0 for all t ∈ I . By (a)(ii), there exists γ ∈ I such that However, F(x) = u(x) – u(x) – 0 = 0 and So, there exists γ ∈ I , such that

F ' (γ ) F (c ) − F ( x ) = . G ' (γ ) G (c ) − G ( x ) 2 ( x − x) G (x ) =

F ' (γ ) F (c ) = . G ' (γ ) G (c )

2

= 0.

Also, F ' (t ) = −u ' (t ) = u ' ' (t )( x − t ) + u ' (t ) = −u ' ' ( x − t ) . Moreover, G ' (t ) = −(x − t ) . Therefore, we have F ' (γ ) = −u ' ' ( x − γ ) and G' (γ ) = −( x − γ ) . So we have

− u ' ' (γ )( x − γ ) u (x ) − u (c ) − u ' (c )( x − c ) = . − (x − γ ) ( x − c )2 2

Thus, we have u ( x ) = u (c ) + u ' (c )(x − c ) +

u ' ' (γ ) ( x − c )2 . 2

Pure Mathematics – Calculus

(c) (i)

p.3

v(0 )

= lim v(x ) x →0

⎛ ⎛ v( x ) ⎞ ⎞ = lim⎜⎜ x⎜ ⎟ ⎟⎟ x →0 ⎝ ⎝ x ⎠⎠

(since v is continuous at 0)

( )

v( x ) ⎞ ⎛ = lim x ⎜ lim ⎟ x →0 x →0 x ⎝ ⎠ =0

v( x ) − v(0) v(x ) = lim = 2006 x →0 x →0 x x−0

v' (0) = lim

(ii) Case 1: x = 0 LHS = v(0) = 0 (by (c)(i)) 2 RHS = (2006)(0) + 0 = 0 So, the inequality holds for x = 0. Case 2: x ≠ 0 Putting u = v and c = 0 in the latter result of (b), there exists d ∈ ℜ such that v( x ) = v(0 ) + v' (0)x +

v' ' (d ) 2 v' ' (d ) 2 x = 2006 x + x ≥ 2006 x + x 2 . 2 2

By combining the above two cases, v( x ) ≥ 2006 x + x 2 for all x ∈ ℜ . Since v' ' (t ) ≥ 2 for all t ∈ ℜ , we have Case 1: x ≥ 0

∫ v' ' (t )dt ≥ ∫ 2dt s

s

0

0

for all s ≥ 0

v' (s ) − v' (0) ≥ 2 s v' (s ) ≥ 2s + 2006

∫

x

0

v' (s )ds ≥ ∫ (2s + 2006 )ds x

0

v( x ) − v(0) ≥ x 2 + 2006 x v( x ) ≥ 2006 x + x 2

Case 2: x < 0 v' ' (t ) ≥ 2 for all t ∈ ℜ

∫ v' ' (t )dt ≥ ∫ 0

0

s

s

2dt for all s ≥ 0

v' (0) − v' (s ) ≥ −2 s − v' (s ) ≥ −(2 s + 2006 ) − ∫ v' (s )ds ≥ − ∫ (2s + 2006 )ds 0

0

x

x

− v(0 ) + v( x ) ≥ x 2 + 2006 x

v( x ) ≥ 2006 x + x 2

By combining the above two cases, v( x ) ≥ 2006 x + x 2 for all x ∈ ℜ .