Solution For 2006-07 Second Term Exam

1. 2000Q6

2. 2001 Q2 (a)

x2 1 −1 ∫ 1 + x 2 dx = ∫ 1 − 1 + x 2 dx = x − tan x + c 2A

3. 2001Q3

4. 2001 Q6

5. 2000 Q3.

6. 2004P1Q2

7. 2005Q1

8. (a)

(i)

f ' ( x) = 1 +

2x − 1 ( x − x + 1) 2 2

x 4 + x 2 + 1 − 2x 3 − 2x + 2x 2 + 2x − 1 = ( x 2 − x + 1) 2 = =

x 4 − 2x 3 + 2x 2 ( x 2 − x + 1) 2 x 2 ( x 2 − 2 x + 3) ( x 2 − x + 1) 2 (b) (i) f ' ( x) > 0 ⇔ x ≠ 0 (ii) f ' ( x) < 0 ⇔ no solution (iii) f " ( x) > 0 ⇔ 0 < x < 1 (iv) f " ( x) < 0 ⇔ x < 0 or x > 1

(c) Point of inflexion (0,-1) and (1,0) (d) x = y

x

x=y y = f ( x)

× (1,0) × (0,-1)

(e)

9. 2003Q9

y

10. (a)

π

1

π

∫ π sin mx sin nxdx = 2 ∫ π (cos(m − n) x − cos(m + n) x)dx −

−

For m=n 1 π (cos 0 − cos 2mx)dx 2 ∫−π 1 sin 2mx π = (x − )| −π 2 2m =π =

For m ≠ n 1 sin( m − n) x sin( m + n) x π − ( )| −π 2 ( m − n) ( m + n) =0 =

π

∫π −

(b)

n

n

(∑ bk sin kx) 2 dx =∑ (bk k =1 n

k =1

2

π

∫ π sin −

2

kx)dx

= π ∑ bk (by (a)) 2

k =1

π

π

0

0

(c) (i) ∫ cos 2 xdx = ∫ (

π cos 2 x + 1 1 xπ π )dx = sin 2 x | + | = 0 2 2 20 2

n kπ 1 1 = (c)(ii) lim ∑ cos 2 n→∞ n π k =1 n

∫

π

0

cos 2 xdx =

1 2

(d) 2

1 π ⎧n ⎛ kπ 1 π ⎧n ⎡ ⎛π ⎞ ⎛π ⎞⎤ ⎫ ⎞⎫ lim ∫ ⎨∑ ⎢sin k ⎜ + x ⎟ − sin k ⎜ − x ⎟⎥ ⎬ dx = lim ∫ ⎨∑ ⎜ 2 cos sin kx ⎟⎬ dx n → ∞ n −π n → ∞ n −π n ⎠⎦ ⎭ ⎝n ⎠ ⎝n ⎠⎭ ⎩ k =1 ⎝ ⎩ k =1 ⎣ 1 π ⎡ n ⎛ kπ ⎞⎤ π ∑ ⎜ 4 cos 2 ⎟ dx (by (b)) = 2π (by (c)(ii)) ⎢ ∫ n →∞ n −π n ⎠⎥⎦ ⎣ k =1 ⎝ 11. 2003Q8 = lim

2

12. 2005Q11