Solution For 2006-07 Second Term Exam

1. 2000Q6

2. 2001 Q2 x2 1 dx = ∫ 1 − dx = x − tan −1 x + c 2A (a) ∫ 2 2 1+ x 1+ x

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 − 2x3 − 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)

(e) × (0,-1)

9. 2003Q9

y

1

π

10. (a)

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

2 (c) (i) ∫0 cos xdx = ∫0 (

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

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

∫

π

0

cos 2 xdx =

1 2 2

2

1 π n  1 π n  kπ π  π    sin k + x − sin k − x dx = lim sin kx  dx       ∑  2 cos (d) lim ∑   ∫ − π n →∞ n ∫−π n → ∞ n  k =1  n n  n     k =1  1 π  n  kπ  π ∑  4 cos 2  dx = 2π (by (c)(ii)) n →∞ n ∫−π  n  (by (b))  k =1 

= lim

11. 2003Q8

12. 2005Q11