# Tag Archives: Os

The diagram shows a river,500m wide between the straight parallel banks where AD = 500m and DB = 2000. A man swims across the river from A to C at 0.5 m/s and then he runs along the banks from C to B at 0.8 m/s

(a)If he swims in the direction AC making an angle of Ɵ and reaches C in t seconds, show that t = 1000cosec Ɵ

(b)If T(in seconds) is the total time taken from A to B, express T in terms of Ɵ

(c)Find the minimum value of T in minutes and seconds, correct to the nearest second.

(a)

Using trigo ratio,

sin Ɵ = 500/AC

AC = 500/sin Ɵ

= 500cosec Ɵ

0.5 = 500cosec Ɵ/t

T = 1000cosec Ɵ

(b)

BC = 2000 – DC

BC = 2000 – 500/tan Ɵ

T_{BC }= BC/s = 2500 – 625/tan Ɵ

Total time, T = t_{AC} + t_{BC}

= 1000cosec Ɵ + 2500 – 625/tan Ɵ

(c)

Given y = 4/√(2x-3), x > 1.5

(a)Find dy/dx

(b)Explain why the graph of y = 4/√(2x – 3) does not have a stationary point

(c)State whether y = 4√(2x – 3) is an increasing or decreasing function. Explain your answer clearly.

y = 4(2x – 3)^{-1/2}

dy/dx = -2(2x – 3)^{-3/2}(2)

= -4(2x – 3)^{-3/2}

(b) Since, dy/dx = -4(2x – 3)^{-3/2} , (2x-3) is always positive

(2x – 3 )^{-3/2} > 0

- dy/dx < 0, dy/dx ≠ 0, therefore graph has no stationary point

Find the minimum gradient of the curve y = 2x^{3} – 9x^{2} + 5x + 3 and the value of x when the minimum gradient occurs.

y = 2x^{3} – 9x^{2} + 5x + 3

gradient, m = 6x^{2} – 18x + 5

dm/dx = 12x – 18

minimum gradient,

dm/dx = 0

12x – 18 =0

X = 3/2 occurs at minimum gradient

= 6(3/2)^{2} – 18(3/2) + 5

= -17/2

A curve with equation in the form of y = ax + b/x

y = ax + b/x^{2} has a stationary point at (3 , 4), where a and b are constants. Find the value of a and of b.

X = 3 y = 4

4 = 3a + b/a —–(1)

Y = ax + bx^{-2}

dy/dx = a – 2bx^{-3}—–(2)

sub dy/dx = 0, x = 3 into (2)

0 = a – 2b(3)^{-3}

0 = a – 2b/27—–(3)

Sub (3) into (1)

4 = 3(2b/27) + b/9

4 = 1/3b

b = 12 , a = 2(12)/27

b = 12 , a = 8/9

Solution:

(a) A=2, B=-5, C=0

(b) 2x/(x^{2}+3)

(c) ln(2x-1) – (5/2)ln(x^{2}+3) + C

(a)6 + 5x – 8x^{2} = A(x^{2} + 3) + (Bx + c)(2x – 1)

Sub x = ½

6 + 5/2 – 8(1/2)^{2} = A(1/4 + 3) + 0

13/2 = (13/4)A —— A=2

When x=0, 6 = 2(3) + c(-1) —— C=0

Compare coefficients of x^{2}

-8 = A + 2B

-10 – 2B —— B = -5

(b)d/dx[ln(x^{2}+3)] = 2x/(x^{2}+3)

=2x/(x^{2}+3)

(c)∫ 2/(2x-1) + ∫(-5x)/(x^{2}+3)

=ln(2x-1) – (5/2)ln(x^{2}+3) + C

Integrate (3-sin^{2}2x)/cos^{2}2x w.r.t x

∫3/cos^{2}2x – sin^{2}x/cos^{2}2x dx

=∫3sec^{2}2x – (sec^{2}2x-1) dx

=∫2sec^{2}2x +1 dx

=2tan2x/2 + x + c

=tan2x + x + c

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