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secondary 4 | A Maths

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I need help for one of either question 10 or 11. Both would be great too!

d/dx ln(x² + 4)

= (d/dx (x² + 4) ) / (x² + 4)

= 2x/(x² + 4)

ii)

if gradient = k,

d/dx ln(x² + 4) = k

2x/(x² + 4) = k

2x = k(x² + 4) = kx² + 4k

kx² - 2x + 4k = 0

iii)

For the equation to have equal roots,

Discriminant = 0

(b² - 4ac = 0)

so

(-2)² - 4(k)(4k) = 0

4 - 16k² = 0

16k² = 4

k² = 4/16 = ¼

k = ±√¼ = ± ½

Or

4 - 16k² = 0

1 - 4k² = 0

(1 - 2k)(1 + 2k) = 0

1 - 2k = 0 or 1 + 2k = 0

2k = 1 or 2k = -1

k = ½ or k = -½

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