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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!
Date Posted: 4 years ago
Views: 510
i)
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
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
See 1 Answer
done
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Q10 and Q11
Date Posted:
4 years ago
Thank you so much! The handwriting was neat and every step was detailed! Could understand it really quickly. Amazing!
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