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Hi – need some help with just question b (i). Thanks in advance!

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6 months ago

```							given ydy/dx= cos2xor ∫ydy= ∫cos2x dxy^2/2= sin2x/2 + cor y^2= sin2x + K. but (pi/4, 2) satisfy this eqn.2^2= sinpi/2 + Kor K= 3hence we have y^2= sin2x + 3or y= ± sqrt(sin2x + 3)but since y is positive when x is pi/4, we drop/reject the negative sign.so, y= g(x)= sqrt(sin2x + 3)clearly sin2x lies in [ – 1, 1].so that sin2x + 3 lies in [2, 4].hence sqrt(sin2x + 3)= g(x) lies in [sqrt2, sqrt4] or [sqrt2, 2], which is the range of g.KINDLY APPROVE :))
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6 months ago
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