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

Question

Aditya Gupta · Answer

given ydy/dx= cos2x or ∫ydy= ∫cos2x dx y^2/2= sin2x/2 + c or y^2= sin2x + K. but (pi/4, 2) satisfy this eqn. 2^2= sinpi/2 + K or K= 3 hence we have y^2= sin2x + 3 or 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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Hi – need some help with just question b (i). Thanks in advance!

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

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