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Grade: 12
        
Hi – need some help with just question b (i). Thanks in advance!
3 months ago

Answers : (1)

Aditya Gupta
2014 Points
							
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 :))
3 months ago
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