Click to Chat
1800-1023-196
+91-120-4616500
CART 0
Use Coupon: CART20 and get 20% off on all online Study Material
Welcome User
OR
LOGIN
Complete Your Registration (Step 2 of 2 )
Hi – need some help with just question b (i). Thanks in advance!
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 :))
Post Question
Dear , Preparing for entrance exams? Register yourself for the free demo class from askiitians.
points won -