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Q.

If u=a2cos2⁡θ+b2sin2⁡θ+a2sin2⁡θ+b2cos2⁡θ  then the difference between the maximum and minimumvalues of u2 is given by

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a

2a2+b2

b

2a2+b2

c

a2+b2

d

a-b2

answer is D.

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Detailed Solution

u2=a2+b2+2a2cos2⁡θ+b2sin2⁡θ×a2sin2⁡θ+b2cos2⁡θ=a2+b2+2sin2⁡θcos2⁡θa4+b4+a2b2sin4⁡θ+cos4⁡θ=a2+b2+2a2b21−2sin2⁡θcos2⁡θ+a4+b4sin2⁡θcos2⁡θ=a2+b2+2a2b2+a2−b22sin2⁡θcos2⁡θ=a2+b2+2a2b2+a2−b224sin2⁡2θ Max. u2=a2+b2+2a2b2+a2−b224 Min. u2=a2+b2+2ab⇒  Difference =2a2b2+a2−b224−2ab=4a2b2+a4+b4−2a2b2−2ab=a2+b22−2ab=a2+b2−2ab=(a−b)2
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If u=a2cos2⁡θ+b2sin2⁡θ+a2sin2⁡θ+b2cos2⁡θ  then the difference between the maximum and minimumvalues of u2 is given by