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Greatdreams (3155)

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Olaaa!! Perrrfect answer. 599  [679 rates]

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Similar to Krishnan sir's work : just a little clearer :

4 sin 4 [ (x - 3 )/ 6 2 ] + 2 cos [( 4x - 3 )/ 3 2]

= [ 2 sin 2 ( 4 x - 3 / 6 2 ) ] 2 + 2 cos [(4x - 3) / 32 ]

= [ 1 - cos ( 4x - 3/  3 2 ) ] 2 + 2 cos [(4x - 3) / 32 ]   

----> [ reason 2 sin 2 x/2 = 1 - cos x ]

= 1 + cos 2 [( 4x - 3 )/ 3 2 ]  ------> [ (a - b) 2 + 2ab  = a 2 + b2 ]

= 3/2 + 1/2 cos [ ( 8x - 6 )/ 3 2]

Now let  be the period ,

So we have f( x + )  = 3/2 + 1/2 cos [ ( 8x + 8 - 6 )/3 2 ]

= f(x)

So we have 8 = 3 2 .  2

So = 3 3 / 4 .......Option a is correct.

Hope it reduced some of Bhatt sir's work



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