If a , b ϵ ( 0 , 2 ) and the equation 2 x 2 + 5 = x − 2 c o s ( a x + b ) has atleast one solution then a + b =
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We can make conclusion from this.
2 x 2 − 2 x + 5 = − 2 cos ( a x + b )
( x − 1 ) 2 + 4 = − 4 cos ( a x + b )
( x − 1 ) 2 = − 4 ( 1 + cos ( a x + b )
( x − 1 ) 2 = − 8 cos 2 2 a x + b
Thus for x ∈ R , the only posibble solution is x = 1 , ⟹ R H S = 0 ⟹ 2 a + b = 2 π
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That mistake only I did while solving , and too writing it , thanks edited
Actually the answer can be any odd multiple of pi.
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Just put x = 1