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$I f$ $0 \leq a \leq 3, 0 \leq b \leq 3$ $a n d t h e e q u a t i o n x^{2} + 4 + 3 \cos (a x + b) = 2 x, h a s a t l e a s t o n e s o l u t i o n, t h e n t h e v a l u e o f$ $a + b$

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Correct option is D

We have x2−2x+4=−3cosax+b⇒x−12+3=−3cosax+bLHS≥3 and RHS≤3 ⇒LHS=RHS=3⇒x=1Also we have 3=-3cos(ax+b)=-3cosa+b ∵x=1cosa+b=-1=cosπ ⇒a+b=π

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If0≤a≤3,0≤b≤3and the equation x2+4+3cosax+b=2x,has at least one solution, then the value ofa+b

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$I f$ $0 \leq a \leq 3, 0 \leq b \leq 3$ $a n d t h e e q u a t i o n x^{2} + 4 + 3 \cos (a x + b) = 2 x, h a s a t l e a s t o n e s o l u t i o n, t h e n t h e v a l u e o f$ $a + b$