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Questions  

If0a3,0b3and the equationx2+4+3cos(ax+b)=2xhas at least solution. Then the values of a+bis

a
0
b
π2
c
π
d
None of these

detailed solution

Correct option is C

We havex2−2x+4=−3cosax+b⇒x−12+3=−3cosax+bAs −1≤cosax+b≤1 and x−12≥0The above is possible only if cosax+b=−1 where x=1⇒cosa+b=-1⇒ a+b=π, 3π, 5π etc; ⇒a+b=π∵a+b≤6

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