Question
If $$0 \leqslant a \leqslant 3,0 \leqslant b \leqslant 3$$ and the equation $${x^2} + 4 + 3\cos \left( {ax + b} \right) = 2x$$ has at least one solution then the value of $$a + b$$ is
A.
$$0$$
B.
$$\frac{\pi }{2}$$
C.
$$\pi $$
D.
None of these
Answer :
$$\pi $$
Solution :
$$\eqalign{
& {x^2} - 2x + 4 = - 3\cos \left( {ax + b} \right) \cr
& \Rightarrow \,\,{\left( {x - 1} \right)^2} + 3 = - 3\cos \left( {ax + b} \right). \cr} $$
$${\text{As }} - 1 \leqslant \cos \left( {ax + b} \right) \leqslant 1\,\,{\text{and }}{\left( {x - 1} \right)^2} \geqslant 0,$$ the above is possible only if $$\cos\left( {ax + b} \right) = - 1\,\,{\text{when }}x = 1.\,{\text{So, }}a + b = \pi ,3\pi ,5\pi ,\,{\text{e}}{\text{.t}}{\text{.c}}{\text{., and }}3\pi > 6.$$