Question

If the equation $x^2+4+3\cos (ax+b)=2x$ has at least one solution where $a,b\in [0,5],$then the value of $(a+b)$ equal to

• A. $5\pi$
• B. $3\pi$
• C. $2\pi$
• D. $\pi$

Answer 1

Answer: (B), (D)

The correct options are
B $3\pi$
D $\pi$

$x^2-2x+4=-3\cos (ax+b)$
$(x-1{)}^2+3=-3\cos (ax+b)$
For above equation to have at least on e solution,
let $f\left(x\right)=(x-1{)}^2+3$ and $f\left(x\right)=-3\cos (ax+b)$
If $x=1$then $LHS=3$ and $RHS=-3\cos (a+b)$
Hence,$\cos (a+b)=-1$
$\therefore a+b=\pi ,3\pi ,5\pi$
But,$0\le a+b\le 10\Rightarrow a+b=\pi$ or $3\pi$
Hence, $\left(b\right)$ and $\left(d\right)$ are the correct answers.