Q113 / 155 · WBJEE 2025 · Mathematics

If $0 \leq a, b \leq 3$ and the equation $x^2+4+3 \cos (a x+b)=2 x$ has real solution, then the value of $(a+b)$ is

A
$\frac{\pi}{4}$
B
$\frac{\pi}{2}$
Correct answer
C
$\pi$
D
$2 \pi$

Solution

$\begin{aligned} & \text { }\left(x^2-2 x+1\right)+3(\cos (a x+b)+1)=0 \\ & \quad \text { or, }(x-1)^2+3(\cos (a x+b)+1)=0 \\ & \quad a+b=\pi\end{aligned}$

Hence, the correct answer is option: 'C'.

If $0 \leq a, b \leq 3$ and the equation $x^2+4+3 \cos (a x+b)=2 x$ has real solution, then the value of $(a+b)$ is

Solution

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