WBJEE · Maths · Trigonometric Ratios & Identities
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 \(\frac{\pi}{2}\)
- C \(\pi\)
- D \(2 \pi\)
Answer & Solution
Correct Answer
(C) \(\pi\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \text { }\left(x^2-2 x+1\right)+3(\cos (a x+b)+1)=0 \\ & \quad \text { or, }(x-1)^2+3(\cos (a x+b)+1)=0 \\ & \quad a+b=\pi\end{aligned}\)
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