If the equation frac{{{x^2} + 5}}{2} = x - 2c | vedclass

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Question

$-\cos (a x+b)=\frac{x^{2}-2 x+5}{4}=\frac{(x-1)^{2}}{4}+1$

$\mathrm{Now},-\cos (\mathrm{ax}+\mathrm{b}) \in[-1,1]$

And $\frac{(x-1)^{2}}{4}+1 \

Vedclass · Accepted Answer

$\pi $

Vedclass · Answer

$-\cos (a x+b)=\frac{x^{2}-2 x+5}{4}=\frac{(x-1)^{2}}{4}+1$

$\mathrm{Now},-\cos (\mathrm{ax}+\mathrm{b}) \in[-1,1]$

And $\frac{(x-1)^{2}}{4}+1 \in[1, \infty)$

$\Rightarrow-\cos (a x+b)=1$ And $\frac{(x-1)^{2}}{4}+1=1$

$\Rightarrow x=1$ And $ a x+b=(2 k+1) \pi, k \in I$

$	herefore a+b=\pi$

If the equation $\frac{{{x^2} + 5}}{2} = x - 2\cos \left( {ax + b} \right)$ has atleast one solution, then $(b + a)$ can be equal to

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