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4-2.Quadratic Equations and Inequations
normal
જો સમીકરણ $\frac{{{x^2} + 5}}{2} = x - 2\cos \left( {ax + b} \right)$ ને ઓછામાં ઓછા એક ઉકેલ મળે તો $(b + a)$ ની કિમત મેળવો
A
$0$
B
$\pi $
C
$2\pi $
D
$4\pi $
Solution
$-\cos (a x+b)=\frac{x^{2}-2 x+5}{4}=\frac{(x-1)^{2}}{4}+1$
$\mathrm{Now},-\cos (\mathrm{ax}+\mathrm{b}) \in[-1,1]$
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$
$\therefore a+b=\pi$
Standard 11
Mathematics
