Question

If $\sqrt{3}\sin \pi x+\cos \pi x=x^2-\frac{2}{3}x+\frac{19}{9}$, then $x$
is equal to

• A. $-\frac{1}{3}$
• B. $\frac{1}{3}$
• C. $\frac{2}{3}$
• D. $\frac{4}{3}$

Answer 1

Answer: (B)

$\frac{1}{3}$

Dividing both sides by 2 we get,

$\frac{\sqrt{3}sin\pi x}{2}+\frac{cos\pi x}{2}=\frac{x^2}{2}-\frac{1}{3}x+\frac{19}{18}\Rightarrow cos\frac{\pi }{6}sin\pi x+sin\frac{\pi }{6}cos\pi x=\frac{9x^2-6x+19}{18}\Rightarrow sin(\frac{\pi }{6}+\pi x)=\frac{(3x-1{)}^2+18}{18}\Rightarrow sin(\frac{\pi }{6}+\pi x)=\frac{(3x-1{)}^2}{18}+1$

Since, Value of Sine Function cannot be Greater than $1$ Therefore,

$\frac{(3x-1{)}^2}{18}=0\Rightarrow (3x-1)=0\Rightarrow x=\frac{1}{3}$

Thererfore, Answer is $\frac{1}{3}$