First slide

Trigonometric equations

Question

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

Moderate

A

$- \frac{1}{3}$
B

$\frac{1}{3}$
C

$\frac{2}{3}$
D

$\frac{4}{3}$

Solution

LHS $= \sqrt{3} \sin π x + \cos π x \leq 2$ and RHS = $x^{2} - \frac{2}{3} x + \frac{19}{9} = {(x - \frac{1}{3})}^{2} + 2 \geq 2$
$∴ L H S = R H S = 2 \Rightarrow x = \frac{1}{3}$

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