Question

Let P(x) be a quadratic in 'x' satisfying P(0) = $co{s}^3{40}^{\circ },P\left(1\right)=cos{40}^{\circ }si{n}^2{40}^{\circ },P\left(2\right)=0$. Then P(3) equals

• A. $cos{80}^{\circ }–2cos{40}^{\circ }$
• B. $\frac{-1}{2}$
• C. $\frac{1}{2}$
• D. $cos{80}^{\circ }+2cos{40}^{\circ }$

Answer 1

The correct option is B $\frac{-1}{2}$

$P\left(x\right)=k(x-2)(x-a)$
When k(2a) $=\frac{-1+6cos{40}^{\circ }}{8}$
$k(a-1)=\frac{2cos{40}^{\circ }+1}{8}$
solve for a, k, $P\left(3\right)=k(3-a)=\frac{-1}{2}$