Question

Let $\sin x$ and $\sin y$ be roots of the quadratic equation $\sin ^2 \theta+b \sin \theta+c=0(a, b, c \in R$ and $a \neq 0)$ such that $\sin x+2 \sin y=1$, then the value of $\left(a^2+2 b^2+3 a b+a c\right)$ equals

• A. 0
• B. 1
• C. 2
• D. 4

Answer 1

Answer: (A)

Given $(\sin x+\sin y)=\frac{-b}{a}$ ....(1)
and $ \sin x \cdot \sin y=\frac{c}{a}$ ....(2)
also $\sin x+2 \sin y=1$ ....(3)
From (1) and (3) $\Rightarrow \sin y=1+\frac{b}{a}=\frac{a+b}{a}$
$\Rightarrow \sin x=\frac{-b}{a}-\frac{(a+b)}{a}=\frac{-a-2 b}{a}$
So, put in (2), we get $\frac{-( a +2 b )}{ a } \cdot \frac{( a + b )}{ a }=\frac{ c }{ a }$
Hence, $a c+(a+2 b)(a+b)=0 \Rightarrow a^2+2 b^2+3 a b+a c=0$