Question

Let $x_1$ and $x_2$ be solutions of the equation ${\sin }^{-1}\left(x^2-3x+\frac{5}{2}\right)=\frac{\pi }{6}$ .Then, the value of $x_1^2+x_2^2$ is:

• A. 4
• B. 5
• C. $\frac{5}{2}$
• D. 6
• E. $\frac{15}{2}$

Answer 1

Answer: (A)

${\sin }^{-1}\left(x^2-3x+\frac{5}{2}\right)=\frac{\pi }{6}$ $\Rightarrow$ $x^2-3x+\frac{5}{2}=\sin \left(\frac{\pi }{6}\right)=\frac{1}{2}$ $\Rightarrow$ $2x^2-6x+5=0$ Since, $x_1{x}_2$ , are the solution set, then $x_1+x_2=\frac{6}{2}=3$ and $x_1{x}_2=\frac{5}{2}$ $\therefore$ $x_1^2+x_2^2={(x_1+x_2)}^2-2x_1{x}_2=9-5=4$