Let x1 and x2 be solutions of the equation sin−1(x2−3x+52)=π6. Then, the value of x12+x22 is

sin−1(x2−3x+52)=π6 ⇒x2−3x+52=sin(π6)=12 ⇒2x2−6x+5=1 ⇒2x2−6x+4=0 x2−3x+2=0 Since, x1, x2 are the roots ∴x1+x2=3 and x1x2=2 ∴x12+x22=(x1+x2)2−2x2x2=9−2(2)=5

Let x1 and x2 be solutions of the equation sin−1(x2−3x+52)=π6. Then, the value of x12+x22 is

sin−1(x2−3x+52)=π6 ⇒x2−3x+52=sin(π6)=12 ⇒2x2−6x+5=1 ⇒2x2−6x+4=0 x2−3x+2=0 Since, x1, x2 are the roots ∴x1+x2=3 and x1x2=2 ∴x12+x22=(x1+x2)2−2x2x2=9−2(2)=5

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Let $x_{1} a n d x_{2}$ be solutions of the equation $\sin^{- 1} (x^{2} - 3 x + \frac{5}{2}) = \frac{π}{6}$ . Then, the value of $x_{1}^{2} + x_{2}^{2}$ is

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$\frac{5}{2}$

answer is B.

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Detailed Solution

$\sin^{- 1} (x^{2} - 3 x + \frac{5}{2}) = \frac{π}{6} \Rightarrow x^{2} - 3 x + \frac{5}{2} = \sin (\frac{π}{6}) = \frac{1}{2} \Rightarrow 2 x^{2} - 6 x + 5 = 1 \Rightarrow 2 x^{2} - 6 x + 4 = 0 x^{2} - 3 x + 2 = 0 S i n c e, x_{1}, x_{2} a r e t h e r o o t s ∴ x_{1} + x_{2} = 3 a n d x_{1} x_{2} = 2 ∴ x_{1}^{2} + x_{2}^{2} = {(x_{1} + x_{2})}^{2} - 2 x_{2} x_{2} = 9 - 2 (2) = 5$