Let (a) and (a) be the roots of the equation ( [3] 1+a -1) x²+( 1+a -1) x+( [6] 1+a -1)=0 where a-1 . Then _ a 0⁺ (a) and _ x 0⁺ (a) are

Question

Let (a) and (a) be the roots of the equation ( [3] 1+a -1) x²+( 1+a -1) x+( [6] 1+a -1)=0 where a>-1 . Then _ a 0⁺ (a) and _ x 0⁺ (a) are

Quantrex Academy · Accepted Answer

Correct answer: B. - 1 2 and -1

( [3] 1+a -1) x²+( 1+a -1) x+( [6] 1+a -1)=0 Let a+1=y , then equation reduces to (y^ 1 / 3 -1 ) x²+ (y^ 1 / 2 -1 ) x+ (y^ 1 / 6 -1 )=0 On dividing both sides by y-1 , we get ( y^ 1 / 3 -1 y-1 ) x²+ ( y^ 1 / 2 -1 y-1 ) x+ ( y^ 1 / 6 -1 y-1 )=0 On taking limit as y 1 i.e. a 0 on both sides, we get 1 3 x²+ 1 2 x+ 1 6 =0 2 x²+3 x+1=0 x=-1,- 1 2 (roots of the equation) _ a 0⁺ (a)=-1, _ a 0⁺ (a)=- 1 2

Let (a) and (a) be the roots of the equation ( [3] 1+a -1) x²+( 1+a -1) x+( [6] 1+a -1)=0 where a>-1 . Then _ a 0⁺ (a) and _ x 0⁺ (a) are

Options

Correct answer

Step-by-step solution

More from Limits