Maximum integral value of a forwhich one root of quadratic equation 2ax? - (a+1x -3 =0 is less than unity and other exceeds unity.
Answers
EXPLANATION.
Maximum integral value of a.
One roots of quadratic equation : 2ax² - (a + 1)x - 3 = 0 is less than unity and other exceeds unity.
As we know that,
Conditions of :
One roots should be less than and other greater than x₀.
⇒ F(x) = ax² + bx + c.
⇒ (a) = D ≥ 0.
⇒ (b) = a. f(x₀) > 0.
Equation : 2ax² - (a + 1)x - 3 = 0.
⇒ (a) = D ≥ 0.
⇒ [(a + 1)² - 4(2a)(-3)] ≥ 0.
⇒ [(a + 1)² + 24a] ≥ 0.
⇒ (a² + 1 + 2a) + 24a ≥ 0.
⇒ a² + 1 + 2a + 24a ≥ 0.
⇒ a² + 26a + 1 ≥ 0.
⇒ a = - 26 ± √(26)² - 4(1)(1)/(2).
⇒ a = - 26 ± √676 - 4/2.
⇒ a = - 26 ± √672/(2).
⇒ a = - 26 ± 4√42/(2).
⇒ a = - 13 ± 2√42.
⇒ a = - 13 + 2√42 and a = - 13 - 2√42.
⇒ a ∈ (- ∞, - 13 - 2√42] ∪ [- 13 + 2√42, ∞). - - - - - (1).
(b) = a. f(1) > 0.
⇒ 2a [2a(1)² - (a + 1)(1) - 3] > 0.
⇒ 2a[2a - a - 1 - 3] > 0.
⇒ 2a[a - 4] > 0.
⇒ a ∈ (- ∞, 0) ∪ (4, ∞). - - - - - (2).
From equation (1) and (2), we get.
(1) ∪ (2).
⇒ a ∈ (- ∞, - 13 - 2√42] ∪ [4, ∞).
Answer:
Given :-
- Maximum integral value for which one root of quadratic equation 2ax?-(a+1x-3=0 is than unity and other exceeds unity. )
□To find :-
- Maximum integral value for quadratic equation.
♧Explanation :-
- Here given,
- Here equation is given ,
- (i)
- a belongs to
- and U (4, inifinity) (2)
♧Here from equation 1 and 2 we get the answer as
- (1)U (2)
Conclusion :-
- If u got any problem in this answer you can refer the above problem a best moderator answer.