Complex Number & Quadratic Equation
Linked Comprehension Type
Quiz-31
Dear Readers,
Complex numbers and quadratic equations is a segment of maths that deals with crucial theorems and concepts along with various formulae. It comprises of linear and quadratic equations along with roots related to the complex number's set (known as complex roots)..
Q1. In the given figure, vertices of ∆ABC lie on y=f(x)=ax^2+bx+c. The ∆ABC is right angled isosceles triangle whose hypotenuse AC=4√2 units
y=f(x) is given by
Solution
(d)
(d)
Q2.Consider the inequality 9x - a3x - a + 3 ≤ 0, where ‘a’ is a real parameter
The given inequality has at least one negative solution for a∈
The given inequality has at least one negative solution for a∈
Solution
(d)
(d)
Q3. Consider the in equation x2+x+a-9< 0
The value of the real parameter ‘a’ so that the given in equation has at least one positive solution:
The value of the real parameter ‘a’ so that the given in equation has at least one positive solution:
Solution
(d)
(d)
Q4. ‘af(μ)< 0’ is the necessary and sufficient condition for a particular real number μ to lie between the roots of a quadratic equation f(x) = 0, where f(x)=ax2+bx+c. Again if f(μ1 )f(μ2 )< 0, then exactly one of the roots will lie between μ1 and μ2
If |b|>|a+c|, then
Solution
(b)
b2 >(a+c)2
⇒ (a+c-b)(a+c+b)< 0
⇒ f(-1)f(1)< 0
So, there is exactly one root in (-1,1)
(b)
b2 >(a+c)2
⇒ (a+c-b)(a+c+b)< 0
⇒ f(-1)f(1)< 0
So, there is exactly one root in (-1,1)
Q5.
The real numbers x1,x2,x3 satisfying the equation x3-x2+βx+γ=0 are in A.P.
All possible values of β are
All possible values of β are
Solution
(a)
From the question, the real roots of x3-x2+βx+γ=0 are x1,x2,x3 and they are in A.P. As x1,x2,x3 are in A.P., let x1=a-d,x2=a,x3=a+d. Now,
x1+x2+x3=-(-1)/1=1
⇒a-d+a+a+d=1
⇒a=1/3 (1)
x1 x2+x2 x3+x3 x1=β/1=β
⇒(a-d)a+a(a+d)+(a+d)(a-d)=β (2)
x1 x2 x3=-γ/1=-γ
⇒(a-d)a(a+d)=-γ (3)
From (1) and (2), we get
3a2-d2=β
⇒ 3 1/9-d2=β, so β=1/3-d2< 1/3
From (1) and (3), we get
1/3 (1/9-d2 )=-γ
⇒ γ=1/3 (d2-1/9)>1/3 (-1/9)=-1/27
γ∈(-1/27,+∞)
(a)
From the question, the real roots of x3-x2+βx+γ=0 are x1,x2,x3 and they are in A.P. As x1,x2,x3 are in A.P., let x1=a-d,x2=a,x3=a+d. Now,
x1+x2+x3=-(-1)/1=1
⇒a-d+a+a+d=1
⇒a=1/3 (1)
x1 x2+x2 x3+x3 x1=β/1=β
⇒(a-d)a+a(a+d)+(a+d)(a-d)=β (2)
x1 x2 x3=-γ/1=-γ
⇒(a-d)a(a+d)=-γ (3)
From (1) and (2), we get
3a2-d2=β
⇒ 3 1/9-d2=β, so β=1/3-d2< 1/3
From (1) and (3), we get
1/3 (1/9-d2 )=-γ
⇒ γ=1/3 (d2-1/9)>1/3 (-1/9)=-1/27
γ∈(-1/27,+∞)
