JEE Advanced Level Test: Quadratic Equation- 2 - JEE MCQ

A = 1
L = 11
Sn = 36
⇒ n( a +L) /2 = 36
⇒ n(1+11)= 72
⇒ n *12 = 72
⇒ n = 6
No.of terms = 6

Let the two numbers be α and β. Given that 

∴ Required equation is x² – 18x + 16 = 0 

a + 3 + 5 = 0, replace a value find roots

The given quadratic equation 5x²−4x+2+k(4x²−2x−1)=0 can be resolved as: 5x²−4x+2+k(4x²−2x−1)=0
⇒5x²−4x+2+4kx²−2kx−k=0
⇒(5+4k)x²−(4+2k)x+(2−k)=0
We know that the sum of the roots of a quadratic equation ax^2+bx+c is − b/a​
 and the product of the roots is c/a
Now, in the equation (5+4k)x²−(4+2k)x+(2−k)=0, we have:
Product of the roots is: c/a = 2−k/(5+4k)
​
Also, it is given that the product of the roots is 2, therefore,
=2−k/(5+4k)=2
⇒2−k=2(5+4k)
⇒2−k=10+8k
⇒−k−8k=10−2
⇒−9k=8
⇒k=−8/9
Hence, k=−8/9

α² + aα + b = 0 and α² + ba + a = 0
⇒ α(a - b) + (b - a) = 0 ⇒ a = 1

 y =−5x²+2x−7
Differentiating the above equation with respect to x, gives us
dy/dx=−10x+2
For maximum, dy/dx=0
2−10x=0
x=1/5​
Hence the above expression reaches its maximum value at x = 1/5

L.H.S. is positive for all x ∈ R

α⋅β = 4
α+β = 2
α²+β² = (α+β)² - 2αβ
= 4−8=−4
(a+b)³ = a³ + b³ + 3ab(a+b)
(α²+β²)³=α⁶ +β⁶+3α²β²(α²+β²)
-64=α6+β6+3*16(-4)
α6+β6=192−64=128

α³ + β³ = (α + β)³ - 3αβ(α + β) 
⇒ αβ = –8

Let original equation be  x² + ax + b = 0 from the given –a = 5, b = –6 

Let α = 2k – 1, β = 2k + 1, k ∈ N
α + β = –b /a = 4k > 0 ⇒ b < 0 and –b = 4ak > 4a ⇒ |b| > 4a

Δ = 4(a² + c² - ac) = 2(a2 + c² + (a - c²)) > 0

(a – 1) x² – (a² – 3a + 2) x + a² – 1 = 0 will have more than two solutions only if it is an identity, i.e., coefficient of x² = coefficient of x = constant term = 0. A – 1 = 0, a² – 3a + 2 = 0 and a² – 1 = 0. Simultaneously. So, a = 1.

Roots of (x - α) (x - β) = λ are γ and δ, 
i.e., (x - α) (x - β) - λ = (x - γ) (x - δ)
So, (x - γ) (x - β) + λ = (x - α) (x - β)

∑a(b - c) = 0, then both roots of equation is 1. 
Product of roots 


Hence, roots are in H.P.

α + β = (a – 2), αβ = –a –1
α² + β² = (α + β)² - 2αβ = (a - 2)² + 2(a + 1)
α² + β² = α² - 2a + 6
ax² + bx+ c takes least value -D/4a, so, least value at of α² + β² is 5.

For ax²+bx+c=0,
if the two roots are of different signs, their product must be –ve => c/a < 0
=> p(p-1)/3 < 0 => p(p-1) < 0 => 0 < p < 1
 = (0,1) 

Let the roots be α, β : β, γ and γ, α then
αβ = b, βγ = c and γα = a 

Let f(x) x² + (a + 2) x – (a + 3)
Diecriminant = (a + 4)² 
So roots are real 
Since 2 lie between the roots so f(2) < 0
⇒ a < –5 ⇒ S = (–∞, -5)

Let f(x) = x² + 2ax + b = (x + a)² + b – a²
So minimum value of f(x) is b – a²
Since hence b - a² > c
i.e., b - c > a²

f(0) = a -1 < 0
⇒ a < 1
f(1) = 1 + 2a + 1 a - 1 < 0
⇒ 3a < -1

D > 0 ⇒ 4a² + 4a + 1 - 4a + 4 > 0
⇒ 4a² + 5 > 0
True for all a ∈ R
From equation (i), (ii) and (iii) 

(x – a)(x – 20) = –1
Then either x – a = 1 and x – 20 = –1 or x – a = –1 and x – 20 = 1 which give a = 18, 22

p + q = –p, pq = q if q = 0 then p = 0
If q ≠ 0 then p = 1 and q = –2. Thus p = 1 or 0.

We can write the given equation as
(x – a)² = 3 – a
This shows that 
Both the roots of the given equation will be less than 3 if the larger of the two roots is less than 3, that is, if 


⇒ 3 - a > 1 or a < 2
Thus a > 3 and a < 2 ⇒ a < 2

Let α, α² be the roots of 3x² + px + 3 = 0. Then 

As α³ = 1 ⇒ α = 1, ω, ω²
We take α = ω so that α² = ω²
Thus - 1 = ω + ω² = α + α² 

((x+b)+(x+a))c =(x+a)(x+b)
⇒ x² +bx+ax-2cx+ab- bc- ca = 0 
Now, let roots are α and β, then
α + β = 0, αβ = ab - bc - ac
and αβ = ab - (b + a) c