JEE Advanced Level Test: Quadratic Equation- 1 - NDA MCQ

Product = 2e^2logk

LHS always positive RHS always negative LHS ≠ RHS

°•° ( x - a ) ( x - b ) + ( x - b ) ( x - c ) + ( x - c ) ( x - a ) = 0 .

==> x² - bx - ax + ab + x² - cx - bx + bc + x² - ax - cx + ac = 0 .

==> 3x² - 2bx - 2ax - 2cx + ab + bc + ca = 0 .

==> 3x² - 2x( a + b + c ) + ( ab + bc + ca ) = 0 .

When equation is compared with Ax² + Bx + C = 0 .

Then , A = 3 .

B = 2( a + b + c ) .

And, C = ( ab + bc + ca ) .

•°• Discriminant ( D ) = b² - 4ac .

= [ 2( a + b + c )]² - 4 × 3 × ( ab + bc + ca ) .

= 4( a + b + c )² - 12( ab + bc + ca ) .

= 4[ ( a + b + c )² - 3( ab + bc + ca ) ] .

= 4( a² + b² + c² + 2ab + 2bc + 2ca - 3ab - 3bc - 3ca ) .

= 4( a² + b² + c² - ab - bc - ca ) .

= 2( 2a² + 2b² + 2c² - 2ab - 2bc - 2ca ) .

= 2[ ( a - b )² + ( b - c )² + ( c - a )² ] ≥ 0 .

[ °•° ( a - b )² ≥ 0, ( b - c )² ≥ 0 and ( c - a )² ≥ 0 ] .

This shows that both the roots of the given equation are real .

For equal roots, we must have : D = 0 .

Now, D = 0 .

==> ( a - b )² + ( b - c )² + ( c - a )² = 0 .

==> ( a - b ) = 0, ( b - c ) = 0 and ( c - a ) = 0 .

aα² + bα + c = 0 and bα² + cα + a = 0 ⇒ α = 1 ∴ a + b + c = 0

We have x² + px + 1 = (x – a)(x – b)
Thus, E = (c – a)(c – b)(-d - a)(-d - b)
= (c² + pc + 1)[(-d²) - pd + 1] = (c² + pc + 1)(d² - pd + 1)       [∴ a + b = -p]
But c² + qc + 1 = 0 and d² + qd + 1 = 0
∴ E = (-qc + pc)(-qd - pd) = cd(q - p)(q + p) = cd(q² - p²) = q² - p² [∴ cd = 1]

For x² + 2ax + 10 - 3a > 0 
D < 0 ⇒ (a + 5) (a - 2) < 0

The discriminant D of the quadratic equation is given by
D = 121(a + b + c)² + 24(a – b)2
As a, b, c are real, 121(a + b + c)² > 0
Also, as a ≠ b ,  (a - b)² > 0 ; Thus, D > 0
Therefore, the equation (1) has real and unequal roots. 

Sum of first n terms of AP, S_n = 2n² + 5n
 
Now choose n =1 and put in the above formula,
First term = 2+5 = 7
 
Now put n=2 to get the sum of first two terms = 2x4 + 5x 2= 8 + 10 = 18
 
This means 
 
first term + second term = 18
 
but first term =7 as calculated above
 
so, second term = 18-7 = 11
 
So common difference becomes, 11-7 = 4
 
So the AP becomes, 7, 11, 15, ....
nth term = a + (n - 1)d = 7 + (n-1)4 = 7 + 4n - 4 = 3 + 4n

given that harmonic mean between the roots of the 
given equation is 4.

α + β = -a,αβ = -b;γ+δ = -a,γδ = b
(α² + aα + b) (β² + aβ +b) = 4b²

 Let α,β be roots of ax³+bx+c=0
γ,δ be roots of px²+2qx+r=0
α/β = γ/δ …………(1) and  
β/α = δ/γ ………….(2)
(1) + (2)
⇒ α/β + β/α = γ/δ + δ/γ
= [(2²+β²)/αβ + 2]= [γ²+δ²]/γδ + 2
⇒ [(α)²+(β)²+2αβ]/αβ ​= [γ²+δ²+2γδ]/γδ
​= [(α+β)²]/αβ = [(γ+δ)²]/γδ
⇒ (4b²/a²)/(c/a) = (4q²/p²)/(r/p)
⇒ b²/ac = q²/pr.

Correct Answer :- A

Explanation : (b-c)x² + 2(c-a)x + (a-b) = 0

b² - 4ac

[2(c-a)² - 4(b-c)(a-b)]

4(c-a)² -4(b-c)(a-b)

4[(c-a)² - (b-c)(a-b)]

4[c² + a² - 2ac -(ab - b² - ca + bc)]

4[c2 + a2 - 2ac -ab + b² + ca - bc)]

4{a² + b² + c² - ab - bc - ca}

4(a² + b² + c²) -2(2ab + 2bc + 2ca)

4(a² + b² + c²) -2{(a+b+c)² - a² - b² - c²}

4(a² + b² + c²) -2(a+b+c)² + 2a² + 2b² + 2c²

4(a² + b² + c²) -2(a+b+c)² > 0 (Real)

(x – a)(x – 10) = -1  
Since a is an integer and x is also integer,
So, (x – a) & (x – 10) can be 1 or –1
For (x – 10) = 1, x = 11
(x – a) = -1, 11 – a = -1, a = 12 
For (x – 10) = -1, x = 9
(x – a) = 1, 9 – a = 1, a = 8 

x² - 2ax + a² -1 = 0 ⇒ (x - a)² = 1
Roots are a - 1 and a + 1
Both roots lie in (-3, 4) ∴ a - 1 > -3, a + 1 < 4
Ie -2 < a < 3 ⇒ [a] = -2, -1, 0, 1, 2 whose sum is zero.

Use extreme value formula

P(x) when divided by x−1, x-2 and x−3 leaves remainder 3,7 and 13 respectively.
From polynomial-remainder theorem, P(1) = 3 and P(2) = 7 P(3) = 13
If the polynomial is divided by (x−1)(x−2)(x-3) then remainder must be of the form ax+b (degree of remainder is less than that of divisor)
 
⇒P(x) = Q(x)(x−1)(x−2)(x-3)+(ax² + bx + c), where Q(x) is some polynomial. 
Substituting for x=1, x=2 and x=3:
P(1) = 3 = a+b+c
P(2) = 7 = 4a+2b+c
P(3) = 13 = 9a+3b+c
=> Subtract P(2) from P(1)
-3a - b = -4.............(1)
Subtract P(3) from P(2)
5a + b = 6..............(2)
From (1) and (2) Solving for a and b, we get 
a = 1 and b = 1, c = 1
⇒Remainder = x² + x + 1

a, b are roots of x² – 5x + 8 = 0 
a + b = 5, ab = 8:

∴ required equation is
x² – (9/8)x + 1 = 0  
⇒  8x² – 9x + 8 = 0 

Write a³x² + abcx + c³ = 0  as 

Let D₁ = b² – 4ac and  D₂ = d²  + 4ac
As ac ≠ 0, either ac < 0 or ac > 0
If ac < 0, then D₁ > 0
In this case P(x) = 0 has distinct two real roots 
If ac > 0, the D₂ > 0
In this case Q(x) = 0 has two distinct real roots.
Thus, in either case P(x)Q(x) = 0 has at least two distinct real roots. 

We have product of the roots = 2e^{4 ln k} - 1 = 31 (given) 

⇒ k⁴ = 16 ⇒ k⁴ - 16 = 0
⇒ (k – 2)(k + 2)(k² + 4) = 0  ⇒ k = 2, -2
as k > 0, we get k = 2.
Sum of the roots = 5k = 10 

But α + β = p and α + 4β = 2q ⇒ β = (1/3)(2q – p) and α = (2/3)(2p – q)

  

Putting y = |x – 2|,we can rewrite the given equation as
y² + y – 2 = 0 or  y² + 2y – y – 2 = 0
⇒ y(y + 2) – (y + 2) = 0  ⇒ (y – 1)(y + 2) = 0
as   y = |x – 2|  > 0,  y + 2 > 2
∴ y – 1 = 0   ⇒ y = 1 ⇒ |x – 2| = 1
⇒ x – 2 = ±1 ⇒ x = 2 ± 1  ⇒ x  = 3 or 1