JEE Advanced (Assertion-Reason Type): Quadratic Equation & Inequalities

Directions: Each question contains Statement - 1 (Assertion) and Statement - 2 (Reason). Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct. So select the correct choice.
Q.1. Consider the equation (a + 2)x² + (a - 3) x = 2a - 1
Statement-1 : Roots of above equation are rational if 'a' is rational and not equal to -2.
Statement-2 : Roots of above equation are rational for all rational values of 'a'.
(a) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement -1.  
(b) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1.  
(c) Statement-1 is True, Statement-2 is False.  
(d) Statement-1 is False, Statement-2 is True. 

Correct Answer is option (c)
Obviously x = 1 is one of the root
∴ Other root =   = rational for all rational a ≠ -2.
(C) is correct option.  

Q.2. Let f(x) = x² = -x² + (a + 1) x + 5
Statement-1 : f(x) is positive for same α < x < β and for all a∈R
Statement-2 : f(x) is always positive for all x∈R and for same real 'a'.  
(a) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement -1.  
(b) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1.  
(c) Statement-1 is True, Statement-2 is False.  
(d) Statement-1 is False, Statement-2 is True.

Correct Answer is option (c)
Here f(x) is a downward parabola
D = (a + 1)² + 20 > 0

 From the graph clearly st (1) is true but st (2) is false

Q.3. Consider f(x) = (x² + x + 1) a² - (x² + 2) a   -3 (2x² + 3x + 1) = 0
Statement-1 : Number of values of 'a' for which f(x) = 0 will be an identity in x is 1.
Statement-2 : a = 3 the only value for which f(x) = 0 will represent an identity.
(a) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement -1.  
(b) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1.  
(c) Statement-1 is True, Statement-2 is False.  
(d) Statement-1 is False, Statement-2 is True.

Correct Answer is option (d)
f(x) = 0 represents an identity if a² - a - 6 = 0  ⇒ a = 3, -2  
⇒ a² - a - 6 = 0  ⇒ a = 3, -2  
⇒ a² - a = 0 ⇒  a = 3, -3  
⇒ a² - 2a -3 =0 ⇒ a = 3, -1  ⇒ a = 3 is the only values.  

Q.4. Let a, b, c be real such that ax² + bx + c = 0 and x² + x + 1= 0 have a common root
Statement-1 : a = b = c
Statement-2 : Two quadratic equations with real coefficients can not have only one imaginary root common.
(a) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement -1.  
(b) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1.  
(c) Statement-1 is True, Statement-2 is False.  
(d) Statement-1 is False, Statement-2 is True.

Correct Answer is option (a)
x² + x + 1 = 0  
D = - 3 < 0
∴ x² + x + 1 = 0 and ax² + bx + c = 0 have both the roots common
⇒ a = b = c.

Q.5. Statement-1 : The number of values of a for which (a² - 3a + 2) x² + (a² - 5a + b) x + a² - 4 = 0 is an identity in x is 1.
Statement-2 : If ax² + bx + c = 0 is an identity in x then a = b = c = 0.
(a) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement -1.  
(b) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1.  
(c) Statement-1 is True, Statement-2 is False.  
(d) Statement-1 is False, Statement-2 is True.

Correct Answer is option (a)
(a² - 3a + 2) x² + (a² - 5a + 6) x + a² - 4 = 0
Clearly only for a = 2, it is an identify.  

Q.6. Let a ∈ (- ∞, 0). 
Statement-1 : ax2 - x + 4 < 0 for all x ∈ R
Statement-2 : If roots of ax² + bx + c = 0, b, c ∈ R are imaginary then signs of ax² + bx + c and a are same for all x ∈ R.
(a) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement -1.  
(b) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1.  
(c) Statement-1 is True, Statement-2 is False.  
(d) Statement-1 is False, Statement-2 is True.

Correct Answer is option (d)
Statement - II is true as if ax² + bx + c = 0 has imaginary roots, then for no real x,  ax2 + bx + c is zero, meaning thereby ax² + bx + c is always of one sign. Further
Statement - I is false, because roots of ax² - x + 4 = 0 are real for any a ∈(- ∞, 0) and hence ax² - x + 4 takes zero, positive and negative values.   Hence (d) is the correct answer.  

Q.7. Let a, b, c ∈ R, a ≠ 0.
Statement-1 : Difference of the roots of the equation ax² + bx + c = 0                          
= Difference of the roots of the equation - ax² + bx - c = 0
Statement-2 : The two quadratic equations over reals have the same difference of roots if product of the coefficient of the two equations are the same.
(a) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement -1.  
(b) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1.  
(c) Statement-1 is True, Statement-2 is False.  
(d) Statement-1 is False, Statement-2 is True.

Correct Answer is option (c)
Statement-I is true, as Difference of the roots of a quadratic equation is always √D , D being the discriminant of the quadratic equation and the two given equations have the same discriminant.
Statement - II is false as if two quadratic equations over reals have the same product of the coefficients, their discriminents need not be same.
Hence (c) is the correct answer.

Q.8. Statement-1 : If the roots of x⁵ - 40x⁴ + Px³ + Qx² + Rx + S = 0 are in G.P. and sum of their reciprocal is 10, then | S |= 32.
Statement-2 : x₁. x₂. x₃.x₄.x5 = S, where x₁, x₂, x₃, x₄, x₅ are the roots of given equation.
(a) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement -1.  
(b) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1.  
(c) Statement-1 is True, Statement-2 is False.  
(d) Statement-1 is False, Statement-2 is True.

Correct Answer is option (c)
Roots of the equation x⁵ - 40x⁴ + px³ + qx² + rx + s = 0 are in G.P., let roots be a, ar, ar², ar³, ar⁴
∴ a + ar + ar² + ar³ + ar⁴ = 40  . . . (i)
and
rom (i) and (ii); ar2 = ± 2  . . . (iii)
Now, - S = product of roots = a⁵r¹⁰ = (ar²)⁵ = ± 32.
∴ | s |= 32 .
∴ Hence (c) is the correct answer.   

Q.9. Statement-1 : If a ≥ 1/2 then α  < 1 < p where α , β are roots of equation -x² + ax + a = 0  
Statement-2 : Roots of quadratic equation are rational if discriminant is perfect square.
(a) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement -1.  
(b) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1.  
(c) Statement-1 is True, Statement-2 is False.  
(d) Statement-1 is False, Statement-2 is True.

Correct Answer is option (b)
x² - ax - a = 0
g(1) < 0
⇒ a > 1/2  

Q.10. Statement-1 : The number of real roots of |x|2 + |x| + 2 = 0 is zero.
Statement-2 : ∀x∈R, |x| ≥ 0.
(a) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement -1.  
(b) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1.  
(c) Statement-1 is True, Statement-2 is False.  
(d) Statement-1 is False, Statement-2 is True.

Correct Answer is option (b)
equation can be written as (2^x)² - (a - 4) 2x - (a - 4) = 0
⇒ 2^x = 1 & 2^x = a - 4
Since x ≤ 0 and 2x = a - 4 [∵ x is non positive]  
∴ 0 < a - 4 ≤ 1
⇒ 4 < a ≤ 5   i.e., a∈ (4, 5]