JEE Advanced (Assertion-Reason Type): Quadratic Equation & Inequalities | Chapter-wise Tests for JEE Main & Advanced PDF Download

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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]