Quadratic Equations (Advanced pattern) (JEE Advanced)

Question 1

The set of all real k for which the number 1 lies strictly between the roots of (k−1)x²−(k+1)x+(k+4)=0 is

Accepted Answer

B. For f(x)=Ax²+Bx+C with A≠0, a number p lies strictly between the roots iff A f(p)<0 (reality is then automatic).
f(1)=(k−1)−(k+1)+(k+4)=k+2.
Condition: (k−1)(k+2)<0, i.e. −2<k<1.
k=1 is excluded since the equation becomes linear.
Trap: using f(1)<0 alone (valid only for a positive leading coefficient) gives k<−2.
Exam pattern: JEE Advanced – location of a point between the roots, a f(k)<0

Question 2

The minimum value of x²−2ax+3 on the interval [0,2] equals −1. Then the set of all possible real values of a is

Accepted Answer

D. The vertex is at x=a. Three cases: a<0: min =f(0)=3. 0≤ a≤2: min =f(a)=3−a². a>2: min =f(2)=7−4a. 3−a²=−1 gives a=±2; only a=2 lies in [0,2]. 7−4a=−1 gives a=2, not >2; the case a<0 gives 3, never −1. Answer {2}. Trap: solving 3−a²=−1 without checking that the vertex lies in the interval gives {2,−2}. Exam pattern: JEE Advanced – extreme values of a quadratic on a closed interval with a parameter

Question 3

The set of all real a for which exactly one root of x²−2ax+a²−4=0 lies in the open interval (1,3) is

Accepted Answer

C. Key insight: x²−2ax+a²−4=(x−a)²−4, so the roots are a−2 and a+2, at distance 4.
The interval (1,3) has length 2, so both roots can never lie in it; we need exactly one inside.
a−2∈(1,3) gives a∈(3,5); a+2∈(1,3) gives a∈(−1,1).
Answer (−1,1)∪(3,5); endpoints put a root at 1 or 3, outside the open interval.
Trap: merging the pieces into (−1,5); for 1≤ a≤3 neither root lies in (1,3).
Exam pattern: JE

Quadratic Equations (Advanced pattern): questions with solutions

Get the full chapter for ₹99