CAT 2018 Question Paper | Quants Slot 2
CAT Previous Year Paper | CAT Inequalities Questions | Question 23
CAT Questions This is a question from the topic: Quadratic Equations. People generally take this topic for granted, however, this question is here to prove you otherwise. If you know the concept of completion of squares, then this is a child's play for you! Try this delightful question which appeared in the CAT 2018 Question Paper, if you aren't able to solve, the detailed video solution is always there to help you. Solve more such questions by clicking here: CAT Question Bank
Question 23 : If a and b are integers such that 2x2 − ax + 2 ＞ 0 and x2 − bx + 8 ≥ 0 for all real numbers x, then the largest possible value of 2a − 6b is [TITA]
🎉 Ace XAT 2025 with 2IIM's Crash Course - Enrol now
Best CAT Online Coaching
Try upto 40 hours for free
Learn from the
best!
2IIM : Best Online CAT Coaching.
Video Explanation
Best CAT Coaching in Chennai
CAT Coaching in Chennai - CAT 2022
Limited Seats Available - Register Now!
Explanatory Answer
Method of solving this CAT Question from Inequalities
Given that a and b are integers such that 2x2 − ax + 2 > 0 and x2 − bx + 8 ≥ 0 for all real numbers x
We have to find the largest possible value of 2a – 6b
Using the completion of squares method
2x2 − ax + 2 > 0 can be written as x2 - \\frac{a}{2})x + 1 > 0
We know that (x – k )2 = x2 – 2kx + k2
Here we have \\frac{a}{2}) = 2k or k = \\frac{a}{4})
This \\frac{a^2}{16}) is added on both sides
⟹ x2 - \\frac{a}{2})x + \\frac{a^2}{16}) > \\frac{a^2}{16}) - 1
⟹ (x - \\frac{a}{4}))2 > \\frac{a^2}{16}) - 1 [ this \\frac{a^2}{16}) - 1 always holds good if it is negative or zero]
\\frac{a^2}{16}) - 1 < 0
a2 < 16
a ∈(−4 , 4)
Since we have to find the largest value as possible and it cannot be 4 so a = 3
Similarly b can be found in the following way
x2 − bx + 8 ≥ 0
x2 − bx + \\frac{b^2}{4}) ≥ \\frac{b^2}{4}) - 8
x2 − bx + \\frac{b^2}{4}) ≥ \\frac{b^2}{4}) - 8 [\\frac{b^2}{4}) = (\\frac{b}{2}))2]
(x - \\frac{b}{2}))2 ≥ \\frac{b^2}{4}) - \\frac{32}{4})
b2 ≤ 32
b2 can go from -5 , -4 , -3 , -2 , -1 , 0 , 1 , 2 , 3 , 4 , 5
Since we have to find the largest possible value of 2a – 6b or b have to be very small so the smallest integer b can take -5
⟹ 2(3) - 6(-5)
⟹ 6 + 30 = 36
The largest possible value of 2a − 6b is 36
The question is "If a and b are integers such that 2x2 − ax + 2 ＞ 0 and x2 − bx + 8 ≥ 0 for all real numbers x, then the largest possible value of 2a − 6b is [TITA]"
Hence, the answer is 36
