JEE Main 2021 · 22 July · evening · Single correct · +4 / −1
Let [x] denote the greatest integer less than or equal to x. Then, the values of x∈R satisfying the equation [eˣ]² + [eˣ + 1] - 3 = 0 lie in the interval :
- (a) [ 0,1/e )
- (b) [logₑ2, logₑ3)
- (c) [1, e)
- (d) [0, logₑ2)
Answer and solution
Answer (d) [0, logₑ2)
Solution
Maths sir · 4 steps
Pull the 1 out of [eˣ + 1], and the whole thing becomes a quadratic in [eˣ].
- Split off the integer. For any integer n, [y + n] = [y] + n. So [eˣ + 1] = [eˣ] + 1. Let t = [eˣ], which is an integer.
t² + (t + 1) - 3 = 0 ⇒ t² + t - 2 = 0 ⇒ (t + 2)(t - 1) = 0
- Reject one root. This gives t = 1 or t = -2. Since eˣ > 0 for every real x, [eˣ] ≥ 0, so t = -2 is impossible. That leaves [eˣ] = 1.
- Back to x. [eˣ] = 1 means 1 ≤ eˣ < 2. Taking logₑ of each part, which keeps the inequality direction, gives 0 ≤ x < logₑ 2.
- So. The solutions are exactly the interval [0, logₑ 2), which is option (d).
The idea to keep. Whenever an integer is added inside the box, pull it out: [y + n] = [y] + n. Then solve for the box as a variable, keep only the integer values it can take, and convert back.
Answer from the NTA final answer key. Solution by PrepWiser.