JEE Main · 2021 · 22 Jul 2021 · Shift 2 · Sets Relations and Functions

Let

[x]

denote the greatest integer less than or equal to

x

. Then, the values of

x \in R

satisfying the equation

[e^{x}]^{2} + [e^{x} + 1] - 3 = 0

lie in the interval :

Answer: (C)

Solution

Let

[e^{x}] = k

, where

k

is an integer. Then the equation becomes

k^{2} + (k + 1) - 3 = 0

, which simplifies to

k^{2} + k - 2 = 0

. Factoring gives

(k + 2) (k - 1) = 0

, so

k = 1

k = - 2

.

Since

e^{x} > 0

for all real

x

, we have

[e^{x}] \geq 0

, so

k = - 2

is not valid. Thus

[e^{x}] = 1

.

This means

1 \leq e^{x} < 2

, which gives

ln 1 \leq x < ln 2

, i.e.,

0 \leq x < lo g_{e} 2

.

Therefore,

x \in [0, lo g_{e} 2)

JEE Main · 2021 · 22 Jul 2021 · Shift 2 · Single correct option