JEE Main · 2022 · 24 Jun 2022 · Shift 2 · Application of Derivatives

The number of distinct real roots of the equation

x^{7} - 7 x - 2 = 0

Answer: (D)

Solution

Let

f (x) = x^{7} - 7 x - 2

. We need to find where

f (x) = 0

, or equivalently where

x^{7} - 7 x = 2

.

Find critical points:

f^{'} (x) = 7 x^{6} - 7 = 7 (x^{6} - 1) = 7 (x^{2} - 1) (x^{4} + x^{2} + 1) = 0

This gives

x = \pm 1

as critical points.

f (- 1) = - 1 + 7 - 2 = 4 > 0

f (1) = 1 - 7 - 2 = - 8 < 0

Since

f

is an odd function and continuous:

For $x < - 1$ : $f (x) \to - \infty$ , and $f (- 1) = 4 > 0$ , so one root in $(- \infty, - 1)$
For $- 1 < x < 1$ : $f (- 1) = 4 > 0$ and $f (1) = - 8 < 0$ , so one root in $(- 1, 1)$
For $x > 1$ : $f (1) = - 8 < 0$ and $f (x) \to \infty$ , so one root in $(1, \infty)$

Therefore, there are 3 distinct real roots.

JEE Main · 2022 · 24 Jun 2022 · Shift 2 · Single correct option