The number of distinct real roots of the equation $x^{5}\left(x^{3}-x^{2}-x+1\right)+x\left(3 x^{3}-4 x^{2}-2 x+4\right)-1=0$ is

JEE Main · 2022 · 26 Jul 2022 · Shift 1 · Quadratic Equations and Inequalities

The number of distinct real roots of the equation

x^{5} (x^{3} - x^{2} - x + 1) + x (3 x^{3} - 4 x^{2} - 2 x + 4) - 1 = 0

Answer: 3

Factor the equation:

x^{5} (x^{3} - x^{2} - x + 1) + x (3 x^{3} - 4 x^{2} - 2 x + 4) - 1 = 0

. Note that

x^{3} - x^{2} - x + 1 = (x - 1) (x^{2} - 1) = (x - 1)^{2} (x + 1)

and

3 x^{3} - 4 x^{2} - 2 x + 4 = (x - 1) (3 x^{2} - x - 4)

. After factoring, the equation becomes

(x - 1)^{2} (x + 1) (x^{5} + 3 x - 1) = 0

. The roots are

x = 1

(double),

x = - 1

, and the roots of

f (x) = x^{5} + 3 x - 1 = 0

. Since

f^{'} (x) = 5 x^{4} + 3 > 0

for all

x

f

is strictly increasing and has exactly one real root. Thus there are 3 distinct real roots:

x = 1, - 1

, and one root from

x^{5} + 3 x - 1 = 0

JEE Main · 2022 · 26 Jul 2022 · Shift 1 · Integer answer