Verified Solution • 2025–26 Session moderate

QJEEM22Mad3fe: JEE Main 2022 Mathematics - Class 11 JEEM22Mad3fe Solution: The number of distinct real roots of the equation is…

The number of distinct real roots of the equation x5(x3−x2−x+1)+x(3x3−4x2−2x+4)−1=0x^{5}\left(x^{3}-x^{2}-x+1\right)+x\left(3 x^{3}-4 x^{2}-2 x+4\right)-1=0 is ______________.

Step-by-Step Verified Solution

x8−x7−x6+x5+3x4−4x3−2x2+4x−1=0{x^8} - {x^7} - {x^6} + {x^5} + 3{x^4} - 4{x^3} - 2{x^2} + 4x - 1 = 0

⇒x7(x−1)−x5(x−1)+3x3(x−1)−x(x2−1)+2x(1−x)+(x−1)=0\Rightarrow {x^7}(x - 1) - {x^5}(x - 1) + 3{x^3}(x - 1) - x({x^2} - 1) + 2x(1 - x) + (x - 1) = 0

⇒(x−1)(x7−x5+3x3−x(x+1)−2x+1)=0\Rightarrow (x - 1)({x^7} - {x^5} + 3{x^3} - x(x + 1) - 2x + 1) = 0

⇒(x−1)(x7−x5+3x3−x2−3x+1)=0\Rightarrow (x - 1)({x^7} - {x^5} + 3{x^3} - {x^2} - 3x + 1) = 0

⇒(x−1)(x5(x2−1)+3x(x2−1)−1(x2−1))=0\Rightarrow (x - 1)({x^5}({x^2} - 1) + 3x({x^2} - 1) - 1({x^2} - 1)) = 0

⇒(x−1)(x2−1)(x5+3x−1)=0\Rightarrow (x - 1)({x^2} - 1)({x^5} + 3x - 1) = 0

∴\therefore x= ± 1x = \, \pm \,1 are roots of above equation and x5+3x−1{x^5} + 3x - 1 is a monotonic term hence vanishs at exactly one value of x other than 1 or −- 1.

∴\therefore 3 real roots.

Related Learning Resources & Exam Preparation