Banner 0
Banner 1
Banner 2
Banner 3
Banner 4
Banner 5

Q.

The number of distinct real roots of the equation x3x2x+1x5+x3x34x22x+41=0 is _________

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away
An Intiative by Sri Chaitanya

answer is 3.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

x5x3x2x+1+x3x34x22x+41=0x8x7x6+x5+3x44x32x2+4x1=0(x1)x7x5+3x3x23x+1=0(x1)x5x21+3xx211x21=0(x1)x21x5+3x1=0(x1)2(x+1)x5+3x1=0 Let f(x)=x5+3x1f'(x)>0xRx5+3x1 is a monotonic function hence vanishes at exactly one value of x other than 1,-1 Hence 3 real distinct roots. 

Best Courses for You

JEE

JEE

NEET

NEET

Foundation JEE

Foundation JEE

Foundation NEET

Foundation NEET

CBSE

CBSE

score_test_img

Get Expert Academic Guidance – Connect with a Counselor Today!

whats app icon