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The number of real roots of the equation e?? - e?? - 2e³? - 12e²? + e? + 1 = 0 is:

Option 1 -

1

Option 2 -

4

Option 3 -

2

Option 4 -

6

0 2 Views | Posted a month ago
Asked by Shiksha User

  • 1 Answer

  • R

    Answered by

    Raj Pandey | Contributor-Level 9

    a month ago
    Correct Option - 3


    Detailed Solution:

    e? - e? - 2e³? - 12e²? + e? + 1 = 0
    e³? - 2 + e? ³? - e? [e³? + 12e? - 1] = 0
    Let y=e? y? -y? -2y³-12y²+y+1=0.
    The solution breaks the equation into (e³? - 4e? - 1) (e³? - 1 + 3e? ) = 0
    Either e³? = 4e? + 1 (One Solution)
    OR e³? = 1 - 3e? (One Solution)
    ∴ the equation has total 2 solutions.

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