The number of real roots of the equation
is:
The number of real roots of the equation is:
Option 1 -
4
Option 2 -
0
Option 3 -
1
Option 4 -
2
Asked by Shiksha User
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1 Answer
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Correct Option - 3
Detailed Solution:Let f (t) = t4 + 2t3 – t – 6

f’ (t) = 4t3 + 6t2 – 1f
Þf’ (0) = -1, f’ (+
= +For t > 0
Þ f’ (t) = 0 has only one root.
One solution of f (t) = 0 is possible
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