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31/01/2022 1:18 pm
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The number of real roots of the equation
e4x + 2e3x – ex – 6 = 0 is
(1) 2
(2) 4
(3) 1
(4) 0
1 Answer
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31/01/2022 1:21 pm
Correct answer: (3) 1
Explanation:
Let ex = t > 0
ƒ(t) = t4 + 2t3 – t – 6 = 0
ƒ'(t) = 4t3 + 6t2 – 1
ƒ"(t) = 12t2 + 12t > 0
ƒ(0) = -6, ƒ(1) = -4, ƒ(2) = 24
⇒ Number of real roots = 1