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Between any two real roots of the equation ex sin x = 1, the equation ex cos x = - 1 has
- a)Atleast one root
- b)Exactly one root
- c)Atmost one root
- d)No root
Correct answer is option 'A'. Can you explain this answer?
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Between any two real roots of the equationexsin x = 1, the equationexc...
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Between any two real roots of the equationexsin x = 1, the equationexc...
Solution:
Let f(x) = exsinx - 1
Let g(x) = excosx + 1
Now, we need to prove that there is atleast one root of g(x) between any two roots of f(x).
Let a and b be two real roots of f(x).
Then, f(a) = exsina - 1 = 0 and f(b) = exsinb - 1 = 0
Also, f(x) is continuous on [a,b] and differentiable on (a,b) (as ex is differentiable everywhere).
Therefore, by Rolle's Theorem, there exists a c ∈ (a,b) such that f'(c) = 0
Now, f'(x) = ex(sin x + cos x)
So, f'(c) = ec(sincos(c) + coscos(c)) = ec cos(c)(sin(c) + 1)
But ec ≠ 0 (as e is a positive constant), so sin(c) + 1 = 0
Therefore, cos(c) ≠ 0 (as sin(c) = -1), which implies that g(c) = excos(c) + 1 ≠ 0
So, g(x) does not have a root at c.
Therefore, there exists a root of g(x) between a and b (as g(a) and g(b) have opposite signs).
Hence, the correct option is (A) - Atleast one root.
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Between any two real roots of the equationexsin x = 1, the equationexcos x = - 1 hasa)Atleast one rootb)Exactly one rootc)Atmost one rootd)No rootCorrect answer is option 'A'. Can you explain this answer?
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