Question
Between any two real roots of the equation ex sin x = 1, the equation ex cos x = - 1 has
Answer: Option A
:
A
Let ∝,β(∝<β) be any two real roots of
f(x) = e - x - sin x
Then, f(∝)=0=f(β)
Moreover, f(x) is continuos and differentiable for xε[∝,β].
Hence, from Rolle's thereom, thereom, there exists atleast one x in ∝,β such t
f′(x)=0⇒−e−x−cosx=0⇒−e−x(1+excosx)=0⇒excosx=−1.
Hence (a) is the correct answer.
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:
A
Let ∝,β(∝<β) be any two real roots of
f(x) = e - x - sin x
Then, f(∝)=0=f(β)
Moreover, f(x) is continuos and differentiable for xε[∝,β].
Hence, from Rolle's thereom, thereom, there exists atleast one x in ∝,β such t
f′(x)=0⇒−e−x−cosx=0⇒−e−x(1+excosx)=0⇒excosx=−1.
Hence (a) is the correct answer.
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