Show that between any two roots of e sinx = 1, there exists at least one root of et cos x + 1 = 0.
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>>Show that between any two roots of the e
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Show that between any two roots of the equation e
x
cosx=1 there exists atleast one root of e
x
sinx−1=0 by continuity and differentiability.
Hard
Solution
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Let a and b be the roots of the equation e
x
cosx=1 then
e
a
cosa=1 and e
b
cosb=1
Let f be a function defined as f(x)=e
−x
−cosx
We observe that:
(i)f(x) is continuous in [a,b] and cosx both are continuous functions.
(ii)f
′
(x)=−e
−x
+sinx, hence function is differentiable in (a,b)
(iii)f(a)=e
−a
−cosa=e
−a
(1−e
a
cosa)=0 and
f(b)=e
−b
−cosb=e
−ab
(1−e
b
cosb)=0
⇒f(a)=f(b)=0
Thus f(x) satisfies all the conditions of Rolle's theorem in [a,b]
hence there exists at least one value of x in (a,b) such that f
′
(c)=0
f
′
(x)=−e
−x
+sinx
f
′
(c)=−e
−c
+sinc
⇒−e
−c
+sinc=0
⇒e
−c
=sinc
⇒e
c
sinx−1=0
Thus, c is a root of the equation e
−x
−sinx=0
Hence between any two roots of the equation e
x
cosx=1 there exists atleast one root of e
x
sinx−1=0