Question

13. In Cauchy's mean value theorem, if f(x)=sin x, g(x)= cos x, show that is independent of both x and h and theta is equal to 1/2

​

Answer 1

Answer:

We have to verify Cauchy's mean value theorem for the given functions

\begin{gathered}f(x) = sinx \\g(x)= cosx\end{gathered}f(x)=sinxg(x)=cosx

for the interval x ∈ (0,\frac{\pi }{2})(0,2π)

Cauchy's mean value theorem states that,

If two functions f(x) and g(x) are continous on closed interval [a.b] and     differentiable on open interval (a,b) and g(x) ≠ 0, ∀ x∈ (a,b) , Then there exist a point c ∈ (a,b) such that

               \frac{f(b) - f(a)}{g(b) - g(a)} = \frac{f'(c)}{g'(c)}g(b)−g(a)f(b)−f(a)=g′(c)f′(c)

Now, verifying cauchy's mean value theorem for the given functions

           \begin{gathered}f(x) = sinx \\g(x)= cosx\end{gathered}f(x)=sinxg(x)=cosx

As sinx and cosx are continous and differential for all x ∈ R , therefore they are also continous for the interval x ∈ [0,\frac{\pi }{2}][0,2π] and differentiable for the interval x ∈ (0,\frac{\pi }{2})(0,2π) .

       Hence cauchy's mean value theorem is valid

Now,

                  \frac{f(\frac{\pi }{2} ) - f(0)}{g(\frac{\pi }{2} ) - g(0)} = \frac{f'(c)}{g'(c)}g(2π)−g(0)f(2π)−f(0)=g′(c)f′(c)

                  \frac{sin\frac{\pi }{2} - sin0}{cos\frac{\pi }{2} - cos0} = \frac{cosc}{-sinc}cos2π−cos0sin2π−sin0=−sinccosc

                  \frac{1-0}{0-1} = -\frac{cosc}{sinc}0−11−0=−sinccosc

                  \begin{gathered}\frac{cosc}{sinc} = 1\\cosc = sinc \\sin(\frac{\pi }{2} -c) = sinc\\\frac{\pi }{2} -c = c\\2c = \frac{\pi }{2} \\\end{gathered}sinccosc=1cosc=sincsin(2π−c)=sinc2π−c=c2c=2π

               ∴  c = \frac{\pi }{4}c=4π ∈ (0,\frac{\pi }{2})(0,2π)

         Hence Cauchy's mean value theorem is verified.