Question

Verify cauchys mean value theorem f(x) =sinx, g(x) =cosx in [0, π/2] answer these question

Answer 1

$\large\underline\purple{\bold{Solution :- }}$

We have Cauchy’s mean value theorem

$\tt \:  \dfrac{f(b) - f(a)}{g(b) - g(a)} = \dfrac{f'(c)}{g'(c)}$

Here,

$\tt \longrightarrow \: f(x) = sinx \\ \tt \longrightarrow \: g(x) \: = cosx \\ \tt \longrightarrow \: f'(x) = cosx \\ \tt \longrightarrow \: g'(x) = - sinx$

☆ Clearly, both f(x) and g(x) are continuous on [ 0, π/2 ]

☆ and both f(x) and g(x) are differentiable on (0, π/2).

☆ Therefore, by Cauchy Mean Value Theorem, there exist atleast one c belongs to (0, π/2) such that

$\tt \:  \dfrac{f(b) - f(a)}{g(b) - g(a)} = \dfrac{f'(c)}{g'(c)}$

$\tt \:  \dfrac{f(\dfrac{\pi}{2} ) - f(0)}{g(\dfrac{\pi}{2} ) - g(0)} = \dfrac{f'(c)}{g'(c)}$

$\tt\implies \:\dfrac{sin\dfrac{\pi}{2} - sin \: 0}{cos \: \dfrac{\pi}{2} - cos \: 0 } = \dfrac{cos \: c}{ - \: sin \: c}$

$\tt\implies \: - \: cot \: c = \dfrac{1 - 0}{0 - ( - 1)}$

$\tt\implies \:cot \: c \: = 1$

$\tt\implies \: \boxed{ \purple{ \bf \: c \: = \: \dfrac{\pi}{4} }}$

$\large{\boxed{\boxed{\bf{Hence, verified}}}}$