Question

If $f\left( x \right) = \cos x,\,0\; \leqslant x\; \leqslant \left( {\dfrac{\pi }{2}} \right)\;$, the real number $c$ of the mean value theorem is:
A. $\dfrac{\pi }{6}$
B. $\dfrac{\pi }{4}$
C. ${\sin ^{ - 1}}\left( {\dfrac{2}{\pi }} \right)$
D. ${\cos ^{ - 1}}\left( {\dfrac{2}{\pi }} \right)$

Answer 1

Hint: As we know that the mean value theorem is sometimes known as the first mean value theorem or Lagrange's Mean Value Theorem. There must be a point along any curve between any two endpoints where the slope of the tangent to the curve equals the slope of the secant between the endpoints. Mathematically, If $f(x)$ is a function defined on $(a,b)$ whereas, $f(x)$ is continuous on $(a,b)$ and $f(x)$is differentiable on $(a,b)$ then Lagrange's mean value theorem holds$(a,b)$. To determine the value of $c$, we need to apply the mean value theorem.

Formula Used:
If $f(x)$ is a function then $f(x)$ is continuous on the closed interval $(a,b)$ and also differentiable on the open interval $(a,b)$. There is a point $c$ in $(a,b)$, that is,$a < c < b$ such that:
$f'(c) = \dfrac{{f(b) - f(a)}}{{b - a}}$.

Complete step by step Solution:
In the question, the function is given by:
$f\left( x \right) = \cos x$for $0\; \leqslant x\; \leqslant \left( {\dfrac{\pi }{2}} \right)\;$
Consider the two functions $f$ and $k$, If $k'(x) \ne 0$ for any $x \in (a,b)$ then at least one point $c \in (a,b)$ is included in such a way that, $\dfrac{{f'(c)}}{{k'(c)}} = \dfrac{{f(b) - f(a)}}{{k(b) - k(a)}}$,
Mean value theorem yields the following result if we take into account $k(x) = x$ for each $x \in (a,b)$:
Then, the mean value theorem is represented as:
$f'(c) = \dfrac{{f(b) - f(a)}}{{b - a}}$
Apply the mean value theorem on the given function, then we have:
$f'(c) = \dfrac{{0 - 1}}{{\left( {\dfrac{\pi }{2}} \right)}} \\$
$\Rightarrow f'(c) = \dfrac{{ - 2}}{\pi }\,\,\,\,....(i) \\$
Now, consider the given function $f\left( x \right) = \cos x$,
As we know that the derivative of $\cos x$ is $ - \sin x$, so:
$f'(x) = - \sin x \\$
$\Rightarrow f'(c) = - \sin c\,\,\,\,....(ii) \\$
From the equation $(i)$ and $(ii)$, we obtain:
$f'(c) = - \sin c = - \dfrac{2}{\pi } \\$
$\Rightarrow c = {\sin ^{ - 1}}\left( {\dfrac{2}{\pi }} \right) \\$

Hence, the correct option is (C).

Note:It should be noted that the initial conditions are crucial when utilizing any of the mean value theorems to solve problems. Every need of the theorem must be satisfied in order for the theorem to be applied correctly; otherwise, the results will be incorrect. Since the equation, in this case, is trigonometric, we have derived the aforesaid function using a derivative; however, if the equation is complex, we should apply the determinant formula.