Question

In which of the following functions Rolle’s theorem is applicable?
(A) f(x) = \left\\{ {\begin{array}{*{20}{c}} {x,0 \leqslant x < 1} \\\ {0,x = 1} \end{array}} \right. on $[0,1]$
(B) f(x) = \left\\{ {\begin{array}{*{20}{c}} {\dfrac{{\sin x}}{x}, - \pi \leqslant x < 0} \\\ {0,x = 0} \end{array}} \right. on $[ - \pi ,0]$
(C) $f(x) = \dfrac{{{x^2} - x - 6}}{{x - 1}}$ on $[ - 2,3]$
(D) f(x) = \left\\{ {\begin{array}{*{20}{c}} {\dfrac{{{x^3} - 2{x^2} - 5x + 6}}{{x - 1}},x \ne 1} \\\ { - 6,x = 1} \end{array}} \right. on $[ - 2,3]$

Answer 1

Hint : According to Rolle’s theorem
If $f(x)$ is a function such that
(a) $f(x)$ is continuous in $[a,b]$
(b) $f(x)$ is differentiable in $(a,b)$
(c) $f(a) = f(b)$
Then there exist at least one $c \in (a,b)$ such that $f'(c) = 0$
Use the theorem above to check if the options given in the question satisfy the Rolle’s theorem or not.

Complete step-by-step answer :
Let us first check the existence of the function.
Clearly, $x - 1 = 0$ when $x = 1$ and $1 \in [ - 2,3]$ . Therefore, (C) does not exist at $x = 1$ as the denominator cannot be equal to zero.
Thus, option (C) is incorrect.
Now we know that every polynomial function is continuous. And every rational function is continuous if the functions in the numerator and denominator are continuous. We also know that every trigonometric function is continuous in their domain.
Thus, all the functions in option (A), (B) and (D) are continuous in their respective intervals.
But we need to check the continuity using the mathematical definition of continuity at the boundary points.
Mathematical definition of continuity is:
A function $f(x)$ is continuous at a point $x = a$ if and only if,
$\mathop {\lim }\limits_{x \to a} f(x) = f(a)$
For option (A)
$\mathop {\lim }\limits_{x \to 1} f(x) = \mathop {\lim }\limits_{x \to 1} x = 1$
But $f(1) = 0$
Thus, $\mathop {\lim }\limits_{x \to a} f(x) \ne f(a)$
Thus, the first condition of Rolle’s theorem is not satisfied at point $x = 1$ .
Therefore, option (A) is incorrect.
For option (B)
$\mathop {\lim }\limits_{x \to 0} f(x) = \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1$
But $f(0) = 0$
$\Rightarrow \mathop {\lim }\limits_{x \to a} f(x) \ne f(a)$
Thus, the first condition of Rolle’s theorem is not satisfied at point $x = 0$ .
Therefore, option (B) is incorrect.
For option (D)
$\mathop {\lim }\limits_{x \to 1} f(x) = \mathop {\lim }\limits_{x \to 1} \dfrac{{{x^3} - 2{x^2} - 5x + 6}}{{x - 1}}$ $\left( {\dfrac{0}{0}form} \right)$
We will solve this limit by L’Hospitals Rule.
According to L’Hospitals Rule.
If $L = \mathop {\lim }\limits_{x \to a} \dfrac{{f(x)}}{{g(x)}}$ is of the form $\dfrac{0}{0}$
Then, $L = \mathop {\lim }\limits_{x \to a} \dfrac{{f'(x)}}{{g'(x)}}$
Since, the above limit satisfied the condition of L’Hospitals Rule. We can write,
$\mathop {\lim }\limits_{x \to 1} f(x) = \mathop {\lim }\limits_{x \to 1} \dfrac{{\dfrac{d}{{dx}}\left( {{x^3} - 2{x^2} - 5x + 6} \right)}}{{\dfrac{d}{{dx}}\left( {x - 1} \right)}}$
$= \mathop {\lim }\limits_{x \to 1} \dfrac{{3{x^2} - 4x - 5}}{1}$
By substituting the value of limit, we get
$= 3 - 4 - 5 = - 6$
And $f(1) = - 6$
Thus the first condition of Rolle’s theorem is satisfied.
Now, we know that every polynomial function is differentiable. And every rational function is differentiable if the functions in the numerator and denominator are differentiable. Therefore, $\dfrac{{{x^3} - 2{x^2} - 5x + 6}}{{x - 1}}$ for all the point on $[ - 2,3] - \\{ 1\\}$.
We also know that every constant function is differentiable, thus $- 6$ is differentiable at $x = 1$ .
Therefore, we can conclude that, $f(x)$ is differentiable on $( - 2,3)$
Thus, the second condition of Rolle’s theorem is satisfied.
Now, $f( - 2) = \dfrac{{{{( - 2)}^3} - 2{{( - 2)}^2} - 5( - 2) + 6}}{{ - 2 - 1}}$
$= \dfrac{{ - 8 - 8 + 10 + 6}}{{ - 3}}$
$\Rightarrow f( - 2) = 0$
$f(3) = \dfrac{{{{(3)}^3} - 2{{(3)}^2} - 5(3) + 6}}{{3 - 1}}$
$= \dfrac{{27 - 18 - 15 + 6}}{2}$
$\Rightarrow f(3) = 0$
$\Rightarrow f( - 2) = f(3)$
Therefore, the third condition of Rolle’s theorem is satisfied.
Hence, Rolle’s theorem is applicable for the function in option (D).
Therefore, from the above explanation, the correct answer is, option (D) f(x) = \left\\{ {\begin{array}{*{20}{c}} {\dfrac{{{x^3} - 2{x^2} - 5x + 6}}{{x - 1}},x \ne 1} \\\ { - 6,x = 1} \end{array}} \right. on $[ - 2,3]$

Note : In this question, it was important to know that the function must exist first. Then we check it for Rolle's theorem. The function in option (C) would have seemed continuous and differentiable both if you would have skipped its existence at $x = 1$ .
We did not need to solve for option (D). We had already proved that options (A), (B) and (C) are incorrect. And since, one of the answers needs to be correct, we could directly mark option (D) as the correct option.