In which of the following functions Rolle’s theorem is applicable… - Vidyadip

MCQ

In which of the following functions Rolle’s theorem is applicable ?

A
$ f(x) =\left\{ \begin{array}{l}x\,\,\,,\,\,0\, \le \,x\, < \,\,1\\0\,\,\,\,,\,\,\,\,\,\,\,\,\,x\,\, = 1\end{array} \right.$ on $[0, 1]$
B
$f(x) = \left\{ \begin{array}{l}\frac{{\sin x}}{x}\,\,,\, - \pi \, \le x\, < 0\\\,0\,\,\,\,\,\,\,\,\,,\,\,\,\,\,\,\,\,\,\,\,\,x\, = \,0\end{array} \right.$ on $[-\pi , 0]$
C
$f(x)= \frac{{{x^2} - x - 6}}{{x - 1}}$ on $[-2,3]$
✓
$f(x) = \left\{ \begin{array}{l}\frac{{{x^3} - 2{x^2} - 5x + 6}}{{x - 1}}\,\,\,if\,\,x\, \ne 1,\,\,on\,[ - 2,3]\\ - 6\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,if\,\,\,\,x\, = 1\end{array} \right.$

✓

Answer

Correct option: D.

$f(x) = \left\{ \begin{array}{l}\frac{{{x^3} - 2{x^2} - 5x + 6}}{{x - 1}}\,\,\,if\,\,x\, \ne 1,\,\,on\,[ - 2,3]\\ - 6\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,if\,\,\,\,x\, = 1\end{array} \right.$

d
$(A)$ discontinuous at $x =1$ not applicable

$(B)$ $f (x)$ is not continuous at $x =0$ hence $(B)$ is incorrect.

$(C)$ discontinuity at $x = 1$ not applicable

$(D)$ Notice that $x^3 - 2x^2 - 5x + 6 = (x-1) (x^2 -x -6)$.

Hence , $f(x) = x^2 - x - 6$ if and $f (1) = - 6$

$==> \,f$ is continuous at $x = 1$.

So $f(x) = x^2 -x - 6$ throughout the interval $[-2,3]$

.Also, note that $f(-2) = f(3) = 0$.

Hence, Rolle’s theorem applies. $f'(x) = 2x -1.$

Setting $f '(x)= 0$ ,

we obtain $x = 1/2$ which lies between $-2$ and $3.$ ]

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