A) \[f(x)=\left\{ \begin{matrix} x,\, \\ 0,\, \\ \end{matrix} \right.\,\begin{matrix} \,\,\,\,\,0\le x<1 \\ x=1 \\ \end{matrix}\] on [0, 1]
B) \[f(x)=\left\{ \begin{matrix} \frac{\sin x}{x},\, \\ 1,\, \\ \end{matrix} \right.\,\,\,\,\begin{matrix} -\pi \le x<0 \\ x=0 \\ \end{matrix}\] on [-\[\pi \],0]
C) \[f(x)=\frac{{{x}^{2}}-x-6}{x-1}\] on [-2, 3]
D) \[f(x)=\left\{ \begin{matrix} \frac{{{x}^{3}}-2{{x}^{2}}-5x+6}{x-1},\,\,\,if\,\,x\ne 1, \\ -\,6,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,if\,x=1 \\ \end{matrix} \right.\]on [-2, 3]
Correct Answer: D
Solution :
[d] (1) Discontinuous at \[x=1\Rightarrow \]not applicable. (2) \[F(x)\]is not continuous (jump discontinuity) at x=0. (3) Discontinuity (missing point) at \[x=1\Rightarrow \]not applicable. (4) Notice that \[{{x}^{3}}-2{{x}^{2}}-5x+6=(x-1)({{x}^{2}}-x-6).\] Hence, \[f(x)={{x}^{2}}-x-6\] if \[x\ne 1\] and \[f(1)=-6.\] Thus, \[f\]is continuous at x=1. So, \[f(x)={{x}^{2}}-x-6\] is continuous in the interval \[\left[ -\,2,\text{ }3 \right]\]. Also, note that \[f(-2)=f(3)=0.\]Hence, Rolle?s Theorem implies \[f'(x)=2x-1.\] Setting \[f'(x)=0\], we obtain x=1/2 which lies between -2 and 3.
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