MCQOPTIONS
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| 1. |
In [0, 1] Lagrange's mean value theorem is NOT applicable to [IIT Screening 2003] |
| A. | \[f(x)=\left\{ \begin{align} & \frac{1}{2}-x,\,\,\,\,\,\,\,x<\frac{1}{2} \\ & {{\left( \frac{1}{2}-x \right)}^{2}},\,\,\,x\ge \frac{1}{2} \\ \end{align} \right.\] |
| B. | \[f(x)=\left\{ \begin{align} & \frac{\sin x}{x},\,\,\,x\ne 0 \\ & \,\,\,\,\,1\,\,\,,\,\,\,x=0 \\ \end{align} \right.\] |
| C. | \[f(x)=x|x|\] |
| D. | \[f(x)=|x|\] |
| Answer» B. \[f(x)=\left\{ \begin{align} & \frac{\sin x}{x},\,\,\,x\ne 0 \\ & \,\,\,\,\,1\,\,\,,\,\,\,x=0 \\ \end{align} \right.\] | |