A) Is equal to \[{{(-1)}^{m}}\]
B) Is equal to \[{{(-1)}^{m+1}}\]
C) Is equal to \[{{(-1)}^{m-1}}\]
D) Does not exist
Correct Answer: D
Solution :
| [d] \[{{A}_{i}}=\frac{x-{{a}_{i}}}{\left| x-{{a}_{i}} \right|},i=1,2,3,...,n\] |
| \[{{a}_{i}}<{{a}_{2}}<{{a}_{3}}<...<{{a}_{n}}.\] |
| If x is in the left neighbourhood of |
| \[{{a}_{1}}<{{a}_{2}}<....{{a}_{m-1}}<x<{{a}_{m}}<{{a}_{m+1}}<...<{{a}_{n}}.\] |
| \[{{A}_{i}}=\frac{x-{{a}_{i}}}{x-{{a}_{i}}}=1,\,\,i=1,\,\,2,...,\,\,m-1;\] |
| \[{{A}_{i}}=\frac{x-{{a}_{i}}}{({{a}_{i}}-x)}=-1\] |
| \[i=m,m-1,...n\] |
| \[\therefore {{A}_{1}}{{A}_{2}}....{{A}_{n}}={{(-1)}^{n-m+1}}\] ? (i) |
| If x is in the right neighbourhood of\[{{a}_{m}}\], |
| \[{{a}_{1}}<{{a}_{2}}<...<{{a}_{m-1}}<{{a}_{m}}<x<{{a}_{m+1}}<...<{{a}_{n}},\] |
| \[{{A}_{i}}=\frac{x-{{a}_{i}}}{x-{{a}_{i}}}=1,i=1,2,...,n.\] |
| \[\therefore {{A}_{1}}{{A}_{2}}...{{A}_{n}}={{(-1)}^{n-m}}\] ?. (ii) |
| \[\therefore \underset{x\to {{a}^{-}}_{m}}{\mathop{\lim }}\,({{A}_{1}}{{A}_{2}}...{{A}_{n}})={{(-1)}^{n-m+1}}\] |
| and \[\underset{x\to {{a}^{+}}_{m}}{\mathop{\lim }}\,({{A}_{1}}{{A}_{2}}...{{A}_{n}})={{(-1)}^{n-m}}\therefore LHL\ne RHL\] |
| Hence, \[\underset{x\to {{a}_{m}}}{\mathop{\lim }}\,({{A}_{1}}{{A}_{2}}...{{A}_{n}})\] does not exist. |
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