Differentiation Question 156
Question: If $ A _{i}=\frac{x-a _{i}}{| x-a _{i} |},i=1,2,3,….,n $ and $ a_1<a_2<a_3….a _{n}, $ then $ \underset{x\to a _{m}}{\mathop{\lim }}(A_1A_2…A _{n}),1\le m\le n $
Options:
A) Is equal to $ {{(-1)}^{m}} $
B) Is equal to $ {{(-1)}^{m+1}} $
C) Is equal to $ {{(-1)}^{m-1}} $
D) Does not exist
Show Answer
Answer:
Correct Answer: D
Solution:
[d] $ A _{i}=\frac{x-a _{i}}{| x-a _{i} |},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_1A_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_1A_2…A _{n}={{(-1)}^{n-m}} $
……. (ii)
$ \therefore \underset{x\to {a^{-}} _{m}}{\mathop{\lim }}(A_1A_2…A _{n})={{(-1)}^{n-m+1}} $ and $ \underset{x\to {a^{+}} _{m}}{\mathop{\lim }}(A_1A_2…A _{n})={{(-1)}^{n-m}}\therefore LHL\ne RHL $ Hence, $ \underset{x\to a _{m}}{\mathop{\lim }}(A_1A_2…A _{n}) $ does not exist.
BETA
