Differentiation Question 474
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
