Functions Question 562
Question: The function $ f(x)=\frac{2x^{2}+7}{x^{3}+3x^{2}-x-3} $ is discontinuous for
[J&K 2005]
Options:
A) $ x=1 $ only
B) $ x=1 $ and $ x=-1 $ only
C) $ x=1,x=-1,x=-3 $ only
D) $ x=1,x=-1,x=-3 $ and other values of x
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Answer:
Correct Answer: C
Solution:
$ f(x)=\frac{2x^{2}+7}{x^{2}(x+3)-1(x+3)}\frac{9x^{2}+7}{(x^{2}-1)(x+3)} $ $ =\frac{2x^{2}+7}{(x-1)(x+1)(x+3)} $ Hence points of discontinuity are $ x=1 $ , $ x=-1 $ and $ x=-3 $ only.
BETA
