A) \[n=1,\]\[m=1\]
B) \[n=1,\]\[m=-\,1\]
C) \[n=2,\]\[m=2\]
D) \[n>2,\]\[m=n\]
Correct Answer: C
Solution :
Given, \[g\left( x \right)=\frac{{{\left( x-1 \right)}^{n}}}{\log {{\cos }^{m}}\left( x-1 \right)}\]P is left bond derivative of \[|x-1|\] |
\[\therefore \,\,\,\,\,\,P=-1\] |
\[\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,g\left( x \right)=P\]\[\Rightarrow \,\,\,\underset{h\to 0}{\mathop{\lim }}\,g\left( 1+h \right)=-1\]\[\Rightarrow \,\,\underset{h\to 0}{\mathop{\lim }}\,\frac{{{h}^{n}}}{\log {{\cos }^{m}}h}=-1\]\[\Rightarrow \,\,\,\underset{h\to 0}{\mathop{\lim }}\,\frac{n{{h}^{n-1}}}{m\left( -\tan \,h \right)}=-1\]\[\Rightarrow \,\,\frac{-n}{n}\underset{h\to 0}{\mathop{\lim }}\,\frac{{{h}^{n-1}}}{\tan \,h}=-1\] |
Which holds of \[n-1=1\,\Rightarrow n=2\]and \[-\frac{n}{m}=-1\,\Rightarrow n=m=2\] |
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