Question
Let $$g\left( x \right) = \frac{{{{\left( {x - 1} \right)}^n}}}{{\log \,{{\cos }^m}\left( {x - 1} \right)}};\,0 < x < 2,$$ $$m$$ and $$n$$ are integers, $$m \ne 0,\,n > 0,$$ and let $$p$$ be the left hand derivative of $$\left| {x - 1} \right|$$ at $$x=1.$$ If $$\mathop {\lim }\limits_{x \to {1^ + }} g\left( x \right) = p,$$ then-
A.
$$n=1,\,\,m=1$$
B.
$$n=1,\,\,m=-1$$
C.
$$n=2,\,\,m=2$$
D.
$$n>2,\,\,m=n$$
Answer :
$$n=2,\,\,m=2$$
Solution :
As per question,
$$p=$$ left hand derivation of $$\left| {x - 1} \right|$$ at $$x = 1\,\, \Rightarrow p = - 1$$
Also $$\mathop {\lim }\limits_{x \to 1^+ } g\left( x \right) = p$$
Where $$g\left( x \right) = \frac{{{{\left( {x - 1} \right)}^n}}}{{\log \,{{\cos }^m}\left( {x - 1} \right)}},\,0 < x < 2,\,m,\,n$$ are integers, $$m \ne 0,\,n > 0$$
$$\therefore $$ We get,
$$\eqalign{
& \mathop {\lim }\limits_{x \to {1^ + }} \frac{{{{\left( {x - 1} \right)}^n}}}{{\log \,{{\cos }^m}\left( {x - 1} \right)}} = - 1 \cr
& \Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{{{h^n}}}{{\log \,{{\cos }^m}h}} = - 1 \cr
& \Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{{{h^n}}}{{m\left( {\log \,\cos h} \right)}} = - 1\,\,\left[ {{\text{Using L'Hospital's rule}}} \right] \cr
& \Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{{n\,{h^{n - 1}}\cos \,h}}{{m\left( { - \sin \,h} \right)}} = - 1\,\,\left[ {{\text{Using L'Hospital's rule}}} \right] \cr
& \Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{{n\,{h^{n - 2}}\cos \,h}}{{m\left( {\frac{{\sin \,h}}{h}} \right)}} = 1 \cr
& \Rightarrow n = 2\,\,{\text{and}}\,\,\,m = 2 \cr} $$