Let g(x)=((x-1)^n)/( ^m(x-1)) ; 0 0 and let p be the left hand derivative of |x-1| at x=1 . If _ x arrow 1^ + g(x)=p , then

Correct answer: C. n=2, m=2 Given, g(x)=((x-1)^n)/( ^m(x-1)) ; 0 x 2, m ≠ 0, n are integers and |x-1|=\ array l x-1 ; x ≥ 1 1-x ; x 1 array . The left hand derivative of |x-1| at x=1 is p=-1 . Also, aligned & _ h arrow 0 ((1+h-1)^n)/( ^m(1+h-1))=-1 _ h arrow 0 (h^n)/( ^m h)=-1 & _ h arrow 0 (h^n)/(m h)=-1 aligned [Using L' Hospital rule] array ll & _ h arrow 0 n · h^ n-1 m (1)/( h)(- h) =-1 & _ h arrow 0 (-(n)/(m)) · h^ n-2 (( h)/(h)) =-1 & ((n)/(m)) _ h arrow 0 h^ n-2 (( h)/(h)) =1 & n=2 and (n)/(m)=1 & m=n=2 array

JEE Advanced 2008Paper 1

Exam: JEE Advanced
Year: 2008
Paper: Paper 1
Subject: Mathematics
Chapter: Limits

JEE Advanced 2008 · Paper 1 · Mathematics · Limits

Let g(x)=((x-1)^n)/( ^m(x-1)) ; 0 0 and let p be the left hand...

Let

g(x)=\frac{(x-1)^n}{\log \cos ^m(x-1)} ; 0 < x < 2, m

and

n

are integers,

m \neq 0, n>0

and let

p

be the left hand derivative of

|x-1|

x=1

. If

\lim _{x \rightarrow 1^{+}} g(x)=p

, then

Options

A $n=1, m=1$
B $n=1, m=-1$
C $n=2, m=2$
D $n>2, m=n$

Correct answer

C. $n=2, m=2$

Step-by-step solution

Given,

g(x)=\frac{(x-1)^n}{\log \cos ^m(x-1)} ; 0 \lt x \lt 2, m \neq 0, n

are integers and

|x-1|=\left\{

\begin{array}{l}x-1 ; x \geq 1 \\ 1-x ; x \lt 1\end{array}

\right.

The left hand derivative of

|x-1|

x=1

p=-1

. Also,

\begin{aligned}& \Rightarrow \quad \lim _{h \rightarrow 0} \frac{(1+h-1)^n}{\log \cos ^m(1+h-1)}=-1 \Rightarrow \lim _{h \rightarrow 0} \frac{h^n}{\log \cos ^m h}=-1 \\ & \Rightarrow \quad \lim _{h \rightarrow 0} \frac{h^n}{m \log \cos h}=-1 \end{aligned}

[Using L' Hospital rule]

\begin{array}{ll} \Rightarrow& \quad \lim _{h \rightarrow 0} \frac{n \cdot h^{n-1}}{m \frac{1}{\cos h}(-\sin h)}=-1 \\ \Rightarrow \quad & \quad \lim _{h \rightarrow 0}\left(-\frac{n}{m}\right) \cdot \frac{h^{n-2}}{\left(\frac{\tan h}{h}\right)}=-1 \\ \Rightarrow \quad & \quad\left(\frac{n}{m}\right) \lim _{h \rightarrow 0} \frac{h^{n-2}}{\left(\frac{\tan h}{h}\right)}=1 \\ \Rightarrow & n=2 \text{ and } \frac{n}{m}=1 \\ \therefore & \quad m=n=2 \end{array}

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