Let $g(x)=\frac{(x-1)^{n}}{\log \cos ^{m}(x-1)} ; 0 0$, and let $p$ be the left hand derivative of $|x-1|$ at $x=1$. If $\lim _{x \rightarrow 1+} g(x)=p$, then

JEE Advanced · 2008 · 13 Apr 2008 · Paper 1 · Limits XII

Let

g (x) = \frac{( x - 1 ) ^{n}}{l o g c o s ^{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 \to 1 +} g (x) = p

, then

Answer: (C)

The left-hand derivative of

∣ x - 1∣

x = 1

p = - 1

(from the graph of

∣ x - 1∣

).

We need:

x \to 1^{+} lim \frac{( x - 1 ) ^{n}}{lo g cos ^{m} ( x - 1 )} = - 1

Let

h = x - 1

, so as

x \to 1^{+}

h \to 0^{+}

h \to 0^{+} lim \frac{h ^{n}}{lo g cos ^{m} h} = h \to 0^{+} lim \frac{h ^{n}}{m lo g cos h}

Using

lo g cos h \approx - \frac{h ^{2}}{2}

for small

h

h \to 0^{+} lim \frac{h ^{n}}{m \cdot ( - \frac{h ^{2}}{2} )} = h \to 0^{+} lim \frac{- 2 h ^{n}}{m h ^{2}} = h \to 0^{+} lim \frac{- 2 h ^{n - 2}}{m}

For this limit to equal

- 1

, we need

n - 2 = 0

(so the limit is finite and non-zero), giving

n = 2

.

Then:

\frac{- 2}{m} = - 1

, so

m = 2

.

Therefore,

n = 2, m = 2

JEE Advanced · 2008 · 13 Apr 2008 · Paper 1 · Single correct option