Interpret the given function

The expression is $g(x)=\frac{(x-1)^n}{\log\left(\cos^m(x-1)\right)}.$

We are also given that $p=\text{left hand derivative of } |x-1| \text{ at } x=1.$

Find }p$

Let $f(x)=|x-1|.$

For $x<1$, $|x-1|=1-x,$ so $f'(x)=-1 \quad (x<1).$

Hence the left hand derivative at $x=1$ is $p=-1.$

Evaluate the limit of }g(x)$as$x\to 1^+$

Put $t=x-1.$ Then as $x\to 1^+$, we have $t\to 0^+$.

So $g(x)=\frac{t^n}{\log(\cos^m t)}.$

Using logarithm properties, $\log(\cos^m t)=m\log(\cos t).$

Therefore $g(x)=\frac{t^n}{m\log(\cos t)}.$

Use the standard expansion of }\log(\cos t)$near$t=0$

We know $\cos t = 1-\frac{t^2}{2}+o(t^2).$

Hence $\log(\cos t) \sim -\frac{t^2}{2} \quad \text{as } t\to 0.$

So $m\log(\cos t) \sim -\frac{m t^2}{2}.$

Thus $g(x)=\frac{t^n}{m\log(\cos t)} \sim \frac{t^n}{-\frac{m t^2}{2}} = -\frac{2}{m}t^{n-2}.$

Match the limit with }p=-1$

We need $\lim_{x\to 1^+} g(x)= -1.$

That is, $\lim_{t\to 0^+} -\frac{2}{m}t^{n-2}=-1.$

Now analyze cases:

• If $n>2$, then $t^{n-2}\to 0$, so limit $=0$, not $-1$.
• If $n<2$, then $t^{n-2}\to \infty$ or undefined behavior, not finite $-1$.
• Therefore, we must have $n=2.$

Then $g(x) \to -\frac{2}{m}.$

For this to equal $-1$, $-\frac{2}{m}=-1 \implies m=2.$

Check options

• A: $n=1,m=1$ gives divergent behavior, not correct.
• B: $n=1,m=-1$ gives divergent behavior, not correct.
• C: $n=2,m=2$ gives limit $-1$, correct.
• D: $n>2,m=n$ gives limit $0$, not correct.

Conclusion

The correct option is $\boxed{\text{C}}.$