Let $g\left( x \right) = \frac{{{{\left( {x - 1} \right)}^n}}}{{\log {{\cos }^m}\left( {x - 1} \right)}},0 0$, 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

JEE MATHEMATICS MCQ: DIFFERENTIATION

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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|

x = 1

. If

\lim\limits_{x \to {1^ + }} g\left( x \right) = p

, then

View answer and explanation

Correct option: C

n = 2,m = 2

Option C (

n = 2,m = 2

) is correct because it satisfies the limit condition

\lim\limits_{{x \to {1^ + }}} g(x) = p = -1