1. Tardigrade
  2. Question
  3. Mathematics
  4. Let g(x) = ((x -1)n/ log cosm (x -1)) ; 0 < x < 2 , m and n are integers, m ≠ 0, n > 0 , and let p be the left hand derivative of |x - 1| at x = 1. If displaystyle limx → 1+ g(x) = p , then

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Q. 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 $|x - 1|$ at $x = 1$. If $\displaystyle \lim_{x \to 1^{+}} \, g(x) = p $ , then

IIT JEEIIT JEE 2008Limits and Derivatives

Solution:

Correct answer is (c) $n = 2, m = 2$