SATHEE: Limit Continuity and Differentiability 7 Question 14

Limit Continuity and Differentiability 7 Question 14

14. 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 $lim_{x \to 1^{+}} g (x) = p$ , then

(a) $n = 1, m = 1$

(b) $n = 1, m = - 1$

(d) $n > 2, m = n$

(2008, 3M)

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Answer:

Correct Answer: 14. (a, b, c, d)

Solution:

Since, $\frac{d x}{d y} = \frac{1}{d y / d x} = {\frac{d y}{d x}}^{- 1}$

$\begin{aligned} \Rightarrow & \frac{d}{d y} \frac{d x}{d y} & = \frac{d}{d x} \frac{d y}{d x}^{- 1} \frac{d x}{d y} \\ \Rightarrow & \frac{d^{2} x}{d y^{2}} & = - {\frac{d^{2} y}{d x^{2}} \frac{d y}{d x}}^{- 2} \frac{d x}{d y} = - {\frac{d^{2} y}{d x^{2}} \frac{d y}{d x}}^{- 3} \end{aligned}$

Limit Continuity and Differentiability 7 Question 14

14. 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 $lim_{x \to 1^{+}} g (x) = p$ , then

Answer:

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Limit Continuity and Differentiability 7 Question 14

14. Let g(x)=(x−1)nlog⁡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 limx→1+g(x)=p, then

Answer:

Engineering Preparation

Medical Preparation

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Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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14. 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 $lim_{x \to 1^{+}} g (x) = p$ , then