Functions Question 269
Question: If $ f(x),={{\cot }^{-1}}( \frac{3x-x^{3}}{1-3x^{2}} ) $ and x $ g(x)={{\cos }^{-1}}( \frac{1-x^{2}}{1+x^{2}} ) $ , then $ \underset{x\to a}{\mathop{\lim }},\frac{f(x)-f(a)}{g(x),-g(a)}, $ $ 0<,a<\frac{1}{2} $ is
[Orissa JEE 2002]
Options:
A) $ \frac{3}{2(1+a^{2})} $
B) $ \frac{3}{2(1+x^{2})} $
C) $ \frac{3}{2} $
D) $ -\frac{3}{2} $
Show Answer
Answer:
Correct Answer: D
Solution:
$ f(x)={{\cot }^{-1}}{ \frac{3x-x^{3}}{1-3x^{2}} } $ and $ g(x)={{\cos }^{-1}}{ \frac{1-x^{2}}{1+x^{2}} } $ Put $ x=\tan \theta $ in both equations $ f(\theta )={{\cot }^{-1}}{ \frac{3\tan \theta -{{\tan }^{3}}\theta }{1-3{{\tan }^{2}}\theta } } $ $ ={{\cot }^{-1}}{ \tan 3\theta } $ $ f(\theta )={{\cot }^{-1}}\cot ( \frac{\pi }{2}-3\theta )=\frac{\pi }{2}-3\theta \Rightarrow {f}’(\theta )=-3 $ .?.(i) and $ g(\theta )={{\cos }^{-1}}{ \frac{1-{{\tan }^{2}}\theta }{1+{{\tan }^{2}}\theta } } $ $ ={{\cos }^{-1}}(\cos 2\theta )=2\theta $
$ \Rightarrow {g}’(\theta )=2 $ ?..(ii) Now $ \underset{x\to a}{\mathop{\lim }},( \frac{f(x)-f(a)}{g(x)-g(a)} )=\underset{x\to a}{\mathop{\lim }},( \frac{f(x)-f(a)}{x-a} )\frac{1}{\underset{x\to a}{\mathop{\lim }},( \frac{g(x)-g(a)}{x-a} )} $ $ ={f}’(x).\frac{1}{{g}’(x)}=-3\times \frac{1}{2}=-\frac{3}{2} $ .
BETA
