Test Index

Manipal Entrance Test 2019 Math Paper

© examsnet.com

Question : 14 of 70

Marks: +1, -0

If

f(x)=\cot^{-1}\left(\frac{3x-x^{3}}{1-3x^{2}}\right)

g(x)=\cos^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right),

\lim\limits_{x\to a}\frac{f(x)-f(a)}{g(x)-g(a)}, 0 < a < \frac{1}{2}

, is

$\frac{3}{2(1+a^{2})}$
$\frac{3}{2(1+x^{2})}$
$\frac{3}{2}$
$-\frac{3}{2}$

Solution:

Given that,

f(x)=\cot^{-1}\left(\frac{3x-x^{3}}{1-3x^{2}}\right)

and

g(x)=\cos^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)

Put

x=\tan\theta

in

f(x),

we get

f(x)=\cot^{-1}\left(\frac{3\tan\theta-\tan^{3}\theta}{1-3\tan^{2}\theta}\right)

\Rightarrow f(x)=\frac{\pi}{2}-3\tan^{-1}x

Again put

x=\tan\theta

in

g(x),

we get

g(x)=\cos^{-1}\left(\frac{1-\tan^{2}\theta}{1+\tan^{2}\theta}\right)

Again put

x=\tan\theta\text{ in }g(x),

we get

g(x)=\cos^{-1}\left(\frac{1-\tan^{2}\theta}{1+\tan^{2}\theta}\right)

g(x)=2\tan^{-1}x

\therefore \lim\limits_{x\to a}\frac{f(x)-f(a)}{g(x)-g(a)}

=\lim\limits_{x\to a}\frac{\left(\frac{\pi}{2}-3\tan^{-1}x\right)-\frac{\pi}{2}+3\tan^{-1}a}{2\tan^{-1}x-2\tan^{-1}a}

=\lim\limits_{x\to a}\frac{-3(\tan^{-1}a-\tan^{-1}x)}{2(\tan^{-1}a-\tan^{-1}x)}=-\frac{3}{2}

© examsnet.com

Go to Question:

Prev Question

Next Question