JEE MATHEMATICS MCQ: INVERSE TRIGONOMETRY

1 Medium

1 of 1

Analyzing +4 | -1

y=\tan ^{-1} \frac{1}{1+x+x^2}+\tan ^{-1} \frac{1}{x^2+3 x+3}+\tan ^{-1} \frac{1}{x^2+5 x+7}+\tan ^{-1} \frac{1}{x^2+7 x+13}+… n \,\,\text{terms,}

then

y^{\prime} (0)

is equal to

View answer and explanation

Correct option: B

-n^2 /\left(n^2+1\right)

Option B (

\frac{{-n^2}}{{(n^2+1)}}

) is correct because the given expression telescopes into a simpler form, allowing for easy differentiation and evaluation at x=0.

More Questions from INVERSE TRIGONOMETRY

Practice All Chapter MCQs →

\cos ^{-1}(\cos 12)-\sin ^{-1}(\sin 12)

\sin \left(2 \tan ^{-1}(1 / 3)\right)+

\cos \left(\tan ^{-1} 2 \sqrt{2}\right)

\cos ^ {-1}\left\{\frac{1}{\sqrt{2}}\left(\cos \frac{9 \pi}{10}-\sin \frac{9 \pi}{10}\right)\right\}=

\sin \left(\sin ^{-1} \frac{1}{5}+\cos ^{-1} x\right)=1

x

{x^2} + {y^2} + {z^2} = {r^2},

{\tan ^{ - 1}}\left( {\frac{{xy}}{{zr}}} \right) + {\tan ^{ - 1}}\left( {\frac{{yz}}{{xr}}} \right) + {\tan ^{ - 1}}\left( {\frac{{xz}}{{yr}}} \right)\;

2{\tan ^{ - 1}}\left( {\frac{1}{3}} \right) + {\tan ^{ - 1}}\left( {\frac{1}{7}} \right)