JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Easy

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Analyzing +4 | -1

y = \left[ {{{\tan }^{ - 1}}\frac{1}{{1 + x + {x^2}}} + {{\tan }^{ - 1}}\frac{1}{{{x^2} + 3x + 3}} + {{\tan }^{ - 1}}\frac{1}{{{x^2} + 5x + 7}} + ... + \,} \right.

upto

n

terms], they

y'\left( 0 \right)

is equal to

View answer and explanation

Correct option: C

\frac{{{n^2}}}{{\left( {{n^2} + 1} \right)}}

The correct answer is C (

\frac{{{n^2}}}{{\left( {{n^2} + 1} \right)}}

) because after simplifying the expression for y using the arctangent subtraction formula and differentiating, y'(0) evaluates to this value.

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