{ If } y= ^{-1}≤ft({1}{1+x+x²})+ ^{-1}≤ft({1}{x²+3 x+3})+ ^{-1}≤ft({1}{x²+5 x+7}) then the value of y^{}(0) is

Correct Answer is : −910-9/10−109

Correct Answer is : −910-9/10−109

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MHTCET 26 Apr 2025 Shift 2 Paper

Question : 147 of 150

Marks: +1, -0

\text { If } y=\tan ^{-1}\left(\frac{1}{1+x+x^{2}}\right)+\tan ^{-1}\left(\frac{1}{x^{2}+3 x+3}\right)+\tan ^{-1}\left(\frac{1}{x^{2}+5 x+7}\right)

y^{\prime}(0)

Solution:

Concept:

Use

\tan^{-1}a - \tan^{-1}b = \tan^{-1}\left(\frac{a-b}{1+ab}\right)

and telescope the series.

Explanation:

Rewrite each term by expressing the numerator as a difference of consecutive terms.

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

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

\tan^{-1}\left(\frac{1}{x^{2}+5x+7}\right) = \tan^{-1}\left(\frac{(x+3)-(x+2)}{1+(x+2)(x+3)}\right) = \tan^{-1}(x+3) - \tan^{-1}(x+2)

Adding these terms, the intermediate terms cancel out:

y = \tan^{-1}(x+3) - \tan^{-1}x

Differentiate with respect to

x

y'(x) = \frac{1}{1+(x+3)^{2}} - \frac{1}{1+x^{2}}

Evaluate at

x=0

y'(0) = \frac{1}{1+9} - \frac{1}{1+0} = \frac{1}{10} - 1 = -\frac{9}{10}

Answer:

y'(0) = -\frac{9}{10}

, hence the correct option is C.

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