If y=[tan-1 1/(1+x+x2) + tan-1 1/(x2+3x+3) + tan-1 1/(x2+5x+7) +....] upto n terms,then y'(0) is equal to:
a)-1/(n2+1)
b)-n2/(n2+1)
c)n2/(n2+1)
d)none of these
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2 Answers
106For such type of questions try to find tr
t_{r} = tan^{-1}\frac{1}{1+(x+r-1)(x+r)}
t_{r} = tan^{-1}\frac{(x+r) - (x+r-1)}{1+(x+r-1)(x+r)}
t_{r} = tan^{-1}\(x+r) - tan^{-1}(x+r-1)
summing this from r=1 to r=n
y = tan^{-1}\(x+n) - tan^{-1}(x)
y = tan^{-1}\frac{n}{1+(x+n)x}
Now can you proceed further?
1yeah,so now on differentiating we get y'(x)=(1/1+(x+n)2) - (1/1+x2).thus y'(0)=-n2/(1+n2) ,that is option b.