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← IIT-JEE 2008 · Paper 1
MathematicsMultiple correct+4 / -1Question 9 of 69

IIT-JEE 2008 · Paper 1, Mathematics question 9

Let
Sn=∑k=1nnn2+kn+k2andTn=∑k=0n−1nn2+kn+k2,S_n=\sum_{k=1}^{n}\frac{n}{n^2+kn+k^2}\quad\text{and}\quad T_n=\sum_{k=0}^{n-1}\frac{n}{n^2+kn+k^2},
for n=1,2,3,…n=1,2,3,\ldots. Then,

Choose one or more options
Answer and worked solution

Official answer: A, D

Riemann sums. Write nn2+kn+k2=1n⋅11+kn+(kn)2\dfrac n{n^2+kn+k^2}=\dfrac1n\cdot\dfrac1{1+\frac kn+\left(\frac kn\right)^2}. With f(x)=11+x+x2f(x)=\dfrac1{1+x+x^2}:
∫01f(x) dx=23[tan⁡−12x+13]01=23(π3−π6)=π33.\int_0^1f(x)\,dx=\frac2{\sqrt3}\left[\tan^{-1}\frac{2x+1}{\sqrt3}\right]_0^1=\frac2{\sqrt3}\left(\frac\pi3-\frac\pi6\right)=\frac\pi{3\sqrt3}.

ff is decreasing on [0,1][0,1], so the right-endpoint sum SnS_n (k=1,…,nk=1,\ldots,n) underestimates the integral and the left-endpoint sum TnT_n (k=0,…,n−1k=0,\ldots,n-1) overestimates it.

(A), (D)