1]If median AD of a triangle ABC makes angle 30 degrees with side BC then \((\cot B-\cot C)^2\) is equal to?\[\]Options:6,9,12,15. \[\] 2]\[T_n=\sum_{r={2n}}^{3n-1}\dfrac{rn}{r^2+n^2},S_n=\sum_{r={2n+1}}^{3n}\dfrac{rn}{r^2+n^2}\].Then,\[T_n,S_n >\ or\ < \dfrac{\ln 2}{2}\].

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TopNewestFor number 1, this is what I thought of:

The way the question is worded implies that the result of \((\cot{B}-\cot{C})^2\) is always the same regardless of what the values of \(m\angle{B}\) and \(m\angle{C}\) are. You can use this to your advantage.

Let \(m\angle{ADC}=30^\circ\), and let \(m\angle{C}=90^\circ\). Now \(\cot{C}=0\), and you can use 30/60/90 triangle relationships to find \(\cot{B}\). I hope this helps! – Andy Hayes · 5 months, 3 weeks ago

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The second one is based on a Jee 2008 question . I'll post the solution next week if you want me to. :) – Keshav Tiwari · 5 months, 3 weeks ago

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– Adarsh Kumar · 5 months, 3 weeks ago

Yeah please do!Log in to reply

– Keshav Tiwari · 5 months, 3 weeks ago

Is the ans. Tn> ln2/2 and Sn<ln2/2 ?. You should see this video https://www.khanacademy.org/math/integral-calculus/indefinite-definite-integrals/riemann-sums/v/simple-riemann-approximation-using-rectanglesLog in to reply

Riemann Sums.... – Rishabh Cool · 5 months, 3 weeks ago

Yep... Its a JEE problem based onLog in to reply

– Adarsh Kumar · 5 months, 3 weeks ago

But how can we use that bcoz we don't know that n tends to infinity? \[\] And yes,best of luck!Log in to reply

– Harsh Shrivastava · 5 months, 3 weeks ago

Best of luck for tmmrw!Log in to reply

– Rishabh Cool · 5 months, 3 weeks ago

Thanks..... :-)Log in to reply