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Question Number 44988 Answers: 1 Comments: 0
$$\boldsymbol{\mathrm{A}}\:\mathrm{tangent}\:\mathrm{to}\:\mathrm{ellipse}\:\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{a}^{\mathrm{2}\:} }+\frac{\mathrm{y}^{\mathrm{2}} }{\mathrm{b}^{\mathrm{2}} }=\mathrm{1}\:\boldsymbol{\mathrm{at}}\:\boldsymbol{\mathrm{point}}\:\boldsymbol{\mathrm{p}}\:\boldsymbol{\mathrm{meets}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{minor}}\:\boldsymbol{\mathrm{axis}} \\ $$$$\boldsymbol{\mathrm{at}}\:\boldsymbol{\mathrm{L}}\:\boldsymbol{\mathrm{if}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{normal}}\:\boldsymbol{\mathrm{at}}\:\boldsymbol{\mathrm{p}}\:\boldsymbol{\mathrm{meets}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{major}}\:\boldsymbol{\mathrm{axis}}\:\boldsymbol{\mathrm{at}}\:\boldsymbol{\mathrm{m}}.\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{locus}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{midpoint}}\:\boldsymbol{\mathrm{LM}} \\ $$
Question Number 44967 Answers: 1 Comments: 0
Question Number 44965 Answers: 0 Comments: 0
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{nth}\:\mathrm{term}\:\mathrm{of}\:\mathrm{the}\:\mathrm{series}:\:\:\:\:\frac{\mathrm{1}}{\mathrm{1}.\mathrm{2}.\mathrm{3}}\:+\:\frac{\mathrm{1}}{\mathrm{4}.\mathrm{5}.\mathrm{6}}\:+\:\frac{\mathrm{1}}{\mathrm{7}.\mathrm{8}.\mathrm{9}}\:\:+\:...\: \\ $$
Question Number 44986 Answers: 1 Comments: 1
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{nth}\:\mathrm{term}\:\mathrm{of}\:\mathrm{the}\:\mathrm{series}:\:\:\:\:\frac{\mathrm{1}}{\mathrm{1}.\mathrm{2}.\mathrm{3}}\:+\:\frac{\mathrm{1}}{\mathrm{4}.\mathrm{5}.\mathrm{6}}\:+\:\frac{\mathrm{1}}{\mathrm{7}.\mathrm{8}.\mathrm{9}}\:+\:... \\ $$
Question Number 44952 Answers: 0 Comments: 1
Question Number 44951 Answers: 1 Comments: 3
Question Number 44950 Answers: 1 Comments: 4
Question Number 44936 Answers: 0 Comments: 0
Question Number 44929 Answers: 2 Comments: 1
Question Number 44928 Answers: 0 Comments: 3
Question Number 44918 Answers: 2 Comments: 5
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{to}\:\mathrm{n}\:\mathrm{terms}\:\mathrm{of}:\:\:\:\:\frac{\mathrm{1}}{\mathrm{1}.\mathrm{2}.\mathrm{3}}\:+\:\frac{\mathrm{3}}{\mathrm{2}.\mathrm{3}.\mathrm{4}}\:+\:\frac{\mathrm{5}}{\mathrm{3}.\mathrm{4}.\mathrm{5}}\:+\:...\: \\ $$
Question Number 44920 Answers: 2 Comments: 0
$$\boldsymbol{\mathrm{solve}}.\mathrm{3}^{\boldsymbol{\mathrm{sin}}\mathrm{2}\boldsymbol{\mathrm{x}}+\mathrm{2}\boldsymbol{\mathrm{cos}}^{\mathrm{2}} \boldsymbol{\mathrm{x}}} +\mathrm{3}^{\mathrm{1}−\boldsymbol{\mathrm{sin}}\mathrm{2}\boldsymbol{\mathrm{x}}+\mathrm{2}\boldsymbol{\mathrm{sin}}^{\mathrm{2}} \boldsymbol{\mathrm{x}}} =\mathrm{28} \\ $$
Question Number 44921 Answers: 1 Comments: 1
$$\int\frac{\mathrm{1}}{\mathrm{1}+\mathrm{ln}\:\mathrm{x}}=? \\ $$
Question Number 44907 Answers: 0 Comments: 1
Question Number 44906 Answers: 0 Comments: 0
Question Number 44901 Answers: 0 Comments: 1
Question Number 44900 Answers: 1 Comments: 0
Question Number 44898 Answers: 1 Comments: 1
$${If}\:\:\:\:{x}^{\mathrm{4}} +{px}^{\mathrm{3}} +{qx}^{\mathrm{2}} +{rx}+\mathrm{5}\:=\:\mathrm{0} \\ $$$${has}\:{four}\:{real}\:{roots},\:{then}\:{find} \\ $$$$\:{the}\:{minimum}\:{value}\:{of}\:\boldsymbol{{pr}}. \\ $$
Question Number 44891 Answers: 1 Comments: 0
Question Number 44892 Answers: 1 Comments: 2
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{formular}\:\mathrm{for}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{first}\:\mathrm{kth}\:\mathrm{power}\:\mathrm{of}\:\mathrm{natural}\:\mathrm{number} \\ $$
Question Number 44876 Answers: 1 Comments: 0
Question Number 44874 Answers: 0 Comments: 2
Question Number 44872 Answers: 1 Comments: 1
Question Number 44858 Answers: 0 Comments: 0
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{formular}\:\mathrm{for}\:\mathrm{the}\:\mathrm{first}\:\mathrm{kth}\:\mathrm{power}\:\mathrm{of}\:\mathrm{natural}\:\mathrm{numbers} \\ $$
Question Number 44826 Answers: 2 Comments: 0
$$\mathrm{Let}\:{A}\:\mathrm{and}\:{B}\:\mathrm{be}\:\mathrm{sets}. \\ $$$$\mathrm{Prove}\:\mathrm{that}\:{A}\:=\:{B}\:\mathrm{if}\:\mathrm{and}\:\mathrm{only}\:\mathrm{if}\:{A}\:\cup\:{B}\:=\:{A}\:\cap\:{B} \\ $$
Question Number 44819 Answers: 0 Comments: 7
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