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Question Number 44918    Answers: 2   Comments: 5

Find the sum to n terms of: (1/(1.2.3)) + (3/(2.3.4)) + (5/(3.4.5)) + ...

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{to}\:\mathrm{n}\:\mathrm{terms}\:\mathrm{of}:\:\:\:\:\frac{\mathrm{1}}{\mathrm{1}.\mathrm{2}.\mathrm{3}}\:+\:\frac{\mathrm{3}}{\mathrm{2}.\mathrm{3}.\mathrm{4}}\:+\:\frac{\mathrm{5}}{\mathrm{3}.\mathrm{4}.\mathrm{5}}\:+\:...\: \\ $$

Question Number 44920    Answers: 2   Comments: 0

solve.3^(sin2x+2cos^2 x) +3^(1−sin2x+2sin^2 x) =28

$$\boldsymbol{\mathrm{solve}}.\mathrm{3}^{\boldsymbol{\mathrm{sin}}\mathrm{2}\boldsymbol{\mathrm{x}}+\mathrm{2}\boldsymbol{\mathrm{cos}}^{\mathrm{2}} \boldsymbol{\mathrm{x}}} +\mathrm{3}^{\mathrm{1}−\boldsymbol{\mathrm{sin}}\mathrm{2}\boldsymbol{\mathrm{x}}+\mathrm{2}\boldsymbol{\mathrm{sin}}^{\mathrm{2}} \boldsymbol{\mathrm{x}}} =\mathrm{28} \\ $$

Question Number 44921    Answers: 1   Comments: 1

∫(1/(1+ln x))=?

$$\int\frac{\mathrm{1}}{\mathrm{1}+\mathrm{ln}\:\mathrm{x}}=? \\ $$

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^4 +px^3 +qx^2 +rx+5 = 0 has four real roots, then find the minimum value of pr.

$${If}\:\:\:\:{x}^{\mathrm{4}} +{px}^{\mathrm{3}} +{qx}^{\mathrm{2}} +{rx}+\mathrm{5}\:=\:\mathrm{0} \\ $$$${has}\:{four}\:{real}\:{roots},\:{then}\:{find} \\ $$$$\:{the}\:{minimum}\:{value}\:{of}\:\boldsymbol{{pr}}. \\ $$

Question Number 44891    Answers: 1   Comments: 0

Question Number 44892    Answers: 1   Comments: 2

Find the formular for the sum of the first kth power of natural number

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{formular}\:\mathrm{for}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{first}\:\mathrm{kth}\:\mathrm{power}\:\mathrm{of}\:\mathrm{natural}\:\mathrm{number} \\ $$

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

Find the formular for the first kth power of natural numbers

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{formular}\:\mathrm{for}\:\mathrm{the}\:\mathrm{first}\:\mathrm{kth}\:\mathrm{power}\:\mathrm{of}\:\mathrm{natural}\:\mathrm{numbers} \\ $$

Question Number 44826    Answers: 2   Comments: 0

Let A and B be sets. Prove that A = B if and only if A ∪ B = A ∩ B

$$\mathrm{Let}\:{A}\:\mathrm{and}\:{B}\:\mathrm{be}\:\mathrm{sets}. \\ $$$$\mathrm{Prove}\:\mathrm{that}\:{A}\:=\:{B}\:\mathrm{if}\:\mathrm{and}\:\mathrm{only}\:\mathrm{if}\:{A}\:\cup\:{B}\:=\:{A}\:\cap\:{B} \\ $$

Question Number 44819    Answers: 0   Comments: 7

Question Number 44801    Answers: 2   Comments: 1

Question Number 44848    Answers: 0   Comments: 2

(1) If a^→ =a_(1i) +a_(2j) +a_(3k) and b^→ =b_(1i) +b_(2j) +b_(3k) show that i. a^→ ×b^→ = determinant ((i,j,k),(a_1 ,a_2 ,a_3 ),(b_1 ,b_2 ,b_3 )) ii. a^→ •b^→ =a_1 b_1 +a_2 b_2 +a_3 b_3 (2) If a^→ =a_(1i) +a_(2j) +a_(3k) , b^→ =b_(1i) +b_(2j) +b_(3k) and c^→ =c_(1i) +c_(2j) +c_(3k) , show that i. a^→ •(b^→ ×c^→ )= determinant ((a_1 ,a_2 ,a_3 ),(b_1 ,b_2 ,b_3 ),(c_1 ,c_2 ,c_3 )) ii. a•(b^→ ×c^→ )=b^→ (a^→ •c^→ )−c^→ (a^→ •b^→ ) iii. (a^→ ×b^→ )×c^→ =b^→ (a^→ •c^→ )−a^→ (b^→ •c^→ ) iv. a^→ ×(b^→ ×c^→ )+b^→ ×(c^→ ×a^→ )+c^→ ×(a^→ ×b^→ )=0

$$\left(\mathrm{1}\right)\:\mathrm{If}\:\overset{\rightarrow} {\boldsymbol{\mathrm{a}}}=\mathrm{a}_{\mathrm{1i}} +\mathrm{a}_{\mathrm{2j}} +\mathrm{a}_{\mathrm{3k}} \:\mathrm{and}\:\overset{\rightarrow} {\boldsymbol{\mathrm{b}}}=\mathrm{b}_{\mathrm{1i}} +\mathrm{b}_{\mathrm{2j}} +\mathrm{b}_{\mathrm{3k}} \:\mathrm{show}\:\mathrm{that} \\ $$$$\mathrm{i}.\:\overset{\rightarrow} {\boldsymbol{\mathrm{a}}}×\overset{\rightarrow} {\boldsymbol{\mathrm{b}}}=\begin{vmatrix}{\mathrm{i}}&{\mathrm{j}}&{\mathrm{k}}\\{\mathrm{a}_{\mathrm{1}} }&{\mathrm{a}_{\mathrm{2}} }&{\mathrm{a}_{\mathrm{3}} }\\{\mathrm{b}_{\mathrm{1}} }&{\mathrm{b}_{\mathrm{2}} }&{\mathrm{b}_{\mathrm{3}} }\end{vmatrix} \\ $$$$\mathrm{ii}.\:\overset{\rightarrow} {\boldsymbol{\mathrm{a}}}\bullet\overset{\rightarrow} {\boldsymbol{\mathrm{b}}}=\mathrm{a}_{\mathrm{1}} \mathrm{b}_{\mathrm{1}} +\mathrm{a}_{\mathrm{2}} \mathrm{b}_{\mathrm{2}} +\mathrm{a}_{\mathrm{3}} \mathrm{b}_{\mathrm{3}} \\ $$$$ \\ $$$$\left(\mathrm{2}\right)\:\mathrm{If}\:\overset{\rightarrow} {\boldsymbol{\mathrm{a}}}=\mathrm{a}_{\mathrm{1i}} +\mathrm{a}_{\mathrm{2j}} +\mathrm{a}_{\mathrm{3k}} ,\:\:\overset{\rightarrow} {\boldsymbol{\mathrm{b}}}=\mathrm{b}_{\mathrm{1i}} +\mathrm{b}_{\mathrm{2j}} +\mathrm{b}_{\mathrm{3k}} \:\mathrm{and}\: \\ $$$$\overset{\rightarrow} {\boldsymbol{\mathrm{c}}}=\mathrm{c}_{\mathrm{1i}} +\mathrm{c}_{\mathrm{2j}} +\mathrm{c}_{\mathrm{3k}} ,\:\mathrm{show}\:\mathrm{that} \\ $$$$\mathrm{i}.\:\overset{\rightarrow} {\boldsymbol{\mathrm{a}}}\bullet\left(\overset{\rightarrow} {\boldsymbol{\mathrm{b}}}×\overset{\rightarrow} {\boldsymbol{\mathrm{c}}}\right)=\begin{vmatrix}{\mathrm{a}_{\mathrm{1}} }&{\mathrm{a}_{\mathrm{2}} }&{\mathrm{a}_{\mathrm{3}} }\\{\mathrm{b}_{\mathrm{1}} }&{\mathrm{b}_{\mathrm{2}} }&{\mathrm{b}_{\mathrm{3}} }\\{\mathrm{c}_{\mathrm{1}} }&{\mathrm{c}_{\mathrm{2}} }&{\mathrm{c}_{\mathrm{3}} }\end{vmatrix} \\ $$$$\mathrm{ii}.\:\boldsymbol{\mathrm{a}}\bullet\left(\overset{\rightarrow} {\boldsymbol{\mathrm{b}}}×\overset{\rightarrow} {\mathrm{c}}\right)=\overset{\rightarrow} {\boldsymbol{\mathrm{b}}}\left(\overset{\rightarrow} {\boldsymbol{\mathrm{a}}}\bullet\overset{\rightarrow} {\boldsymbol{\mathrm{c}}}\right)−\overset{\rightarrow} {\boldsymbol{\mathrm{c}}}\left(\overset{\rightarrow} {\boldsymbol{\mathrm{a}}}\bullet\overset{\rightarrow} {\boldsymbol{\mathrm{b}}}\right) \\ $$$$\mathrm{iii}.\:\left(\overset{\rightarrow} {\boldsymbol{\mathrm{a}}}×\overset{\rightarrow} {\boldsymbol{\mathrm{b}}}\right)×\overset{\rightarrow} {\boldsymbol{\mathrm{c}}}=\overset{\rightarrow} {\boldsymbol{\mathrm{b}}}\left(\overset{\rightarrow} {\boldsymbol{\mathrm{a}}}\bullet\overset{\rightarrow} {\boldsymbol{\mathrm{c}}}\right)−\overset{\rightarrow} {\boldsymbol{\mathrm{a}}}\left(\overset{\rightarrow} {\boldsymbol{\mathrm{b}}}\bullet\overset{\rightarrow} {\boldsymbol{\mathrm{c}}}\right) \\ $$$$\mathrm{iv}.\:\overset{\rightarrow} {\boldsymbol{\mathrm{a}}}×\left(\overset{\rightarrow} {\boldsymbol{\mathrm{b}}}×\overset{\rightarrow} {\boldsymbol{\mathrm{c}}}\right)+\overset{\rightarrow} {\boldsymbol{\mathrm{b}}}×\left(\overset{\rightarrow} {\boldsymbol{\mathrm{c}}}×\overset{\rightarrow} {\boldsymbol{\mathrm{a}}}\right)+\overset{\rightarrow} {\boldsymbol{\mathrm{c}}}×\left(\overset{\rightarrow} {\boldsymbol{\mathrm{a}}}×\overset{\rightarrow} {\boldsymbol{\mathrm{b}}}\right)=\mathrm{0} \\ $$

Question Number 44812    Answers: 0   Comments: 1

Question Number 44811    Answers: 0   Comments: 1

Question Number 44810    Answers: 1   Comments: 0

Question Number 44809    Answers: 0   Comments: 1

Question Number 44808    Answers: 1   Comments: 1

Question Number 44807    Answers: 1   Comments: 0

Question Number 44806    Answers: 1   Comments: 0

Question Number 44805    Answers: 1   Comments: 0

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