physics chemistry maths science forums
become expert I help I sign up I login
refer a friend - earn nickels!!   
 advanced
 
Home
Ask & Discuss Questions
Study Material
Experts Zone
Hang Out!
sboosy   sboosy is offline sboosy's messages in the community
Message
Catalogs Discussion Forums -> Differential Calculus -> integration -> Go to message
This Post 0 points    (0 Olaaa!! Perrrfect answer.   in 0 votes )   [?]

f
Catalogs Discussion Forums -> Integral Calculus -> find f(x) -> Go to message
This Post 7 points    (1 Olaaa!! Perrrfect answer.   in 2 votes )   [?]

\mbox{Consider} \ f(x) = ax^3+bx^2+cx+d \\ \\ f(1) = 2 \ \Rightarrow a+b+c+d = 2 \\ \\ f(2) = 5 \ \Rightarrow 8a+4b+2c+d = 5 \\ \\ f(3) = 9 \ \Rightarrow 27a+9b+3c+d=9 \\ \\ f(5)= 32 \ \Rightarrow 125a+25b+5c+d = 32 \\ \\ \mbox{Four equations, four unknowns} \\ \\ \mbox{Solving we get} \ a=\frac{1}{2},b=\frac{-5}{2},c=7,d=-3 \\ \\ \Rightarrow f(x) = \frac{1}{2}x^3 - \frac{5}{2}x^2+7x-3 \\ \\ \Rightarrow f(4) = 32-40+28-3 =17

Catalogs Discussion Forums -> Trignometry -> TRIANGLES -> Go to message
This Post 0 points    (0 Olaaa!! Perrrfect answer.   in 0 votes )   [?]

The second question has the answer in the following link ..


http://www.goiit.com/posts/list/trignometry-sot-challenge-46259.htm#228738
Catalogs Discussion Forums -> Trignometry -> Summation -> Go to message
This Post 15 points    (3 Olaaa!! Perrrfect answer.   in 3 votes )   [?]

\mbox{Continuing from where sandeep has left} \\ \\ \tan^{-1}\left(\frac{1}{2r^2}\right) = \tan^{-1}\left(\frac{(2r+1)-(2r-1)}{1+(2r+1)(2r-1)}\right) \\ \\ = \tan^{-1}(2r+1) - \tan^{-1} (2r-1) \\ \\ \mbox{Summing from 1 to} \ \infty \ \mbox{we get} \\ \\ \tan^{-1}(3) - \tan^{-1}(1) +\tan^{-1}(5) - \tan^{-1}(3)+......\tan^{1}(\infty) \\ \\ = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}


 


:gu
Catalogs Discussion Forums -> Algebra -> question -> Go to message
This Post 5 points    (1 Olaaa!! Perrrfect answer.   in 1 votes )   [?]

(1-2a)x^2 -6ax-1=0 \ \mbox{Let the roots be} \ \alpha,\beta \\ \\ ax^2-x+1=0 \ \mbox{Let the roots be} \ \alpha,\gamma<br/>\\ \\ \alpha+\beta = \frac{6a}{1-2a}...(1) \\ \\ \alpha\beta = \frac{1}{2a-1}....(2) \\ \\ \alpha+\gamma = \frac{1}{a}....(3) \\ \\ \alpha\gamma = \frac{1}{a}....(4) \\ \\ \mbox{Using 1 in 2 and 3 in 4 we get} \\ \\ \alpha\left(\frac{6a}{1-2a}\right) - (\alpha)^2 = \frac{1}{2a-1} ....(5) \\ \\ \alpha\left(\frac{1}{a}\right) - (\alpha)^2 = \frac{1}{a} ....(6) \\ \\ \mbox{Subtract these 2 equation to get rid of} \ (\alpha)^2 \\ \\ \Rightarrow \alpha = \frac{a-1}{6a^2+2a-1} \\ \\ \Rightarrow \gamma = \frac{6a^2+2a-1}{a(a-1)} \\ \\ \mbox{Now using 3 we get to solve} \ \frac{a-1}{6a^2+2a-1}+\frac{6a^2+2a-1}{a(a-1)} = \frac{1}{a} \\ \\ \mbox{which gives} \ a= \frac{2}{9}


 


:gu
Catalogs Discussion Forums -> Integral Calculus -> integrate [dx/sin(x-a)cos(x-b)] -> Go to message
This Post 0 points    (0 Olaaa!! Perrrfect answer.   in 0 votes )   [?]

\mbox{Better it would be if we multiplied and divided by} \ \cos(b-a) \\ \\ \frac{1}{\cos(b-a)}\int \frac{\cos((x-a)-(x-b))}{\sin(x-a)\cos(x-b)} \,dx \\ \\ = \frac{1}{\cos(b-a)}\left(\int \cot(x-a) \,dx + \int \tan(x-b) \,dx \right) \\ \\ = \frac{1}{\cos(b-a)}\left(\log|\sin(x-a)|+\log|\sec(x-b)|\right) + C


:gu
Catalogs Discussion Forums -> Integral Calculus -> integration -> Go to message
This Post 5 points    (1 Olaaa!! Perrrfect answer.   in 1 votes )   [?]

\int_0^\infty \frac{\log(x+\frac{1}{x})}{1+x^2}\,dx \\ \\ x=\tan(\theta) \\ \\ \int_0^\frac{\pi}{2} -\log(\sin(\theta)\cos(\theta))\,d\theta \\ \\ = -2\times\frac{-\pi}{2}\log(2) = \pi\log(2)


:gu
Catalogs Discussion Forums -> Analytical Geometry -> check this -> Go to message
This Post 12 points    (2 Olaaa!! Perrrfect answer.   in 3 votes )   [?]

A(1,1),B(4,-2),C(5,5) \ \mbox{are the vertices} \\ \\ \mbox{Thus the equations of the sides are} \ x+y=2,x=y,7x-y=30 \\ \\ \mbox{Also it is easy to note now that angle A is} \ 90^{\circ} \\ \\ \mbox{Thus the bisector of A is parallel to x axis} \ \mbox{and its equation is} \ y=1 \\ \\ \mbox{Thus the perpendicular equation is} \ x=k \\ \\ \mbox{But since it has to pass through} \ (5,5) \\ \\ \mbox{Its equation is} \ x=5


:gu
Catalogs Discussion Forums -> Algebra -> determinent -> Go to message
This Post 14 points    (2 Olaaa!! Perrrfect answer.   in 4 votes )   [?]

\mbox{Determinant of A} \ = abc-a-b-c+2>0 \ \mbox{(Determinant given to be positive)} \\ \\ abc+2>a+b+c\geq3(abc)^{\frac{1}{3}} \ \mbox{(Second part using AM-GM)} \\ \\ \Rightarrow abc+2>3(abc)^{\frac{1}{3}} \\ \\ \mbox{Let} \ abc=k \ \Rightarrow k^3+6k^2-15k+8>0 \\ \\ \Rightarrow (k-1)^2(k+8) >0 \\ \\ \Rightarrow abc>-8
Catalogs Discussion Forums -> Mechanics -> A particle starts frm rest wid accn 2m/s2. accn of the particle decreases -> Go to message
This Post 5 points    (1 Olaaa!! Perrrfect answer.   in 1 votes )   [?]


:gu
Catalogs Discussion Forums -> Mechanics -> problem in mechanics... -> Go to message
This Post 0 points    (0 Olaaa!! Perrrfect answer.   in 0 votes )   [?]

Since the lift is accelerating up .mark a pseudo force down on the block = ma = 2m (in this case)


Thus in equilibrium we have     


 


In case a one kg block is also added we have



Further elongation is 0.12 m :gu
Catalogs Discussion Forums -> Mechanics -> problem in mechanics... -> Go to message
This Post 0 points    (0 Olaaa!! Perrrfect answer.   in 0 votes )   [?]



 


:gu
Catalogs Discussion Forums -> Algebra -> probability... -> Go to message
This Post 0 points    (0 Olaaa!! Perrrfect answer.   in 0 votes )   [?]

When we throw thrice , sum of 9 can come on the first 2 throws or first and last or last 2 throws (3 cases)


Thus the multiplication by 3 is done :gu:gu:gu
Catalogs Discussion Forums -> Algebra -> probability... -> Go to message
This Post 2 points    (0 Olaaa!! Perrrfect answer.   in 1 votes )   [?]


 


:gu:gu:gu
Catalogs Discussion Forums -> Algebra -> question -> Go to message
This Post 0 points    (0 Olaaa!! Perrrfect answer.   in 0 votes )   [?]

Ya ur right abt it ...The new editor is really nice :d :d


:bo :bo :bo :gu:gu
 
 
Go to:   

Top Offers for goIITians
Correspondence Courses
Brilliant Tutorials
Narayana Institute
Quest Tutorials
Aakash Institute
Classroom/Crash Courses
Narayana - Kota , Delhi , Others
Brilliant Tutorials - Class , Crash
Aakash Institute - Medical , Engg
Online Test Series
Brilliant Tutorials
Narayana Institute
Aakash Institute
Mahesh Tutorials
AMITY      Sri Chaitanya