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

 


 In the first place excepting 0 all the 9 numbers can come ..in the next place excepting the number which appeared to its left (1st number) , other 9 numbers can come ...similarly all n places 9 numbers can come ..


Thus, the answer is 9*9*9*9....n times = 9n 


 
Catalogs Discussion Forums -> Algebra -> An interesting result -> Go to message
This Post 12 points    (2 Olaaa!! Perrrfect answer.   in 3 votes )   [?]
f(x)=(1+x)^2 = x^2+2x+1 , f(x)=0 \ \mbox{has repeated root} \ -1 \\ \\ \mbox{Also} \ x^2+3x+1=0,x^2+4x+1=0 ...,x^2+ax+1 (a\geq2) \\ \\ \mbox{always have real roots.Also} \ x^2+ax+1(a\geq2)\geq x^2+2x+1 = (1+x)^2(x \ \mbox{positive}) \\ \\ \mbox{Now just extending this concept to n from power 2} \\ \\ x^n+a_1x^{n-1}+a_2x^{n-2}+....1 \geq x^n+nC_1x^{n-1}+nC_2x^{n-2}+....1 = (1+x)^n \\ \\ \mbox{Thus} \ f(2)\geq (1+2)^n = 3^n
Catalogs Discussion Forums -> Algebra -> quadratic.. -> Go to message
This Post 0 points    (0 Olaaa!! Perrrfect answer.   in 0 votes )   [?]
ax^3+bx^2+cx+d=0 \\ \\ \mbox{Let the roots of the equation be} \ \alpha,\beta,\gamma \\ \\ \mbox{Sum of the roots} = \frac{\mbox{-coeff.of} \ x^2}{\mbox{coeff.of} \ x^3} \Rightarrow \alpha+\beta+\gamma = \frac{-b}{a} \\ \\ \mbox{Sum taken two at a time} = \frac{\mbox{coeff.of} \ x}{\mbox{coeff.of} \ x^3} \ \Rightarrow \alpha\beta+\beta\gamma+\gamma\alpha = \frac{c}{a} \\ \\ \mbox{Product of the roots} \ = \frac{-\mbox{Constant term}}{\mbox{coeff.of} \ x^3} \ \Rightarrow \alpha\beta\gamma = \frac{-d}{a} \\
Catalogs Discussion Forums -> Algebra -> question -> Go to message
This Post 27 points    (5 Olaaa!! Perrrfect answer.   in 6 votes )   [?]
10^1+10^2+10^3+10^4+....10^{1000} \ \mbox{mod} \ 7 = ? \\ \\ \mbox{Final remainder} = \mbox{Remainder obtained when S is divided by 7} \\ \\ \mbox{where S is the sum of the remainders when each number is individually divided by 7} \\ \\ 10 \ \mbox{mod} \ 7 = 3 ,\ 100 \ \mbox{mod} \ 7 = 2 ....\\ \\ \mbox{The pattern is} \ (3,2,6,4,5,1),(3,2,6,4,5,1),(3,2,6,4,5,1) .....166 \ \mbox{times} ,(3,2,6,4) \\ \\ \mbox{Sum of each bracket is} \ 21 \ \mbox{and each bracket} \\ \\ \mbox{appears 166 times(i.e) upto} \ 10^{996} \ \mbox{and for the last four} \\ \\ \mbox{we have} \ (3,2,6,4) \\ \\ \mbox{Thus the final remainder is} \ 166*21+15 \ \mbox{mod} \ 7 = 1
Catalogs Discussion Forums -> Algebra -> Summation -> Go to message
This Post 15 points    (3 Olaaa!! Perrrfect answer.   in 3 votes )   [?]
1^2-2^2+3^2-4^2+....-2002^2+2003^2 \\ \\ \mbox{Considering in pairs}\\ \\ \ (1-2)(1+2)+(3-4)(3+4)+....(2001-2002)(2001+2002)+2003^2 \\ \\ \mbox{Taking} \ -1 \ \mbox{common in each bracket,we get} \\ \\ -1[1+2+3+....2002]+2003^2 \\ \\ = -1\left(\frac{(2002)(2003)}{2}\right)+2003^2 \\ \\ = 2003(2003-1001) = (2003)(1002) = 2007006
Catalogs Discussion Forums -> Trignometry -> find min value -> Go to message
This Post 0 points    (0 Olaaa!! Perrrfect answer.   in 0 votes )   [?]
\frac{2^{\sin(x)}+2^{\cos(x)}}{2} \geq 2^{\frac{\sin(x)+\cos(x)}{2}} \ \mbox{(Using AM-GM)} \\ \\ \sin(x)+\cos(x) \geq -\sqrt{2} \ \Rightarrow \frac{\sin(x)+\cos(x)}{2} \geq -\frac{1}{\sqrt{2}} \\ \\ \frac{2^{\sin(x)}+2^{\cos(x)}}{2} \geq 2^{\frac{-1}{\sqrt{2}}} \\ \\ 2^{\sin(x)}+2^{\cos(x)} \geq 2^{1-\frac{1}{\sqrt{2}}}
Catalogs Discussion Forums -> Analytical Geometry -> Prime number OR NOT: is 91 digits number with 1 as all digits (111....111) -> Go to message
This Post 10 points    (2 Olaaa!! Perrrfect answer.   in 2 votes )   [?]
Now 1111...91 times is not a prime number ...this is so becuase ..consider 1111111(7 times 1) ...
Now 111..91 times is nothing but 1111111,1111111, ...like this into 13 such groups ...that is it is in other words 1111...91 times is divisible by 1111111(7 times) ...
Hence it is not a prime number ,,,
 
In other words 111...(n times) cannot be a prime number if n is itself non prime (91 = 13*7) ..but n being a prime may not always be a sufficient condition to know if 1111..(that many times) is prime or not ..because 111(3 times) is not a prime number(obviously div. by 3) although 3 is prime ....
Catalogs Discussion Forums -> Differential Calculus -> Domain -> Go to message
This Post 5 points    (1 Olaaa!! Perrrfect answer.   in 1 votes )   [?]
y=\sqrt{\cos^{-1}(\frac{1-|x|}{2})} \\ \\ -1\leq\frac{1-|x|}{2}\leq1 \\ \\ 1-|x|\geq-2 \ \Rightarrow |x|\leq3 \\ \\ \Rightarrow -3\leq x \leq3 \\ \\ |x|\geq -1 \ \mbox{is always true} \\ \\ \mbox{Thus} \ -3 \leq x \leq 3
Catalogs Discussion Forums -> Trignometry -> problem -> Go to message
This Post 10 points    (2 Olaaa!! Perrrfect answer.   in 2 votes )   [?]
8^{1+|\cos(x)|+|\cos^2(x)|+|\cos^3(x)|+....} = 4^3 = 2^6 \\ \\ 8^{\frac{1}{1-|\cos(x)|}} = 2^6 \\ \\ \Rightarrow 2^{\frac{3}{1-|\cos(x)|}}=2^6 \\ \\ \Rightarrow 1-|\cos(x)| = \frac{1}{2} \\ \\ \Rightarrow |\cos(x)| = \frac{1}{2} \\ \\ \Rightarrow \cos(x) = \pm \frac{1}{2} \\ \\ \cos(x) = \frac{1}{2} \ \Rightarrow x=2n\pi \pm \frac{\pi}{3} \\ \\ \cos(x) = \frac{-1}{2} \ \Rightarrow x = 2n\pi \pm \frac{2\pi}{3} \\ \\ \mbox{Their union is the general solution}
 
 
Go to:   

Top Offers for goIITians
Correspondence Courses
Brilliant Tutorials
Narayana Institute
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