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Catalogs Discussion Forums -> Algebra -> An interesting result -> Go to message
This Post 17 points    (3 Olaaa!! Perrrfect answer.   in 4 votes )   [?]
3 replies   
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 )   [?]
10 replies   
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 )   [?]
1 replies   
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 )   [?]
2 replies   
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 )   [?]
1 replies   
\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 )   [?]
1 replies   
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 )   [?]
9 replies   
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 )   [?]
1 replies   
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}
Catalogs Discussion Forums -> Algebra -> quad. equation -> Go to message
This Post 2 points    (0 Olaaa!! Perrrfect answer.   in 1 votes )   [?]
2 replies   
Since it is an upward drawn parabola ...the value of f for those values between the roots (-2 and 2 minimum) are negative
so f(1)<0 =>a+b+c<0
f(-1)<0 =>a-b+c<0
Thus a+|b|+c<0
Catalogs Discussion Forums -> Mechanics -> Hei Guys! Lend me ur attention..rates assured..OSCILLATIONS -> Go to message
This Post 5 points    (1 Olaaa!! Perrrfect answer.   in 1 votes )   [?]
5 replies   
Assume it is displacd by x straight wrt to one spring...then correspondingly the extensions in the other two springs will be xcos(60) each
The force eqn (which is not much different,would be)
 
F = kx+(k2x2cos2(60)+k2x2cos2(60)+2k2x2cos2(60)cos(120))
(Using the formula of vectors .a+b(vectors)(magnitude) = (a2+b2+2abcos(alpha)
where alpha is the angle between the vectors a and b
 
Thus F = kx+kx/2 =3kx/2
T = 2(m/keff) where keff in this case is 3k/2
Thus T = 2(2m/3k)
Catalogs Discussion Forums -> Algebra -> quadratic equation -> Go to message
This Post 0 points    (0 Olaaa!! Perrrfect answer.   in 0 votes )   [?]
8 replies   
sorry ..didnt read the question properly ...i took it to be
ax^2+bx+c ...sorry for the miss ..mayb the time is a factor ...
Catalogs Discussion Forums -> Algebra -> quadratic equation -> Go to message
This Post 0 points    (0 Olaaa!! Perrrfect answer.   in 0 votes )   [?]
8 replies   
I think it is true ....infact this is the condition u ll usually apply in problems
Catalogs Discussion Forums -> Algebra -> no of solutions.........find plz......... -> Go to message
This Post 0 points    (0 Olaaa!! Perrrfect answer.   in 0 votes )   [?]
3 replies   
I suppose everybody is very serious about VITEEE because the same question has come thrice in 3 days
 
Here is the link
 
Catalogs Discussion Forums -> Mechanics -> WORK AND ENERGY -> Go to message
This Post 5 points    (1 Olaaa!! Perrrfect answer.   in 1 votes )   [?]
2 replies   
The important point is that it is a rod and not a string ..a string can slack but a rod cannot
Thus the least condition is that the topmost pt ..the velocity can be 0 .
thus mg(2l) = 1/2 m(v2)
which means v = 4gl
Catalogs Discussion Forums -> Algebra -> do fast -> Go to message
This Post 0 points    (0 Olaaa!! Perrrfect answer.   in 0 votes )   [?]
3 replies   
Ryu ..i guess u've made a small mistake...
Question says ...each cat has 7 kittens ...
so u actually have 7 cats +7*7 kittens = 56 cats!
 
 
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