[chandan.mnnit]

Originally Posted by maxcute12

please solve

Q)Both the roots of the equation x^2 - 6ax + 2 - 2a + 9a^2 = 0 exceeds 3,then

a) a<-7/9 & a >1
b)a<10
c)a>11/9
d)a>1/9

Solution-

As both roots are greater than 3 so ...

sum of roots = -(b/a) >6

or 6a>6

or a>1 ----(1)

also product of roots = 2 - 2a + 9a^2 >9

or 9a^2 - 2a -7>0

or solving we get a< -7/9 or a>1 ---(2)

hence the answer (a)

[DON@IIM]

Originally Posted by maxcute12

please solve

Q)Both the roots of the equation x^2 - 6ax + 2 - 2a + 9a^2 = 0 exceeds 3,then

a) a<-7/9 & a >1
b)a<10
c)a>11/9
d)a>1/9

Hi,

Sum of the 2 roots > 6
ie, 6a>6......using sum of roots=-b/a for the quadratic eqn. ax^2+bx+c=0.
ie, a>1............................(1)

Now, product of roots > 9,
ie, 9a^2-2a+2 > 9
ie, 9a^2-2a-7 > 0
ie, (a-1)(9a+7) > 0
Using the no. line: a>1 or a<-7/9.......................(2)

Now combining (1) & (2) we get a>1.

But none of the options match....

Do you have the answer?

Cheers.......

[DON@IIM]

Originally Posted by first_timer

Product of the roots = 2 - 2a + 9a^2
Both roots exceed 3 so
2 - 2a + 9a^2 > 9
(a-1)(9a+7)>0

So a<-7/9 and a>1

But for a < -7/9 the sum of the roots is not greater than 6.

Only a>1 is satisfying both the conditions.

Isn't it?

[aryan12o]

A function f(x) is a polinomial of deg 2002 such that f(1)= 1, f(2)=3, f(3)=5 , f(4)=7................f(2002)=4003. also ,the co-efficient of x^2002 in f(x) is 1.
then f(2002)- f(2002) is given by.
1. (2003)! 2. 4005 3. (2002)! 4. none

[kamra.abhinav]

Originally Posted by maxcute12

please solve

Q)Both the roots of the equation x^2 - 6ax + 2 - 2a + 9a^2 = 0 exceeds 3,then

a) a<-7/9 & a >1
b)a<10
c)a>11/9
d)a>1/9

here we hve got 2 cases
case1:sum>6
i.e. 6a>6 ie a>1.......(1)
case2roduct>9
i.e. 9a^2-2a-7>9
which on solving gives a<-9/7 and a>1but for a<-9/7 the condition of sum of roots dunn satisfy so a>1

[DON@IIM]

Originally Posted by aryan12o

A function f(x) is a polinomial of deg 2002 such that f(1)= 1, f(2)=3, f(3)=5 , f(4)=7................f(2002)=4003. also ,the co-efficient of x^2002 in f(x) is 1.
then f(2002)- f(2002) is given by.
1. (2003)! 2. 4005 3. (2002)! 4. none

Hi,

We are asked to find f(2002)- f(2002)?????

[aryan12o]

f(2003)-f(2002)........................ correction of previous Q

[justtj]

Originally Posted by aryan12o

A function f(x) is a polynomial of deg 2002 such that f(1)= 1, f(2)=3, f(3)=5 , f(4)=7................f(2002)=4003. also ,the co-efficient of x^2002 in f(x) is 1.
then f(2003)-f(2002) is given by.
1. (2003)! 2. 4005 3. (2002)! 4. none

nth term can be written as 2n -1

so f(2003) = f(2002) + 2

f(2003) - f(2002) = 2

this was to be the answer if the statement function f(x) is a polynomial of deg 2002, so i think I would mark the answer as None...

am I correct?

[prateek7563]

Originally Posted by aryan12o

A function f(x) is a polinomial of deg 2002 such that f(1)= 1, f(2)=3, f(3)=5 , f(4)=7................f(2002)=4003. also ,the co-efficient of x^2002 in f(x) is 1.
then f(2002)- f(2002) is given by.
1. (2003)! 2. 4005 3. (2002)! 4. none

is the answer none of these

[aryan12o]

U R WRONG JUSTTJ,
let g(x)=f(x)-(2x-1)
then,g(1)=g(2)=.................=g(2002)=0
g(x)=(x-1)(x-2)...........(x-2002)
f(x)=(x-1)(x-2).............(x-2002)+2x-1
f(2003)=(2002)!+4005=(2002!+4003+2
f(2003)-f(2002)=(2002)!+2
option 4.