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algebra 400 ques doubts
by gaurika gupta - Friday, 22 August 2014, 02:01 AM
  Sir
I have doubts in the following questions.Please see if you could help.

1.how many real numbers are solutions to the equation x^4+4|x|=10?

2.if 2^2^x+4^2^x=56 then what is the value of 2^2^2^x?

3.let f(x) be a polynomial of degree 2006 satisfying
f(k)=1/k,1<=k<=2007
what is the value of f(2008)?

4.suppose real numbers x and y satisfy x^2+9y^2-4x+6y+4=0.what is the maximum value of 4x-9y?
Re: algebra 400 ques doubts
by TG Team - Friday, 22 August 2014, 01:00 PM
  Hi Gaurika smile

Starting from last one..

4. x2 + 9y2 - 4x + 6y + 4 = 0
i.e. {x2 - 2(x)(2) + 22} + {(3y)2 + 2(3y)(1) + 1} = 1
i.e. (x - 2)2 + (3y + 1)2 = 1

So either x - 2 = 0 i.e. x = 2 and |3y + 1| = 1 i.e. y = 0 or y = -2/3
OR |x - 2| = 1 i.e. x = 3 or 1 and 3y + 1 = 0 i.e. y = -1/3.

So possible values of 4x - 9y = 4(2) - 9(-2/3) = 14 OR 4(3) - 9(-1/3) = 15.

Clearly maximum value of 4x - 9y = 15. smile

Kamal Lohia
Re: algebra 400 ques doubts
by TG Team - Friday, 22 August 2014, 04:31 PM
  3. Let xf(x) - 1 be a new polynomial of degree 1 more than that of f(x) i.e. 2007. So it'll have 2007 roots and all of those are known to us as 1, 2, 3, ...., 2007.

So we get that xf(x) - 1 = A(x - 1)(x - 2)(x - 3).....(x - 2007)

Putting x = 0, we get that -1 = A(-1)(-2)(-3)...(-2007)
i.e. -1 = -A(2007!)
i.e. A = 1/2007!

Thus xf(x) - 1 = [
(x - 1)(x - 2)(x - 3).....(x - 2007)]/2007!
and 2008f(2008) - 1 = [
(2008 - 1)(2008 - 2)(2008 - 3).....(2008 - 2007)]/2007! = 1
i.e. f(2008) = 2/2008 = 1/1004. smile

Kamal Lohia
 
Re: algebra 400 ques doubts
by TG Team - Friday, 22 August 2014, 04:41 PM
  2. 22^x + 42^x = 56
Let 22^x = t

So given equation becomes t + t2 = 56
i.e. (t - 7)(t + 8) = 0
i.e. t = 7 as t is positive.

Now we need to calculate 2t = 27 = 128. smile

Kamal Lohia
Re: algebra 400 ques doubts
by gaurika gupta - Sunday, 24 August 2014, 12:41 PM
  Sir

the answer given is 16.Is it incorrect?
what if say (x-2)^2=1/2
and (3y+1)^2= 1/2 and similar cases? why are we considering only integral values to them?

and can you please help with the first question also?
Thank You :)
Re: algebra 400 ques doubts
by TG Team - Tuesday, 26 August 2014, 03:51 PM
  Thanks for pointing out the error in solution of 4th one Gaurika. Here it is again for you in .... two different ways. smile

4. x2 + 9y2 - 4x + 6y + 4 = 0
i.e. {x2 - 2(x)(2) + 22} + {(3y)2 + 2(3y)(1) + 1} = 1
i.e. (x - 2)2 + (3y + 1)2 = 1

Now we want to maximise 4x - 9y. Let 4x - 9y = k, i.e. x = (k + 9y)/4

So eliminating x, above equation becomes: (k + 9y - 8)2 + 16(3y + 1)2 = 16
i.e. 225y2 + {18(k - 8) + 96}y + (k - 8)2 = 0
i.e. 225y2 + (18k - 48)y + (k - 8)2 = 0

Now as y is a real number, so discriminant of the above equation should be greater than or equal to zero.
So, (18k - 48)2 - 4(225)(k - 8)2 ≥ 0
i.e. (18k - 48)2 - (30k - 240)2 ≥ 0
i.e. (3k - 8)2 - (5k - 40)2 ≥ 0
i.e. (-2k + 32)(8k - 48) ≥ 0
i.e. (k - 6)(k - 16) ≤ 0
i.e. 6 ≤ k ≤ 16.

Thus required maximum value of 'k' is 16. smile

Alternate approach is geometric one smile
We know that
(x - 2)2 + (3y + 1)2 = 1 and we need to maximise 4x - 9y.

Now, let's replace 3y by Y, so that know equation turns into a circle

(x - 2)2 + (Y + 1)2 = 1 with center at (2, -1) and radius 1.
And now we need to find the maximum value of 4x - 3Y which will be equal to 4x - 3Y = k (say) and will be achieved when this line is tangential to the circle.

Rest part is simple. I hope you know how to find the perpendicular distance of a point from a line.

Using the relation for the distance, we get that |{4(2) - 3(-1) - k}/√(42 + 33)| = 1
i.e. |11 - k| = 5
i.e. k = 6 or 16.

Thus maximum value of k = 16 as found earlier too. smile

Kamal Lohia
Re: algebra 400 ques doubts
by TG Team - Tuesday, 26 August 2014, 04:02 PM
  Hi Gaurika smile

1. x4 + 4|x| = 10
i.e. x4 = 10 - 4|x|
i.e. we just need to find the number of intersections of y = x4 and y = 10 - 4|x| which is simply 2 as first curve draws a steeper parabola closer to Y=axis and the second curve draws an inverted V having a vertex at (0, 10).

Kamal Lohia
Re: algebra 400 ques doubts
by soumya sawarn - Thursday, 4 September 2014, 12:14 PM
  Sir why do we take the polynomial as xf(x)-1
Re: algebra 400 ques doubts
by TG Team - Thursday, 4 September 2014, 12:42 PM
  Hi Soumya smile

It is given that f(k) = 1/k for some particular values of x = k.

So for those values of x = k, we have k*f(k) = 1 i.e. k*f(k) - 1 = 0

i.e. for some particular values of x, we get x*f(x) - 1 as zero. And this x*f(x) - 1 is also a polynomial of one higher degree than that of f(x). And the values of x at which this new polynomial becomes zero are its roots.

I hope it helps. smile

Kamal Lohia