On Tue, Feb 9, 2010 at 3:18 PM, Ronit Sen wrote:
I am sending you some.Please try for it.
They are as follows:
1)Prove that for all real values of x, the expression 3x^2-6x+10 is always greater that 7.
2)If a^2+b^2+c^2=1 then prove that bc+ca+ab≥-1/2.
3)Find the equation to the straight line which passes through the point (-4,3)
and is such that the portion of the line between the axes is divided at the
point internally in the ratio 5:3.
I did the height and distance sum while giving you the sum.
Please send it to me as fast as possible..
Sir please send it fast.
Find the solutions as below:
1) we can write the expression as 3(x-1)^2+7, since (x-1)^2 will
always be positive, the minimum value of the expression will always be
7, even if x=1, ie, (x-1)^2 is zero.
So, your solution is:
Since a square is always positive, the minimum value of the exprssion
will always be 7.
Now since (a+b+c) is a square, the value has to be greater than zero.
3) The formula to be used here is :
since it is dividing internally,
so the equation is: 5x-3y+29=0
The Target ICSE team