Q10: IIT LEVEL Solved I

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A particular algebraic knowledge is required to solve this problem. The algebra is shown in the next diagram. Partially it is treated herein too. 

Within the equation given, in step 2, we have divided with 1+x+x^2. Than can be done only if this expression is not equal to zero. If you see x^2+x+1, the D=1-4=-3<0 which means that this quadratic expression has no roots i.e. it is above the x axis and is thus y or the expression is always positive, so we can divide our equation with it. 

Do you see how quadratic expressions pop up everywhere and each time the diagram of the quadratic must pop up in your mind too letting you correctly arrive at the interval in which the relevant expression is greater than or lesser than 0 i.e. positive or negative. 


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The algebra used in the above question has been shown again here. Just read it and you will know it. It is useful to remember a few unusual algebraic combinations. You never know when they come useful.

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Since the expression is given as being greater than or equal to zero that means the entire quadratic will be either completely above the x axis making all the y values positive which implies D<0 or
​the quadratic will touch the x axis at one point which implies D=0. 

Since it is stated in the question that the expression is greater than or equal to zero FOR ALL X which are REAL, it means D cannot be greater than zero because if D>0, there would be two roots and in the segment between the roots, the y values will be negative which is not permitted as per the question. 


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As shown in the diagram, if 2 is greater than both the roots, then f(2)> has to be greater than zero. 

The student must get this very very clearly. That the f values or the y values which are the same thing are negative between the roots and are positive beyond the roots. This is true whenever a>0 and a is the coefficient of x^2 which is 4 in this question. 

So, once again
if a>0, 
y<0 between the roots &
y>0 beyond the roots on both sides

The other important point is as you calculate D or f(d) and you arrive at another quadratic, you have to treat that quadratic independently and apply the same idea there itself to find when that expression is greater or lesser than zero. And then apply to the conditions in the question to arrive at the answer being asked. 


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This is a simple examination of possibilities. The examiners who set your question papers must be employing their faculties to investigate into different properties of the quadratic in order to come up with creative questions that test your understanding. 

Here we test what if the coefficients b and c of the quadratic ax^2+bx+c
are square of the other.
​Two cases confront us: when b is square of c i.e. Case II in the diagram and when c is square of b i.e. Case I. 

We have shown that when c is square of b, there are no roots. 

And when b is square of c and if the values p are between 0 to 1.6, D<0 i.e there are no roots. The only integer between 0 and 1.6 is 1 and you can test it: the quadratic expression for x^2+p^2.x +p=0 will become x^2+x+1 which has negative D and no roots. 
And any integer less than 0 or more than 2 will have D>0 and thus would have positive D and real roots. 

We have also shown a simple method to find the cube root of 4 which is the value of p when D=0


Here we first take when b and c are decreasing consecutive integers and find that D>0 for all p and thus there are real roots.
In the second case b and c are increasing integers and we find that the D is equal to 
p^2-4p-4=0 whose D is 16+16>0 and the roots of this equation is 2+-sqrt 32/2 i.e. 2+-2sqrt2 i.e. between 2-2sqrt2 and 2+2sqrt2, the expression will be negative and beyond them the expression will be positive. 
Sqrt 2 is 1.4 implying 2-2*1.4=-0.8 and 2 +2.8=4.8 i.e. values of p from – 0.8 to 1.8 includes the integers 0,1,2,3&4 as shown. This has also been shown clearly in the diagram to the right

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in this problem the concept of tangent has been introduced. A quadratic can have only one tangent which is parallel to the x axis and that is the one at its vertex. For U shaped curves with a>0, the vertex is the lowest point of the quadratic and we know from our analysis of axis which is given by x=- b/2a that the vertex has the coordinates [- b/2a, – D/4a]

​So, the tangent has to have an equation as y = – D/4a. 

Thus, we would do well to remember that 

AXIS:                    x=-b/2a 
TANGENT:          y=-D/4a



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this is basically a logarithm problem with a quadratic component thrown in. You will find later that a quadratic expression can crop up anywhere in Physics and other chapters of mathematics. In fact quadratics is ubiquitous and you cannot do without it. 

Since we see log (2) x repeated twice in the question we take it as y and proceed accordingly. Logarithm is a very playful chapter and you must learn to play along with its many colourful properties. See the Logarithm page for more entertaining examples.

Additionally we have a cubic equation here and we have shown a simple method to solve the cubic.