Find the number of real solutions to the equation r = 1 ∑ 1 0 0 x − r r = x
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Nice one! If someone is unable to find the solution in the third quadrant, you could also reason it this way -
There are a total of 1 0 1 roots to the whole polynomial as the 1 0 0 degree denominator on the left will get multiplied by x on the right. And 1 0 0 real roots are clearly visible due to the asymptotes. So, the 1 0 1 th root also has to be real as complex roots to polynomials with real coefficients occur as conjugates.
Nice graph, Deepanshu. Is this Geogebra? WolframAlpha wasn't able to duplicate it.
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Thanks Sir ! Actually I used Paint to draw this graph :)
Did it same way! :)
Nice solution..
There are a total of 101 roots to the whole polynomial as the100 degree denominator on the left will get multiplied by x on the right. And 100 real roots are clearly visible due to the asymptotes. So, the101th root also has to be real as complex roots to polynomials with real coefficients occur as conjugates.
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Draw approximate graph of this question :
hence no: of solution is 101
q.e.d