Don't think of Vieta's theorem
Let a , b , c , d and e be the roots of the equation x 5 − 4 x 2 + x − 1 3 = 0 . Find the value of 2 − a 1 + 2 − b 1 + 2 − c 1 + 2 − d 1 + 2 − e 1 .
The answer is 13.
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3 solutions
I solved it using transformation of equations to get the given expression as sum of the roots of equation.
Can you show your working
I have posted a solution, check that out..
Has this question come in jee? Which years paper it was??
By the way I never knew that such a short method existed, I am currently preparing for JMO to be held in the sept. and calculus is not a part if the syllabus. Hence, the solution became so long...Thanks for the question..
The solution is as follows:
Don't you think that's too long for a jee problem.Calculus makes it straightforward
You can combine the fractions into one fraction and expand the numerator more easily using combinatorics/symmetry. 2 5 − 4 ( 2 ) 2 + 2 − 1 3 5 ⋅ 2 4 − 5 ⋅ 2 3 ⋅ 5 C 1 4 C 1 ( 0 ) + 5 ⋅ 2 2 5 C 2 4 C 2 ( 0 ) − 5 ⋅ 2 ⋅ 5 C 3 4 C 3 ( 4 ) + 5 ⋅ 5 C 4 4 C 4 ( 1 ) = 5 8 0 − 1 6 + 1 = 1 3
For the given function f(x) calculate f'(x) then find the ratio f'(x)/f(x) which gives us the required expression if we put x = 2.