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Question

Let a be an integer such that all the real roots of the polynomial 2x5+5x4+10x3+10x2 +10x+10 lie in the interval (a,a+1). Then, ais equal to


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Solution

Explanation for the correct answer:

Given :

f(x)=2x5+5x4+10x3+10x2+10x+10.f(x)=x5+x5+5x4+10x3+10x2+10x+1+91

we have,
(x+1)5=c05x0+c15x+c25x2+c35x3+c45x4+c55x5(x+1)5=1+5x+10x2+10x3+5x4+x5

Equation (1) becomes,

f(x)=x5+9+(x+1)5

Now differentiating with respect to 'x'

f'(x)=5x4+5(x+1)4·1

f'(x)0
f(x) is an increasing function.

Let x=-2, the f(-2)
f(-2)=(-2+1)5+(-2)5+9=-1-32+9=-24

Let x=-1
f(-1)=(-1+1)5+(-1)5+9=0-1+9=8

Therefore, the roots lies between -2 to -1.

Since we have given,

a,a+1,a=-2a=2

Therefore, the value of ais 2.


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