204
Question
If x1,x2,x3,..,xn are the roots of the equation xn+ax+b=0, then the value of (x1−x2)(x1−x3)(x1−x4)...(x1−xn) is equal to
If x1,x2,x3,..,xn are the roots of the equation xn+ax+b=0, then the value of (x1−x2)(x1−x3)(x1−x4)...(x1−xn) is equal to
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Solution
The correct option is B nx1n−1+a
Given:x1,x2,x3.....xn are the roots of xn+ax+b=0To find: (x1−x2)(x1−x3)......(x1−xn)=?Sol: (x1−x2)(x1−x3)......(x1−xn)=(x1−x1)(x1−x2)(x1−x1)........(x1−x2)=limx→x1(x−x1)(x−x2).........(x−xn)(x−x1)=limx→x1xn+ax+b(x−x1)Using l'Hospital rule, we getlimx→x1n.xn−1+a1=nxn−11+aHence, correct answer is nxn−11+a
Given:x1,x2,x3.....xn are the roots of xn+ax+b=0
To find: (x1−x2)(x1−x3)......(x1−xn)=?
Sol: (x1−x2)(x1−x3)......(x1−xn)
=(x1−x1)(x1−x2)(x1−x1)........(x1−x2)
=limx→x1(x−x1)(x−x2).........(x−xn)(x−x1)
=limx→x1xn+ax+b(x−x1)
Using l'Hospital rule, we get
limx→x1n.xn−1+a1=nxn−11+a
Hence, correct answer is nxn−11+a
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