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Question

Consider the function f:(-,)(-,) defined byf(x)=[x2-ax+1][x2+ax+1];0<a<2. Which of the following is true


A

(2+a)2f''(1)+(2a)2f''(-1)=0

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B

(2a)2f''(1)(2+a)2f''(-1)=0

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C

f'(1)f'(-1)=(2a)2

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D

f'(1)f'(-1)=-(2+a)2

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Solution

The correct option is A

(2+a)2f''(1)+(2a)2f''(-1)=0


Explanation of the correct option :

Step1. Given function :

f(x)=[x2-ax+1][x2+ax+1];0<a<2

Discriminant of denominator a24<0

Denominator 0,xR

Step 2. Differentiate with respect to x :

f'(x)=[(x2+ax+1)(-2a)+2ax(2x+a)][x2+ax+1]2=[2ax22a][x2+ax+1]2=[2a(x21)][x2+ax+1]2f'(1)=f'(-1)=0f''(x)=[2a(x2+ax+1)2(2x)2a(x21)2(x2+ax+1)(2x+a)](x2+ax+1)4f''(1)=2a[2(2+a)](2+a)3=4a(2+a)2f''(-1)=-4a(2-a)2f''(1)(2+a)2+f''(-1)(2a)2=0

Hence, the correct option is A.


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