Q.

Let tangent at  (α1,β1)  on the curve  y=x42x2x touch the curve again at  (α2,β2), then the value of  |α1|+|α2|+|β1|+|β2|  is equal to

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answer is 4.

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Detailed Solution

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Let  y=mx+c be the tangent  x42x2x(mx+c)=(x2+ax+b)2
 a=0,b=1,m=1,c=1
x+y+1=0  is tangent to the curve at  (1,0)  and  (1,2)
|α1|+|α2|+|β1|+|β2|=4

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