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Question : 68 of 160
Marks:
+1,
-0
Solution:
Equation of given curve is
y=x4−6x3+13x2−10x+5 So,
=4x3−18x2+26x−10 As the slope of tangent at point
P(x1,y1) and
Q(x2,y2) on curve is given as 2, so put
=2 ⇒4x3−18x2+26x−10=2 ⇒2x3−9x2+13x−6=0 ⇒(x−1)(2x2−7x+6)=0 ⇒(x−1)(2x2−4x−3x+6)=0 ⇒(x−1)[2x(x−2)−3(x−2)]=0 ⇒(x−1)(x−2)(2x−3)=0 ∵x1,x2∈N⇒y1=3 and
y2=5 Therefore,
x1x2+y1y2=(1×2)+(3×5) =2+15=17
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