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JEE Advanced2009MathematicsApplication of DerivativesActual

Let p(x) be a polynomial of degree 4 having extremum at x=1,2 and _ x 0 [1+ p(x) x^2 ]=2 . Then, the value of p(2) is

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Step-by-step solution

Let p(x)=a x^4+b x^3+c x^2+d x+e p^ (x)=4 a x^3+3 b x^2+2 c x+d p^ (1)=4 a+3 b+2 c+d=0 and p^ (2)=32 a+12 b+4 c+d=0 Since, _ x 0 (1+ p(x) x^2 )=2 [given] _ x 0 a x^4+b x^3+(c+1) x^2+d x+e x^2 =2 array rlrl & & x+1=2, d & =0, e=0 & c & =1 array From Eqs. (i) and (ii) array rlrl & & 4 a+3 b=-2 & and 32 a+12 b=-4 & & a= 1 4 and b=-1 & & p(x)= x^4 4 -x^3+x^2 & & p(2)= 16 4 -8+4 array

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