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← IIT-JEE 2011 · Paper 1
MathematicsInteger+4Question 66 of 69

IIT-JEE 2011 · Paper 1, Mathematics question 66

Let f:[1,∞)→[2,∞)f:[1,\infty)\to[2,\infty) be a differentiable function such that f(1)=2f(1)=2. If
6∫1xf(t) dt=3xf(x)−x36\int_1^xf(t)\,dt=3xf(x)-x^3
for all x≥1x\ge1, then the value of f(2)f(2) is

Answer and worked solution

Official answer: Bonus: full marks to everyone

Official key: marks to all.

The data are inconsistent. At x=1x=1 the equation gives 0=3f(1)−10=3f(1)-1, i.e. f(1)=13f(1)=\frac13, contradicting f(1)=2f(1)=2.

Formal working. Differentiating, 6f=3f+3xf′−3x26f=3f+3xf'-3x^2, so f′−fx=xf'-\frac fx=x and f=x2+Cxf=x^2+Cx. Neither f(1)=2f(1)=2 (C=1C=1, f(2)=6f(2)=6) nor the integral equation itself (C=−23C=-\frac23, which also violates f≥2f\ge2) gives a consistent answer.