JEE Advanced 2016 · Paper 1 · Mathematics Q43

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MathematicsMultiple correct+4 / -2Question 43 of 54

Let $f:(0,\infty)\to\mathbb R$ be a differentiable function such that $f'(x)=2-\dfrac{f(x)}{x}$ for all $x\in(0,\infty)$ and $f(1)\ne1$ . Then

\displaystyle\lim_{x\to0+}f'\left(\frac1x\right)=1

\displaystyle\lim_{x\to0+}xf\left(\frac1x\right)=2

\displaystyle\lim_{x\to0+}x^2f'(x)=0

|f(x)|\le2

for all

x\in(0,2)

Answer and worked solution

Official answer: A

Solve. The equation is $xf'+f=2x$ , i.e. $(xf)'=2x$ . So
$xf=x^2+c\implies f(x)=x+\frac cx,$
with $c\ne0$ because $f(1)\ne1$ . Then $f'(x)=1-\dfrac c{x^2}$ .

Check each option.

(A) $f'(1/x)=1-cx^2\to1$ . True.
(B) $xf(1/x)=1+cx^2\to1$ , not 2. False.
(C) $x^2f'(x)=x^2-c\to-c\ne0$ . False.
(D) $f$ is unbounded near 0. False.

(A)

Topics:Linear first-order equations Limits

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JEE Advanced 2016 · Paper 1, Mathematics question 43