JEE Main 2016 · 9 Apr · Shift 1 · Mathematics Q77

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MathematicsSingle correct+4 / -1Question 77 of 90

If f(x) is a differentiable function in the interval (0, ∞) such that f(1) = 1 and $\displaystyle\lim_{t\to x}\frac{t^2f(x)-x^2f(t)}{t-x}=1$ , for each x > 0, then $f\left(\dfrac32\right)$ is equal to :

\dfrac{13}6

\dfrac{23}{18}

\dfrac{25}9

\dfrac{31}{18}

Answer and worked solution

Official answer: D

The limit is $2xf(x)-x^2f'(x)=1$ , i.e. $\left(\frac f{x^2}\right)'=-\frac1{x^4}$ . So $f=\frac1{3x}+cx^2$ with $c=\frac23$ from f(1) = 1, and $f\left(\frac32\right)=\frac29+\frac32=\frac{31}{18}$ .

Topics:Linear first-order equations Limits

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JEE Main 2016 · 9 Apr · Shift 1, Mathematics question 77