IIT-JEE 2010 · Paper 1 · Mathematics Q34

MathematicsSingle correct+3 / -1Question 34 of 84

The value of $\displaystyle\lim_{x\to0}\frac1{x^3}\int_0^x\frac{t\ln(1+t)}{t^4+4}\,dt$ is

\dfrac1{12}

\dfrac1{24}

\dfrac1{64}

Answer and worked solution

Official answer: B

Small $t$ . $\dfrac{t\ln(1+t)}{t^4+4}=\dfrac{t^2}4+O(t^3)$ , so
$\int_0^x\frac{t\ln(1+t)}{t^4+4}dt=\frac{x^3}{12}+O(x^4).$

Limit. Dividing by $x^3$ : $\dfrac1{12}$ . (Equivalently, L'Hôpital: $\lim\dfrac{x\ln(1+x)}{3x^2(x^4+4)}=\dfrac1{12}$ .)

(B)

Topics:L'Hospital's rule Fundamental theorem of calculus

IIT-JEE 2010 · Paper 1, Mathematics question 34