JEE Advanced 2015 · Paper 2 · Mathematics Q47

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MathematicsInteger+4Question 47 of 60

Let $f:\mathbb R\to\mathbb R$ be a continuous odd function, which vanishes exactly at one point and $f(1)=\dfrac12$ . Suppose that $F(x)=\displaystyle\int_{-1}^xf(t)\,dt$ for all $x\in[-1,2]$ and $G(x)=\displaystyle\int_{-1}^xt|f(f(t))|\,dt$ for all $x\in[-1,2]$ . If $\displaystyle\lim_{x\to1}\frac{F(x)}{G(x)}=\frac1{14}$ , then the value of $f\left(\dfrac12\right)$ is

Your answer

Answer and worked solution

Official answer: 7

Both vanish at 1.

$F(1)=\int_{-1}^1f=0$ , since $f$ is odd.
$t\,|f(f(t))|$ is odd ( $f\circ f$ is odd, so its absolute value is even), so $G(1)=0$ too.

L'Hôpital.
$\lim_{x\to1}\frac{F}{G}=\frac{f(1)}{1\cdot|f(f(1))|}=\frac{1/2}{|f(1/2)|}=\frac1{14}\implies|f(\tfrac12)|=7.$

Sign. $f$ is continuous, vanishes only at 0 and $f(1)>0$ , so $f>0$ on $(0,\infty)$ . Hence $f(\frac12)=7$ .

Answer: 7

Topics:L'Hospital's rule Leibniz rule

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JEE Advanced 2015 · Paper 2, Mathematics question 47