JEE Main 2015 · 11 Apr · Shift 1 · Mathematics Q70

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MathematicsSingle correct+4 / -1Question 70 of 90

Let k be a non-zero real number. If
$f(x)=\begin{cases}\dfrac{(e^x-1)^2}{\sin\left(\dfrac xk\right)\log\left(1+\dfrac x4\right)},&x\ne0\\12,&x=0\end{cases}$
is a continuous function, then the value of k is :

Answer and worked solution

Official answer: C

Near 0: $(e^x-1)^2\approx x^2$ and $\sin\frac xk\log\left(1+\frac x4\right)\approx\frac{x^2}{4k}$ , so the limit is $4k=12$ and $k=3$ .

Topics:Continuity Limits

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JEE Main 2015 · 11 Apr · Shift 1, Mathematics question 70