JEE Main 2017 · 9 Apr · Shift 1 · Mathematics Q70

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The value of k for which the function
$f(x)=\begin{cases}\left(\dfrac45\right)^{\frac{\tan4x}{\tan5x}}, & 0<x<\dfrac\pi2\\[2mm] k+\dfrac25, & x=\dfrac\pi2\end{cases}$
is continuous at $x=\dfrac\pi2$ , is :

\dfrac{17}{20}

\dfrac25

\dfrac35

-\dfrac25

Answer and worked solution

Official answer: C

As $x\to\frac\pi2$ , $\tan4x\to0$ and $\tan5x\to\pm\infty$ , so the exponent tends to 0 and $f\to\left(\frac45\right)^0=1$ . Continuity needs $k+\frac25=1$ : $k=\frac35$ .

Topics:Continuity Limits

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JEE Main 2017 · 9 Apr · Shift 1, Mathematics question 70