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If $f(x)=\left\{\begin{array}{ll}k, & \text { for } x=1 \\ \frac{(9 x-1)(\sqrt{x}-1)}{3 x^2+2 x-5}, & \text { for } x \neq 1\end{array}\right.$ is continuous on $[0, \infty)$, then $k=$
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Verified Answer
The correct answer is:
$\frac{1}{2}$
We have,
$f(x)$ is continuous at $x=1$
$\therefore \quad k=\lim _{x \rightarrow 1} \frac{(9 x-1)(\sqrt{x}-1)}{3 x^2+2 x-5}$
$\begin{array}{ll}\Rightarrow & k=\lim _{x \rightarrow 1} \frac{9 x^{3 / 2}-9 x-\sqrt{x}+1}{3 x^2+2 x-5} \\ \Rightarrow & k=\lim _{x \rightarrow 1} \frac{9\left(\frac{3}{2} x^{1 / 2}\right)-9-\frac{1}{2 \sqrt{x}}}{6 x+2} \\ \Rightarrow & k=\frac{\frac{27}{2}-9-\frac{1}{2}}{6+2}=\frac{13-9}{8}=\frac{4}{8}=\frac{1}{2}\end{array}$
$f(x)$ is continuous at $x=1$
$\therefore \quad k=\lim _{x \rightarrow 1} \frac{(9 x-1)(\sqrt{x}-1)}{3 x^2+2 x-5}$
$\begin{array}{ll}\Rightarrow & k=\lim _{x \rightarrow 1} \frac{9 x^{3 / 2}-9 x-\sqrt{x}+1}{3 x^2+2 x-5} \\ \Rightarrow & k=\lim _{x \rightarrow 1} \frac{9\left(\frac{3}{2} x^{1 / 2}\right)-9-\frac{1}{2 \sqrt{x}}}{6 x+2} \\ \Rightarrow & k=\frac{\frac{27}{2}-9-\frac{1}{2}}{6+2}=\frac{13-9}{8}=\frac{4}{8}=\frac{1}{2}\end{array}$
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