Question

# If $f\left(x\right)=\left\{\begin{array}{ll}\frac{\mathrm{log}\left(1+ax\right)-\mathrm{log}\left(1-bx\right)}{x},& x\ne 0\\ k,& x=0\end{array}\right\$ and f (x) is continuous at x = 0, then the value of k is (a) a − b (b) a + b (c) log a + log b (d) none of these

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Solution

## $\left(b\right)a+b$ Given: $f\left(x\right)=\left\{\begin{array}{l}\frac{\mathrm{log}\left(1+ax\right)-\mathrm{log}\left(1-bx\right)}{x},x\ne 0\\ k,x=0\end{array}\right\$ If f(x) is continuous at x = 0, then $\underset{x\to 0}{\mathrm{lim}}f\left(x\right)=f\left(0\right)$ $⇒\underset{x\to 0}{\mathrm{lim}}\left(\frac{\mathrm{log}\left(1+ax\right)-\mathrm{log}\left(1-bx\right)}{x}\right)=k$ $⇒\underset{x\to 0}{\mathrm{lim}}\left(\frac{a\mathrm{log}\left(1+ax\right)}{ax}-\frac{b\mathrm{log}\left(1-bx\right)}{bx}\right)=k\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}⇒a\underset{x\to 0}{\mathrm{lim}}\left(\frac{\mathrm{log}\left(1+ax\right)}{ax}\right)-b\underset{x\to 0}{\mathrm{lim}}\left(\frac{\mathrm{log}\left(1-bx\right)}{bx}\right)=k\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}⇒a\underset{x\to 0}{\mathrm{lim}}\left(\frac{\mathrm{log}\left(1+ax\right)}{ax}\right)+b\underset{x\to 0}{\mathrm{lim}}\left(\frac{\mathrm{log}\left(1-bx\right)}{-bx}\right)=k\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}⇒a×1+b×1=k\left[\because \underset{x\to 0}{\mathrm{lim}}\frac{\mathrm{log}\left(1+x\right)}{x}=1\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}⇒k=a+b$

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