Question

If $f\left(x\right)=\{\begin{array}{c}\frac{\log (1+ax)-\log (1-bx)}{x},x\ne 0\\ k,x=0\end{array}$, is contiuous at $x=0$, then $k$ is equal to-

• A. $2a+b$
• B. $2a-b$
• C. $b-2a$
• D. $a+b$

Answer 1

Answer: (D)

$a+b$

If$f\left(x\right)=\{\begin{array}{c}\frac{\log (1+ax)-\log (1-bx)}{x},x\ne 0\\ k,x=0\end{array}$,is continuous at $x=0$

We know that $log(1+x)=x-\frac{x^2}{2}$+Higher order terms $log(1+ax)=ax+$Terms containing $\left(x^2\right)$

$log(1-bx)=-bx+$Higher order

$log(1+ax)-log(1-bx)=(a+b)x+$ terms containing $\left(x^2\right)$ or higher

$f\left(x\right)=\frac{(a+b)x+x^2\left(Q\right(x\left)\right)}{x}$ where $Q\left(x\right)$ is any function of $x$

$\underset{x⟶0}{lim}f\left(x\right)=a+b$

As $f\left(x\right)$ is continuous so $a+b=k$