Question

Let $f\left(a\right)=g\left(a\right)=k$ and their $n^{th}$ derivatives $f^n\left(a\right)$, $g^n\left(a\right)$ exist and are not equal for some $n$. Further if $\underset{x\to a}{lim}\frac{f\left(a\right)g\left(x\right)-f\left(a\right)-g\left(a\right)f\left(x\right)+g\left(a\right)}{g\left(x\right)-f\left(x\right)}=4$, then the value of $k$ is:

• A. 4
• B. 2
• C. 1
• D. 0

Answer 1

The correct option is A

4

$\underset{x\to a}{lim}\frac{f\left(a\right)g\left(x\right)-f\left(a\right)-g\left(a\right)f\left(x\right)+g\left(a\right)}{g\left(x\right)-f\left(x\right)}=4\text{Applying L' Hospital rule}\underset{x\to a}{lim}\frac{f\left(a\right).g^{\prime }\left(x\right)-g\left(a\right).f^{\prime }\left(x\right)}{g^{\prime }\left(x\right)-f^{\prime }\left(x\right)}=4\Rightarrow \underset{x\to a}{lim}\frac{k{g}^{\prime }\left(x\right)-k{f}^{\prime }\left(x\right)}{g^{\prime }\left(x\right)-f^{\prime }\left(x\right)}=4\Rightarrow \underset{x\to a}{lim}\frac{k\left[g^{\prime }\right(x)-f^{\prime }(x\left)\right]}{g^{\prime }\left(x\right)-f^{\prime }\left(x\right)}=4\Rightarrow k=4$