Question

Let $f\left(a\right)=g\left(a\right)=k$ and their $n^{th}$ order derivatives $f^n\left(a\right),g^n$ exist and are not equal for some n $\in$ 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. $0$
• B. $4$
• C. $2$
• D. $1$

Answer 1

Answer: (B)

$4$

We have,
$\underset{x\to a}{lim}\frac{f\left(a\right)g\left(x\right)-f\left(a\right)-g\left(a\right)f\left(x\right)+g\left(a\right)}{g\left(x\right)-f\left(x\right)}=4$

$\Rightarrow \underset{x\to a}{lim}\frac{f\left(a\right)g^{\prime }\left(x\right)-0-g\left(a\right)f^{\prime }\left(x\right)+0}{g^{\prime }\left(x\right)-f^{\prime }\left(x\right)}=4$

$\Rightarrow \frac{f\left(a\right)g^{\prime }\left(a\right)-g\left(a\right)f^{\prime }\left(a\right)}{g^{\prime }\left(a\right)-f^{\prime }\left(a\right)}=4$ [Using L' Hospital's Rule]

$\Rightarrow \frac{k\left\{g^{\prime }\right(a)-f^{\prime }(a\left)\right\}}{\left\{g^{\prime }\right(a)-f^{\prime }(a\left)\right\}}=4$ $[\because f(a)=g(a)=k]$

$\Rightarrow k=4$