Question

Let $f\left(a\right)=g\left(a\right)=k$ and their $nth$ derivatives $f^{\left(n\right)}\left(a\right),g^{\left(n\right)}\left(a\right)$ 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. $2$
• B. $1$
• C. $0$
• D. $4$

Answer 1

Answer: (D)

$4$

As it is given, $f\left(a\right)=g\left(a\right)=k$
Thus, on applying the limit, we observe that this is of the $\frac{0}{0}$ form
So we apply L-Hospital's Rule and differentiate the numerator and the denominator individually.
$\underset{x\to a}{lim}\frac{f\left(a\right)g^{\prime }\left(x\right)-g\left(a\right)f^{\prime }\left(x\right)}{g^{\prime }\left(x\right)-f^{\prime }\left(x\right)}$

$\underset{x\to a}{lim}\frac{f\left(a\right)g^{\prime }\left(x\right)-g\left(a\right)f^{\prime }\left(x\right)}{g^{\prime }\left(x\right)-f^{\prime }\left(x\right)}$

$\frac{k\left(g^{\prime }\right(x)-f^{\prime }(x\left)\right)}{g^{\prime }\left(x\right)-f^{\prime }\left(x\right)}$

$=k\times 1=4$
Or $k=4$