Question

Let $S(x)=\int\limits_{x^2}^{x^3} \ln t d t(x>0)$ and $H(x)=\frac{S^{\prime}(x)}{x}$. Then $H(x)$ is :

• A. continuous but not derivable in its domain
• B. derivable and continuous in its domain
• C. neither derivable nor continuous in its domain
• D. derivable but not continuous in its domain.

Answer 1

Answer: (B)

$ S^{\prime}(x)=\ln x^3 \cdot 3 x^2-\ln x^2 \cdot 2 x =9 x^2 \ln x-4 x \ln x$ $=x \ln x(9 x-4)$.
Hence $\frac{S^{\prime}(x)}{x}=\ln x(9 x-4)$.
Now it is obvious that $\frac{ S ^{\prime}( x )}{ x }$ is continuous and derivable in its domain.