Let S(x)=∫ limitsx2x3 ln t d t(x0) and H(x)=(S prime(x)/x). Then H(x) is :

Question

Let S(x)=∫ limitsx2x3 ln t d t(x>0) and H(x)=(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.

Tardigrade Admin · Accepted Answer

S prime(x)= ln x3 ⋅ 3 x2- ln x2 ⋅ 2 x =9 x2 ln x-4 x ln x =x ln x(9 x-4). Hence (S prime(x)/x)= ln x(9 x-4). Now it is obvious that ( S prime( x )/ x ) is continuous and derivable in its domain.

Q. Let $S (x) = x^{2} \int x^{3} ln t d t (x > 0)$ and $H (x) = \frac{S ^{'} ( x )}{x}$ . Then $H (x)$ is :

continuous but not derivable in its domain

derivable and continuous in its domain

neither derivable nor continuous in its domain

derivable but not continuous in its domain.

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

Q. Let S(x)=x2∫x3​lntdt(x>0) and H(x)=xS′(x)​. Then H(x) is :