Question:
Let a function $f:[0,5] \rightarrow \mathbf{R}$ be continuous, $f(1)=3$ and $\mathrm{F}$ be defined as:
$\mathrm{F}(x)=\int_{1}^{x} t^{2} g(t) d t$, where $g(t)=\int_{1}^{x} f(u) d u$
Then for the function $\mathrm{F}$, the point $x=1$ is :
Correct Option: 1
Solution:
$F(x)=\int_{1}^{x} t^{2} g(t) d t$
Differentiate by using Leibnitz's rule, we get
$F^{\prime}(x)=x^{2} g(x)=x^{2} \int_{1}^{x} f(u) d u$ .........(1)
At $x=1$,
$F^{\prime}(1)=1 \int_{1}^{1} f(u) d u=0$
Now, differentiate eqn (i)
$F^{\prime \prime}(x)=x^{2} f(x)-2 x \int_{1}^{x} f(u) d u$
At $x=1$
$F^{\prime \prime}(1)=1 . f(1)-2 \times 1 . \int_{1}^{1} f(u) d u$
$=f(1)-2 \times 0=f(1)$
Then, for $F^{\prime}(1)=0, F^{\prime \prime}(1)=3>0$
Hence, $x=1$ is a point of local minima.