Let a function f:[0,5]

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 :


  1. (1) a point of local minima.

  2. (2) not a critical point.

  3. (3) a point of local maxima.

  4. (4) a point of inflection.

Correct Option: 1


$F(x)=\int_{1}^{x} t^{2} g(t) d t$

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.



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