The maximum area of a right-angled triangle with hypotenuse $h$ is

In the interval $(0, 1)$,the maximum value of the function $f(x) = |x \ln x|$ is:

Let $f(x)$ be a polynomial of degree four having extreme values at $x=1$ and $x=2$. If $\mathop {\lim }\limits_{x \to 0} \left[ {1 + \frac{{f(x)}}{{{x^2}}}} \right] = 3$,then $f(2)$ is equal to:

Let $p(x)$ be a polynomial of degree $4$ having extrema at $x=1$ and $x=2$,and $\lim_{x \rightarrow 0} \left(1+\frac{p(x)}{x^2}\right) = 2$. Then the value of $p(2)$ is

The sufficient conditions for the function $f:R \to R$ to have a local maximum at $x = a$ are:

Let $f: R \rightarrow R$ be given by $f(x)=(x-1)(x-2)(x-5)$. Define $F(x)=\int_0^x f(t) dt$ for $x>0$. Which of the following options is/are correct?
$(1)$ $F$ has a local minimum at $x=1$
$(2)$ $F$ has a local maximum at $x=2$
$(3)$ $F(x) \neq 0$ for all $x \in (0,5)$
$(4)$ $F$ has two local maxima and one local minimum in $(0, \infty)$