The number of polynomials $p: \mathbb{R} \rightarrow \mathbb{R}$ satisfying $p(0)=0$,$p(x) > x^2$ for all $x \neq 0$,and $p^{\prime \prime}(0) = \frac{1}{2}$ is

  • A
    $0$
  • B
    $1$
  • C
    more than $1$,but finite
  • D
    infinite

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$(i)$ $f(x)$ is continuous and defined for all real numbers.
$(ii)$ $f'(-5) = 0$; $f'(2)$ is not defined and $f'(4) = 0$.
$(iii)$ $(-5, 12)$ is a point which lies on the graph of $f(x)$.
$(iv)$ $f''(2)$ is undefined,but $f''(x)$ is negative everywhere else.
$(v)$ The signs of $f'(x)$ are given below:
$f'(x)$ sign chart:
- For $x < -5$,$f'(x) > 0$
- For $-5 < x < 2$,$f'(x) < 0$
- For $2 < x < 4$,$f'(x) > 0$
- For $x > 4$,$f'(x) < 0$
From the possible graph of $y = f(x)$,we can say that:

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Let $h$ be a twice continuously differentiable positive function on an open interval $J.$ Let $g(x) = \ln(h(x))$ for each $x \in J$. Suppose $(h'(x))^2 > h''(x) h(x)$ for each $x \in J$. Then

Let $a_1, a_2, \ldots, a_{100}$ be non-zero real numbers such that $a_1+a_2+\ldots+a_{100}=0$. Then,

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