If $2a + 3b + 6c = 0$ and $a, b, c \in \mathbb{R}$,then the equation $ax^2 + bx + c = 0$ has at least one root between $0$ and $1$.

  • A
    $ax + bx + c = 0$
  • B
    $ax^2 - bx + c = 0$
  • C
    $ax^2 + bx + c = 0$
  • D
    $ax^2 - bx - c = 0$

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Let $\psi_1:[0, \infty) \rightarrow \mathbb{R}$,$\psi_2:[0, \infty) \rightarrow \mathbb{R}$,$f:[0, \infty) \rightarrow \mathbb{R}$,and $g:[0, \infty) \rightarrow \mathbb{R}$ be functions such that $f(0)=g(0)=0$,$\psi_1(x)=e^{-x}+x$ for $x \geq 0$,$\psi_2(x)=x^2-2x-2e^{-x}+2$ for $x \geq 0$,$f(x)=\int_{-x}^{x}(|t|-t^2)e^{-t^2} dt$ for $x>0$,and $g(x)=\int_0^{x^2} \sqrt{t} e^{-t} dt$ for $x>0$.
$(1)$ Which of the following statements is $TRUE$?
$(A)$ $f(\sqrt{\ln 3})+g(\sqrt{\ln 3})=\frac{1}{3}$
$(B)$ For every $x>1$,there exists an $\alpha \in(1, x)$ such that $\psi_1(x)=1+\alpha x$
$(C)$ For every $x>0$,there exists a $\beta \in(0, x)$ such that $\psi_2(x)=2x(\psi_1(\beta)-1)$
$(D)$ $f$ is an increasing function on the interval $[0, \frac{3}{2}]$
$(2)$ Which of the following statements is $TRUE$?
$(A)$ $\psi_1(x) \leq 1$,for all $x>0$
$(B)$ $\psi_2(x) \leq 0$,for all $x>0$
$(C)$ $f(x) \geq 1-e^{-x^2}-\frac{2}{3}x^3+\frac{2}{5}x^5$,for all $x \in(0, \frac{1}{2})$
$(D)$ $g(x) \leq \frac{2}{3}x^3-\frac{2}{5}x^5+\frac{1}{7}x^7$,for all $x \in(0, \frac{1}{2})$

Given that $f(x)$ is continuously differentiable on $a \le x \le b$ where $a < b, f(a) < 0$ and $f(b) > 0$,which of the following are always true?
$(i)$ $f(x)$ is bounded on $a \le x \le b$.
$(ii)$ The equation $f(x) = 0$ has at least one solution in $a < x < b$.
$(iii)$ The maximum and minimum values of $f(x)$ on $a \le x \le b$ occur at points where $f'(c) = 0$.
$(iv)$ There is at least one point $c$ with $a < c < b$ where $f'(c) > 0$.
$(v)$ There is at least one point $d$ with $a < d < b$ where $f'(d) < 0$.

Rolle's theorem is not applicable to the function $f(x) = |x|$ defined on $[-1, 1]$ because

If $f(x)=|x-2|, x \in[0,4]$ then the Rolle's theorem cannot be applied to the function because

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a twice continuously differentiable function such that $f(0)=f(1)=f^{\prime}(0)=0$. Then:

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