If $f:[-5,5] \rightarrow \mathrm{R}$ is a differentiable function and if $f^{\prime}(x)$ does not vanish anywhere, then prove that $f(-5) \neq f(5).$

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It is given that $f:[-5,5] \rightarrow R$ is a differentiable function.

Since every differentiable function is a continuous function, we obtain

a) $f$ is continuous on $[-5,5].$

b) $f$ is continuous on $(-5,5).$

Therefore, by the Mean Value Theorem, there exists $c \in(-5,5)$ such that

$f^{\prime}(c)=\frac{f(5)-f(-5)}{5-(-5)}$

$\Rightarrow 10 f^{\prime}(c)=f(5)-f(-5)$

It is also given that $f^{\prime}(x)$ does not vanish anywhere.

$\therefore f^{\prime}(c) \neq 0$

$\Rightarrow 10 f^{\prime}(c) \neq 0$

$\Rightarrow f(5)-f(-5) \neq 0$

$\Rightarrow f(5) \neq f(-5)$

Hence, proved.

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and

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($1$) 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 $\psi_1(x)=1+\alpha x$

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($2$) 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 $x \in\left(0, \frac{1}{2}\right)$

$(D)$ $g(x) \leq \frac{2}{3} x^3-\frac{2}{5} x^5+\frac{1}{7} x^7$, for all $x \in\left(0, \frac{1}{2}\right)$

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