Application of Derivatives 2 Question 11
11. For all $x \in(0,1)$
(2000, 1M)
(a) $e^{x}<1+x$
(b) $\log _e(1+x)<x$
(c) $\sin x>x$
(d) $\log _{\varepsilon} x>x$
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Solution:
- PLAN Inequation based upon uncompatible function. This type of inequation can be solved by calculus only.
Option (a) Let $f(x)=e^{x}-1-x$.
then $\quad f^{\prime}(x)=e^{x}-1>0, \forall x \in(0,1)$
$\Rightarrow f(x)$ increase in $(0,1)$
$\Rightarrow f(x)>f(0)$ for $0<x<1$
$\Rightarrow e^{x}-1-x>0$ or $e^{x}>1+x$ for $0<x<1$ Option (b) Let $g(x)=\log _e(1+x)-x, 0<x<1$
$$ \begin{array}{lll} & g^{\prime}(x)=\frac{1}{1+x}-1=-\frac{x}{1+x}<0 \text { for } 0<x<1 \\ \Rightarrow & g(x) \text { decreases } & \text { for } 0<x<1 \\ \Rightarrow & g(x)<g(0) & \text { for } 0<x<1 \\ \Rightarrow & \log _e(1+x)-x<0 & \text { for } 0<x<1 \\ \text { or } & \log _e(1+x)<x & \text { for } 0<x<1 \end{array} $$
Therefore, option (b) is the answer.
Option $\sin x>x$
$$ \text { Let } \begin{aligned} h(x) & =\sin x-x \\ h^{\prime}(x) & =\cos x-1 \end{aligned} $$
For $x \in(0,1), \cos x-1<0$
$\Rightarrow h(x)$ is decreasing function.
$\Rightarrow$ $h(x)<h(0)$
$\Rightarrow \quad \sin x-x<0$
$$ \Rightarrow \quad \sin x<x, \text { which is not true. } $$
Option (d) $\quad p(x)=\log x-x$
$$ p^{\prime}(x)=\frac{1}{x}-1>0, \forall x \in(0,1) $$
Therefore, $p^{\prime}(x)$ is an increasing function.
$$ \begin{array}{lr} \Rightarrow & p(0)<p(x)<p(1) \\ \Rightarrow & -\infty<\log x-x<-1 \\ \Rightarrow & \log x-x<0 \\ \Rightarrow & \log x<x \end{array} $$
Therefore, option (d) is not the answer.
