SATHEE: Application of Derivatives 2 Question 11

Application of Derivatives 2 Question 11

####11. For all $x \in (0, 1)$

(2000, 1M)

(a) $e^{x} < 1 + x$

(b) $\log_{e} (1 + x) < x$

(d) $\log_{ε} x > x$

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Answer:

(c)

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 $f^{'} (x) = e^{x} - 1 > 0, \forall x \in (0, 1)$

$\Rightarrow f (x)$ 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 $g (x) = \log_{e} (1 + x) - x, 0 < x < 1$

$\begin{array}{lll} g^{'} (x) = \frac{1}{1 + x} - 1 = - \frac{x}{1 + x} < 0 for 0 < x < 1 \\ \Rightarrow & g (x) decreases & for 0 < x < 1 \\ \Rightarrow & g (x) < g (0) & for 0 < x < 1 \\ \Rightarrow & \log_{e} (1 + x) - x < 0 & for 0 < x < 1 \\ or & \log_{e} (1 + x) < x & for 0 < x < 1 \end{array}$

Therefore, option (b) is the answer.

Option $\sin x > x$

$Let \begin{aligned} h (x) & = \sin x - x \\ h^{'} (x) & = \cos x - 1 \end{aligned}$

For $x \in (0, 1), \cos x - 1 < 0$

$\Rightarrow h (x)$ is decreasing function.

$\Rightarrow$ $h (x) < h (0)$

$\Rightarrow \sin x - x < 0$

$\Rightarrow \sin x < x, which is not true.$

Option (d) $p (x) = \log x - x$

$p^{'} (x) = \frac{1}{x} - 1 > 0, \forall x \in (0, 1)$

Therefore, $p^{'} (x)$ 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.

Application of Derivatives 2 Question 11

Answer:

Solution:

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Application of Derivatives 2 Question 11

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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