1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

# The function $\mathrm{f}\left(\mathrm{x}\right)=\mathrm{log}\left(1+\mathrm{x}\right)-\left(\frac{2\mathrm{x}}{\left(2+\mathrm{x}\right)}\right)$ is increasing for all values of $\mathrm{x}$, then

A

$\left(0,\infty \right)$

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B

$\left(-\infty ,0\right)$

No worries! We‘ve got your back. Try BYJU‘S free classes today!
C

$\left(-\infty ,\infty \right)$

No worries! We‘ve got your back. Try BYJU‘S free classes today!
D

None of these

No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

## The correct option is A $\left(0,\infty \right)$Explanation for correct optionStep 1. Determine the interval of the function The given function is, $\mathrm{f}\left(\mathrm{x}\right)=\mathrm{log}\left(1+\mathrm{x}\right)-\left(\frac{2\mathrm{x}}{\left(2+\mathrm{x}\right)}\right)$Differentiate with respect to $\mathrm{x}$.$⇒$ $\mathrm{f}\text{'}\left(\mathrm{x}\right)=\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{log}\left(1+\mathrm{x}\right)-\left(\frac{2}{\left(2+\mathrm{x}\right)}\right)\right)$Use sum/difference rule $\left(\mathrm{f}±\mathrm{g}\right)\text{'}=\mathrm{f}\text{'}±\mathrm{g}\text{'}$, where $\mathrm{f}$ and $\mathrm{g}$ are the function.$⇒$ $\mathrm{f}\text{'}\left(\mathrm{x}\right)=\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{ln}\left(1+\mathrm{x}\right)\right)-\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{2\mathrm{x}}{2+\mathrm{x}}\right)$Use chain rule$⇒$ $\mathrm{f}\text{'}\left(\mathrm{x}\right)=\frac{1}{1+\mathrm{x}}\frac{\mathrm{d}}{\mathrm{dx}}\left(1+\mathrm{x}\right)-\frac{\frac{1}{2\mathrm{x}}}{2+\mathrm{x}}\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{2\mathrm{x}}{2+\mathrm{x}}\right)$Derivative of constant $\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{a}\right)=0$ and common derivative $\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}\right)=1$.$⇒$ $\mathrm{f}\text{'}\left(\mathrm{x}\right)=\left(\frac{1}{1+\mathrm{x}}×1\right)-2\frac{1\left(2+\mathrm{x}\right)-1×\mathrm{x}}{{\left(2+\mathrm{x}\right)}^{2}}$ $=\frac{1}{1+\mathrm{x}}-\frac{4}{{\left(2+\mathrm{x}\right)}^{2}}$Step 2 simplify the equationWe will simplify the equation $\frac{1}{1+\mathrm{x}}-\frac{4}{{\left(2+\mathrm{x}\right)}^{2}}$.$⇒$ $\mathrm{f}\text{'}\left(\mathrm{x}\right)=\frac{{\left(\mathrm{x}+2\right)}^{2}-4\left(\mathrm{x}+1\right)}{\left(\mathrm{x}+1\right){\left(\mathrm{x}+2\right)}^{2}}$ $=\frac{\left({\mathrm{x}}^{2}+4\mathrm{x}+4\right)}{\left(\mathrm{x}+1\right){\left(\mathrm{x}+2\right)}^{2}}$ $=\frac{\left({\mathrm{x}}^{2}+4\mathrm{x}-4\mathrm{x}+4-4\right)}{\left(\mathrm{x}+1\right){\left(\mathrm{x}+2\right)}^{2}}$ $=\frac{{\mathrm{x}}^{2}}{\left(\mathrm{x}+1\right){\left(\mathrm{x}+2\right)}^{2}}$ Therefore, $\mathrm{f}\text{'}\left(\mathrm{x}\right)>0$By substituting the values,$\frac{{\mathrm{x}}^{2}}{\left(\mathrm{x}+1\right){\left(\mathrm{x}+2\right)}^{2}}>0$, where $\left(1+\mathrm{x}\right)>0$Thus, $\mathrm{f}\left(\mathrm{x}\right)$ is increasing function on $\left(0,\infty \right)$ for all $\mathrm{x}>-1$.Hence, option $\mathrm{A}$ is correct answer.

Suggest Corrections
19
Join BYJU'S Learning Program
Related Videos
Single Point Continuity
MATHEMATICS
Watch in App
Explore more
Join BYJU'S Learning Program