Differentiation and its Application: Let f:(-1,infty)tomathbfR be defined by f(0)=1 and… | JEE Main 2020

JEE Main 2020 (Online) 2nd September Evening ShiftQuestion 1

Let $f:(-1,\infty)\to\mathbf{R}$ be defined by $f(0)=1$ and $f(x)=\frac1x\log_e(1+x)$, $x\ne0$. Then the function $f$ :

Differentiation and its Application

decreases in $(-1,0)$ and increases in $(0,\infty)$.increases in $(-1,0)$ and decreases in $(0,\infty)$.increases in $(-1,\infty)$.decreases in $(-1,\infty)$.✓ Correct answer

Solution

Concept. The numerator $h(x)=\frac{x}{1+x}-\ln(1+x)$ of $f'$ has its maximum $0$ at $x=0$. Formulas used. (1) $f'(x)=\dfrac{h(x)}{x^2}$, $\ h'(x)=\dfrac{1}{(1+x)^2}-\dfrac{1}{1+x}=-\dfrac{x}{(1+x)^2}$Step 1 - $h\le0$. By (1), $h$ increases on $(-1,0)$ and decreases on $(0,\infty)$, with $h(0)=0$. So $h<0$ for $x\ne0$. Step 2 - $f'$. $f'<0$ for $x\ne0$, and $f$ is continuous at $0$ ($\lim\frac{\ln(1+x)}{x}=1=f(0)$). Step 3 - Conclude. $f$ decreases on $(-1,\infty)$. Answer: (D)

Differentiation and its Application - JEE Mathematics question (JEE Main 2020 (Online) 2nd September Evening Shift)

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