JEE Main 2026 (Online) 24th January Morning ShiftQuestion 1
Let $(2\alpha,\alpha)$ be the largest interval in which the function $f(t)=\frac{|t+1|}{t^2}$, $t<0$, is strictly decreasing. Then the local maximum value of the function $g(x)=2\log_e(x-2)+\alpha x^2+4x-\alpha$, $x>2$, is $\_\_\_\_\_\_$ .
Differentiation and its Application
Answer$4$
Solution
Concept. Find where $f$ decreases to get $\alpha$, then maximise $g$. Formulas used. (1) $\dfrac{d}{dt}\left(\dfrac{t+1}{t^2}\right)=-\dfrac{t+2}{t^3}$Step 1 - $t<-1$. $f=-\frac{t+1}{t^2}$, so by (1) $f'=\frac{t+2}{t^3}$. With $t^3<0$: $f'<0$ iff $t>-2$. So $f$ decreases on $(-2,-1)$ and increases on $(-\infty,-2)$. Step 2 - $-1<t<0$. $f=\frac{t+1}{t^2}$, $f'=-\frac{t+2}{t^3}>0$, so $f$ increases. Step 3 - $\alpha$. The largest decreasing interval is $(-2,-1)=(2\alpha,\alpha)$, so $\alpha=-1$. Step 4 - Maximise $g$. $g(x)=2\ln(x-2)-x^2+4x+1$: $$g'(x)=\frac{2}{x-2}-2(x-2)=0\ \Longrightarrow\ (x-2)^2=1\ \Longrightarrow\ x=3.$$ $g'$ changes from $+$ to $-$ at $x=3$, so it is a local maximum. Step 5 - Value. $g(3)=0-9+12+1=4$. Answer: 4