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JEE Main 202624 JanuaryMorning Shift
Exam
JEE Main
Year
2026
Date
24 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Application of Derivatives

JEE Main 2026 · 24 January · Morning Shift · Mathematics · Application of Derivatives

Let (2 α, α) be the largest interval in which the function f(t)=...

Let (2α,α)(2 \alpha, \alpha) be the largest interval in which the function f(t)=∣t+1∣t2,t<0f(t)=\frac{|t+1|}{t^{2}}, t<0, is strictly decreasing. Then the local maximum value of the function g(x)=2log⁡e(x−2)+αx2+4x−α,x>2g(x)=2 \log _{\mathrm{e}}(x-2)+\alpha x^{2}+4 x-\alpha, x>2, is ____\_\_\_\_

Correct answer

4

Step-by-step solution

For t<0t < 0, analyze f(t)=∣t+1∣t2f(t) = \frac{|t+1|}{t^2} in two regions.
For −1<t<0-1 < t < 0:
f(t)=t+1t2f(t) = \frac{t+1}{t^2}, so f′(t)=−(t+2)t3>0f'(t) = \frac{-(t+2)}{t^3} > 0 (increasing).
For t<−1t < -1:
f(t)=−t−1t2f(t) = \frac{-t-1}{t^2}, so f′(t)=t+2t3f'(t) = \frac{t+2}{t^3}.
For −2<t<−1-2 < t < -1:
f′(t)<0f'(t) < 0 (decreasing).
For t<−2t < -2:
f′(t)>0f'(t) > 0 (increasing).
Largest decreasing interval is (−2,−1)=(2α,α)(-2, -1) = (2\alpha, \alpha), so α=−1\alpha = -1.
For g(x)=2ln⁡(x−2)−x2+4x+1g(x) = 2\ln(x-2) - x^2 + 4x + 1:
g′(x)=2x−2−2x+4=0g'(x) = \frac{2}{x-2} - 2x + 4 = 0 gives (x−2)2=1(x-2)^2 = 1, so x=3x = 3.
Since g′′(x)=−2(x−2)2−2<0g''(x) = -\frac{2}{(x-2)^2} - 2 < 0, x=3x=3 is maximum.
g(3)=2ln⁡(1)−9+12+1=4g(3) = 2\ln(1) - 9 + 12 + 1 = 4
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