Let $(2\alpha, \alpha)$ be the largest interval in which the function $f(t)=\frac{|t+1|}{t^{2}}, t 2$, is ______

JEE Main · 2026 · 24 Jan 2026 · Shift 1 · Application of Derivatives

Let

(2 α, α)

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 lo g_{e} (x - 2) + α x^{2} + 4 x - α, x > 2

, is ______

Answer: 4

For

t < 0

f (t) = \frac{∣ t + 1∣}{t ^{2}}

. On

t < - 1

f (t) = - \frac{t + 1}{t ^{2}}

with

f^{'} (t) = \frac{t + 2}{t ^{3}}

, which is negative on

(- 2, - 1)

; on

- 1 \leq t < 0

f^{'} (t) = \frac{- t - 2}{t ^{3}} > 0

. So the largest interval of strict decrease is

(- 2, - 1)

, giving

(2 α, α) = (- 2, - 1)

and

α = - 1

. Then

g (x) = 2 lo g_{e} (x - 2) - x^{2} + 4 x + 1

for

x > 2

, with

g^{'} (x) = \frac{2}{x - 2} - 2 x + 4 = \frac{2 ( 3 - x ) ( x - 1 )}{x - 2}

. For

x > 2

g^{'}

changes from positive to negative at

x = 3

, a local maximum, and

g (3) = 2 lo g_{e} 1 - 9 + 12 + 1 = 4

. The local maximum value is

4

JEE Main · 2026 · 24 Jan 2026 · Shift 1 · Integer answer