Let be the largest interval in which the function is strictly decreasing. Then the local maximum value of the function is ________ . [2026]

Question

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 $g (x) = 2 \log_{e} (x - 2) + α x^{2} + 4 x - α, x > 2,$ is ________ . [2026]

Smart Achievers · Accepted Answer

(4)

$Drawing graph of f (t) for t < 0$

$g (x) = \log_{e} (x - 2) - x^{2} + 4 x + 1, x > 2$

$g' (x) = \frac{2}{x - 2} - 2 (x - 2), x > 2$

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

$as x > 2$

$maxima occur at x = 3$

$g (3) = 2 \log_{e} 1 - 9 + 12 + 1 = 4$

Solution

Q.
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 $g (x) = 2 \log_{e} (x - 2) + α x^{2} + 4 x - α, x > 2,$ is ________ . [2026]

Ans.
(4)

$Drawing graph of f (t) for t < 0$

$g (x) = \log_{e} (x - 2) - x^{2} + 4 x + 1, x > 2$

$g' (x) = \frac{2}{x - 2} - 2 (x - 2), x > 2$

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

$as x > 2$

$maxima occur at x = 3$

$g (3) = 2 \log_{e} 1 - 9 + 12 + 1 = 4$

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Solution

Q. Let (2α,α) be the largest interval in which the function f(t)=|t+1|t2, t<0, is strictly decreasing. Then the local maximum value of the function g(x)=2loge(x-2)+αx2+4x-α, x>2, is ________ . [2026]

Ans. (4) Drawing graph of f(t) for t<0 g(x)=loge(x-2)-x2+4x+1, x>2 g'(x)=2x-2-2(x-2), x>2 g'(x)=1-(x-2)2x-2=-(x-3)(x-1)x-2 as x>2 maxima occur at x=3 g(3)=2loge1-9+12+1=4

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Q.
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 $g (x) = 2 \log_{e} (x - 2) + α x^{2} + 4 x - α, x > 2,$ is ________ . [2026]