Let (2, 3) be the largest open interval in which the function is strictly increasing and (b, c) be the largest open interval, in which the function is strictly decreasing. Then 100 (a + b – c) is equal to: [2025]

Question

Let (2, 3) be the largest open interval in which the function $f (x) = 2 \log_{e} (x - 2) - x^{2} + a x + 1$ is strictly increasing and (b, c) be the largest open interval, in which the function $g (x) = {(x - 1)}^{3} {(x + 2 - a)}^{2}$ is strictly decreasing. Then 100 (a + b – c) is equal to: [2025]

Smart Achievers · Accepted Answer

(1)

Given : $f (x) = 2 \log_{e} (x - 2) - x^{2} + a x + 1$ is strictly increasing on (2, 3)

$g (x) = {(x - 1)}^{3} {(x + 2 - a)}^{2}$ is strictly decreasing on (b, c).

Using f(x),

$f' (x) = \frac{2}{x - 2} - 2 x + a$

$\Rightarrow f'' (x) = \frac{- 2}{{(x - 2)}^{2}} - 2 < 0$

SInce f(x) is strictly increasing, so $f' (x) > 0$ .

But we have given that (2, 3) is the largest open interval where f(x) is strictly increasing.

$\Rightarrow f' (3) = 0 \Rightarrow 2 - 6 + a = 0 \Rightarrow a = 4$

Taking, $g (x) = {(x - 1)}^{3} {(x + 2 - a)}^{2}$

$= {(x - 1)}^{3} {(x - 2)}^{2}$ [ $∵$ a = 4]

$\Rightarrow g' (x) = {(x - 1)}^{2} (x - 2) (2 x - 2 + 3 x - 6)$

$= {(x - 1)}^{2} (x - 2) (5 x - 8)$

Here, $g' (x) < 0$ [ $∵$ g(x) is strictly decreasing]

$\Rightarrow {(x - 1)}^{2} (x - 2) (5 x - 8) < 0$

$\Rightarrow x \in (\frac{8}{5}, 2)$

$\Rightarrow b = \frac{8}{5} and c = 2$

Finally, we get 100 (a + b – c) = $100 (4 + \frac{8}{5} - 2) = 360$ .

Solution

1 360

2 160

3 280

4 420

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