Let (2alpha, alpha)$(2\alpha, \alpha)$ be the largest interval in which the function f(t) = frac|t + 1|t^2$f(t) = \frac{|t + 1|}{t^2}$, t < 0$t < 0$, 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$g(x) = 2 \log_e(x - 2) + \alpha x^2 + 4x - \alpha, x > 2$, is

### Related Formula g'(x) = f'(h(x)) cdot h'(x)$$g'(x) = f'(h(x)) \cdot h'(x)$$ For monotonicity, g'(x) > 0 implies$g'(x) > 0 \implies$ increasing, and g'(x) < 0 implies$g'(x) < 0 \implies$ decreasing. ### Core Logic Given g(x) = f((tan x - 1)^2 + a - 1)$g(x) = f((\tan x - 1)^2 + a - 1)$. Differentiating w.r.t. x$x$: g'(x) = f'((tan x - 1)^2 + a - 1) cdot 2(tan x - 1)sec^2 x$$g'(x) = f'((\tan x - 1)^2 + a - 1) \cdot 2(\tan x - 1)\sec^2 x$$ ### Step 1: Analyze the Sign of the Derivative We know f''(x) > 0 implies f'(x)$f''(x) > 0 \implies f'(x)$ is strictly increasing. Since f'(a-1) = 0$f'(a-1) = 0$, for any input X > a-1$X > a-1$, f'(X) > 0$f'(X) > 0$. Here, the input to f'$f'$ is (tan x - 1)^2 + a - 1$(\tan x - 1)^2 + a - 1$. Since (tan x - 1)^2 geq 0$(\tan x - 1)^2 \geq 0$, it is strictly positive for x neq fracpi4$x \neq \frac{\pi}{4}$ in the given interval. Thus, (tan x - 1)^2 + a - 1 geq a - 1 implies f'((tan x - 1)^2 + a - 1) > 0$(\tan x - 1)^2 + a - 1 \geq a - 1 \implies f'((\tan x - 1)^2 + a - 1) > 0$ for all valid x$x$. ### Step 2: Determine Intervals of Monotonicity Now, the sign of g'(x)$g'(x)$ depends solely on the term (tan x - 1)$(\tan x - 1)$ because 2sec^2 x > 0$2\sec^2 x > 0$. Case 1: x in left(0, fracpi4right)$x \in \left(0, \frac{\pi}{4}\right)$ Here tan x < 1 implies tan x - 1 < 0$\tan x < 1 \implies \tan x - 1 < 0$. So, g'(x) < 0 implies g(x)$g'(x) < 0 \implies g(x)$ is decreasing. Case 2: x in left(fracpi4, fracpi2right)$x \in \left(\frac{\pi}{4}, \frac{\pi}{2}\right)$ Here tan x > 1 implies tan x - 1 > 0$\tan x > 1 \implies \tan x - 1 > 0$. So, g'(x) > 0 implies g(x)$g'(x) > 0 \implies g(x)$ is increasing. Therefore, neither statement (I) nor (II) is true. ### Pattern Recognition When a function wraps a quadratic, f((u-k)^2 + c)$f((u-k)^2 + c)$, the critical points match the inner function's extrema. Evaluate the sign directly from the inner derivative u'$u'$ and (u-k)$(u-k)$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Application of Derivatives

### Related Formula f'(x) = 0 text gives the critical points for extreme values.$$f'(x) = 0 \text{ gives the critical points for extreme values.}$$ ### Core Logic Given f(x) = x^2025 - x^2000$f(x) = x^{2025} - x^{2000}$ on x in [0,1]$x \in [0,1]$. Differentiate to find critical points: f'(x) = 2025 x^2024 - 2000 x^1999 = 0$$f'(x) = 2025 x^{2024} - 2000 x^{1999} = 0$$ x^1999(2025x^25 - 2000) = 0$$x^{1999}(2025x^{25} - 2000) = 0$$ ### Step 1: Identifying the Minima The critical points are x = 0$x = 0$ and x = left(frac20002025right)^frac125 = left(frac8081right)^frac125$x = \left(\frac{2000}{2025}\right)^{\frac{1}{25}} = \left(\frac{80}{81}\right)^{\frac{1}{25}}$. Let this root be alpha$\alpha$. Evaluating the function at endpoints and the critical point: - f(0) = 0$f(0) = 0$ - f(1) = 1 - 1 = 0$f(1) = 1 - 1 = 0$ - f(alpha) = alpha^2025 - alpha^2000 = alpha^2000(alpha^25 - 1)$f(\alpha) = \alpha^{2025} - \alpha^{2000} = \alpha^{2000}(\alpha^{25} - 1)$ Substitute alpha^25 = frac8081$\alpha^{25} = \frac{80}{81}$: f(alpha) = left(left(frac8081right)^frac125right)^2000 cdot left(frac8081 - 1right)$$f(\alpha) = \left(\left(\frac{80}{81}\right)^{\frac{1}{25}}\right)^{2000} \cdot \left(\frac{80}{81} - 1\right)$$ f(alpha) = left(frac8081right)^80 cdot left(-frac181right)$$f(\alpha) = \left(\frac{80}{81}\right)^{80} \cdot \left(-\frac{1}{81}\right)$$ ### Step 2: Structuring into the Given Format f(alpha) = frac80^8081^80 cdot (-81)^-1 = 80^80 cdot (-81)^-81$$f(\alpha) = \frac{80^{80}}{81^{80}} \cdot (-81)^{-1} = 80^{80} \cdot (-81)^{-81}$$ Comparing this with (80)^80(n)^-81$(80)^{80}(n)^{-81}$, we immediately get n = -81$n = -81$. ### Pattern Recognition When evaluating exponents in the form x^A - x^B$x^A - x^B$, the stationary point occurs at x = (B/A)^1/(A-B)$x = (B/A)^{1/(A-B)}$. Plugging this back into the original function seamlessly factors out (B/A)^B cdot (B/A - 1)$(B/A)^B \cdot (B/A - 1)$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Applications of Derivatives

### Related Formula For local extrema of a differentiable function, set the first derivative to zero: f'(x) = 0$f'(x) = 0$ ### Core Logic Differentiate f(x)$f(x)$ to obtain its critical points p$p$ and q$q$ in terms of a$a$, use the constraint p^2 = q$p^2 = q$ to fix a$a$, and evaluate f(3)$f(3)$. ### Step 1: Differentiate and Find Critical Points f'(x) = 6x^2 - 18ax + 12a^2 $ f'(x) = 6x^2 - 18ax + 12a^2 $ Set $ Set $f'(x) = 0: $: $6(x^2 - 3ax + 2a^2) = 0 implies 6(x-a)(x-2a) = 0$6(x^2 - 3ax + 2a^2) = 0 \implies 6(x-a)(x-2a) = 0$ The critical points are $ The critical points are $x = a and $ and $x = 2a. Since $. Since $a > 0, checking the sign change of $, checking the sign change of $f'(x) shows that the local maximum occurs at the smaller root ($ shows that the local maximum occurs at the smaller root ($p = a) and the local minimum at the larger root ($) and the local minimum at the larger root ($q = 2a). ### Step 2: Apply Root Constraint Given $). ### Step 2: Apply Root Constraint Given $p^2 = q: $: $a^2 = 2a implies a(a-2) = 0$a^2 = 2a \implies a(a-2) = 0$ Since $ Since $a > 0, we get $, we get $a = 2. ### Step 3: Evaluate function at x = 3 Substitute $. ### Step 3: Evaluate function at x = 3 Substitute $a = 2 back into the definition of $ back into the definition of $f(x): $: $f(x) = 2x^3 - 18x^2 + 48x + 1$f(x) = 2x^3 - 18x^2 + 48x + 1$ Now compute $ Now compute $f(3): $: $f(3) = 2(3)^3 - 18(3)^2 + 48(3) + 1 = 54 - 162 + 144 + 1 = 37$f(3) = 2(3)^3 - 18(3)^2 + 48(3) + 1 = 54 - 162 + 144 + 1 = 37$ ### Pattern Recognition For cubic equations with two distinct real critical roots, the smaller root is always the local maximum if the leading coefficient is positive ($ ### Pattern Recognition For cubic equations with two distinct real critical roots, the smaller root is always the local maximum if the leading coefficient is positive ($2 > 0). This ensures $). This ensures $p=a and $ and $q=2a$ immediately without relying on secondary derivative checks. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Application of Derivatives

### Related Formula Critical points occur where f'(x) = 0$f'(x) = 0$. For absolute maximum and minimum on an interval [c, d]$[c, d]$, evaluate function values at the boundaries and at any local critical points falling inside the domain. ### Core Logic Given f(x) = x^3 + ax^2 + bln|x| + 1$f(x) = x^3 + ax^2 + b\ln|x| + 1$ Differentiating with respect to x$x$: f'(x) = 3x^2 + 2ax + fracbx$$f'(x) = 3x^2 + 2ax + \frac{b}{x}$$ Since x = -1$x = -1$ and x = 2$x = 2$ are critical points: f'(-1) = 3(-1)^2 + 2a(-1) + fracb-1 = 3 - 2a - b = 0 implies 2a + b = 3$$f'(-1) = 3(-1)^2 + 2a(-1) + \frac{b}{-1} = 3 - 2a - b = 0 \implies 2a + b = 3$$ f'(2) = 3(2)^2 + 2a(2) + fracb2 = 12 + 4a + fracb2 = 0 implies 8a + b = -24$$f'(2) = 3(2)^2 + 2a(2) + \frac{b}{2} = 12 + 4a + \frac{b}{2} = 0 \implies 8a + b = -24$$ ### Step 1: Solve for Coefficients Subtracting the first simplified derivative equation from the second: (8a + b) - (2a + b) = -24 - 3$$(8a + b) - (2a + b) = -24 - 3$$ 6a = -27 implies a = -frac92$$6a = -27 \implies a = -\frac{9}{2}$$ Substituting a$a$ back to get b$b$: 2left(-frac92right) + b = 3 implies -9 + b = 3 implies b = 12$$2\left(-\frac{9}{2}\right) + b = 3 \implies -9 + b = 3 \implies b = 12$$ Thus, the function is: f(x) = x^3 - frac92x^2 + 12ln|x| + 1$$f(x) = x^3 - \frac{9}{2}x^2 + 12\ln|x| + 1$$ ### Step 2: Check Critical Points in Target Interval The given interval is [-2, -1/2]$[-2, -1/2]$. Inside this interval, the relevant critical point is x = -1$x = -1$ (since x = 2$x = 2$ lies outside). Evaluate the function values at x = -2, -1, -1/2$x = -2, -1, -1/2$: 1. At x = -1$x = -1$: f(-1) = (-1)^3 - frac92(-1)^2 + 12ln|-1| + 1 = -1 - 4.5 + 0 + 1 = -4.5$$f(-1) = (-1)^3 - \frac{9}{2}(-1)^2 + 12\ln|-1| + 1 = -1 - 4.5 + 0 + 1 = -4.5$$ 2. At x = -2$x = -2$: f(-2) = (-2)^3 - frac92(-2)^2 + 12ln|-2| + 1 = -8 - 18 + 12(0.7) + 1 = -25 + 8.4 = -16.6$$f(-2) = (-2)^3 - \frac{9}{2}(-2)^2 + 12\ln|-2| + 1 = -8 - 18 + 12(0.7) + 1 = -25 + 8.4 = -16.6$$ 3. At x = -1/2$x = -1/2$: f(-1/2) = left(-frac12right)^3 - frac92left(-frac12right)^2 + 12lnleft|-frac12right| + 1 = -0.125 - 1.125 - 12(0.7) + 1 = -1.25 - 8.4 + 1 = -8.65$$f(-1/2) = \left(-\frac{1}{2}\right)^3 - \frac{9}{2}\left(-\frac{1}{2}\right)^2 + 12\ln\left|-\frac{1}{2}\right| + 1 = -0.125 - 1.125 - 12(0.7) + 1 = -1.25 - 8.4 + 1 = -8.65$$ ### Step 3: Calculate Absolute Sum Comparing the calculated values: M = textAbsolute Maximum = -4.5 quad (textat x = -1)$$M = \text{Absolute Maximum} = -4.5 \quad (\text{at } x = -1)$$ m = textAbsolute Minimum = -16.6 quad (textat x = -2)$$m = \text{Absolute Minimum} = -16.6 \quad (\text{at } x = -2)$$ Therefore: |M + m| = |-4.5 + (-16.6)| = |-21.1| = 21.1$$|M + m| = |-4.5 + (-16.6)| = |-21.1| = 21.1$$ ### Pattern Recognition Always verify whether the critical points lie inside the requested boundary interval before blindly testing all values. Here, x=2$x=2$ was an irrelevant trap for the interval valuation phase. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Application of Derivatives