### 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

### Related Formula f'(x) > 0 implies textIncreasing$f'(x) > 0 \implies \text{Increasing}$ f'(x) < 0 implies textDecreasing$f'(x) < 0 \implies \text{Decreasing}$ ### Core Logic Differentiate the rational function and determine the critical intervals by assessing where the derivative flips signs around critical points and domain boundaries. ### Step 1: Derivative Assessment Given f(x) = fracx3 + frac3x + 3$f(x) = \frac{x}{3} + \frac{3}{x} + 3$ f'(x) = frac13 - frac3x^2 = fracx^2 - 93x^2$f'(x) = \frac{1}{3} - \frac{3}{x^2} = \frac{x^2 - 9}{3x^2}$ Critical points occur at x = pm 3$x = \pm 3$, and a domain discontinuity sits at x = 0$x = 0$. ### Step 2: Sign Scheme Mapping Analyzing interval signs: * Strictly Increasing (f'(x) > 0$f'(x) > 0$): x in (-infty, -3) cup (3, infty)$x \in (-\infty, -3) \cup (3, \infty)$ * Strictly Decreasing (f'(x) < 0$f'(x) < 0$): x in (-3, 0) cup (0, 3)$x \in (-3, 0) \cup (0, 3)$ ### Step 3: Interval Summation Comparing bounds with assigned symbols: alpha_1 = -3, quad alpha_2 = 3, quad alpha_3 = -3, quad alpha_4 = 0, quad alpha_5 = 3$\alpha_1 = -3, \quad \alpha_2 = 3, \quad \alpha_3 = -3, \quad \alpha_4 = 0, \quad \alpha_5 = 3$ sum_i=1^5 alpha_i^2 = (-3)^2 + (3)^2 + (-3)^2 + (0)^2 + (3)^2 = 9 + 9 + 9 + 0 + 9 = 36$\sum_{i=1}^{5} \alpha_i^2 = (-3)^2 + (3)^2 + (-3)^2 + (0)^2 + (3)^2 = 9 + 9 + 9 + 0 + 9 = 36$ ### Pattern Recognition Functions of the form x + frackx$x + \frac{k}{x}$ always present localized extrema turning symmetric zones at pmsqrtk$\pm\sqrt{k}$. Always include the asymptotes (x=0$x=0$) when stating precise disjoint decreasing intervals. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Application of Derivatives

### Related Formula Leibniz Rule for differentiation under integral sign: fracddx left( int_0^x g(t) \, dt right) = g(x)$\frac{d}{dx} \left( \int_{0}^{x} g(t) \, dt \right) = g(x)$ ### Core Logic Find critical points inside the interval by finding f'(x) = 0$f'(x) = 0$: f'(x) = x(x^2 - 9x + 20) = x(x - 4)(x - 5) = 0$f'(x) = x(x^2 - 9x + 20) = x(x - 4)(x - 5) = 0$

Critical points inside the interval [1, 5]$[1, 5]$ are x = 4$x = 4$ and x = 5$x = 5$. ### Step 1: Perform Integration Integrate the function to evaluate boundary metrics: f(x) = int_0^x (t^3 - 9t^2 + 20t) \, dt = left[ fract^44 - 3t^3 + 10t^2 right]_0^x$f(x) = \int_{0}^{x} (t^3 - 9t^2 + 20t) \, dt = \left[ \frac{t^4}{4} - 3t^3 + 10t^2 \right]_{0}^{x}$ f(x) = fracx^44 - 3x^3 + 10x^2$f(x) = \frac{x^4}{4} - 3x^3 + 10x^2$ ### Step 2: Evaluate Points and Sum up Range Boundaries Evaluate at endpoints and critical points: f(1) = frac14 - 3 + 10 = frac294 = 7.25$f(1) = \frac{1}{4} - 3 + 10 = \frac{29}{4} = 7.25$ f(4) = frac2564 - 3(64) + 10(16) = 64 - 192 + 160 = 32$f(4) = \frac{256}{4} - 3(64) + 10(16) = 64 - 192 + 160 = 32$ f(5) = frac6254 - 3(125) + 10(25) = 156.25 - 375 + 250 = 31.25$f(5) = \frac{625}{4} - 3(125) + 10(25) = 156.25 - 375 + 250 = 31.25$ Thus, the global minimum is alpha = f(1) = frac294$\alpha = f(1) = \frac{29}{4}$, and the global maximum is \beta = f(4) = 32. 4(alpha + beta) = 4 left( frac294 + 32 right) = 29 + 128 = 157$4(\alpha + \beta) = 4 \left( \frac{29}{4} + 32 \right) = 29 + 128 = 157$ ### Pattern Recognition Always check boundary points along with interior critical values when finding the absolute range on a closed interval. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Application of Derivatives Class 12 Mathematics: Integral Calculus

### Related Formula Local minima occur at points where the derivative changes sign from negative to positive, or at sharp corners where the function reaches a local low point. ### Core Logic Analyze the piecewise sections across all transition domains:

For x in [-1, 2]$x \in [-1, 2]$, f(x) = frac13(7 + 2|x|)$f(x) = \frac{1}{3}(7 + 2|x|)$. This curve has a sharp structural minimum at x = 0$x = 0$ since |x|$|x|$ decreases then increases. Value at x = 0 implies f(0) = frac73$x = 0 \implies f(0) = \frac{7}{3}$. ### Step 1: Analyzing the Quadratic Domain Boundaries For x > 2$x > 2$, f(x) = frac1118(x^2 - 9x + 20)$f(x) = \frac{11}{18}(x^2 - 9x + 20)$. Finding the vertex point by setting the derivative to zero: f^prime(x) = frac1118(2x - 9) = 0 implies x = frac92 = 4.5$f^\prime(x) = \frac{11}{18}(2x - 9) = 0 \implies x = \frac{9}{2} = 4.5$ Since 4.5 > 2$4.5 > 2$, this vertex is within the valid domain and represents a local minimum. Value at x = 4.5 implies fleft(frac92right) = frac1118left(frac12right)left(-frac12right) = -frac1172$x = 4.5 \implies f\left(\frac{9}{2}\right) = \frac{11}{18}\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right) = -\frac{11}{72}$. ### Step 2: Calculating Final Combined Summation Summing both local minima: textTotal Minima Sum = frac73 + left(-frac1172right) = frac168 - 1172 = frac15772$\text{Total Minima Sum} = \frac{7}{3} + \left(-\frac{11}{72}\right) = \frac{168 - 11}{72} = \frac{157}{72}$ ### Pattern Recognition For piecewise function systems, always verify if the quadratic vertices fall inside their restricted intervals before adding them to your solution path. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Application of Derivatives