### Related Formula For a function f(x)$f(x)$ to have a local maximum or minimum at a point, its first derivative must vanish at that point: f'(x) = 0$f'(x) = 0$ ### Core Logic Differentiating the given function f(x) = 6x^3 - 45ax^2 + 108a^2x + 1$f(x) = 6x^3 - 45ax^2 + 108a^2x + 1$ with respect to x$x$: f'(x) = 18x^2 - 90ax + 108a^2 = 0$f'(x) = 18x^2 - 90ax + 108a^2 = 0$ Dividing the entire equation by 18$18$: x^2 - 5ax + 6a^2 = 0$x^2 - 5ax + 6a^2 = 0$ Factoring the quadratic equation: (x - 2a)(x - 3a) = 0$(x - 2a)(x - 3a) = 0$ Thus, the critical points are x = 2a$x = 2a$ and x = 3a$x = 3a$. Since a > 0$a > 0$, we assign x_1 = 2a$x_1 = 2a$ and x_2 = 3a$x_2 = 3a$. ### Step 1: Finding the value of a Given that the product of the roots x_1x_2 = 54$x_1x_2 = 54$: (2a)(3a) = 54$(2a)(3a) = 54$ 6a^2 = 54 implies a^2 = 9$6a^2 = 54 \implies a^2 = 9$ Since a > 0$a > 0$, we have a = 3$a = 3$. ### Step 2: Calculating the final expression Substituting a = 3$a = 3$ back to find x_1$x_1$ and x_2$x_2$: x_1 = 2(3) = 6$x_1 = 2(3) = 6$ x_2 = 3(3) = 9$x_2 = 3(3) = 9$ Now, evaluating a + x_1 + x_2$a + x_1 + x_2$: a + x_1 + x_2 = 3 + 6 + 9 = 18$a + x_1 + x_2 = 3 + 6 + 9 = 18$ ### Pattern Recognition When critical points are expressed in terms of a parameter, relate the given root condition (x_1x_2 = 54$x_1x_2 = 54$) directly to the product of roots formula (fracca$\frac{c}{a}$) of the simplified quadratic equation to save factoring time. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Application of Derivatives

### Related Formula A differentiable function is strictly increasing where its first derivative is positive (f'(x) ge 0$f'(x) \ge 0$) and strictly decreasing where its first derivative is negative (g'(x) le 0$g'(x) \le 0$). ### Step 1: Differentiate f(x)$f(x)$ to solve for a$a$ Find f'(x)$f'(x)$ : f'(x) = frac2x-2 - 2x + a ge 0$f'(x) = \frac{2}{x-2} - 2x + a \ge 0$ Since (2,3)$(2,3)$ is the largest open interval of increasing behavior [cite: 3335, 3336], the transition root occurs at the upper boundary x=3$x=3$ : f'(3) = 0 Rightarrow frac23-2 - 2(3) + a = 0 Rightarrow 2 - 6 + a = 0 Rightarrow a = 4$f'(3) = 0 \Rightarrow \frac{2}{3-2} - 2(3) + a = 0 \Rightarrow 2 - 6 + a = 0 \Rightarrow a = 4$ [cite: 3989, 3990, 3991] ### Step 2: Differentiate g(x)$g(x)$ to solve for interval (b, c)$(b, c)$ Substitute a = 4$a = 4$ into g(x)$g(x)$ [cite: 4002, 4003]: g(x) = (x-1)^3(x + 2 - 4)^2 = (x-1)^3(x-2)^2$g(x) = (x-1)^3(x + 2 - 4)^2 = (x-1)^3(x-2)^2$ Compute g'(x)$g'(x)$ using the product rule : g'(x) = 3(x-1)^2(x-2)^2 + (x-1)^3 cdot 2(x-2)$g'(x) = 3(x-1)^2(x-2)^2 + (x-1)^3 \cdot 2(x-2)$ g'(x) = (x-1)^2(x-2)left[3(x-2) + 2(x-1)right] = (x-1)^2(x-2)(5x - 8)$g'(x) = (x-1)^2(x-2)\left[3(x-2) + 2(x-1)\right] = (x-1)^2(x-2)(5x - 8)$ [cite: 4005, 4006] For g(x)$g(x)$ to be strictly decreasing, set g'(x) < 0$g'(x) < 0$ : Since (x-1)^2 ge 0$(x-1)^2 \ge 0$, we require (x-2)(5x-8) < 0 Rightarrow x in left(frac85, 2right)$(x-2)(5x-8) < 0 \Rightarrow x \in \left(\frac{8}{5}, 2\right)$ [cite: 4006, 4007]. Thus, the interval is (b, c) = left(frac85, 2right)$(b, c) = \left(\frac{8}{5}, 2\right)$, yielding b = frac85 = 1.6$b = \frac{8}{5} = 1.6$ and c = 2$c = 2$[cite: 3336, 4007]. ### Step 3: Compute Target Value Evaluate the objective expression [cite: 3337, 4008]: 100(a + b - c) = 100left(4 + frac85 - 2right) = 100(3.6) = 360$100(a + b - c) = 100\left(4 + \frac{8}{5} - 2\right) = 100(3.6) = 360$ ### Pattern Recognition Boundary parameters of maximal monotonic intervals are always the exact zero-crossings of the derivative function. Setting f'(3) = 0$f'(3) = 0$ immediately establishes a=4$a=4$ without secondary algebraic transformations. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Application of Derivatives

### Related Formula The area of a rectangle bounded between an upper function curve y_2(x)$y_2(x)$ and lower function curve y_1(x)$y_1(x)$ spanning width x = t$x = t$ is formulated as: A(t) = t cdot [y_2(t) - y_1(t)]$A(t) = t \cdot [y_2(t) - y_1(t)]$ ### Core Logic The region is bounded below by the line y = x$y = x$ and above by the downward opening parabola y = 9 - frac113x^2$y = 9 - \frac{11}{3}x^2$ in the first quadrant:

Let a vertex of the rectangle lie on the upper parabolic boundary at x = t$x = t$. The corresponding height span of the rectangle is bounded by the line y = t$y = t$ at the base:

Thus, the area equation as a function of variable parameter t$t$ is: A(t) = t cdot left( 9 - frac113t^2 - t right) = 9t - t^2 - frac113t^3$A(t) = t \cdot \left( 9 - \frac{11}{3}t^2 - t \right) = 9t - t^2 - \frac{11}{3}t^3$ ### Step 1: Differentiate to Identify Critical Values Differentiate the area function with respect to t$t$ and equate to zero: fracdAdt = 9 - 2t - 11t^2 = 0$\frac{dA}{dt} = 9 - 2t - 11t^2 = 0$ 11t^2 + 2t - 9 = 0$11t^2 + 2t - 9 = 0$ 11t^2 + 11t - 9t - 9 = 0$11t^2 + 11t - 9t - 9 = 0$ +(11t - 9)(t + 1) = 0$+(11t - 9)(t + 1) = 0$ Since $ Since $x \geq 0, we reject the negative root $, we reject the negative root $t = -1. This isolates the physical critical point at: $. This isolates the physical critical point at: $t = frac911$t = \frac{9}{11}$ ### Step 2: Evaluate Maximum Inscribed Area Substitute $ ### Step 2: Evaluate Maximum Inscribed Area Substitute $t = \frac{9}{11} back into the factored area equation formulation: $ back into the factored area equation formulation: $A_textmax = frac911 cdot left( 9 - frac911 - frac113left(frac911right)^2 right)$A_{\text{\max}} = \frac{9}{11} \cdot \left( 9 - \frac{9}{11} - \frac{11}{3}\left(\frac{9}{11}\right)^2 \right)$ $ $A_textmax = frac911 cdot left( 9 - frac911 - frac2711 right)$A_{\text{\max}} = \frac{9}{11} \cdot \left( 9 - \frac{9}{11} - \frac{27}{11} \right)$ $ $A_textmax = frac911 cdot left( 9 - frac3611 right) = frac911 cdot frac6311 = frac567121$A_{\text{\max}} = \frac{9}{11} \cdot \left( 9 - \frac{36}{11} \right) = \frac{9}{11} \cdot \frac{63}{11} = \frac{567}{121}$$ ### Pattern Recognition For optimized area allocation inside custom functional borders, setting the optimization parameter strictly to the horizontal coordinates simplifies high-degree polynomials down to standard derivative templates. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Application of Derivatives

### Related Formula 1. Functional equations containing x$x$ and frac1x$\frac{1}{x}$ can be solved by substituting x to frac1x$x \to \frac{1}{x}$ to create a system of simultaneous equations. 2. A function is strictly increasing in intervals where its first derivative is positive: fracdydx > 0$\frac{dy}{dx} > 0$. ### Core Logic Given the functional equation equation: 5f(x) + 4fleft(frac1xright) = x^2 - 2 quad implies (1)$5f(x) + 4f\left(\frac{1}{x}\right) = x^2 - 2 \quad \implies (1)$ Substitute x to frac1x$x \to \frac{1}{x}$: 5fleft(frac1xright) + 4f(x) = frac1x^2 - 2 quad implies (2)$5f\left(\frac{1}{x}\right) + 4f(x) = \frac{1}{x^2} - 2 \quad \implies (2)$ ### Step 1: Solve for f(x) To eliminate fleft(frac1xright)$f\left(\frac{1}{x}\right)$, multiply equation (1) by 5 and equation (2) by 4: 25f(x) + 20fleft(frac1xright) = 5x^2 - 10$25f(x) + 20f\left(\frac{1}{x}\right) = 5x^2 - 10$ 16f(x) + 20fleft(frac1xright) = frac4x^2 - 8$16f(x) + 20f\left(\frac{1}{x}\right) = \frac{4}{x^2} - 8$ Subtract the second equation from the first: 9f(x) = 5x^2 - frac4x^2 - 2$9f(x) = 5x^2 - \frac{4}{x^2} - 2$ ### Step 2: Differentiate the Target Equation We are given y = 9x^2 f(x) = x^2 [9f(x)]$y = 9x^2 f(x) = x^2 [9f(x)]$. Substituting our equation for 9f(x)$9f(x)$: y = x^2 left( 5x^2 - frac4x^2 - 2 right)$y = x^2 \left( 5x^2 - \frac{4}{x^2} - 2 \right)$ y = 5x^4 - 2x^2 - 4$y = 5x^4 - 2x^2 - 4$ Differentiating with respect to x$x$: fracdydx = 20x^3 - 4x = 4x(5x^2 - 1)$\frac{dy}{dx} = 20x^3 - 4x = 4x(5x^2 - 1)$ ### Step 3: Analyze the Monotonicity Intervals For the function to be strictly increasing, set fracdydx > 0$\frac{dy}{dx} > 0$: 4x(5x^2 - 1) > 0 implies xleft(x - frac1sqrt5right)left(x + frac1sqrt5right) > 0$4x(5x^2 - 1) > 0 \implies x\left(x - \frac{1}{\sqrt{5}}\right)\left(x + \frac{1}{\sqrt{5}}\right) > 0$ Using the wavy curve method with critical roots -frac1sqrt5, \, 0, \, frac1sqrt5$-\frac{1}{\sqrt{5}}, \, 0, \, \frac{1}{\sqrt{5}}$: - Positive regions occur where x in left(-frac1sqrt5, 0right) cup left(frac1sqrt5, inftyright)$x \in \left(-\frac{1}{\sqrt{5}}, 0\right) \cup \left(\frac{1}{\sqrt{5}}, \infty\right)$. ### Pattern Recognition Sees: Symmetric reciprocal arguments inside a functional format. Shortcut: Recognizing that y = 9x^2 f(x)$y = 9x^2 f(x)$ simplifies directly to an even polynomial (5x^4 - 2x^2 - 4$5x^4 - 2x^2 - 4$) means the final derivative will be odd, ensuring a symmetric layout across the origin. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Application of Derivatives Class 11 Mathematics: Relations and Functions

### Related Formula For local maxima or minima, check the sign flip of f'(x)$f'(x)$ across critical points. ### Core Logic Given f(x) = 2x + 3x^2/3$f(x) = 2x + 3x^{2/3}$. Differentiating with respect to x$x$: f'(x) = 2 + 3 cdot frac23 x^-1/3 = 2 + 2x^-1/3 = 2 left(1 + frac1x^1/3right) = 2 left(fracx^1/3 + 1x^1/3right)$f'(x) = 2 + 3 \cdot \frac{2}{3} x^{-1/3} = 2 + 2x^{-1/3} = 2 \left(1 + \frac{1}{x^{1/3}}\right) = 2 \left(\frac{x^{1/3} + 1}{x^{1/3}}\right)$ Critical points occur where f'(x) = 0$f'(x) = 0$ or where f'(x)$f'(x)$ is undefined: * f'(x) = 0 implies x^1/3 + 1 = 0 implies x = -1$f'(x) = 0 \implies x^{1/3} + 1 = 0 \implies x = -1$ * f'(x)$f'(x)$ is undefined at x = 0$x = 0$ ### Step 1: Sign Scheme Analysis Let us check the sign changes of f'(x)$f'(x)$: * For x < -1$x < -1$: x^1/3+1 < 0$x^{1/3}+1 < 0$ and x^1/3 < 0 implies f'(x) > 0$x^{1/3} < 0 \implies f'(x) > 0$ (+) * For -1 < x < 0$-1 < x < 0$: x^1/3+1 > 0$x^{1/3}+1 > 0$ and x^1/3 < 0 implies f'(x) < 0$x^{1/3} < 0 \implies f'(x) < 0$ (-) * For x > 0$x > 0$: x^1/3+1 > 0$x^{1/3}+1 > 0$ and x^1/3 > 0 implies f'(x) > 0$x^{1/3} > 0 \implies f'(x) > 0$ (+) Sign changes: * At x = -1$x = -1$: positive to negative rightarrow$\rightarrow$ Local Maxima * At x = 0$x = 0$: negative to positive rightarrow$\rightarrow$ Local Minima Thus, there is exactly one point of local maxima and exactly one point of local minima. ### Pattern Recognition Points where the derivative is undefined (cusps or vertical tangents) are equally valid critical point candidates for local extrema. Always include them in your sign scheme layout. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Application of Derivatives