Let the function f(x) = fracx3 + frac3x + 3, x neq 0$f(x) = \frac{x}{3} + \frac{3}{x} + 3, x \neq 0$ be strictly increasing in (-infty, alpha_1) cup (alpha_2, infty)$(-\infty, \alpha_1) \cup (\alpha_2, \infty)$ and strictly decreasing in (alpha_3, alpha_4) cup (alpha_4, alpha_5)$(\alpha_3, \alpha_4) \cup (\alpha_4, \alpha_5)$. Then sum_i=1^5 alpha_i^2$\sum_{i=1}^{5} \alpha_i^2$ is equal to:

### Related Formula textTo find absolute extrema, evaluate f(x) text at critical points and domain boundaries.$$\text{To find absolute extrema, evaluate } f(x) \text{ at critical points and domain boundaries.}$$ ### Core Logic Calculate the derivative of f(x)$f(x)$ using the product rule: f'(x) = fracddx [(x + 3)^2] cdot (x - 2)^3 + (x + 3)^2 cdot fracddx [(x - 2)^3]$$f'(x) = \frac{d}{dx} [(x + 3)^2] \cdot (x - 2)^3 + (x + 3)^2 \cdot \frac{d}{dx} [(x - 2)^3]$$ f'(x) = 2(x + 3)(x - 2)^3 + 3(x + 3)^2(x - 2)^2$$f'(x) = 2(x + 3)(x - 2)^3 + 3(x + 3)^2(x - 2)^2$$ Factor out the common terms (x+3)(x-2)^2$(x+3)(x-2)^2$: f'(x) = (x + 3)(x - 2)^2 [2(x - 2) + 3(x + 3)]$$f'(x) = (x + 3)(x - 2)^2 [2(x - 2) + 3(x + 3)]$$ f'(x) = (x + 3)(x - 2)^2 [2x - 4 + 3x + 9]$$f'(x) = (x + 3)(x - 2)^2 [2x - 4 + 3x + 9]$$ f'(x) = 5(x + 3)(x - 2)^2 (x + 1)$$f'(x) = 5(x + 3)(x - 2)^2 (x + 1)$$ Setting f'(x) = 0$f'(x) = 0$ gives the critical points: x = -3, -1, 2$x = -3, -1, 2$ ### Step 1: Evaluating at Critical Points and Boundaries Evaluate f(x)$f(x)$ at x = -4, -3, -1, 2, 4$x = -4, -3, -1, 2, 4$:

For x = -4$x = -4$: f(-4) = (-4 + 3)^2(-4 - 2)^3 = (1)(-216) = -216$f(-4) = (-4 + 3)^2(-4 - 2)^3 = (1)(-216) = -216$ For x = -3$x = -3$: f(-3) = 0$f(-3) = 0$ For x = -1$x = -1$: f(-1) = (-1 + 3)^2(-1 - 2)^3 = (4)(-27) = -108$f(-1) = (-1 + 3)^2(-1 - 2)^3 = (4)(-27) = -108$ For x = 2$x = 2$: f(2) = 0$f(2) = 0$ For x = 4$x = 4$: f(4) = (4 + 3)^2(4 - 2)^3 = 49 times 8 = 392$f(4) = (4 + 3)^2(4 - 2)^3 = 49 \times 8 = 392$ ### Step 2: Finding M and m From the evaluated values, the maximum M = 392$M = 392$ and the minimum m = -216$m = -216$. M - m = 392 - (-216) = 392 + 216 = 608$$M - m = 392 - (-216) = 392 + 216 = 608$$ ### Pattern Recognition For polynomials on a closed interval, critical points combined with boundary values reliably expose the absolute max and min. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Application of Derivatives

### Related Formula textArea of Delta = frac12 times textbase times textheight$$\text{Area of } \Delta = \frac{1}{2} \times \text{base} \times \text{height}$$ ### Core Logic

The vertices of the triangle are (0, 0)$(0, 0)$, (x, y)$(x, y)$, and (-x, y)$(-x, y)$. The base of the triangle lies along the horizontal line segment joining (-x, y)$(-x, y)$ and (x, y)$(x, y)$. Length of base = 2x$= 2x$. The height of the triangle from (0,0)$(0,0)$ to the line segment is y$y$. So, Area Delta = frac12 (2x) (y) = xy$\Delta = \frac{1}{2} (2x) (y) = xy$ (Assuming x > 0$x > 0$). ### Step 1: Setting up the area function Substitute y = -2x^2 + 54$y = -2x^2 + 54$ into the area function: textArea(Delta) = A(x) = x(-2x^2 + 54) = -2x^3 + 54x$$\text{Area}(\Delta) = A(x) = x(-2x^2 + 54) = -2x^3 + 54x$$ ### Step 2: Differentiating to find maximum area To find the maximum area, take the derivative with respect to x$x$ and set it to zero: A'(x) = fracdAdx = -6x^2 + 54$$A'(x) = \frac{dA}{dx} = -6x^2 + 54$$ Setting A'(x) = 0$A'(x) = 0$: -6x^2 + 54 = 0 Rightarrow 6x^2 = 54 Rightarrow x^2 = 9$$-6x^2 + 54 = 0 \Rightarrow 6x^2 = 54 \Rightarrow x^2 = 9$$ Since x$x$ represents half the base length, x = 3$x = 3$. ### Step 3: Calculating maximum area Substitute x = 3$x = 3$ back into the area function: A_textmax = 3(-2(3)^2 + 54) = 3(-18 + 54) = 3(36) = 108$$A_{\text{max}} = 3(-2(3)^2 + 54) = 3(-18 + 54) = 3(36) = 108$$ ### Pattern Recognition For symmetric figures inscribed under a parabolic arch, the area function A(x) = x cdot y(x)$A(x) = x \cdot y(x)$ is standard. Differentiate and find the critical point directly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Application of Derivatives

### Related Formula Rolle's Theorem: If f(x)$f(x)$ is continuous on [a, b]$[a, b]$, differentiable on (a, b)$(a, b)$, and f(a) = f(b)$f(a) = f(b)$, then there exists at least one c in (a, b)$c \in (a, b)$ such that f'(c) = 0$f'(c) = 0$. ### Core Logic Given f(x) = frac12left[g(x) + g(2 - x)right]$f(x) = \frac{1}{2}\left[g(x) + g(2 - x)\right]$. Differentiating f(x)$f(x)$ with respect to x$x$: f'(x) = fracg'(x) - g'(2 - x)2$$f'(x) = \frac{g'(x) - g'(2 - x)}{2}$$ ### Step 1: Evaluating the derivative at given points Evaluate f'left(frac32right)$f'\left(\frac{3}{2}\right)$: f'left(frac32right) = fracg'left(frac32right) - g'left(frac12right)2$$f'\left(\frac{3}{2}\right) = \frac{g'\left(\frac{3}{2}\right) - g'\left(\frac{1}{2}\right)}{2}$$ Since g'left(frac12right) = g'left(frac32right)$g'\left(\frac{1}{2}\right) = g'\left(\frac{3}{2}\right)$, we get: f'left(frac32right) = 0$$f'\left(\frac{3}{2}\right) = 0$$ Also evaluate f'left(frac12right)$f'\left(\frac{1}{2}\right)$: f'left(frac12right) = fracg'left(frac12right) - g'left(frac32right)2 = 0$$f'\left(\frac{1}{2}\right) = \frac{g'\left(\frac{1}{2}\right) - g'\left(\frac{3}{2}\right)}{2} = 0$$ Additionally, observe that f'(1) = fracg'(1) - g'(1)2 = 0$f'(1) = \frac{g'(1) - g'(1)}{2} = 0$. ### Step 2: Applying Rolle's Theorem We have f'left(frac12right) = 0$f'\left(\frac{1}{2}\right) = 0$, f'(1) = 0$f'(1) = 0$, and f'left(frac32right) = 0$f'\left(\frac{3}{2}\right) = 0$. By Rolle's Theorem on f'(x)$f'(x)$ (though we are asked about f'(x)=0$f'(x)=0$ directly), we already found three distinct roots for f'(x) = 0$f'(x) = 0$: x = frac12, 1, frac32$x = \frac{1}{2}, 1, \frac{3}{2}$. Notice that x = 1/2$x = 1/2$ and x=1$x=1$ are in (0, 2)$(0, 2)$, and x=3/2$x=3/2$ is in (0, 2)$(0, 2)$. Thus f'(x) = 0$f'(x) = 0$ for at least three values in (0, 2)$(0, 2)$, which satisfies the condition "at least two x$x$ in (0, 2)$(0, 2)$". ### Pattern Recognition Symmetry in function definition g(x) + g(a-x)$g(x) + g(a-x)$ guarantees a critical point at x = a/2$x = a/2$. Additional critical points arise due to the given condition on g'(x)$g'(x)$, creating a scenario perfectly suited for Rolle's implications. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Application of Derivatives

### Related Formula textDistance of point (x_1, y_1) text from line Ax+By+C=0 text is d = frac|Ax_1 + By_1 + C|sqrtA^2 + B^2$$\text{Distance of point } (x_1, y_1) \text{ from line } Ax+By+C=0 \text{ is } d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$$ ### Core Logic

Given f(x) = e^x^3 - 3x + 1$f(x) = e^{x^3 - 3x + 1}$. Check monotonicity on (-infty, -1]$(-\infty, -1]$: f'(x) = e^x^3 - 3x + 1 cdot (3x^2 - 3) = 3(x-1)(x+1)e^x^3 - 3x + 1$$f'(x) = e^{x^3 - 3x + 1} \cdot (3x^2 - 3) = 3(x-1)(x+1)e^{x^3 - 3x + 1}$$ For x le -1$x \le -1$, (x+1) le 0$(x+1) \le 0$ and (x-1) < 0$(x-1) < 0$, making f'(x) ge 0$f'(x) \ge 0$. Hence f(x)$f(x)$ is strictly increasing. Since f$f$ is onto (a,b]$(a,b]$, the range is dictated by the domain boundaries: a = lim_xto-infty f(x) = e^-infty = 0$$a = \lim_{x\to-\infty} f(x) = e^{-\infty} = 0$$ b = f(-1) = e^-1 + 3 + 1 = e^3$$b = f(-1) = e^{-1 + 3 + 1} = e^3$$ Point P(2b + 4, a + 2) equiv P(2e^3 + 4, 2)$P(2b + 4, a + 2) \equiv P(2e^3 + 4, 2)$. Find distance from line x + e^-3y - 4 = 0$x + e^{-3}y - 4 = 0$: d = frac|2e^3 + 4 + e^-3(2) - 4|sqrt1^2 + (e^-3)^2 = frac2e^3 + 2e^-3sqrt1 + e^-6$$d = \frac{|2e^3 + 4 + e^{-3}(2) - 4|}{\sqrt{1^2 + (e^{-3})^2}} = \frac{2e^3 + 2e^{-3}}{\sqrt{1 + e^{-6}}}$$ d = frac2e^3(1 + e^-6)sqrt1 + e^-6 = 2e^3 sqrt1 + e^-6 = 2sqrte^6 + 1$$d = \frac{2e^3(1 + e^{-6})}{\sqrt{1 + e^{-6}}} = 2e^3 \sqrt{1 + e^{-6}} = 2\sqrt{e^6 + 1}$$ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Applications of Derivatives Class 11 Maths: Straight Lines

### Related Formula f'(x) = 0 text for critical points. Change from + to - text means local maxima, - to + text means local minima.$$f'(x) = 0 \text{ for critical points. Change from } + \to - \text{ means local maxima, } - \to + \text{ means local minima.}$$ ### Core Logic Using Newton-Leibniz formula: f'(x) = (e^x - 1)^11 (2x - 1)^5 (x - 2)^7 (x - 3)^12 (2x - 10)^61$$f'(x) = (e^x - 1)^{11} (2x - 1)^5 (x - 2)^7 (x - 3)^{12} (2x - 10)^{61}$$ The critical points are roots of f'(x) = 0$f'(x) = 0$: x = 0, frac12, 2, 3, 5$x = 0, \frac{1}{2}, 2, 3, 5$.

### Step 1: Sign Change Analysis Perform the Wavy Curve Method for x in (-1, infty)$x \in (-1, \infty)$: At x=5$x=5$ (odd power 61): sign changes - to +$- \to +$ (Local Minima) At x=3$x=3$ (even power 12): sign doesn't change (Neither) At x=2$x=2$ (odd power 7): sign changes + to -$+ \to -$ (Local Maxima) At x=1/2$x=1/2$ (odd power 5): sign changes - to +$- \to +$ (Local Minima) At x=0$x=0$ (from e^x - 1$e^x - 1$, odd power 11): sign changes + to -$+ \to -$ (Local Maxima) ### Step 2: Compute p and q Local minima at x = 5, frac12$x = 5, \frac{1}{2}$. Local maxima at x = 2, 0$x = 2, 0$. p = 0^2 + 2^2 = 4$$p = 0^2 + 2^2 = 4$$ q = 5 + frac12 = frac112$$q = 5 + \frac{1}{2} = \frac{11}{2}$$ ### Step 3: Final Computation p^2 + 2q = 16 + 2left(frac112right) = 16 + 11 = 27$$p^2 + 2q = 16 + 2\left(\frac{11}{2}\right) = 16 + 11 = 27$$ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Application of Derivatives Class 12 Maths: Integrals