### Related Formula textTaylor Series expansion of cos u = 1 - fracu^22 + fracu^424 + O(u^6)$\text{Taylor Series expansion of } \cos u = 1 - \frac{u^2}{2} + \frac{u^4}{24} + O(u^6)$ ### Core Logic Since the denominator has x^4$x^4$, we expand the numerator using Taylor series up to x^4$x^4$. For the limit to exist and be finite, the coefficients of lower powers of x$x$ (specifically x^0$x^0$ and x^2$x^2$) must be zero. ### Step 1: Write down series expansions Expand cos(2x)$\cos(2x)$ and cos(4x)$\cos(4x)$: cos 2x = 1 - frac4x^22 + frac16x^424 + O(x^6) = 1 - 2x^2 + frac23x^4 + O(x^6)$\cos 2x = 1 - \frac{4x^2}{2} + \frac{16x^4}{24} + O(x^6) = 1 - 2x^2 + \frac{2}{3}x^4 + O(x^6)$ cos 4x = 1 - frac16x^22 + frac256x^424 + O(x^6) = 1 - 8x^2 + frac323x^4 + O(x^6)$\cos 4x = 1 - \frac{16x^2}{2} + \frac{256x^4}{24} + O(x^6) = 1 - 8x^2 + \frac{32}{3}x^4 + O(x^6)$ ### Step 2: Collect coefficients in the numerator The numerator of the limit is: cos(2x) + acos(4x) - b = left( 1 - 2x^2 + frac23x^4 right) + aleft( 1 - 8x^2 + frac323x^4 right) - b$\cos(2x) + a\cos(4x) - b = \left( 1 - 2x^2 + \frac{2}{3}x^4 \right) + a\left( 1 - 8x^2 + \frac{32}{3}x^4 \right) - b$ = (1 + a - b) - x^2(2 + 8a) + x^4left(frac23 + frac323aright) + O(x^6)$= (1 + a - b) - x^2(2 + 8a) + x^4\left(\frac{2}{3} + \frac{32}{3}a\right) + O(x^6)$ ### Step 3: Set lower order coefficients to zero For the limit to be finite, the coefficients of x^0$x^0$ and x^2$x^2$ must vanish: - From x^2$x^2$ coefficient: 2 + 8a = 0 implies a = -frac14$2 + 8a = 0 \implies a = -\frac{1}{4}$ - From constant term: 1 + a - b = 0 implies b = a + 1 = -frac14 + 1 = frac34$1 + a - b = 0 \implies b = a + 1 = -\frac{1}{4} + 1 = \frac{3}{4}$ Now calculate the sum: a + b = -frac14 + frac34 = frac12$a + b = -\frac{1}{4} + \frac{3}{4} = \frac{1}{2}$ ### Pattern Recognition Finiteness condition: When a limit is finite with a denominator of x^n$x^n$, it implies that the numerator is a function of order O(x^n)$O(x^n)$ near zero. Taylor expansions allow you to quickly extract the necessary values of unknown coefficients. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Limits, Continuity and Differentiability

### Related Formula Trigonometric Sine expansion difference: sin(x-y) = sin x cos y - cos x sin y$\sin(x-y) = \sin x \cos y - \cos x \sin y$ sin(x+y) = sin x cos y + cos x sin y$\sin(x+y) = \sin x \cos y + \cos x \sin y$ ### Core Logic Rearrange the given expression to isolate the variables symmetrically: f(2x+2y)(sin x cos y - cos x sin y) = f(2x-2y)(sin x cos y + cos x sin y)$f(2x+2y)(\sin x \cos y - \cos x \sin y) = f(2x-2y)(\sin x \cos y + \cos x \sin y)$ f(2x+2y)sin(x-y) = f(2x-2y)sin(x+y)$f(2x+2y)\sin(x-y) = f(2x-2y)\sin(x+y)$ fracf(2x+2y)sin(x+y) = fracf(2x-2y)sin(x-y)$\frac{f(2x+2y)}{\sin(x+y)} = \frac{f(2x-2y)}{\sin(x-y)}$ ### Step 1: Convert to Single Variable Let 2x+2y = m$2x+2y = m$ and 2x-2y = n$2x-2y = n$. Then x+y = fracm2$x+y = \frac{m}{2}$ and x-y = fracn2$x-y = \frac{n}{2}$. fracf(m)sinleft(fracm2right) = fracf(n)sinleft(fracn2right) = K implies f(x) = K sinleft(fracx2right)$\frac{f(m)}{\sin\left(\frac{m}{2}\right)} = \frac{f(n)}{\sin\left(\frac{n}{2}\right)} = K \implies f(x) = K \sin\left(\frac{x}{2}\right)$ ### Step 2: Find K using First Derivative Differentiating f(x)$f(x)$: f'(x) = fracK2 cosleft(fracx2right)$f'(x) = \frac{K}{2} \cos\left(\frac{x}{2}\right)$ Given f'(0) = frac12$f'(0) = \frac{1}{2}$: frac12 = fracK2(1) implies K = 1$\frac{1}{2} = \frac{K}{2}(1) \implies K = 1$ Thus, f(x) = sinleft(fracx2right)$f(x) = \sin\left(\frac{x}{2}\right)$, f'(x) = frac12cosleft(fracx2right)$f'(x) = \frac{1}{2}\cos\left(\frac{x}{2}\right)$, and f''(x) = -frac14sinleft(fracx2right)$f''(x) = -\frac{1}{4}\sin\left(\frac{x}{2}\right)$. ### Step 3: Evaluate Second Derivative Value For x = frac5pi3$x = \frac{5\pi}{3}$: f''left(frac5pi3right) = -frac14 sinleft(frac5pi6right) = -frac14 left(frac12right) = -frac18$f''\left(\frac{5\pi}{3}\right) = -\frac{1}{4} \sin\left(\frac{5\pi}{6}\right) = -\frac{1}{4} \left(\frac{1}{2}\right) = -\frac{1}{8}$ Multiply by 24$24$: 24 f''left(frac5pi3right) = 24 left(-frac18right) = -3$24 f''\left(\frac{5\pi}{3}\right) = 24 \left(-\frac{1}{8}\right) = -3$ ### Pattern Recognition Grouping terms containing f(2x+2y)$f(2x+2y)$ and f(2x-2y)$f(2x-2y)$ directly creates standard sine difference/sum structures, simplifying the equation into a separable form matching a classic sine function template. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Limits, Continuity and Differentiability Class 11 Mathematics: Trigonometric Functions

### Related Formula Standard Taylor series expansions centered at x=0$x=0$: sin x = x - fracx^36 + dots$\sin x = x - \frac{x^3}{6} + \dots$ e^x = 1 + x + fracx^22 + dots$e^x = 1 + x + \frac{x^2}{2} + \dots$ ### Core Logic Since the limit evaluates to a finite value (3$3$) while the denominator goes to zero when x to 0$x \to 0$ (if 2-beta=0$2-\beta=0$), the numerator coefficients of lower-degree terms must vanish to resolve the indetermination. ### Step 1: Substitute Expansions Substitute series expansions into numerator and denominator: textNumerator = x^2(alpha x) + (gamma - 1)left(1 + x^2 + fracx^42 + dotsright)$\text{Numerator} = x^2(\alpha x) + (\gamma - 1)\left(1 + x^2 + \frac{x^4}{2} + \dots\right)$ textDenominator = left(2x - frac8x^36 + dotsright) - beta x = (2 - beta)x - frac43x^3 + dots$\text{Denominator} = \left(2x - \frac{8x^3}{6} + \dots\right) - \beta x = (2 - \beta)x - \frac{4}{3}x^3 + \dots$ ### Step 2: Equate Coefficients to Avoid Infinity Combine terms by degree: lim_x to 0 frac(gamma - 1) + (gamma - 1)x^2 + alpha x^3(2 - beta)x - frac43x^3 = 3$\lim_{x \to 0} \frac{(\gamma - 1) + (\gamma - 1)x^2 + \alpha x^3}{(2 - \beta)x - \frac{4}{3}x^3} = 3$ For a valid finite limit, the lowest power in the numerator cannot be smaller than the lowest power in the denominator. * Constraining constant term to zero: gamma - 1 = 0 implies gamma = 1$\gamma - 1 = 0 \implies \gamma = 1$ * This also forces the x^2$x^2$ coefficient to vanish: (gamma - 1) = 0$(\gamma - 1) = 0$. * To balance the remaining leading x^3$x^3$ terms, the x$x$ term in the denominator must vanish: 2 - beta = 0 implies beta = 2$2 - \beta = 0 \implies \beta = 2$. ### Step 3: Evaluate Remaining Limit Value Now compute the remaining simplified limit of x^3$x^3$ variables: lim_x to 0 fracalpha x^3-frac43x^3 = frac-3alpha4 = 3 implies alpha = -4$\lim_{x \to 0} \frac{\alpha x^3}{-\frac{4}{3}x^3} = \frac{-3\alpha}{4} = 3 \implies \alpha = -4$ ### Step 4: Final Expression Calculation Substitute the found parameters into beta + gamma - alpha$\beta + \gamma - \alpha$: beta + gamma - alpha = 2 + 1 - (-4) = 7$\beta + \gamma - \alpha = 2 + 1 - (-4) = 7$ ### Pattern Recognition Taylor expansions are far safer than consecutive L'Hopital iterations here because multiple variables are spread across distinct polynomial powers, isolating components explicitly by structural degree. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Limits, Continuity and Differentiability

### Related Formula For a continuous function f(x)$f(x)$: - A point x = c$x = c$ is a local minimum if f(x)$f(x)$ changes from decreasing to increasing as x$x$ passes through c$c$ (i.e., a trough in the graph). - A point x = c$x = c$ is a local maximum if f(x)$f(x)$ changes from increasing to decreasing as x$x$ passes through c$c$ (i.e., a peak in the graph). ### Core Logic Let's analyze g(x) = |x+2| - 2|x|$g(x) = |x+2| - 2|x|$ first to construct f(x) = |g(x)|$f(x) = |g(x)|$. 1. If x le -2$x \le -2$: g(x) = -(x+2) - 2(-x) = x - 2$g(x) = -(x+2) - 2(-x) = x - 2$ 2. If -2 < x le 0$-2 < x \le 0$: g(x) = (x+2) - 2(-x) = 3x + 2$g(x) = (x+2) - 2(-x) = 3x + 2$ 3. If x > 0$x > 0$: g(x) = (x+2) - 2x = -x + 2$g(x) = (x+2) - 2x = -x + 2$ Now, we find the critical transition points where g(x) = 0$g(x) = 0$: - x - 2 = 0 implies x = 2$x - 2 = 0 \implies x = 2$ (not in x le -2$x \le -2$) - 3x + 2 = 0 implies x = -2/3$3x + 2 = 0 \implies x = -2/3$ (lies in -2 < x le 0$-2 < x \le 0$) - -x + 2 = 0 implies x = 2$-x + 2 = 0 \implies x = 2$ (lies in x > 0$x > 0$) Thus, the absolute value function f(x) = |g(x)|$f(x) = |g(x)|$ transitions at x = -2$x = -2$, x = -2/3$x = -2/3$, x = 0$x = 0$, and x = 2$x = 2$. ### Step 1: Graph Reconstruction and Critical Points Analysis Evaluating values at these boundaries: - f(-2) = | |-2+2| - 2|-2| | = |0 - 4| = 4$f(-2) = | |-2+2| - 2|-2| | = |0 - 4| = 4$ - f(-2/3) = 0$f(-2/3) = 0$ (trough, local minimum) - f(0) = | |2| - 0 | = 2$f(0) = | |2| - 0 | = 2$ (peak, local maximum) - f(2) = 0$f(2) = 0$ (trough, local minimum) Let's trace the graph: - For x < -2$x < -2$, f(x) = |x-2| = 2-x$f(x) = |x-2| = 2-x$, which decreases towards 4$4$ as x to -2$x \to -2$. - At x = -2$x = -2$, there is a corner point (-2,4)$(-2,4)$, but it is not an extremum because the function continues to decrease to 0$0$ on the right. - At x = -2/3$x = -2/3$, it reaches 0$0$ and turns upwards (local minimum). - At x = 0$x = 0$, it reaches a local peak of 2$2$ and turns downwards (local maximum). - At x = 2$x = 2$, it reaches 0$0$ and turns upwards (local minimum).

### Step 2: Calculating m + n$m + n$ From the verified graph and transition behaviors: - Local minima points (m$m$): x = -2/3$x = -2/3$ and x = 2 implies m = 2$x = 2 \implies m = 2$ - Local maxima points (n$n$): x = 0 implies n = 1$x = 0 \implies n = 1$ textSum m + n = 2 + 1 = 3$\text{Sum } m + n = 2 + 1 = 3$ ### Pattern Recognition Whenever you have f(x) = |g(x)|$f(x) = |g(x)|$ where g(x)$g(x)$ is continuous: - Any point where g(x) = 0$g(x) = 0$ becomes a local minimum with value 0$0$ (unless it was a tangent point already, which still remains a minimum). - Corners of the original absolute segments like x=0, -2$x=0, -2$ must be checked sequentially for slope changes. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Limits, Continuity and Differentiability Class 11 Mathematics: Relations and Functions

### Related Formula For a limit of 1^infty$1^{\infty}$ form, where lim f(x) = 1$\lim f(x) = 1$ and lim g(x) = infty$\lim g(x) = \infty$: lim [f(x)]^g(x) = e^lim g(x)(f(x) - 1)$\lim [f(x)]^{g(x)} = e^{\lim g(x)(f(x) - 1)}$ Taylor expansion of tan x$\tan x$: tan x = x + fracx^33 + frac2x^515 + dots$\tan x = x + \frac{x^3}{3} + \frac{2x^5}{15} + \dots$ ### Core Logic Here, lim_x to 0 fractan xx = 1$\lim_{x \to 0} \frac{\tan x}{x} = 1$ and lim_x to 0 frac1x^2 = infty$\lim_{x \to 0} \frac{1}{x^2} = \infty$. p = e^lim_x to 0 frac1x^2left(fractan xx - 1right) = e^lim_x to 0 fractan x - xx^3$p = e^{\lim_{x \to 0} \frac{1}{x^2}\left(\frac{\tan x}{x} - 1\right)} = e^{\lim_{x \to 0} \frac{\tan x - x}{x^3}}$ ### Step 1: Finding limit exponent Using expansion of tan x$\tan x$: lim_x to 0 fracleft(x + fracx^33 + dotsright) - xx^3 = lim_x to 0 fracfracx^33 + O(x^5)x^3 = frac13$\lim_{x \to 0} \frac{\left(x + \frac{x^3}{3} + \dots\right) - x}{x^3} = \lim_{x \to 0} \frac{\frac{x^3}{3} + O(x^5)}{x^3} = \frac{1}{3}$ Thus: p = e^1/3$p = e^{1/3}$ 96log_e p = 96left(frac13right) = 32$96\log_e p = 96\left(\frac{1}{3}\right) = 32$ ### Pattern Recognition Limits of forms like 1^infty$1^{\infty}$ always boil down to evaluating standard polynomials in the exponent. Using Taylor series expansion rather than L'Hopital's rule directly avoids taking multiple heavy derivatives. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Limits, Continuity and Differentiability