If lim_xto 0fraccos(2x) + acos(4x) - bx^4$\lim_{x\to 0}\frac{\cos(2x) + a\cos(4x) - b}{x^4}$ is finite, then (a + b)$(a + b)$ is equal to:

### Related Formula For a limit of 1^infty$1^{\infty}$ form, lim_x to a [g(x)]^h(x)$\lim_{x \to a} [g(x)]^{h(x)}$ equals: e^lim_x to a h(x)[g(x) - 1]$$e^{\lim_{x \to a} h(x)[g(x) - 1]}$$ ### Core Logic Let T = lim_x to 1 left( fracf(x + 2)f(3) right)^frac18(x - 1)^2$T = \lim_{x \to 1} \left( \frac{f(x + 2)}{f(3)} \right)^{\frac{18}{(x - 1)^2}}$. As x to 1$x \to 1$, fracf(x+2)f(3) to fracf(3)f(3) = 1$\frac{f(x+2)}{f(3)} \to \frac{f(3)}{f(3)} = 1$. The exponent goes to infty$\infty$. This is a standard 1^infty$1^{\infty}$ form. T = e^lim_x to 1 frac18(x - 1)^2 left( fracf(x + 2) - f(3)f(3) right)$$T = e^{\lim_{x \to 1} \frac{18}{(x - 1)^2} \left( \frac{f(x + 2) - f(3)}{f(3)} \right)}$$ ### Step 1: Simplify Exponent Limit Given f(3) = 18$f(3) = 18$: T = e^lim_x to 1 frac18(x - 1)^2 left( fracf(x + 2) - f(3)18 right)$$T = e^{\lim_{x \to 1} \frac{18}{(x - 1)^2} \left( \frac{f(x + 2) - f(3)}{18} \right)}$$ T = e^lim_x to 1 fracf(x + 2) - f(3)(x - 1)^2$$T = e^{\lim_{x \to 1} \frac{f(x + 2) - f(3)}{(x - 1)^2}}$$ This is a frac00$\frac{0}{0}$ form limit. ### Step 2: Apply L'Hopital's Rule Differentiate numerator and denominator w.r.t x$x$: T = e^lim_x to 1 fracf'(x + 2)2(x - 1)$$T = e^{\lim_{x \to 1} \frac{f'(x + 2)}{2(x - 1)}}$$ This is still a frac00$\frac{0}{0}$ form since f'(3) = 0$f'(3) = 0$. Apply L'Hopital's Rule again: T = e^lim_x to 1 fracf''(x + 2)2$$T = e^{\lim_{x \to 1} \frac{f''(x + 2)}{2}}$$ Substitute x = 1$x = 1$: T = e^fracf''(3)2$$T = e^{\frac{f''(3)}{2}}$$ ### Step 3: Final Calculation Given f''(3) = 4$f''(3) = 4$: T = e^frac42 = e^2$$T = e^{\frac{4}{2}} = e^2$$ The question asks for log_e(T)$\log_e(T)$: log_e(T) = log_e(e^2) = 2$$\log_e(T) = \log_e(e^2) = 2$$ ### Pattern Recognition When expanding f(x)$f(x)$ around an extrema (f'(a)=0$f'(a)=0$) inside a 1^infty$1^{\infty}$ limit form, double L'Hopital isolates the second derivative divided by the Taylor factorial (2! = 2$2! = 2$). The limit elegantly shrinks directly to f''(a)/2$f''(a)/2$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Limits and Derivatives Class 12 Maths: 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