Let f: mathbfR to mathbfR$f: \mathbf{R} \to \mathbf{R}$ be a polynomial function of degree four having extreme values at x = 4$x = 4$ and x = 5$x = 5$. If lim_mathbfxto 0fracf(mathbfx)mathbfx^2 = 5$\lim_{\mathbf{x}\to 0}\frac{f(\mathbf{x})}{\mathbf{x}^2} = 5$, then f(2)$f(2)$ is equal to :

### Related Formula A function is non-differentiable at sharp corner transition points where left-hand and right-hand derivatives do not match. ### Core Logic Analyze the behavior of powers of x$x$ across significant transition domains: For x < -1$x < -1$: x$x$ is the largest because higher odd powers of negative fractions decrease rapidly (x > x^3 > x^5...$x > x^3 > x^5...$). For -1 le x < 0$-1 \le x < 0$: x^21$x^{21}$ is largest (closest to zero from below). For 0 le x < 1$0 \le x < 1$: x$x$ is largest. For x ge 1$x \ge 1$: x^21$x^{21}$ is largest. f(x) = begincases x, & x < -1 \\ x^21, & -1 le x < 0 \\ x, & 0 le x < 1 \\ x^21, & x ge 1 endcases$$f(x) = \begin{cases} x, & x < -1 \\ x^{21}, & -1 \le x < 0 \\ x, & 0 \le x < 1 \\ x^{21}, & x \ge 1 \end{cases}$$ ### Step 1: Continuity and Differentiability Checks At critical intersection boundaries x = -1, 0, 1$x = -1, 0, 1$, f(x)$f(x)$ matches continuous values perfectly, so n = 0$n = 0$. Now check derivative transitions f'(x)$f'(x)$: f'(x) = begincases 1, & x < -1 \\ 21x^20, & -1 < x < 0 \\ 1, & 0 < x < 1 \\ 21x^20, & x > 1 endcases$$f'(x) = \begin{cases} 1, & x < -1 \\ 21x^{20}, & -1 < x < 0 \\ 1, & 0 < x < 1 \\ 21x^{20}, & x > 1 \end{cases}$$ At x = -1$x = -1$: textLHD = 1$\text{LHD} = 1$, textRHD = 21(-1)^20 = 21 implies textNon-differentiable$\text{RHD} = 21(-1)^{20} = 21 \implies \text{Non-differentiable}$. At x = 0$x = 0$: textLHD = 0$\text{LHD} = 0$, textRHD = 1 implies textNon-differentiable$\text{RHD} = 1 \implies \text{Non-differentiable}$. At x = 1$x = 1$: textLHD = 1$\text{LHD} = 1$, textRHD = 21(1)^20 = 21 implies textNon-differentiable$\text{RHD} = 21(1)^{20} = 21 \implies \text{Non-differentiable}$. ### Step 2: Conclusion Thus, the function is non-differentiable at exactly 3 points (x = -1, 0, 1$x = -1, 0, 1$), so m = 3$m = 3$. Since n = 0$n = 0$: m + n = 3 + 0 = 3$$m + n = 3 + 0 = 3$$ ### Pattern Recognition Maximum boundary tracking curves for standard power elements always form continuous shapes but introduce non-differentiable sharp corners at every intersection crossover point. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Limits, Continuity and Differentiability

### Related Formula For continuity at x=0$x=0$, f(0) = lim_x to 0 f(x)$f(0) = \lim_{x \to 0} f(x)$. ### Core Logic We need to evaluate the limit: lim_x rightarrow 0 fractan(tan x) - sin(sin x)tan x - sin x$$\lim_{x \rightarrow 0} \frac{\tan(\tan x) - \sin(\sin x)}{\tan x - \sin x}$$ Adding and subtracting tan x$\tan x$ inside the numerator: lim_x rightarrow 0 frac(tan(tan x) - tan x) + (tan x - sin x) + (sin x - sin(sin x))tan x - sin x$$\lim_{x \rightarrow 0} \frac{(\tan(\tan x) - \tan x) + (\tan x - \sin x) + (\sin x - \sin(\sin x))}{\tan x - \sin x}$$ Divide individual parts by x^3$x^3$ across standard series layouts directly yields the combined fractional evaluation equal to 2. ### Step 1: Final Resolution The limit evaluates cleanly to 2. Therefore, for continuity, f(0) = 2$f(0) = 2$. ### Pattern Recognition Expansion of expansion functions like tan(tan x)$\tan(\tan x)$ simplifies smoothly when paired strategically with basic structural Taylor series expansions. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Limits, Continuity and Differentiability

### Related Formula Sum of roots for a quadratic equation Ax^2 + Bx + C = 0$Ax^2 + Bx + C = 0$ satisfies: alpha + beta = -fracBA$$\alpha + \beta = -\frac{B}{A}$$ ### Core Logic We need to find lim_t to -1 (alpha_t + beta_t) = a + b$\lim_{t \to -1} (\alpha_t + \beta_t) = a + b$: a + b = lim_t to -1 -frac(t+2)^1/6 - 1(t+2)^1/7 - 1$$a + b = \lim_{t \to -1} -\frac{(t+2)^{1/6} - 1}{(t+2)^{1/7} - 1}$$ Let y = t+2$y = t+2$. As t to -1$t \to -1$, y to 1$y \to 1$. a + b = lim_y to 1 -fracy^1/6 - 1y^1/7 - 1$$a + b = \lim_{y \to 1} -\frac{y^{1/6} - 1}{y^{1/7} - 1}$$ ### Step 1: Evaluate Limit Applying L'Hopital's Rule or standard limit templates: a + b = -fracfrac16frac17 = -frac76$$a + b = -\frac{\frac{1}{6}}{\frac{1}{7}} = -\frac{7}{6}$$ Squaring the sum alignment: (a + b)^2 = frac4936$$(a + b)^2 = \frac{49}{36}$$ 72(a + b)^2 = 72 cdot frac4936 = 98$$72(a + b)^2 = 72 \cdot \frac{49}{36} = 98$$ ### Pattern Recognition Treating lim(alpha + beta)$\lim(\alpha + \beta)$ collectively allows direct evaluation via standard root identities without solving for individual root entities. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Limits, Continuity and Differentiability Class 11 Mathematics: Quadratic Equations

### Related Formula The greatest integer function [x]$[x]$ is discontinuous at all integer points. The absolute value function |x-x_0|$|x-x_0|$ is continuous everywhere but non-differentiable at its corner tip x = x_0$x = x_0$. ### Core Logic Break down the function f(x) = [x] + |x-2|$f(x) = [x] + |x-2|$ in the open domain (-2, 3)$(-2, 3)$ across sub-intervals between integers [cite: 3278, 3951]: f(x) = begincases -2 - (x-2) = -x & -2 < x < -1 \\ -1 - (x-2) = -x+1 & -1 le x < 0 \\ 0 - (x-2) = -x+2 & 0 le x < 1 \\ 1 - (x-2) = -x+3 & 1 le x < 2 \\ 2 + (x-2) = x & 2 le x < 3 endcases$$f(x) = \begin{cases} -2 - (x-2) = -x & -2 < x < -1 \\ -1 - (x-2) = -x+1 & -1 \le x < 0 \\ 0 - (x-2) = -x+2 & 0 \le x < 1 \\ 1 - (x-2) = -x+3 & 1 \le x < 2 \\ 2 + (x-2) = x & 2 \le x < 3 \end{cases}$$ ### Step 1: Count Discontinuity Points (m$m$) Evaluate the limits at internal integers \-1, 0, 1, 2\$\{-1, 0, 1, 2\}$: - At x = -1$x = -1$: textLHL = 1$\text{LHL} = 1$, textRHL = 2 Rightarrow$\text{RHL} = 2 \Rightarrow$ Discontinuous. - At x = 0$x = 0$: textLHL = 1$\text{LHL} = 1$, textRHL = 2 Rightarrow$\text{RHL} = 2 \Rightarrow$ Discontinuous. - At x = 1$x = 1$: textLHL = 1$\text{LHL} = 1$, textRHL = 2 Rightarrow$\text{RHL} = 2 \Rightarrow$ Discontinuous. - At x = 2$x = 2$: textLHL = 1$\text{LHL} = 1$, textRHL = 2 Rightarrow$\text{RHL} = 2 \Rightarrow$ Discontinuous. Thus, f(x)$f(x)$ is discontinuous at exactly 4 integer locations , meaning m = 4$m = 4$. ### Step 2: Count Non-Differentiability Points (n$n$) Since discontinuity automatically implies non-differentiability, the points \-1, 0, 1, 2\$\{-1, 0, 1, 2\}$ are non-differentiable. Let\'s check if there are other sharp corners. The modulus part |x-2|$|x-2|$ turns sharp at x=2$x=2$, which is already covered in our discontinuity list. Hence, there are no additional non-differentiable points. Thus, n = 4$n = 4$. ### Step 3: Total Evaluation Calculate the \sum requested : m + n = 4 + 4 = 8$$m + n = 4 + 4 = 8$$ ### Pattern Recognition For expressions containing [x]$[x]$, the discontinuity at integers usually drives the overall non-differentiability tally, making any coincidental sharp points from continuous elements redundant if they happen at the exact same integers. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Limits, Continuity and Differentiability