### Related Formula For a finite limit lim_xto 0 fracf(x)x^n = L$\lim_{x\to 0} \frac{f(x)}{x^n} = L$, the lowest powers of x$x$ below degree n$n$ in the polynomial f(x)$f(x)$ must vanish. ### Core Logic Let the 4th-degree polynomial be: f(x) = ax^4 + bx^3 + cx^2 + dx + e$f(x) = ax^4 + bx^3 + cx^2 + dx + e$ Given: lim_xrightarrow 0 fracax^4 + bx^3 + cx^2 + dx + ex^2 = 5$\lim_{x\rightarrow 0} \frac{ax^4 + bx^3 + cx^2 + dx + e}{x^2} = 5$ For the limit to exist and equal 5, the terms dx$dx$ and e$e$ must be 0$0$, and the coefficient of x^2$x^2$ must be equal to 5: c = 5, quad d = 0, quad e = 0$c = 5, \quad d = 0, \quad e = 0$ Thus, the polynomial simplifies to: f(x) = ax^4 + bx^3 + 5x^2$f(x) = ax^4 + bx^3 + 5x^2$ ### Step 1: Use Extrema Conditions Differentiating f(x)$f(x)$ with respect to x$x$: f'(x) = 4ax^3 + 3bx^2 + 10x = x(4ax^2 + 3bx + 10)$f'(x) = 4ax^3 + 3bx^2 + 10x = x(4ax^2 + 3bx + 10)$ Since f(x)$f(x)$ has extreme values at x=4$x=4$ and x=5$x=5$, f'(4) = 0$f'(4) = 0$ and f'(5) = 0$f'(5) = 0$. This means 4$4$ and 5$5$ are roots of the quadratic factor 4ax^2 + 3bx + 10 = 0$4ax^2 + 3bx + 10 = 0$. ### Step 2: Solve Coefficients Using properties of roots for 4ax^2 + 3bx + 10 = 0$4ax^2 + 3bx + 10 = 0$: textProduct of roots = 4 cdot 5 = 20 = frac104a implies 4a = frac1020 = frac12 implies a = frac18$\text{Product of roots} = 4 \cdot 5 = 20 = \frac{10}{4a} \implies 4a = \frac{10}{20} = \frac{1}{2} \implies a = \frac{1}{8}$ textSum of roots = 4 + 5 = 9 = -frac3b4a$\text{Sum of roots} = 4 + 5 = 9 = -\frac{3b}{4a}$ Substituting 4a = frac12$4a = \frac{1}{2}$: 9 = -frac3b1/2 = -6b implies b = -frac96 = -frac32$9 = -\frac{3b}{1/2} = -6b \implies b = -\frac{9}{6} = -\frac{3}{2}$ Our full polynomial is: f(x) = frac18x^4 - frac32x^3 + 5x^2$f(x) = \frac{1}{8}x^4 - \frac{3}{2}x^3 + 5x^2$ ### Step 3: Calculate f(2) Evaluate at x = 2$x = 2$: f(2) = frac18(2^4) - frac32(2^3) + 5(2^2)$f(2) = \frac{1}{8}(2^4) - \frac{3}{2}(2^3) + 5(2^2)$ = frac168 - frac242 + 20 = 2 - 12 + 20 = 10$= \frac{16}{8} - \frac{24}{2} + 20 = 2 - 12 + 20 = 10$ ### Pattern Recognition Whenever a limit explicitly matches a denominator power x^n$x^n$, it directly yields both the lower-order coefficients as zeroes and the x^n$x^n$ coefficient as the limit value. ### 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

### Related Formula The function min\f(x), g(x)\$\min\{f(x), g(x)\}$ chooses the lower vertical path between the two curves at any coordinate x$x$. ### Core Logic Analyze the conditions for the \right-hand function min\|x-3|, |x+2|\$\min\{|x-3|, |x+2|\}$: - The intersection of |x-3| = |x+2|$|x-3| = |x+2|$ happens at x - 3 = -(x + 2) Rightarrow 2x = 1 Rightarrow x = 0.5$x - 3 = -(x + 2) \Rightarrow 2x = 1 \Rightarrow x = 0.5$. - For x le 0.5$x \le 0.5$, |x+2| le |x-3| Rightarrow min = |x+2|$|x+2| \le |x-3| \Rightarrow \min = |x+2|$. - For x > 0.5$x > 0.5$, |x-3| le |x+2| Rightarrow min = |x-3|$|x-3| \le |x+2| \Rightarrow \min = |x-3|$.

### Step 1: Check Interval x le -2$x \le -2$ Here, |x+2| = -(x+2) = -x-2$|x+2| = -(x+2) = -x-2$: x^2 + 3x + 2 = -x - 2 Rightarrow x^2 + 4x + 4 = 0$x^2 + 3x + 2 = -x - 2 \Rightarrow x^2 + 4x + 4 = 0$ (x+2)^2 = 0 Rightarrow x = -2$(x+2)^2 = 0 \Rightarrow x = -2$ This is a valid solution as it lies precisely within the interval condition boundary. ### Step 2: Check Interval -2 < x le 0.5$-2 < x \le 0.5$ Here, |x+2| = x+2$|x+2| = x+2$: x^2 + 3x + 2 = x + 2 Rightarrow x^2 + 2x = 0$x^2 + 3x + 2 = x + 2 \Rightarrow x^2 + 2x = 0$ x(x+2) = 0 Rightarrow x = 0 quad textor quad x = -2$x(x+2) = 0 \Rightarrow x = 0 \quad \text{or} \quad x = -2$ Only x = 0$x = 0$ fits inside this interval. ### Step 3: Check Interval x > 0.5$x > 0.5$ Here, min = |x-3| = 3-x$\min = |x-3| = 3-x$: x^2 + 3x + 2 = 3 - x Rightarrow x^2 + 4x - 1 = 0$x^2 + 3x + 2 = 3 - x \Rightarrow x^2 + 4x - 1 = 0$ x = frac-4 pm sqrt16 - 4(1)(-1)2 = -2 pm sqrt5$x = \frac{-4 \pm \sqrt{16 - 4(1)(-1)}}{2} = -2 \pm \sqrt{5}$ Evaluating values: -2 + sqrt5 approx 0.236$-2 + \sqrt{5} \approx 0.236$, which does not satisfy x > 0.5$x > 0.5$. Thus, no real roots occur in this span. Combining valid points, we find exactly 2 distinct real solutions (x = -2, 0$x = -2, 0$). ### Pattern Recognition Sketching a rough visualization showing the parabola crossing below the sharp wedge of the combined absolute values makes it visually clear that there are exactly two crossing points, confirming the algebraic count. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Quadratic Equations Class 12 Mathematics: Limits, Continuity and Differentiability