### Related Formula For a limit of the form lim_x to 0 [g(x)]^h(x)$\lim_{x \to 0} [g(x)]^{h(x)}$ where g(x) to 1$g(x) \to 1$ and h(x) to infty$h(x) \to \infty$, the limit evaluates to: e^lim_x to 0 h(x)[g(x) - 1]$e^{\lim_{x \to 0} h(x)[g(x) - 1]}$ ### Core Logic Given the limit expression: lim_x to 0 (f(2 + x))^3/x = e^alpha$\lim_{x \to 0} (f(2 + x))^{3/x} = e^{\alpha}$ Using the standard form as f(2)=1$f(2)=1$, this transforms to: e^lim_x to 0 frac3x (f(2 + x) - 1) = e^alpha$e^{\lim_{x \to 0} \frac{3}{x} (f(2 + x) - 1)} = e^{\alpha}$ Recognizing the definition of the derivative f'(2) = lim_x to 0 fracf(2+x)-1x$f'(2) = \lim_{x \to 0} \frac{f(2+x)-1}{x}$: e^3 f'(2) = e^alpha$e^{3 f'(2)} = e^{\alpha}$ Given f'(2) = 4$f'(2) = 4$: e^3(4) = e^12 = e^alpha implies alpha = 12$e^{3(4)} = e^{12} = e^{\alpha} \implies \alpha = 12$ ### Step 1: Finding Intersection points with x-axis Substitute alpha = 12$\alpha = 12$ into the equation of the curve: y = 4x^3 - 4x^2 - 4(12 - 7)x - 12$y = 4x^3 - 4x^2 - 4(12 - 7)x - 12$ y = 4x^3 - 4x^2 - 20x - 12$y = 4x^3 - 4x^2 - 20x - 12$ To find where it meets the x-axis, set y = 0$y = 0$: 4x^3 - 4x^2 - 20x - 12 = 0 implies x^3 - x^2 - 5x - 3 = 0$4x^3 - 4x^2 - 20x - 12 = 0 \implies x^3 - x^2 - 5x - 3 = 0$ Testing for rational roots, x = -1$x = -1$ is a root because (-1)^3 - (-1)^2 - 5(-1) - 3 = -1 - 1 + 5 - 3 = 0$(-1)^3 - (-1)^2 - 5(-1) - 3 = -1 - 1 + 5 - 3 = 0$. ### Step 2: Factoring the cubic polynomial Dividing x^3 - x^2 - 5x - 3$x^3 - x^2 - 5x - 3$ by (x+1)$(x+1)$ gives: (x + 1)(x^2 - 2x - 3) = 0$(x + 1)(x^2 - 2x - 3) = 0$ (x + 1)(x + 1)(x - 3) = 0 implies (x + 1)^2(x - 3) = 0$(x + 1)(x + 1)(x - 3) = 0 \implies (x + 1)^2(x - 3) = 0$ The roots are x = -1$x = -1$ (repeated root) and x = 3$x = 3$. Therefore, the distinct real values of x$x$ where the curve intersects the x-axis are -1$-1$ and 3$3$, meaning it meets the x-axis exactly 2$2$ times. ### Pattern Recognition A repeated root like (x+1)^2$(x+1)^2$ means the curve is tangent to the x-axis at that point, but it still counts as a meeting point. Always count distinct real roots when determining the number of meeting points with the coordinate axes. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Limits, Continuity and Differentiability Class 11 Mathematics: Theory of Equations

### Related Formula Standard Taylor expansions near zero: cos h = 1 - frach^22! + frach^44! - dots$\cos h = 1 - \frac{h^2}{2!} + \frac{h^4}{4!} - \dots$ sin h = h - frach^33! + frach^55! - dots$\sin h = h - \frac{h^3}{3!} + \frac{h^5}{5!} - \dots$ ### Core Logic Let x - 1 = h$x - 1 = h$, where h to 0^+$h \to 0^{+}$. The expression transforms into: lim_h to 0frach(6 + lambdacos h) - musin hh^3 = -1$\lim_{h \to 0}\frac{h(6 + \lambda\cos h) - \mu\sin h}{h^3} = -1$ Substitute the expansions into the numerator: lim_h to 0frachleft[6 + lambdaleft(1 - frach^22right)right] - muleft(h - frach^36right)h^3 = -1$\lim_{h \to 0}\frac{h\left[6 + \lambda\left(1 - \frac{h^2}{2}\right)\right] - \mu\left(h - \frac{h^3}{6}\right)}{h^3} = -1$ lim_h to 0frac(6 + lambda - mu)h + left(-fraclambda2 + fracmu6right)h^3h^3 = -1$\lim_{h \to 0}\frac{(6 + \lambda - \mu)h + \left(-\frac{\lambda}{2} + \frac{\mu}{6}\right)h^3}{h^3} = -1$ ### Step 1: Match Coefficients for Existence For the limit to be finite, the coefficient of h$h$ must vanish: 6 + lambda - mu = 0 implies mu - lambda = 6 quad dots (1)$6 + \lambda - \mu = 0 \implies \mu - \lambda = 6 \quad \dots (1)$ Equating the h^3$h^3$ term to the given limit value: -fraclambda2 + fracmu6 = -1 implies -3lambda + mu = -6 quad dots (2)$-\frac{\lambda}{2} + \frac{\mu}{6} = -1 \implies -3\lambda + \mu = -6 \quad \dots (2)$ ### Step 2: Solve System of Equations Subtract equation (1) from (2): (-3lambda + mu) - (mu - lambda) = -6 - 6 implies -2lambda = -12 implies lambda = 6$(-3\lambda + \mu) - (\mu - \lambda) = -6 - 6 \implies -2\lambda = -12 \implies \lambda = 6$ From (1), mu = 6 + 6 = 12$\mu = 6 + 6 = 12$. lambda + mu = 6 + 12 = 18$\lambda + \mu = 6 + 12 = 18$ ### Pattern Recognition When dealing with indeterminate form limits involving mixed trigonometric expressions with a non-zero denominator power, polynomial substitution using Taylor series is much cleaner and less prone to differentiation tracking mistakes compared to multiple L'Hôpital cycles. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Limits, Continuity and Differentiability

### 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 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