### Related Formula To evaluate limits of indeterminate types containing radical forms, rationalize the numerator directly by multiplying by its conjugate element matching: (sqrtA - sqrtB)(sqrtA + sqrtB) = A - B$(\sqrt{A} - \sqrt{B})(\sqrt{A} + \sqrt{B}) = A - B$ ### Core Logic Rewrite the expression as a fraction with sin x$\sin x$ in the denominator and rationalize the numerator: lim_x to 0 frac(2cos^2 x + 3cos x) - (cos^2 x + sin x + 4)sin x cdot left(sqrt2cos^2 x + 3cos x + sqrtcos^2 x + sin x + 4right)$\lim_{x \to 0} \frac{(2\cos^2 x + 3\cos x) - (\cos^2 x + \sin x + 4)}{\sin x \cdot \left(\sqrt{2\cos^2 x + 3\cos x} + \sqrt{\cos^2 x + \sin x + 4}\right)}$ = lim_x to 0 fraccos^2 x + 3cos x - sin x - 4sin x cdot left(sqrt2cos^2 x + 3cos x + sqrtcos^2 x + sin x + 4right)$= \lim_{x \to 0} \frac{\cos^2 x + 3\cos x - \sin x - 4}{\sin x \cdot \left(\sqrt{2\cos^2 x + 3\cos x} + \sqrt{\cos^2 x + \sin x + 4}\right)}$ ### Step 1: Simplify Numerator and Group Terms Express the numerator terms to isolate algebraic patterns: cos^2 x + 3cos x - 4 - sin x = (cos x - 1)(cos x + 4) - sin x$\cos^2 x + 3\cos x - 4 - \sin x = (\cos x - 1)(\cos x + 4) - \sin x$ Substitute this back into our rationalized limit format: = lim_x to 0 frac(cos x - 1)(cos x + 4) - sin xsin x cdot left(sqrt2cos^2 x + 3cos x + sqrtcos^2 x + sin x + 4right)$= \lim_{x \to 0} \frac{(\cos x - 1)(\cos x + 4) - \sin x}{\sin x \cdot \left(\sqrt{2\cos^2 x + 3\cos x} + \sqrt{\cos^2 x + \sin x + 4}\right)}$ ### Step 2: Distribute sin x$\sin x$ in Denominator Split the limit across the two separated numerator expressions: = lim_x to 0 left[ fraccos x - 1sin x cdot (cos x + 4) - 1 right] cdot frac1sqrt2(1)+3 + sqrt1+0+4$= \lim_{x \to 0} \left[ \frac{\cos x - 1}{\sin x} \cdot (\cos x + 4) - 1 \right] \cdot \frac{1}{\sqrt{2(1)+3} + \sqrt{1+0+4}}$ Evaluate the limit component values: lim_x to 0 fraccos x - 1sin x = lim_x to 0 frac-2sin^2(x/2)2sin(x/2)cos(x/2) = lim_x to 0 [-tan(x/2)] = 0$\lim_{x \to 0} \frac{\cos x - 1}{\sin x} = \lim_{x \to 0} \frac{-2\sin^2(x/2)}{2\sin(x/2)\cos(x/2)} = \lim_{x \to 0} [-\tan(x/2)] = 0$ Substituting this zero value simplifies the numerator expression directly: = left[ 0 cdot (1 + 4) - 1 right] cdot frac1sqrt5 + sqrt5 = frac-12sqrt5$= \left[ 0 \cdot (1 + 4) - 1 \right] \cdot \frac{1}{\sqrt{5} + \sqrt{5}} = \frac{-1}{2\sqrt{5}}$ ### Pattern Recognition Recognizing that fraccos x - 1sin x to 0$\frac{\cos x - 1}{\sin x} \to 0$ as x to 0$x \to 0$ isolates the non-vanishing trigonometric components without needing full multi-stage application of L'Hôpital's rule. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Limits and Derivatives

### Related Formula Trigonometric identity: fractantheta + tan^3theta1-tan^2theta = tantheta left( frac1+tan^2theta1-tan^2theta right) = fractanthetacos 2theta$\frac{\tan\theta + \tan^3\theta}{1-\tan^2\theta} = \tan\theta \left( \frac{1+\tan^2\theta}{1-\tan^2\theta} \right) = \frac{\tan\theta}{\cos 2\theta}$ Also note standard telescopic identity: tan 2phi - tanphi = fractanphicos 2phi$\tan 2\phi - \tan\phi = \frac{\tan\phi}{\cos 2\phi}$ ### Core Logic Let theta = fracx2^r+1$\theta = \frac{x}{2^{r+1}}$. The term inside the summation simplifies to: tanleft(fracx2^rright) - tanleft(fracx2^r+1right)$\tan\left(\frac{x}{2^r}\right) - \tan\left(\frac{x}{2^{r+1}}\right)$ Now, evaluating the summation: sum_r=0^n left[ tanleft(fracx2^rright) - tanleft(fracx2^r+1right) right] = tan x - tanleft(fracx2^n+1right)$\sum_{r=0}^{n} \left[ \tan\left(\frac{x}{2^r}\right) - \tan\left(\frac{x}{2^{r+1}}\right) \right] = \tan x - \tan\left(\frac{x}{2^{n+1}}\right)$ Taking the limit as n to infty$n \to \infty$, tanleft(fracx2^n+1right) to tan(0) = 0$\tan\left(\frac{x}{2^{n+1}}\right) \to \tan(0) = 0$. Therefore, f(x) = tan x$f(x) = \tan x$. ### Step 1: Evaluate the Limit We need to find: lim_xrightarrow0frace^x-e^tan xx-tan x$\lim_{x\rightarrow0}\frac{e^{x}-e^{\tan x}}{x-\tan x}$ Factor out e^tan x$e^{\tan x}$ from the numerator: lim_xrightarrow0 e^tan x cdot left[ frace^x-tan x - 1x-tan x right]$\lim_{x\rightarrow0} e^{\tan x} \cdot \left[ \frac{e^{x-\tan x} - 1}{x-\tan x} \right]$ Let u = x - tan x$u = x - \tan x$. As x to 0$x \to 0$, u to 0$u \to 0$. The limit becomes: lim_urightarrow0 e^0 cdot left[ frace^u - 1u right] = 1 times 1 = 1$\lim_{u\rightarrow0} e^0 \cdot \left[ \frac{e^u - 1}{u} \right] = 1 \times 1 = 1$ ### Pattern Recognition Standard limit substitution lim_y to 0 frace^y - 1y = 1$\lim_{y \to 0} \frac{e^y - 1}{y} = 1$ applies cleanly whenever the argument in the exponent matches the entire denominator layout. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Trigonometry Class 12 Mathematics: Limits, Continuity and Differentiability

### Related Formula sum_k=1^infty left( frac1k! - frac1(k+3)! right) implies textTelescoping Series simplification$\sum_{k=1}^{\infty} \left( \frac{1}{k!} - \frac{1}{(k+3)!} \right) \implies \text{Telescoping Series simplification}$ ### Core Logic Rewrite the numerator polynomial to establish factor terms matching the factorial expansion base (k+3)$(k+3)$: k^3 + 6k^2 + 11k + 5 = (k^3 + 6k^2 + 11k + 6) - 1 = (k+1)(k+2)(k+3) - 1$k^3 + 6k^2 + 11k + 5 = (k^3 + 6k^2 + 11k + 6) - 1 = (k+1)(k+2)(k+3) - 1$ ### Step 1: Simplify General Term T_k = frac(k+1)(k+2)(k+3)(k+3)! - frac1(k+3)!$T_k = \frac{(k+1)(k+2)(k+3)}{(k+3)!} - \frac{1}{(k+3)!}$ T_k = frac1k! - frac1(k+3)!$T_k = \frac{1}{k!} - \frac{1}{(k+3)!}$ This creates a clean telescoping layout format structure. ### Step 2: Sum the Series Writing out expanded \partial sums up to infinity: S = left( frac11! + frac12! + frac13! + frac14! + dots right) - left( frac14! + frac15! + frac16! + dots right)$S = \left( \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \dots \right) - \left( \frac{1}{4!} + \frac{1}{5!} + \frac{1}{6!} + \dots \right)$ All higher terms cancel out systematically, leaving exactly the leading remaining fragments: S = frac11! + frac12! + frac13! = 1 + frac12 + frac16 = frac106 = frac53$S = \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} = 1 + \frac{1}{2} + \frac{1}{6} = \frac{10}{6} = \frac{5}{3}$ ### Pattern Recognition Whenever factorials dominate fraction denominators, manipulate structural terms to align components via Telescoping sums (V_n - V_n-k$V_n - V_{n-k}$). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Limits, Continuity and Differentiability Class 11 Mathematics: Sequences and Series

### Related Formula lim_x to 0^+ x left[ frackx right] = k$\lim_{x \to 0^+} x \left[ \frac{k}{x} \right] = k$ sum_k=1^n k = fracn(n+1)2, quad sum_k=1^n k^2 = fracn(n+1)(2n+1)6$\sum_{k=1}^{n} k = \frac{n(n+1)}{2}, \quad \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$ ### Core Logic Using properties of Greatest Integer Function limits, as x to 0^+$x \to 0^+$, fraction values diverge cleanly to continuous variable distributions: lim_x to 0^+ x left[ frackx right] = k implies sum_k=1^p k = fracp(p+1)2$\lim_{x \to 0^+} x \left[ \frac{k}{x} \right] = k \implies \sum_{k=1}^{p} k = \frac{p(p+1)}{2}$ Similarly, for the second block component part: lim_x to 0^+ x^2 left[ frack^2x^2 right] = k^2 implies sum_k=1^9 k^2 = frac9 times 10 times 196 = 285$\lim_{x \to 0^+} x^2 \left[ \frac{k^2}{x^2} \right] = k^2 \implies \sum_{k=1}^{9} k^2 = \frac{9 \times 10 \times 19}{6} = 285$ ### Step 1: Setup Inequality Formulation Combine evaluated limits component parts: fracp(p+1)2 - 285 geq 1$\frac{p(p+1)}{2} - 285 \geq 1$ fracp(p+1)2 geq 286 implies p(p+1) geq 572$\frac{p(p+1)}{2} \geq 286 \implies p(p+1) \geq 572$ ### Step 2: Solve for least natural number Evaluate product bounds of adjacent integers: If p = 23 implies 23 times 24 = 552$p = 23 \implies 23 \times 24 = 552$ (False) If p = 24 implies 24 times 25 = 600$p = 24 \implies 24 \times 25 = 600$ (True) Therefore, the least natural value of p$p$ is 24$24$. ### Pattern Recognition Greatest Integer fractions simplify directly to standard scalar values inside limits evaluated at infinity or zero, letting you drop brackets and treat them as arithmetic sequences. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Limits

### Related Formula For a function to be continuous at a boundary point x = x_0$x = x_0$, the left-hand limit, right-hand limit, and exact function value must all match: lim_x to x_0^- f(x) = lim_x to x_0^+ f(x) = f(x_0)$\lim_{x \to x_0^-} f(x) = \lim_{x \to x_0^+} f(x) = f(x_0)$ ### Core Logic Let's enforce continuity at the critical boundaries, x = 1$x = 1$ and x = 0$x = 0$: 1. **Continuity at x = 1$x = 1$:** f(1^-) = f(1) = 1^2 + c(1) + 2 = 3 + c$f(1^-) = f(1) = 1^2 + c(1) + 2 = 3 + c$ f(1^+) = 2(1) + 1 = 3$f(1^+) = 2(1) + 1 = 3$ Equating both configurations: 3 + c = 3 implies c = 0$3 + c = 3 \implies c = 0$. 2. **Continuity at x = 0$x = 0$:** f(0^+) = f(0) = 0^2 + 0 + 2 = 2$f(0^+) = f(0) = 0^2 + 0 + 2 = 2$ f(0^-) = lim_h to 0 fraca - b cos(2h)h^2$f(0^-) = \lim_{h \to 0} \frac{a - b \cos(2h)}{h^2}$ Using the Taylor expansion cos(2h) = 1 - frac4h^22! + frac16h^44! - dots = 1 - 2h^2 + frac23h^4 - dots$\cos(2h) = 1 - \frac{4h^2}{2!} + \frac{16h^4}{4!} - \dots = 1 - 2h^2 + \frac{2}{3}h^4 - \dots$ lim_h to 0 fraca - bleft(1 - 2h^2 + frac23h^4 - dotsright)h^2 = lim_h to 0 frac(a-b) + 2bh^2 - frac23bh^4 + dotsh^2$\lim_{h \to 0} \frac{a - b\left(1 - 2h^2 + \frac{2}{3}h^4 - \dots\right)}{h^2} = \lim_{h \to 0} \frac{(a-b) + 2bh^2 - \frac{2}{3}bh^4 + \dots}{h^2}$ For the limit to exist and remain finite, the constant term must vanish: a - b = 0 implies a = b$a - b = 0 \implies a = b$. The value of the limit is then equal to 2b$2b$. To satisfy continuity: 2b = 2 implies b = 1 implies a = 1$2b = 2 \implies b = 1 \implies a = 1$. ### Step 1: Checking Differentiability at x = 0 Evaluating the Left-Hand Derivative (LHD) at x = 0$x = 0$ using values a=1, b=1$a=1, b=1$: textLHD = lim_h to 0 fracf(-h) - f(0)-h = lim_h to 0 fracfrac1 - cos(2h)h^2 - 2-h$\text{LHD} = \lim_{h \to 0} \frac{f(-h) - f(0)}{-h} = \lim_{h \to 0} \frac{\frac{1 - \cos(2h)}{h^2} - 2}{-h}$ textLHD = lim_h to 0 fracleft(2 - frac23h^2 + dotsright) - 2-h = lim_h to 0 frac23h = 0$\text{LHD} = \lim_{h \to 0} \frac{\left(2 - \frac{2}{3}h^2 + \dots\right) - 2}{-h} = \lim_{h \to 0} \frac{2}{3}h = 0$ Evaluating the Right-Hand Derivative (RHD) at x = 0$x = 0$: textRHD = lim_h to 0 fracf(h) - f(0)h = lim_h to 0 frac(h^2 + 2) - 2h = lim_h to 0 h = 0$\text{RHD} = \lim_{h \to 0} \frac{f(h) - f(0)}{h} = \lim_{h \to 0} \frac{(h^2 + 2) - 2}{h} = \lim_{h \to 0} h = 0$ Since textLHD = textRHD = 0$\text{LHD} = \text{RHD} = 0$, the function is fully differentiable at x = 0$x = 0$. ### Step 2: Checking Differentiability at x = 1 Evaluating derivatives at x = 1$x = 1$ with parameter c = 0$c = 0$: - For 0 le x le 1$0 \le x \le 1$, f(x) = x^2 + 2 implies f'(x) = 2x implies f'(1^-) = 2$f(x) = x^2 + 2 \implies f'(x) = 2x \implies f'(1^-) = 2$. - For x > 1$x > 1$, f(x) = 2x + 1 implies f'(x) = 2 implies f'(1^+) = 2$f(x) = 2x + 1 \implies f'(x) = 2 \implies f'(1^+) = 2$. Since the left derivative equals the right derivative at x = 1$x = 1$, the function is differentiable at x = 1$x = 1$. Thus, the function is differentiable everywhere, giving m = 0$m = 0$ points of non-differentiability. ### Step 3: Finding the Requested Evaluation Sum Now substitute the values m=0, a=1, b=1, c=0$m=0, a=1, b=1, c=0$ into the target equation: m + a + b + c = 0 + 1 + 1 + 0 = 2$m + a + b + c = 0 + 1 + 1 + 0 = 2$ ### Pattern Recognition Sees: Continuity conditions paired with rational surd trigonometric expansion. Shortcut: When tracking indeterminate limits like fraca-bcos 2xx^2$\frac{a-b\cos 2x}{x^2}$, matching expansions row by row prevents typical computation errors encountered with standard L'Hopital differentiation loops. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Limits, Continuity and Differentiability