Given below are two statements : Statement I : lim _ mathrm x rightarrow 0 left(frac tan^ - 1 mathrm x + log_ mathrm e sqrt frac 1 + mathrm x1 - mathrm x - 2 mathrm xmathrm x ^ 5right) = frac 25$$\lim _ {\mathrm {x} \rightarrow 0} \left(\frac {\tan^ {- 1} \mathrm {x} + \log_ {\mathrm {e}} \sqrt {\frac {1 + \mathrm {x}}{1 - \mathrm {x}}} - 2 \mathrm {x}}{\mathrm {x} ^ {5}}\right) = \frac {2}{5}$$ Statement II: lim_xto 1left(x^frac21 - xright) = frac1e^2$\lim_{x\to 1}\left(x^{\frac{2}{1 - x}}\right) = \frac{1}{e^{2}}$ In the light of the above statements, choose the correct answer from the options given below:

### 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

### 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