Let [t] be the greatest integer less than or equal to t. Then the least value of mathbfp in mathbbN$\mathbf{p} \in \mathbb{N}$ for which lim_mathrmx to 0^+ left( mathrmx left( left[ frac1mathrmx right] + left[ frac2mathrmx right] + dots + left[ fracmathrmpmathrmx right] right) - mathrmx^2 left( left[ frac1mathrmx^2 right] + left[ frac2^2mathrmx^2 right] + dots + left[ frac9^2mathrmx^2 right] right) right) geq 1$\lim_{\mathrm{x} \to 0^{+}} \left( \mathrm{x} \left( \left[ \frac{1}{\mathrm{x}} \right] + \left[ \frac{2}{\mathrm{x}} \right] + \dots + \left[ \frac{\mathrm{p}}{\mathrm{x}} \right] \right) - \mathrm{x}^{2} \left( \left[ \frac{1}{\mathrm{x}^{2}} \right] + \left[ \frac{2^{2}}{\mathrm{x}^{2}} \right] + \dots + \left[ \frac{9^{2}}{\mathrm{x}^{2}} \right] \right) \right) \geq 1$ is equal to

### Core Logic Let g(x) = ax + b$g(x) = ax + b$. Since f(x)$f(x)$ is continuous at x = 0$x = 0$: lim_x to 0^+ f(x) = f(0)$$\lim_{x \to 0^+} f(x) = f(0)$$ lim_x to 0 left(frac1+x2+xright)^frac1x = b$$\lim_{x \to 0} \left(\frac{1+x}{2+x}\right)^{\frac{1}{x}} = b$$ As x to 0$x \to 0$, the base approaches frac12$\frac{1}{2}$, and exponent approaches infty$\infty$. Thus, left(frac12right)^infty = 0$\left(\frac{1}{2}\right)^{\infty} = 0$. So, b = 0$b = 0$. Thus, g(x) = ax$g(x) = ax$. ### Step 1: Calculate Derivative For x > 0$x > 0$, f(x) = left(frac1+x2+xright)^frac1x$f(x) = \left(\frac{1+x}{2+x}\right)^{\frac{1}{x}}$. Let y = f(x)$y = f(x)$. ln y = frac1x lnleft(frac1+x2+xright)$$\ln y = \frac{1}{x} \ln\left(\frac{1+x}{2+x}\right)$$ Differentiating both sides w.r.t x$x$: frac1y y' = -frac1x^2 lnleft(frac1+x2+xright) + frac1x cdot frac2+x1+x cdot frac1(2+x) - (1+x)1(2+x)^2$$\frac{1}{y} y' = -\frac{1}{x^2} \ln\left(\frac{1+x}{2+x}\right) + \frac{1}{x} \cdot \frac{2+x}{1+x} \cdot \frac{1(2+x) - (1+x)1}{(2+x)^2}$$ y' = y left[ -frac1x^2 lnleft(frac1+x2+xright) + frac1x(1+x)(2+x) right]$$y' = y \left[ -\frac{1}{x^2} \ln\left(\frac{1+x}{2+x}\right) + \frac{1}{x(1+x)(2+x)} \right]$$ ### Step 2: Apply Condition At x=1$x=1$, y = f(1) = frac23$y = f(1) = \frac{2}{3}$. f'(1) = frac23 left[ -1 lnleft(frac23right) + frac16 right] = -frac23 lnleft(frac23right) + frac19$$f'(1) = \frac{2}{3} \left[ -1 \ln\left(\frac{2}{3}\right) + \frac{1}{6} \right] = -\frac{2}{3} \ln\left(\frac{2}{3}\right) + \frac{1}{9}$$ Also f(-1) = g(-1) = -a$f(-1) = g(-1) = -a$. Given f'(1) = f(-1) implies -a = -frac23 lnleft(frac23right) + frac19$f'(1) = f(-1) \implies -a = -\frac{2}{3} \ln\left(\frac{2}{3}\right) + \frac{1}{9}$. a = frac23 lnleft(frac23right) - frac19$$a = \frac{2}{3} \ln\left(\frac{2}{3}\right) - \frac{1}{9}$$ ### Step 3: Evaluate g(3) g(3) = 3a = 2 lnleft(frac23right) - frac13$$g(3) = 3a = 2 \ln\left(\frac{2}{3}\right) - \frac{1}{3}$$ g(3) = lnleft(frac49right) - frac13 = lnleft(frac49right) - ln(e^1/3) = lnleft(frac49e^1/3right)$$g(3) = \ln\left(\frac{4}{9}\right) - \frac{1}{3} = \ln\left(\frac{4}{9}\right) - \ln(e^{1/3}) = \ln\left(\frac{4}{9e^{1/3}}\right)$$ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Continuity and Differentiability Class 12 Maths: Application of Derivatives