If f(x) = begincases fraca | x | + x^2 - 2(sin | x |)(cos | x |)x, & x neq 0 \\ b, & x = 0 endcases$f(x) = \begin{cases} \frac{a | x | + x^2 - 2(\sin | x |)(\cos | x |)}{x}, & x \neq 0 \\ b, & x = 0 \end{cases}$ is continuous at x = 0$x = 0$, then a + b$a + b$ is equal to :

### Related Formula For a limit of 1^infty$1^{\infty}$ form, lim_x to a [g(x)]^h(x)$\lim_{x \to a} [g(x)]^{h(x)}$ equals: e^lim_x to a h(x)[g(x) - 1]$$e^{\lim_{x \to a} h(x)[g(x) - 1]}$$ ### Core Logic Let T = lim_x to 1 left( fracf(x + 2)f(3) right)^frac18(x - 1)^2$T = \lim_{x \to 1} \left( \frac{f(x + 2)}{f(3)} \right)^{\frac{18}{(x - 1)^2}}$. As x to 1$x \to 1$, fracf(x+2)f(3) to fracf(3)f(3) = 1$\frac{f(x+2)}{f(3)} \to \frac{f(3)}{f(3)} = 1$. The exponent goes to infty$\infty$. This is a standard 1^infty$1^{\infty}$ form. T = e^lim_x to 1 frac18(x - 1)^2 left( fracf(x + 2) - f(3)f(3) right)$$T = e^{\lim_{x \to 1} \frac{18}{(x - 1)^2} \left( \frac{f(x + 2) - f(3)}{f(3)} \right)}$$ ### Step 1: Simplify Exponent Limit Given f(3) = 18$f(3) = 18$: T = e^lim_x to 1 frac18(x - 1)^2 left( fracf(x + 2) - f(3)18 right)$$T = e^{\lim_{x \to 1} \frac{18}{(x - 1)^2} \left( \frac{f(x + 2) - f(3)}{18} \right)}$$ T = e^lim_x to 1 fracf(x + 2) - f(3)(x - 1)^2$$T = e^{\lim_{x \to 1} \frac{f(x + 2) - f(3)}{(x - 1)^2}}$$ This is a frac00$\frac{0}{0}$ form limit. ### Step 2: Apply L'Hopital's Rule Differentiate numerator and denominator w.r.t x$x$: T = e^lim_x to 1 fracf'(x + 2)2(x - 1)$$T = e^{\lim_{x \to 1} \frac{f'(x + 2)}{2(x - 1)}}$$ This is still a frac00$\frac{0}{0}$ form since f'(3) = 0$f'(3) = 0$. Apply L'Hopital's Rule again: T = e^lim_x to 1 fracf''(x + 2)2$$T = e^{\lim_{x \to 1} \frac{f''(x + 2)}{2}}$$ Substitute x = 1$x = 1$: T = e^fracf''(3)2$$T = e^{\frac{f''(3)}{2}}$$ ### Step 3: Final Calculation Given f''(3) = 4$f''(3) = 4$: T = e^frac42 = e^2$$T = e^{\frac{4}{2}} = e^2$$ The question asks for log_e(T)$\log_e(T)$: log_e(T) = log_e(e^2) = 2$$\log_e(T) = \log_e(e^2) = 2$$ ### Pattern Recognition When expanding f(x)$f(x)$ around an extrema (f'(a)=0$f'(a)=0$) inside a 1^infty$1^{\infty}$ limit form, double L'Hopital isolates the second derivative divided by the Taylor factorial (2! = 2$2! = 2$). The limit elegantly shrinks directly to f''(a)/2$f''(a)/2$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Limits and Derivatives Class 12 Maths: Limits, Continuity and Differentiability

### Related Formula textSandwich Theorem for greatest integer function: x - 1 < [x] leq x$$\text{Sandwich Theorem for greatest integer function: } x - 1 < [x] \leq x$$ sum_k=1^n k^2 = fracn(n+1)(2n+1)6$$\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$$ ### Core Logic Evaluate f(x)$f(x)$ using Squeeze Theorem on the summation bounds due to the Greatest Integer Function. frack^23^x - 1 < left[frack^23^xright] leq frack^23^x$$\frac{k^2}{3^x} - 1 < \left[\frac{k^2}{3^x}\right] \leq \frac{k^2}{3^x}$$ ### Step 1: Evaluate Limit f(x) Sum over bounds: sum_k=1^n left( frack^23^x - 1 right) < sum_k=1^n left[ frack^23^x right] leq sum_k=1^n frack^23^x$$\sum_{k=1}^n \left( \frac{k^2}{3^x} - 1 \right) < \sum_{k=1}^n \left[ \frac{k^2}{3^x} \right] \leq \sum_{k=1}^n \frac{k^2}{3^x}$$ frac13^x fracn(n+1)(2n+1)6 - n < sum_k=1^n left[ frack^23^x right] leq frac13^x fracn(n+1)(2n+1)6$$\frac{1}{3^x} \frac{n(n+1)(2n+1)}{6} - n < \sum_{k=1}^n \left[ \frac{k^2}{3^x} \right] \leq \frac{1}{3^x} \frac{n(n+1)(2n+1)}{6}$$ Divide by n^3$n^3$ and apply limit n to infty$n \to \infty$: lim_n to infty left( frac2n^3 + 3n^2 + n6n^3 cdot 3^x - frac1n^2 right) < f(x) leq lim_n to infty frac2n^3 + 3n^2 + n6n^3 cdot 3^x$$\lim_{n \to \infty} \left( \frac{2n^3 + 3n^2 + n}{6n^3 \cdot 3^x} - \frac{1}{n^2} \right) < f(x) \leq \lim_{n \to \infty} \frac{2n^3 + 3n^2 + n}{6n^3 \cdot 3^x}$$ f(x) = frac26 cdot 3^x = frac13 cdot 3^x = frac13^x+1$$f(x) = \frac{2}{6 \cdot 3^x} = \frac{1}{3 \cdot 3^x} = \frac{1}{3^{x+1}}$$ ### Step 2: Evaluate Final Summation We need 12sum_j=1^infty f(j)$12\sum_{j=1}^{\infty} f(j)$: 12 sum_j=1^infty frac13^j+1 = 12 left( frac13^2 + frac13^3 + ldots right)$$12 \sum_{j=1}^{\infty} \frac{1}{3^{j+1}} = 12 \left( \frac{1}{3^2} + \frac{1}{3^3} + \ldots \right)$$ This is an infinite geometric progression with a = frac19$a = \frac{1}{9}$ and r = frac13$r = \frac{1}{3}$. textSum = fraca1 - r = frac1/91 - 1/3 = frac1/92/3 = frac16$$\text{Sum} = \frac{a}{1 - r} = \frac{1/9}{1 - 1/3} = \frac{1/9}{2/3} = \frac{1}{6}$$ Finally, 12 times frac16 = 2$12 \times \frac{1}{6} = 2$. ### Pattern Recognition When evaluating infinite limits over greatest integer sums lim_n to infty frac1n^p+1 sum [k^p/C]$\lim_{n \to \infty} \frac{1}{n^{p+1}} \sum [k^p/C]$, the G.I.F brackets can be safely replaced by exact values due to Sandwich theorem bounding the -1$-1$ residual to 0$0$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Limits, Continuity and Differentiability Class 11 Maths: Sequence and Series

### Related Formula Standard Taylor series expansions: e^x = 1 + x + fracx^22! + dots, quad cos x = 1 - fracx^22! + dots, quad log_e(1+x) = x - fracx^22 + dots$$e^x = 1 + x + \frac{x^2}{2!} + \dots, \quad \cos x = 1 - \frac{x^2}{2!} + \dots, \quad \log_e(1+x) = x - \frac{x^2}{2} + \dots$$ ### Core Logic Expand numerator and denominator around x = 0$x = 0$: Denominator: xleft(1 - fracx^22right) - left(x - fracx^22right) = fracx^22 + O(x^3)$x\left(1 - \frac{x^2}{2}\right) - \left(x - \frac{x^2}{2}\right) = \frac{x^2}{2} + O(x^3)$. For limit to be finite, coefficients of x^0$x^0$ and x^1$x^1$ in numerator must be zero: - Coefficient of x^0$x^0$: 1 + 2 + c - 2 = 0 implies c = -1$1 + 2 + c - 2 = 0 \implies c = -1$ - Coefficient of x^1$x^1$: (a-1) - (c-2) = 0 implies a - 1 + 3 = 0 implies a = -2$(a-1) - (c-2) = 0 \implies a - 1 + 3 = 0 \implies a = -2$ ### Step 1: Coefficient of x^2 Numerator coefficient of x^2$x^2$ is frac(a-1)^22 - b^2 + fracc-22$\frac{(a-1)^2}{2} - b^2 + \frac{c-2}{2}$. Given limit value is 2$2$: fracfrac(a-1)^22 - b^2 + fracc-221/2 = 2 implies frac92 - b^2 - frac32 = 1 implies b^2 = 2$$\frac{\frac{(a-1)^2}{2} - b^2 + \frac{c-2}{2}}{1/2} = 2 \implies \frac{9}{2} - b^2 - \frac{3}{2} = 1 \implies b^2 = 2$$ ### Step 2: Final Calculation a^2 + b^2 + c^2 = (-2)^2 + 2 + (-1)^2 = 4 + 2 + 1 = 7$$a^2 + b^2 + c^2 = (-2)^2 + 2 + (-1)^2 = 4 + 2 + 1 = 7$$ ### Pattern Recognition Match powers of x$x$ in Taylor series to resolve indeterminate limit form frac00$\frac{0}{0}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Limits, Continuity and Differentiability

### Related Formula Greatest integer function [x^2]$[x^2]$ is discontinuous where x^2$x^2$ takes integer values, except possibly where |x| = 0$|x| = 0$. ### Core Logic In (-2, 2)$(-2, 2)$, x^2 in [0, 4)$x^2 \in [0, 4)$. Integer values occur at x = 0, pm 1, pmsqrt2, pmsqrt3$x = 0, \pm 1, \pm\sqrt{2}, \pm\sqrt{3}$. At x = 0$x = 0$, g(0) = 0$g(0) = 0$ and lim_x to 0 g(x) = 0$\lim_{x \to 0} g(x) = 0$, so g(x)$g(x)$ is continuous at x = 0$x = 0$. Points of discontinuity: S = \-1, 1, -sqrt2, sqrt2, -sqrt3, sqrt3\$S = \{-1, 1, -\sqrt{2}, \sqrt{2}, -\sqrt{3}, \sqrt{3}\}$. ### Step 1: Evaluate f(x) for x in S For f(x) = min\sqrt2x, x^2\$f(x) = \min\{\sqrt{2}x, x^2\}$: - f(-1) = min\-sqrt2, 1\ = -sqrt2$f(-1) = \min\{-\sqrt{2}, 1\} = -\sqrt{2}$ - f(1) = min\sqrt2, 1\ = 1$f(1) = \min\{\sqrt{2}, 1\} = 1$ - f(-sqrt2) = min\-2, 2\ = -2$f(-\sqrt{2}) = \min\{-2, 2\} = -2$ - f(sqrt2) = min\2, 2\ = 2$f(\sqrt{2}) = \min\{2, 2\} = 2$ - f(-sqrt3) = min\-sqrt6, 3\ = -sqrt6$f(-\sqrt{3}) = \min\{-\sqrt{6}, 3\} = -\sqrt{6}$ - f(sqrt3) = min\sqrt6, 3\ = sqrt6$f(\sqrt{3}) = \min\{\sqrt{6}, 3\} = \sqrt{6}$ ### Step 2: Summation sum_x in S f(x) = -sqrt2 + 1 - 2 + 2 - sqrt6 + sqrt6 = 1 - sqrt2$$\sum_{x \in S} f(x) = -\sqrt{2} + 1 - 2 + 2 - \sqrt{6} + \sqrt{6} = 1 - \sqrt{2}$$ ### Pattern Recognition Check origin continuity explicitly for |x|[x^2]$|x|[x^2]$; evaluate min\sqrt2x, x^2\$\min\{\sqrt{2}x, x^2\}$ case-by-case on set S$S$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Limits, Continuity and Differentiability

### Core Logic For f(x)$f(x)$ to be continuous at x = -frac32$x = -\frac{3}{2}$, the limit as x to -frac32$x \to -\frac{3}{2}$ must exist and equal fleft(-frac32right) = b$f\left(-\frac{3}{2}\right) = b$. lim_x to -3/2 fracax^2 + 2ax + 3(2x - 1)(2x + 3)$$\lim_{x \to -3/2} \frac{ax^2 + 2ax + 3}{(2x - 1)(2x + 3)}$$ Since the denominator is zero at x = -frac32$x = -\frac{3}{2}$, for the limit to exist, the numerator must also be zero at x = -frac32$x = -\frac{3}{2}$. ### Step 1: Determine 'a' Set the numerator to 0$0$ at x = -frac32$x = -\frac{3}{2}$: aleft(-frac32right)^2 + 2aleft(-frac32right) + 3 = 0$$a\left(-\frac{3}{2}\right)^2 + 2a\left(-\frac{3}{2}\right) + 3 = 0$$ frac9a4 - 3a + 3 = 0$$\frac{9a}{4} - 3a + 3 = 0$$ frac-3a4 + 3 = 0 Rightarrow frac3a4 = 3 Rightarrow a = 4$$\frac{-3a}{4} + 3 = 0 \Rightarrow \frac{3a}{4} = 3 \Rightarrow a = 4$$ ### Step 2: Simplify f(x) Substitute a = 4$a = 4$ into f(x)$f(x)$ for x neq -frac32, frac12$x \neq -\frac{3}{2}, \frac{1}{2}$: f(x) = frac4x^2 + 8x + 3(2x - 1)(2x + 3)$$f(x) = \frac{4x^2 + 8x + 3}{(2x - 1)(2x + 3)}$$ Factorizing the numerator: 4x^2 + 8x + 3 = (2x + 1)(2x + 3)$$4x^2 + 8x + 3 = (2x + 1)(2x + 3)$$ Thus, f(x) = frac(2x + 1)(2x + 3)(2x - 1)(2x + 3) = frac2x + 12x - 1$f(x) = \frac{(2x + 1)(2x + 3)}{(2x - 1)(2x + 3)} = \frac{2x + 1}{2x - 1}$ for x neq -frac32$x \neq -\frac{3}{2}$. ### Step 3: Solve f(f(x)) = 7/5 Evaluate fof(x)$fof(x)$: f(f(x)) = fleft(frac2x + 12x - 1right) = frac2left(frac2x + 12x - 1right) + 12left(frac2x + 12x - 1right) - 1$$f(f(x)) = f\left(\frac{2x + 1}{2x - 1}\right) = \frac{2\left(\frac{2x + 1}{2x - 1}\right) + 1}{2\left(\frac{2x + 1}{2x - 1}\right) - 1}$$ = frac2(2x + 1) + (2x - 1)2(2x + 1) - (2x - 1) = frac4x + 2 + 2x - 14x + 2 - 2x + 1 = frac6x + 12x + 3$$= \frac{2(2x + 1) + (2x - 1)}{2(2x + 1) - (2x - 1)} = \frac{4x + 2 + 2x - 1}{4x + 2 - 2x + 1} = \frac{6x + 1}{2x + 3}$$ Equate to frac75$\frac{7}{5}$: frac6x + 12x + 3 = frac75 Rightarrow 5(6x + 1) = 7(2x + 3)$$\frac{6x + 1}{2x + 3} = \frac{7}{5} \Rightarrow 5(6x + 1) = 7(2x + 3)$$ 30x + 5 = 14x + 21 Rightarrow 16x = 16 Rightarrow x = 1$$30x + 5 = 14x + 21 \Rightarrow 16x = 16 \Rightarrow x = 1$$ ### Pattern Recognition Indeterminate forms at points of continuity explicitly lock polynomial coefficients. Always resolve the 0/0$0/0$ form to extract missing variables before addressing composite functions. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Limits, Continuity and Differentiability