For alpha, beta, gamma in mathbbR$\alpha, \beta, \gamma \in \mathbb{R}$, if lim_x to 0 fracx^2 sin alpha x + (gamma - 1) e^x^2sin 2x - beta x = 3$\lim_{x \to 0} \frac{x^2 \sin \alpha x + (\gamma - 1) e^{x^2}}{\sin 2x - \beta x} = 3$, then beta + gamma - alpha$\beta + \gamma - \alpha$ is equal to:

### 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

### Related Formula Definition of fractional part function: - For x to 0^+$x \to 0^+$, \x\ = x - 0 = x$\{x\} = x - 0 = x$. - For x to 0^-$x \to 0^-$, \x\ = x - (-1) = x + 1$\{x\} = x - (-1) = x + 1$. ### Core Logic Let's evaluate the left-hand limit (L$L$) and right-hand limit (R$R$) separately by setting up substitution parameters around the point x=0$x=0$. ### Step 1: Evaluate Right Hand Limit (R) As x to 0^+$x \to 0^+$, substitute \x\ = h$\{x\} = h$ where h to 0$h \to 0$: R = lim_h to 0 fraccos^-1(1-h^2)sin^-1(1-h)h(1-h^2) = lim_h to 0 fraccos^-1(1-h^2)h cdot left(fracsin^-111right) = fracpi2 lim_h to 0 fraccos^-1(1-h^2)h$$R = \lim_{h \to 0} \frac{\cos^{-1}(1-h^2)\sin^{-1}(1-h)}{h(1-h^2)} = \lim_{h \to 0} \frac{\cos^{-1}(1-h^2)}{h} \cdot \left(\frac{\sin^{-1}1}{1}\right) = \frac{\pi}{2} \lim_{h \to 0} \frac{\cos^{-1}(1-h^2)}{h}$$ Let cos^-1(1-h^2) = theta implies 1-h^2 = costheta implies h^2 = 1 - costheta = 2sin^2(theta/2)$\cos^{-1}(1-h^2) = \theta \implies 1-h^2 = \cos\theta \implies h^2 = 1 - \cos\theta = 2\sin^2(\theta/2)$. As h to 0, theta to 0$h \to 0, \theta \to 0$, so h approx fracthetasqrt2$h \approx \frac{\theta}{\sqrt{2}}$: R = fracpi2 lim_theta to 0 fracthetafracthetasqrt2 = fracpisqrt2$$R = \frac{\pi}{2} \lim_{\theta \to 0} \frac{\theta}{\frac{\theta}{\sqrt{2}}} = \frac{\pi}{\sqrt{2}}$$ ### Step 2: Evaluate Left Hand Limit (L) As x to 0^-$x \to 0^-$, let x = -h implies \x\ = 1-h$x = -h \implies \{x\} = 1-h$ where h to 0$h \to 0$: L = lim_h to 0 fraccos^-1(1-(1-h)^2)sin^-1(1-(1-h))(1-h) - (1-h)^3$$L = \lim_{h \to 0} \frac{\cos^{-1}(1-(1-h)^2)\sin^{-1}(1-(1-h))}{(1-h) - (1-h)^3}$$ L = lim_h to 0 fraccos^-1(2h-h^2)sin^-1h(1-h)[1 - (1-h)^2] = lim_h to 0 fraccos^-1(0)sin^-1h1 cdot (2h-h^2)$$L = \lim_{h \to 0} \frac{\cos^{-1}(2h-h^2)\sin^{-1}h}{(1-h)[1 - (1-h)^2]} = \lim_{h \to 0} \frac{\cos^{-1}(0)\sin^{-1}h}{1 \cdot (2h-h^2)}$$ L = fracpi2 lim_h to 0 left( fracsin^-1hh cdot frac12-h right) = fracpi2 cdot 1 cdot frac12 = fracpi4$$L = \frac{\pi}{2} \lim_{h \to 0} \left( \frac{\sin^{-1}h}{h} \cdot \frac{1}{2-h} \right) = \frac{\pi}{2} \cdot 1 \cdot \frac{1}{2} = \frac{\pi}{4}$$ ### Step 3: Calculate the Target Value Substituting the computed limits L = fracpi4$L = \frac{\pi}{4}$ and R = fracpisqrt2$R = \frac{\pi}{\sqrt{2}}$ into the target expression: frac32pi^2(L^2+R^2) = frac32pi^2 left( fracpi^216 + fracpi^22 right) = 32 left( frac116 + frac12 right) = 2 + 16 = 18$$\frac{32}{\pi^{2}}(L^{2}+R^{2}) = \frac{32}{\pi^2} \left( \frac{\pi^2}{16} + \frac{\pi^2}{2} \right) = 32 \left( \frac{1}{16} + \frac{1}{2} \right) = 2 + 16 = 18$$ ### Pattern Recognition Sees: Discontinuous fractional part function framing an indeterminate limit form. Trap: Be extremely careful when managing fractional limits below zero: \x\ to 1$\{x\} \to 1$ when x to 0^-$x \to 0^-$, transforming expressions significantly compared to right-hand approaches. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Limits, Continuity and Differentiability Class 11 Mathematics: Relations and Functions

### Related Formula y = ln(1 - x^2) - ln(1 + x^2)$$y = \ln(1 - x^2) - \ln(1 + x^2)$$ ### Core Logic Let us differentiate the simplified logarithm form: y' = frac-2x1 - x^2 - frac2x1 + x^2 = -2x left( frac1 + x^2 + 1 - x^21 - x^4 right) = frac-4x1 - x^4$$y' = \frac{-2x}{1 - x^2} - \frac{2x}{1 + x^2} = -2x \left( \frac{1 + x^2 + 1 - x^2}{1 - x^4} \right) = \frac{-4x}{1 - x^4}$$ Now, computing the second derivative y''$y''$ using the quotient rule: y'' = frac-4(1 - x^4) - (-4x)(-4x^3)(1 - x^4)^2 = frac-4 + 4x^4 - 16x^4(1 - x^4)^2 = frac-4(1 + 3x^4)(1 - x^4)^2$$y'' = \frac{-4(1 - x^4) - (-4x)(-4x^3)}{(1 - x^4)^2} = \frac{-4 + 4x^4 - 16x^4}{(1 - x^4)^2} = \frac{-4(1 + 3x^4)}{(1 - x^4)^2}$$ ### Step 1: Finding the Combined Value Let us substitute x = frac12$x = \frac{1}{2}$ into the expressions: 1 - x^4 = 1 - frac116 = frac1516$$1 - x^4 = 1 - \frac{1}{16} = \frac{15}{16}$$ y' = frac-4(1/2)15/16 = frac-215/16 = -frac3215$$y' = \frac{-4(1/2)}{15/16} = \frac{-2}{15/16} = -\frac{32}{15}$$ y'' = frac-4(1 + 3/16)(15/16)^2 = frac-4(19/16)225/256 = -frac194 times frac256225 = -frac19 times 64225 = -frac1216225$$y'' = \frac{-4(1 + 3/16)}{(15/16)^2} = \frac{-4(19/16)}{225/256} = -\frac{19}{4} \times \frac{256}{225} = -\frac{19 \times 64}{225} = -\frac{1216}{225}$$ ### Step 2: Resolving the Target Multiplier Compute y' - y''$y' - y''$: y' - y'' = -frac3215 - left(-frac1216225right) = -frac480225 + frac1216225 = frac736225$$y' - y'' = -\frac{32}{15} - \left(-\frac{1216}{225}\right) = -\frac{480}{225} + \frac{1216}{225} = \frac{736}{225}$$ Multiplying this by 225: 225(y' - y'') = 225 times frac736225 = 736$$225(y' - y'') = 225 \times \frac{736}{225} = 736$$ ### Pattern Recognition Always break log quotient blocks into independent terms before differentiating (lnfracab = ln a - ln b$\ln\frac{a}{b} = \ln a - \ln b$). Differentiating fractions directly invites errors. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Continuity and Differentiability

### Related Formula Using the Newton-Leibniz formula for differentiating an integral: fracddx left( int_a^x f(t)\, dt right) = f(x)$$\frac{d}{dx} \left( \int_{a}^{x} f(t)\, dt \right) = f(x)$$ ### Core Logic The line equation is 45x + 5y + 3 = 0 implies y = -9x - frac35$45x + 5y + 3 = 0 \implies y = -9x - \frac{3}{5}$. Its slope is -9$-9$. Equating the slope expressions: 27r_1 + frac9r_22 = -9 implies 3r_1 + fracr_22 = -1 quad dots (i)$$27r_1 + \frac{9r_2}{2} = -9 \implies 3r_1 + \frac{r_2}{2} = -1 \quad \dots (i)$$ ### Step 1: Applying L'Hopital's Rule to the Limit The limit is in the frac00$\frac{0}{0}$ form as x rightarrow 3$x \rightarrow 3$. Differentiating the numerator and denominator using L'Hopital's Rule: textLimit = lim_x rightarrow 3 frac8x^2frac3r_22 - 2r_2 x - 3r_1 x^2 - 3$$\text{Limit} = \lim_{x \rightarrow 3} \frac{8x^2}{\frac{3r_2}{2} - 2r_2 x - 3r_1 x^2 - 3}$$ ### Step 2: Evaluating the Target Denominator Value Substitute x = 3$x = 3$ into the differentiated structure: textDenominator = frac3r_22 - 6r_2 - 27r_1 - 3 = -frac9r_22 - 27r_1 - 3 = -9left(3r_1 + fracr_22right) - 3$$\text{Denominator} = \frac{3r_2}{2} - 6r_2 - 27r_1 - 3 = -\frac{9r_2}{2} - 27r_1 - 3 = -9\left(3r_1 + \frac{r_2}{2}\right) - 3$$ From equation (i), we substitute 3r_1 + fracr_22 = -1$3r_1 + \frac{r_2}{2} = -1$: textDenominator = -9(-1) - 3 = 9 - 3 = 6$$\text{Denominator} = -9(-1) - 3 = 9 - 3 = 6$$ Evaluating the full limit: textLimit = frac8(3)^26 = frac726 = 12$$\text{Limit} = \frac{8(3)^2}{6} = \frac{72}{6} = 12$$ ### Pattern Recognition L'Hopital transformations reduce parameter sets back into exact multiples of the initial constraint formula. This avoids solving for r_1$r_1$ and r_2$r_2$ individually. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Limits, Continuity and Differentiability

### Related Formula textFor continuity at x=a, lim_x to a^- f(x) = lim_x to a^+ f(x) = f(a)$$\text{For continuity at } x=a, \lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = f(a)$$ lim_h to 0 fracsin hh = 1$$\lim_{h \to 0} \frac{\sin h}{h} = 1$$ ### Core Logic We need to evaluate the Left Hand Limit (LHL) and Right Hand Limit (RHL) at x = 3$x = 3$. For LHL (x < 3$x < 3$): f(x) = fraca(7x-12-x^2)b(x^2-7x+12)$$f(x) = \frac{a(7x-12-x^2)}{b(x^2-7x+12)}$$ Factor the polynomials: Numerator quadratic: -(x^2 - 7x + 12)$-(x^2 - 7x + 12)$ f(x) = frac-a(x^2-7x+12)b(x^2-7x+12) = frac-ab$$f(x) = \frac{-a(x^2-7x+12)}{b(x^2-7x+12)} = \frac{-a}{b}$$ Thus, lim_x to 3^- f(x) = frac-ab$\lim_{x \to 3^-} f(x) = \frac{-a}{b}$. ### Step 1: Evaluating Right Hand Limit For RHL (x > 3$x > 3$), as x to 3^+$x \to 3^+$, the value of the greatest integer function [x] = 3$[x] = 3$. f(x) = frac2sin(x-3)x-[x]$$f(x) = \frac{2\sin(x-3)}{x-[x]}$$ Substituting [x] = 3$[x] = 3$: lim_x to 3^+ f(x) = lim_x to 3^+ frac2sin(x-3)x-3$$\lim_{x \to 3^+} f(x) = \lim_{x \to 3^+} \frac{2\sin(x-3)}{x-3}$$ Applying the standard limit lim_theta to 0 fracsin thetatheta = 1$\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$: textRHL = 2(1) = 2$$\text{RHL} = 2(1) = 2$$ ### Step 2: Equating Limits For the function to be continuous at x=3$x=3$, LHL = RHL = f(3)$f(3)$. We are given f(3) = b$f(3) = b$. Therefore: frac-ab = 2 = b$$\frac{-a}{b} = 2 = b$$ From the right equation, b = 2$b = 2$. Substitute b$b$ into the left equation: frac-a2 = 2 Rightarrow a = -4$$\frac{-a}{2} = 2 \Rightarrow a = -4$$ ### Step 3: Final Conclusion The only ordered pair (a, b)$(a, b)$ that makes the function continuous is (-4, 2)$(-4, 2)$. The number of elements in the set S$S$ is 1. ### Pattern Recognition For limits involving [x]$[x]$ as x to k^+$x \to k^+$, you can immediately replace [x]$[x]$ with k$k$. When evaluating algebraic limits where the numerator is the exact negative of the denominator, they cancel out natively leaving just the constant ratio. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Continuity and Differentiability Class 11 Maths: Limits and Derivatives