Let a$a$ and b$b$ be real constants such that the function f$f$ defined by f(x) = begincases x^2 + 3x + a, & x le 1 \\ bx + 2, & x gt 1 endcases$$f(x) = \begin{cases} x^2 + 3x + a, & x \le 1 \\ bx + 2, & x \gt 1 \end{cases}$$ be differentiable on mathbbR$\mathbb{R}$. Then, the value of int_-2^2f(x)dx$\int_{-2}^{2}f(x)dx$ equals

### Related Formula textContinuity at x=c: lim_x to c^- f(x) = lim_x to c^+ f(x)$$\text{Continuity at } x=c: \lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x)$$ textDifferentiability at x=c: lim_x to c^- f'(x) = lim_x to c^+ f'(x)$$\text{Differentiability at } x=c: \lim_{x \to c^-} f'(x) = \lim_{x \to c^+} f'(x)$$ ### Core Logic Rewrite the piecewise function without absolute values: f(x) = begincases frac1x & , x geq 2 \\ ax^2 + 2b & , -2 < x < 2 \\ -frac1x & , x leq -2 endcases$$f(x) = \begin{cases} \frac{1}{x} & , x \geq 2 \\ ax^2 + 2b & , -2 < x < 2 \\ -\frac{1}{x} & , x \leq -2 \end{cases}$$ ### Step 1: Applying Continuity For f(x)$f(x)$ to be continuous at x = 2$x = 2$: lim_x to 2^- (ax^2 + 2b) = lim_x to 2^+ frac1x$$\lim_{x \to 2^-} (ax^2 + 2b) = \lim_{x \to 2^+} \frac{1}{x}$$ a(2)^2 + 2b = frac12 Rightarrow 4a + 2b = frac12 quad dots (1)$$a(2)^2 + 2b = \frac{1}{2} \Rightarrow 4a + 2b = \frac{1}{2} \quad \dots (1)$$ Because the function is even, continuity at x = -2$x = -2$ yields the exact same equation: 4a + 2b = 1/2$4a + 2b = 1/2$. ### Step 2: Applying Differentiability Find the derivative f'(x)$f'(x)$ for piecewise sections: f'(x) = begincases -frac1x^2 & , x > 2 \\ 2ax & , -2 < x < 2 \\ frac1x^2 & , x < -2 endcases$$f'(x) = \begin{cases} -\frac{1}{x^2} & , x > 2 \\ 2ax & , -2 < x < 2 \\ \frac{1}{x^2} & , x < -2 \end{cases}$$ For f(x)$f(x)$ to be differentiable at x = 2$x = 2$: lim_x to 2^- (2ax) = lim_x to 2^+ left(-frac1x^2right)$$\lim_{x \to 2^-} (2ax) = \lim_{x \to 2^+} \left(-\frac{1}{x^2}\right)$$ 4a = -frac14 Rightarrow a = -frac116$$4a = -\frac{1}{4} \Rightarrow a = -\frac{1}{16}$$ ### Step 3: Finding variables and final target Substitute a$a$ back into equation (1): 4left(-frac116right) + 2b = frac12$$4\left(-\frac{1}{16}\right) + 2b = \frac{1}{2}$$ -frac14 + 2b = frac12 Rightarrow 2b = frac34 Rightarrow b = frac38$$-\frac{1}{4} + 2b = \frac{1}{2} \Rightarrow 2b = \frac{3}{4} \Rightarrow b = \frac{3}{8}$$ We need to evaluate 48(a + b)$48(a + b)$: 48left(-frac116 + frac38right) = 48left(frac-1 + 616right) = 48left(frac516right) = 3 times 5 = 15$$48\left(-\frac{1}{16} + \frac{3}{8}\right) = 48\left(\frac{-1 + 6}{16}\right) = 48\left(\frac{5}{16}\right) = 3 \times 5 = 15$$ ### Pattern Recognition Piecewise differentiability forces simultaneous linear equations matching function values and their first derivatives at boundary limits. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Continuity and Differentiability

### Related Formula textSandwich / Squeeze Theorem: If g(x) le h(x) le k(x) text and lim g(x) = lim k(x) = L, text then lim h(x) = L$$\text{Sandwich / Squeeze Theorem: If } g(x) \le h(x) \le k(x) \text{ and } \lim g(x) = \lim k(x) = L, \text{ then } \lim h(x) = L$$ ### Core Logic Since f$f$ is a strictly increasing function mapping to (0,infty)$(0,\infty)$: For x > 0$x > 0$, we have x < 5x < 7x$x < 5x < 7x$. Thus, f(x) < f(5x) < f(7x)$f(x) < f(5x) < f(7x)$. Divide everything by f(x)$f(x)$ (which is strictly positive): 1 < fracf(5x)f(x) < fracf(7x)f(x)$$1 < \frac{f(5x)}{f(x)} < \frac{f(7x)}{f(x)}$$ Take the limit as x to infty$x \to \infty$: lim_xtoinfty 1 le lim_xtoinfty fracf(5x)f(x) le lim_xtoinfty fracf(7x)f(x)$$\lim_{x\to\infty} 1 \le \lim_{x\to\infty} \frac{f(5x)}{f(x)} \le \lim_{x\to\infty} \frac{f(7x)}{f(x)}$$ 1 le lim_xtoinfty fracf(5x)f(x) le 1$$1 \le \lim_{x\to\infty} \frac{f(5x)}{f(x)} \le 1$$ Therefore, lim_xtoinfty fracf(5x)f(x) = 1$\lim_{x\to\infty} \frac{f(5x)}{f(x)} = 1$. The required value is: lim_x rightarrow infty left[ fracf(5x)f(x) - 1 right] = 1 - 1 = 0$$\lim_{x \rightarrow \infty} \left[ \frac{f(5x)}{f(x)} - 1 \right] = 1 - 1 = 0$$ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Limits and Derivatives Class 12 Maths: Continuity and Differentiability

### Core Logic

The function is f(x) = e^-|ln x|$f(x) = e^{-|\ln x|}$. Rewrite piecewise for (0, infty)$(0, \infty)$: f(x) = begincases e^-(-ln x) & textif 0 < x < 1 \\ e^-ln x & textif x ge 1 endcases$$f(x) = \begin{cases} e^{-(-\ln x)} & \text{if } 0 < x < 1 \\ e^{-\ln x} & \text{if } x \ge 1 \end{cases}$$ f(x) = begincases e^ln x = x & textif 0 < x < 1 \\ frac1e^ln x = frac1x & textif x ge 1 endcases$$f(x) = \begin{cases} e^{\ln x} = x & \text{if } 0 < x < 1 \\ \frac{1}{e^{\ln x}} = \frac{1}{x} & \text{if } x \ge 1 \end{cases}$$ Check continuity at x = 1$x = 1$: lim_x to 1^- f(x) = lim_x to 1^- x = 1$$\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} x = 1$$ lim_x to 1^+ f(x) = lim_x to 1^+ frac1x = 1$$\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} \frac{1}{x} = 1$$ f(1) = 1$f(1) = 1$. The function is continuous everywhere on (0, infty)$(0, \infty)$. Thus, m = 0$m = 0$. Check differentiability at x = 1$x = 1$: LHD = lim_x to 1^- f'(x) = 1$$LHD = \lim_{x \to 1^-} f'(x) = 1$$ RHD = lim_x to 1^+ f'(x) = -frac1x^2Big|_x=1 = -1$$RHD = \lim_{x \to 1^+} f'(x) = -\frac{1}{x^2}\Big|_{x=1} = -1$$ Since LHD neq RHD$LHD \neq RHD$, the function is not differentiable at x = 1$x = 1$. Thus, n = 1$n = 1$. Finally, m + n = 0 + 1 = 1$m + n = 0 + 1 = 1$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Continuity and Differentiability

### Related Formula e^x = 1 + x + fracx^22! + dots$$e^x = 1 + x + \frac{x^2}{2!} + \dots$$ ln(1+x) = x - fracx^22 + fracx^33 - dots$$\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \dots$$ sin x approx x implies x^2sin x approx x^3$$\sin x \approx x \implies x^2\sin x \approx x^3$$ ### Core Logic Expand the numerator terms using Maclaurin series around x=0$x=0$: ax^2 left(1 + x + fracx^22 + dotsright) - b left(x - fracx^22 + fracx^33 - dotsright) + cx left(1 - x + fracx^22 - fracx^36 + dotsright)$$ax^2 \left(1 + x + \frac{x^2}{2} + \dots\right) - b \left(x - \frac{x^2}{2} + \frac{x^3}{3} - \dots\right) + cx \left(1 - x + \frac{x^2}{2} - \frac{x^3}{6} + \dots\right)$$ Denominator behavior is x^3$x^3$. Group by powers of x$x$: Coefficient of x$x$: -b + c = 0 implies c = b$-b + c = 0 \implies c = b$ Coefficient of x^2$x^2$: a + fracb2 - c = 0 implies a = c - fracb2 = fracb2$a + \frac{b}{2} - c = 0 \implies a = c - \frac{b}{2} = \frac{b}{2}$ Coefficient of x^3$x^3$: a - fracb3 + fracc2 = 1$a - \frac{b}{3} + \frac{c}{2} = 1$ Substitute a = b/2$a = b/2$ and c = b$c = b$ into the x^3$x^3$ equation: fracb2 - fracb3 + fracb2 = 1$$\frac{b}{2} - \frac{b}{3} + \frac{b}{2} = 1$$ b - fracb3 = 1 implies frac2b3 = 1 implies b = frac32$$b - \frac{b}{3} = 1 \implies \frac{2b}{3} = 1 \implies b = \frac{3}{2}$$ This gives c = frac32$c = \frac{3}{2}$ and a = frac34$a = \frac{3}{4}$. Calculate the required value: 16(a^2 + b^2 + c^2) = 16left(frac916 + frac94 + frac94right)$$16(a^2 + b^2 + c^2) = 16\left(\frac{9}{16} + \frac{9}{4} + \frac{9}{4}\right)$$ = 9 + 36 + 36 = 81$$= 9 + 36 + 36 = 81$$ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Limits and Derivatives

### Core Logic Evaluate lim_x to 0 frace^2|sin x| - 2|sin x| - 1x^2$\lim_{x \to 0} \frac{e^{2|\sin x|} - 2|\sin x| - 1}{x^2}$. Multiply and divide by |sin x|^2$|\sin x|^2$: = lim_x to 0 frace^2|sin x| - 2|sin x| - 1|sin x|^2 times fracsin^2 xx^2$$= \lim_{x \to 0} \frac{e^{2|\sin x|} - 2|\sin x| - 1}{|\sin x|^2} \times \frac{\sin^2 x}{x^2}$$ ### Step 1: Substitution and L'Hôpital Let |sin x| = t$|\sin x| = t$. As x to 0, t to 0$x \to 0, t \to 0$. lim_t to 0 frace^2t - 2t - 1t^2 times lim_x to 0 fracsin^2 xx^2$$\lim_{t \to 0} \frac{e^{2t} - 2t - 1}{t^2} \times \lim_{x \to 0} \frac{\sin^2 x}{x^2}$$ The second limit evaluates to 1$1$. Using L'Hôpital's Rule on the first limit: = lim_t to 0 frac2e^2t - 22t = lim_t to 0 frac4e^2t2 = 2$$= \lim_{t \to 0} \frac{2e^{2t} - 2}{2t} = \lim_{t \to 0} \frac{4e^{2t}}{2} = 2$$ ### Step 2: Final Result 2 times 1 = 2$2 \times 1 = 2$ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Limits and Derivatives