### Related Formula textNewton-Leibniz Formula: fracddx int_h(x)^g(x) f(t)dt = f(g(x)) cdot g'(x) - f(h(x)) cdot h'(x)$\text{Newton-Leibniz Formula: } \frac{d}{dx} \int_{h(x)}^{g(x)} f(t)dt = f(g(x)) \cdot g'(x) - f(h(x)) \cdot h'(x)$ lim_x to a fracf(x)g(x) = lim_x to a fracf'(x)g'(x) quad text(L'Hopital's Rule for frac00 text forms)$\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)} \quad \text{(L'Hopital's Rule for } \frac{0}{0} \text{ forms)}$ ### Core Logic Evaluate the limit L = lim_xrightarrowfracpi2 fracint_x^3^(pi/2)^3 cos(t^1/3) dt(x-fracpi2)^2$L = \lim_{x\rightarrow\frac{\pi}{2}} \frac{\int_{x^3}^{(\pi/2)^3} \cos(t^{1/3}) dt}{(x-\frac{\pi}{2})^2}$. When x to fracpi2$x \to \frac{\pi}{2}$, the integral limits become from (fracpi2)^3$(\frac{\pi}{2})^3$ to (fracpi2)^3$(\frac{\pi}{2})^3$, so the numerator is 0. The denominator evaluates to 0^2 = 0$0^2 = 0$. This is a frac00$\frac{0}{0}$ form, meaning L'Hopital's rule must be applied. Differentiate the numerator using Newton-Leibniz theorem: N'(x) = fracddx left[ int_x^3^(pi/2)^3 cos(t^1/3) dt right]$N'(x) = \frac{d}{dx} \left[ \int_{x^3}^{(\pi/2)^3} \cos(t^{1/3}) dt \right]$ = cosleft(left((pi/2)^3right)^1/3right) cdot 0 - cosleft((x^3)^1/3right) cdot fracddx(x^3)$= \cos\left(\left((\pi/2)^3\right)^{1/3}\right) \cdot 0 - \cos\left((x^3)^{1/3}\right) \cdot \frac{d}{dx}(x^3)$ = 0 - cos(x) cdot 3x^2 = -3x^2 cos(x)$= 0 - \cos(x) \cdot 3x^2 = -3x^2 \cos(x)$ Differentiate the denominator: D'(x) = fracddxleft[ (x-fracpi2)^2 right] = 2(x-fracpi2)$D'(x) = \frac{d}{dx}\left[ (x-\frac{\pi}{2})^2 \right] = 2(x-\frac{\pi}{2})$ ### Step 1: Simplify and Re-evaluate Limit Substitute the derivatives back into the limit expression: L = lim_xrightarrowfracpi2 frac-3x^2 cos x2(x-fracpi2)$L = \lim_{x\rightarrow\frac{\pi}{2}} \frac{-3x^2 \cos x}{2(x-\frac{\pi}{2})}$ Notice that cos(x) = sin(fracpi2 - x) = -sin(x - fracpi2)$\cos(x) = \sin(\frac{\pi}{2} - x) = -\sin(x - \frac{\pi}{2})$. Substituting this equivalence: L = lim_xrightarrowfracpi2 frac-3x^2 cdot (-sin(x - fracpi2))2(x-fracpi2)$L = \lim_{x\rightarrow\frac{\pi}{2}} \frac{-3x^2 \cdot (-\sin(x - \frac{\pi}{2}))}{2(x-\frac{\pi}{2})}$ L = lim_xrightarrowfracpi2 left[ fracsin(x-fracpi2)x-fracpi2 right] times left[ frac3x^22 right]$L = \lim_{x\rightarrow\frac{\pi}{2}} \left[ \frac{\sin(x-\frac{\pi}{2})}{x-\frac{\pi}{2}} \right] \times \left[ \frac{3x^2}{2} \right]$ ### Step 2: Apply Standard Limit Since lim_theta to 0 fracsin thetatheta = 1$\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$, where theta = x - fracpi2$\theta = x - \frac{\pi}{2}$: L = 1 times frac3(pi/2)^22$L = 1 \times \frac{3(\pi/2)^2}{2}$ L = frac3 cdot fracpi^242 = frac3pi^28$L = \frac{3 \cdot \frac{\pi^2}{4}}{2} = \frac{3\pi^2}{8}$ ### Pattern Recognition Integral over a variable boundary over a 0-yielding polynomial denominator is the classic signal for the Newton-Leibniz differentiation combined with L'Hopital's rule. Watch out for shifting cos x$\cos x$ to -sin(x - fracpi2)$-\sin(x - \frac{\pi}{2})$ to match the denominator structure for standard trigonometric limits. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Limit and Continuity Class 12 Mathematics: Integral Calculus

### Related Formula f'(0) = lim_h to 0 fracf(h) - f(0)h$f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h}$ Standard Limits: lim_h to 0 fractan hh = 1$\lim_{h \to 0} \frac{\tan h}{h} = 1$ ### Core Logic First, evaluate f(0)$f(0)$ to ensure the first principle approach simplifies: f(0) = frac(2^0 + 2^-0)tan(0)sqrttan^-1(0-0+1)(0+0+1)^3$f(0) = \frac{(2^0 + 2^{-0})\tan(0)\sqrt{\tan^{-1}(0-0+1)}}{(0+0+1)^3}$ Since tan(0) = 0$\tan(0) = 0$, the entire numerator collapses, giving f(0) = 0$f(0) = 0$. Set up the limit definition of the derivative at x = 0$x = 0$: f'(0) = lim_h to 0 fracf(h) - 0h$f'(0) = \lim_{h \to 0} \frac{f(h) - 0}{h}$ f'(0) = lim_h to 0 frac1h left( frac(2^h + 2^-h)tan hsqrttan^-1(h^2-h+1)(7h^2+3h+1)^3 right)$f'(0) = \lim_{h \to 0} \frac{1}{h} \left( \frac{(2^h + 2^{-h})\tan h\sqrt{\tan^{-1}(h^2-h+1)}}{(7h^2+3h+1)^3} \right)$ ### Step 1: Group Standard Limit Forms Regroup the expression to isolate the known limit forms: f'(0) = lim_h to 0 left( fractan hh right) times left( 2^h + 2^-h right) times fracsqrttan^-1(h^2-h+1)(7h^2+3h+1)^3$f'(0) = \lim_{h \to 0} \left( \frac{\tan h}{h} \right) \times \left( 2^h + 2^{-h} \right) \times \frac{\sqrt{\tan^{-1}(h^2-h+1)}}{(7h^2+3h+1)^3}$ Now, evaluate the limit of each independent non-zero segment as h to 0$h \to 0$: 1. lim_h to 0 fractan hh = 1$\lim_{h \to 0} \frac{\tan h}{h} = 1$ 2. lim_h to 0 (2^h + 2^-h) = 2^0 + 2^-0 = 1 + 1 = 2$\lim_{h \to 0} (2^h + 2^{-h}) = 2^0 + 2^{-0} = 1 + 1 = 2$ 3. lim_h to 0 sqrttan^-1(h^2-h+1) = sqrttan^-1(1) = sqrtfracpi4 = fracsqrtpi2$\lim_{h \to 0} \sqrt{\tan^{-1}(h^2-h+1)} = \sqrt{\tan^{-1}(1)} = \sqrt{\frac{\pi}{4}} = \frac{\sqrt{\pi}}{2}$ 4. lim_h to 0 (7h^2+3h+1)^3 = (0+0+1)^3 = 1$\lim_{h \to 0} (7h^2+3h+1)^3 = (0+0+1)^3 = 1$ ### Step 2: Combine Limits Multiply the evaluated continuous components together: f'(0) = 1 times 2 times fracfracsqrtpi21$f'(0) = 1 \times 2 \times \frac{\frac{\sqrt{\pi}}{2}}{1}$ f'(0) = sqrtpi$f'(0) = \sqrt{\pi}$ ### Pattern Recognition If you are asked to find f'(0)$f'(0)$ for a massive, horrifying fraction where f(0)=0$f(0)=0$ (usually due to a rogue sin x$\sin x$, tan x$\tan x$, or x$x$ term), completely ignore the quotient rule. Use the first principle formula lim_hto0 f(h)/h$\lim_{h\to0} f(h)/h$ to instantly isolate standard limit identities and plug 0 into everything else. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Limit and Continuity Class 11 Mathematics: Derivatives

### Related Formula lim_y to 0 frace^y - 1y = 1$\lim_{y \to 0} \frac{e^y - 1}{y} = 1$ Leibniz Integral Rule: fracddx int_0^x f(t) dt = f(x)$\frac{d}{dx} \int_0^x f(t) dt = f(x)$ ### Core Logic Given limit is: alpha = lim_x to 0 fracx int_0^x f(t) dte^x^2 - 1$\alpha = \lim_{x \to 0} \frac{x \int_{0}^{x} f(t) dt}{e^{x^2} - 1}$ Multiply and divide the denominator by x^2$x^2$ to use standard exponential limit: alpha = lim_x to 0 fracx int_0^x f(t) dtleft(frace^x^2 - 1x^2right) cdot x^2$\alpha = \lim_{x \to 0} \frac{x \int_{0}^{x} f(t) dt}{\left(\frac{e^{x^2} - 1}{x^2}\right) \cdot x^2}$ Since lim_xto 0 frace^x^2 - 1x^2 = 1$\lim_{x\to 0} \frac{e^{x^2} - 1}{x^2} = 1$, the expression simplifies to: alpha = lim_x to 0 fracx int_0^x f(t) dt1 cdot x^2 = lim_x to 0 fracint_0^x f(t) dtx$\alpha = \lim_{x \to 0} \frac{x \int_{0}^{x} f(t) dt}{1 \cdot x^2} = \lim_{x \to 0} \frac{\int_{0}^{x} f(t) dt}{x}$ ### Step 1: Applying L'Hôpital's Rule This is a 0/0$0/0$ form. Apply L'Hôpital's Rule by differentiating numerator and denominator w.r.t x$x$: alpha = lim_x to 0 fracfracddx int_0^x f(t) dtfracddx(x) = lim_x to 0 fracf(x)1$\alpha = \lim_{x \to 0} \frac{\frac{d}{dx} \int_{0}^{x} f(t) dt}{\frac{d}{dx}(x)} = \lim_{x \to 0} \frac{f(x)}{1}$ By continuity of differentiable function f$f$ at 0$0$: alpha = f(0)$\alpha = f(0)$ ### Step 2: Final Calculation We are given f(0) = frac12$f(0) = \frac{1}{2}$, so alpha = frac12$\alpha = \frac{1}{2}$. We need to find 8alpha^2$8\alpha^2$: 8alpha^2 = 8 left(frac12right)^2 = 8 left(frac14right) = 2$8\alpha^2 = 8 \left(\frac{1}{2}\right)^2 = 8 \left(\frac{1}{4}\right) = 2$ ### Pattern Recognition Standard expansion/limits on isolated terms in denominators immediately reduce the power of x$x$, setting up a trivial Leibniz derivative application. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Limits and Derivatives Class 12 Maths: Integrals

### 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