### Related Formula Sum of roots for a quadratic equation Ax^2 + Bx + C = 0$Ax^2 + Bx + C = 0$ satisfies: alpha + beta = -fracBA$\alpha + \beta = -\frac{B}{A}$ ### Core Logic We need to find lim_t to -1 (alpha_t + beta_t) = a + b$\lim_{t \to -1} (\alpha_t + \beta_t) = a + b$: a + b = lim_t to -1 -frac(t+2)^1/6 - 1(t+2)^1/7 - 1$a + b = \lim_{t \to -1} -\frac{(t+2)^{1/6} - 1}{(t+2)^{1/7} - 1}$ Let y = t+2$y = t+2$. As t to -1$t \to -1$, y to 1$y \to 1$. a + b = lim_y to 1 -fracy^1/6 - 1y^1/7 - 1$a + b = \lim_{y \to 1} -\frac{y^{1/6} - 1}{y^{1/7} - 1}$ ### Step 1: Evaluate Limit Applying L'Hopital's Rule or standard limit templates: a + b = -fracfrac16frac17 = -frac76$a + b = -\frac{\frac{1}{6}}{\frac{1}{7}} = -\frac{7}{6}$ Squaring the sum alignment: (a + b)^2 = frac4936$(a + b)^2 = \frac{49}{36}$ 72(a + b)^2 = 72 cdot frac4936 = 98$72(a + b)^2 = 72 \cdot \frac{49}{36} = 98$ ### Pattern Recognition Treating lim(alpha + beta)$\lim(\alpha + \beta)$ collectively allows direct evaluation via standard root identities without solving for individual root entities. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Limits, Continuity and Differentiability Class 11 Mathematics: Quadratic Equations

### Related Formula The greatest integer function [x]$[x]$ is discontinuous at all integer points. The absolute value function |x-x_0|$|x-x_0|$ is continuous everywhere but non-differentiable at its corner tip x = x_0$x = x_0$. ### Core Logic Break down the function f(x) = [x] + |x-2|$f(x) = [x] + |x-2|$ in the open domain (-2, 3)$(-2, 3)$ across sub-intervals between integers [cite: 3278, 3951]: f(x) = begincases -2 - (x-2) = -x & -2 < x < -1 \\ -1 - (x-2) = -x+1 & -1 le x < 0 \\ 0 - (x-2) = -x+2 & 0 le x < 1 \\ 1 - (x-2) = -x+3 & 1 le x < 2 \\ 2 + (x-2) = x & 2 le x < 3 endcases$f(x) = \begin{cases} -2 - (x-2) = -x & -2 < x < -1 \\ -1 - (x-2) = -x+1 & -1 \le x < 0 \\ 0 - (x-2) = -x+2 & 0 \le x < 1 \\ 1 - (x-2) = -x+3 & 1 \le x < 2 \\ 2 + (x-2) = x & 2 \le x < 3 \end{cases}$ ### Step 1: Count Discontinuity Points (m$m$) Evaluate the limits at internal integers \-1, 0, 1, 2\$\{-1, 0, 1, 2\}$: - At x = -1$x = -1$: textLHL = 1$\text{LHL} = 1$, textRHL = 2 Rightarrow$\text{RHL} = 2 \Rightarrow$ Discontinuous. - At x = 0$x = 0$: textLHL = 1$\text{LHL} = 1$, textRHL = 2 Rightarrow$\text{RHL} = 2 \Rightarrow$ Discontinuous. - At x = 1$x = 1$: textLHL = 1$\text{LHL} = 1$, textRHL = 2 Rightarrow$\text{RHL} = 2 \Rightarrow$ Discontinuous. - At x = 2$x = 2$: textLHL = 1$\text{LHL} = 1$, textRHL = 2 Rightarrow$\text{RHL} = 2 \Rightarrow$ Discontinuous. Thus, f(x)$f(x)$ is discontinuous at exactly 4 integer locations , meaning m = 4$m = 4$. ### Step 2: Count Non-Differentiability Points (n$n$) Since discontinuity automatically implies non-differentiability, the points \-1, 0, 1, 2\$\{-1, 0, 1, 2\}$ are non-differentiable. Let\'s check if there are other sharp corners. The modulus part |x-2|$|x-2|$ turns sharp at x=2$x=2$, which is already covered in our discontinuity list. Hence, there are no additional non-differentiable points. Thus, n = 4$n = 4$. ### Step 3: Total Evaluation Calculate the \sum requested : m + n = 4 + 4 = 8$m + n = 4 + 4 = 8$ ### Pattern Recognition For expressions containing [x]$[x]$, the discontinuity at integers usually drives the overall non-differentiability tally, making any coincidental sharp points from continuous elements redundant if they happen at the exact same integers. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Limits, Continuity and Differentiability

### Related Formula The function min\f(x), g(x)\$\min\{f(x), g(x)\}$ chooses the lower vertical path between the two curves at any coordinate x$x$. ### Core Logic Analyze the conditions for the \right-hand function min\|x-3|, |x+2|\$\min\{|x-3|, |x+2|\}$: - The intersection of |x-3| = |x+2|$|x-3| = |x+2|$ happens at x - 3 = -(x + 2) Rightarrow 2x = 1 Rightarrow x = 0.5$x - 3 = -(x + 2) \Rightarrow 2x = 1 \Rightarrow x = 0.5$. - For x le 0.5$x \le 0.5$, |x+2| le |x-3| Rightarrow min = |x+2|$|x+2| \le |x-3| \Rightarrow \min = |x+2|$. - For x > 0.5$x > 0.5$, |x-3| le |x+2| Rightarrow min = |x-3|$|x-3| \le |x+2| \Rightarrow \min = |x-3|$.

### Step 1: Check Interval x le -2$x \le -2$ Here, |x+2| = -(x+2) = -x-2$|x+2| = -(x+2) = -x-2$: x^2 + 3x + 2 = -x - 2 Rightarrow x^2 + 4x + 4 = 0$x^2 + 3x + 2 = -x - 2 \Rightarrow x^2 + 4x + 4 = 0$ (x+2)^2 = 0 Rightarrow x = -2$(x+2)^2 = 0 \Rightarrow x = -2$ This is a valid solution as it lies precisely within the interval condition boundary. ### Step 2: Check Interval -2 < x le 0.5$-2 < x \le 0.5$ Here, |x+2| = x+2$|x+2| = x+2$: x^2 + 3x + 2 = x + 2 Rightarrow x^2 + 2x = 0$x^2 + 3x + 2 = x + 2 \Rightarrow x^2 + 2x = 0$ x(x+2) = 0 Rightarrow x = 0 quad textor quad x = -2$x(x+2) = 0 \Rightarrow x = 0 \quad \text{or} \quad x = -2$ Only x = 0$x = 0$ fits inside this interval. ### Step 3: Check Interval x > 0.5$x > 0.5$ Here, min = |x-3| = 3-x$\min = |x-3| = 3-x$: x^2 + 3x + 2 = 3 - x Rightarrow x^2 + 4x - 1 = 0$x^2 + 3x + 2 = 3 - x \Rightarrow x^2 + 4x - 1 = 0$ x = frac-4 pm sqrt16 - 4(1)(-1)2 = -2 pm sqrt5$x = \frac{-4 \pm \sqrt{16 - 4(1)(-1)}}{2} = -2 \pm \sqrt{5}$ Evaluating values: -2 + sqrt5 approx 0.236$-2 + \sqrt{5} \approx 0.236$, which does not satisfy x > 0.5$x > 0.5$. Thus, no real roots occur in this span. Combining valid points, we find exactly 2 distinct real solutions (x = -2, 0$x = -2, 0$). ### Pattern Recognition Sketching a rough visualization showing the parabola crossing below the sharp wedge of the combined absolute values makes it visually clear that there are exactly two crossing points, confirming the algebraic count. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Quadratic Equations Class 12 Mathematics: Limits, Continuity and Differentiability

### Related Formula To evaluate limits of indeterminate types containing radical forms, rationalize the numerator directly by multiplying by its conjugate element matching: (sqrtA - sqrtB)(sqrtA + sqrtB) = A - B$(\sqrt{A} - \sqrt{B})(\sqrt{A} + \sqrt{B}) = A - B$ ### Core Logic Rewrite the expression as a fraction with sin x$\sin x$ in the denominator and rationalize the numerator: lim_x to 0 frac(2cos^2 x + 3cos x) - (cos^2 x + sin x + 4)sin x cdot left(sqrt2cos^2 x + 3cos x + sqrtcos^2 x + sin x + 4right)$\lim_{x \to 0} \frac{(2\cos^2 x + 3\cos x) - (\cos^2 x + \sin x + 4)}{\sin x \cdot \left(\sqrt{2\cos^2 x + 3\cos x} + \sqrt{\cos^2 x + \sin x + 4}\right)}$ = lim_x to 0 fraccos^2 x + 3cos x - sin x - 4sin x cdot left(sqrt2cos^2 x + 3cos x + sqrtcos^2 x + sin x + 4right)$= \lim_{x \to 0} \frac{\cos^2 x + 3\cos x - \sin x - 4}{\sin x \cdot \left(\sqrt{2\cos^2 x + 3\cos x} + \sqrt{\cos^2 x + \sin x + 4}\right)}$ ### Step 1: Simplify Numerator and Group Terms Express the numerator terms to isolate algebraic patterns: cos^2 x + 3cos x - 4 - sin x = (cos x - 1)(cos x + 4) - sin x$\cos^2 x + 3\cos x - 4 - \sin x = (\cos x - 1)(\cos x + 4) - \sin x$ Substitute this back into our rationalized limit format: = lim_x to 0 frac(cos x - 1)(cos x + 4) - sin xsin x cdot left(sqrt2cos^2 x + 3cos x + sqrtcos^2 x + sin x + 4right)$= \lim_{x \to 0} \frac{(\cos x - 1)(\cos x + 4) - \sin x}{\sin x \cdot \left(\sqrt{2\cos^2 x + 3\cos x} + \sqrt{\cos^2 x + \sin x + 4}\right)}$ ### Step 2: Distribute sin x$\sin x$ in Denominator Split the limit across the two separated numerator expressions: = lim_x to 0 left[ fraccos x - 1sin x cdot (cos x + 4) - 1 right] cdot frac1sqrt2(1)+3 + sqrt1+0+4$= \lim_{x \to 0} \left[ \frac{\cos x - 1}{\sin x} \cdot (\cos x + 4) - 1 \right] \cdot \frac{1}{\sqrt{2(1)+3} + \sqrt{1+0+4}}$ Evaluate the limit component values: lim_x to 0 fraccos x - 1sin x = lim_x to 0 frac-2sin^2(x/2)2sin(x/2)cos(x/2) = lim_x to 0 [-tan(x/2)] = 0$\lim_{x \to 0} \frac{\cos x - 1}{\sin x} = \lim_{x \to 0} \frac{-2\sin^2(x/2)}{2\sin(x/2)\cos(x/2)} = \lim_{x \to 0} [-\tan(x/2)] = 0$ Substituting this zero value simplifies the numerator expression directly: = left[ 0 cdot (1 + 4) - 1 right] cdot frac1sqrt5 + sqrt5 = frac-12sqrt5$= \left[ 0 \cdot (1 + 4) - 1 \right] \cdot \frac{1}{\sqrt{5} + \sqrt{5}} = \frac{-1}{2\sqrt{5}}$ ### Pattern Recognition Recognizing that fraccos x - 1sin x to 0$\frac{\cos x - 1}{\sin x} \to 0$ as x to 0$x \to 0$ isolates the non-vanishing trigonometric components without needing full multi-stage application of L'Hôpital's rule. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Limits and Derivatives

### Related Formula Trigonometric identity: fractantheta + tan^3theta1-tan^2theta = tantheta left( frac1+tan^2theta1-tan^2theta right) = fractanthetacos 2theta$\frac{\tan\theta + \tan^3\theta}{1-\tan^2\theta} = \tan\theta \left( \frac{1+\tan^2\theta}{1-\tan^2\theta} \right) = \frac{\tan\theta}{\cos 2\theta}$ Also note standard telescopic identity: tan 2phi - tanphi = fractanphicos 2phi$\tan 2\phi - \tan\phi = \frac{\tan\phi}{\cos 2\phi}$ ### Core Logic Let theta = fracx2^r+1$\theta = \frac{x}{2^{r+1}}$. The term inside the summation simplifies to: tanleft(fracx2^rright) - tanleft(fracx2^r+1right)$\tan\left(\frac{x}{2^r}\right) - \tan\left(\frac{x}{2^{r+1}}\right)$ Now, evaluating the summation: sum_r=0^n left[ tanleft(fracx2^rright) - tanleft(fracx2^r+1right) right] = tan x - tanleft(fracx2^n+1right)$\sum_{r=0}^{n} \left[ \tan\left(\frac{x}{2^r}\right) - \tan\left(\frac{x}{2^{r+1}}\right) \right] = \tan x - \tan\left(\frac{x}{2^{n+1}}\right)$ Taking the limit as n to infty$n \to \infty$, tanleft(fracx2^n+1right) to tan(0) = 0$\tan\left(\frac{x}{2^{n+1}}\right) \to \tan(0) = 0$. Therefore, f(x) = tan x$f(x) = \tan x$. ### Step 1: Evaluate the Limit We need to find: lim_xrightarrow0frace^x-e^tan xx-tan x$\lim_{x\rightarrow0}\frac{e^{x}-e^{\tan x}}{x-\tan x}$ Factor out e^tan x$e^{\tan x}$ from the numerator: lim_xrightarrow0 e^tan x cdot left[ frace^x-tan x - 1x-tan x right]$\lim_{x\rightarrow0} e^{\tan x} \cdot \left[ \frac{e^{x-\tan x} - 1}{x-\tan x} \right]$ Let u = x - tan x$u = x - \tan x$. As x to 0$x \to 0$, u to 0$u \to 0$. The limit becomes: lim_urightarrow0 e^0 cdot left[ frace^u - 1u right] = 1 times 1 = 1$\lim_{u\rightarrow0} e^0 \cdot \left[ \frac{e^u - 1}{u} \right] = 1 \times 1 = 1$ ### Pattern Recognition Standard limit substitution lim_y to 0 frace^y - 1y = 1$\lim_{y \to 0} \frac{e^y - 1}{y} = 1$ applies cleanly whenever the argument in the exponent matches the entire denominator layout. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Trigonometry Class 12 Mathematics: Limits, Continuity and Differentiability