Let f$f$ be a differentiable function satisfying f(x) = 1 - 2x + int_0^x e^(x-t) f(t) dt, x in R$f(x) = 1 - 2x + \int_{0}^{x} e^{(x-t)} f(t) dt, x \in R$ and let g(x) = int_0^x (f(t) + 2)^15 (t - 4)^6 (t + 12)^17 dt, x in R$g(x) = \int_{0}^{x} (f(t) + 2)^{15} (t - 4)^{6} (t + 12)^{17} dt, x \in R$. If p$p$ and q$q$ are respectively the points of local minima and local maxima of g$g$, then the value of |p + q|$|p + q|$ is equal to

### Related Formula g'(x) = f'(h(x)) cdot h'(x)$$g'(x) = f'(h(x)) \cdot h'(x)$$ For monotonicity, g'(x) > 0 implies$g'(x) > 0 \implies$ increasing, and g'(x) < 0 implies$g'(x) < 0 \implies$ decreasing. ### Core Logic Given g(x) = f((tan x - 1)^2 + a - 1)$g(x) = f((\tan x - 1)^2 + a - 1)$. Differentiating w.r.t. x$x$: g'(x) = f'((tan x - 1)^2 + a - 1) cdot 2(tan x - 1)sec^2 x$$g'(x) = f'((\tan x - 1)^2 + a - 1) \cdot 2(\tan x - 1)\sec^2 x$$ ### Step 1: Analyze the Sign of the Derivative We know f''(x) > 0 implies f'(x)$f''(x) > 0 \implies f'(x)$ is strictly increasing. Since f'(a-1) = 0$f'(a-1) = 0$, for any input X > a-1$X > a-1$, f'(X) > 0$f'(X) > 0$. Here, the input to f'$f'$ is (tan x - 1)^2 + a - 1$(\tan x - 1)^2 + a - 1$. Since (tan x - 1)^2 geq 0$(\tan x - 1)^2 \geq 0$, it is strictly positive for x neq fracpi4$x \neq \frac{\pi}{4}$ in the given interval. Thus, (tan x - 1)^2 + a - 1 geq a - 1 implies f'((tan x - 1)^2 + a - 1) > 0$(\tan x - 1)^2 + a - 1 \geq a - 1 \implies f'((\tan x - 1)^2 + a - 1) > 0$ for all valid x$x$. ### Step 2: Determine Intervals of Monotonicity Now, the sign of g'(x)$g'(x)$ depends solely on the term (tan x - 1)$(\tan x - 1)$ because 2sec^2 x > 0$2\sec^2 x > 0$. Case 1: x in left(0, fracpi4right)$x \in \left(0, \frac{\pi}{4}\right)$ Here tan x < 1 implies tan x - 1 < 0$\tan x < 1 \implies \tan x - 1 < 0$. So, g'(x) < 0 implies g(x)$g'(x) < 0 \implies g(x)$ is decreasing. Case 2: x in left(fracpi4, fracpi2right)$x \in \left(\frac{\pi}{4}, \frac{\pi}{2}\right)$ Here tan x > 1 implies tan x - 1 > 0$\tan x > 1 \implies \tan x - 1 > 0$. So, g'(x) > 0 implies g(x)$g'(x) > 0 \implies g(x)$ is increasing. Therefore, neither statement (I) nor (II) is true. ### Pattern Recognition When a function wraps a quadratic, f((u-k)^2 + c)$f((u-k)^2 + c)$, the critical points match the inner function's extrema. Evaluate the sign directly from the inner derivative u'$u'$ and (u-k)$(u-k)$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Application of Derivatives

### Related Formula f'(x) = 0 text gives the critical points for extreme values.$$f'(x) = 0 \text{ gives the critical points for extreme values.}$$ ### Core Logic Given f(x) = x^2025 - x^2000$f(x) = x^{2025} - x^{2000}$ on x in [0,1]$x \in [0,1]$. Differentiate to find critical points: f'(x) = 2025 x^2024 - 2000 x^1999 = 0$$f'(x) = 2025 x^{2024} - 2000 x^{1999} = 0$$ x^1999(2025x^25 - 2000) = 0$$x^{1999}(2025x^{25} - 2000) = 0$$ ### Step 1: Identifying the Minima The critical points are x = 0$x = 0$ and x = left(frac20002025right)^frac125 = left(frac8081right)^frac125$x = \left(\frac{2000}{2025}\right)^{\frac{1}{25}} = \left(\frac{80}{81}\right)^{\frac{1}{25}}$. Let this root be alpha$\alpha$. Evaluating the function at endpoints and the critical point: - f(0) = 0$f(0) = 0$ - f(1) = 1 - 1 = 0$f(1) = 1 - 1 = 0$ - f(alpha) = alpha^2025 - alpha^2000 = alpha^2000(alpha^25 - 1)$f(\alpha) = \alpha^{2025} - \alpha^{2000} = \alpha^{2000}(\alpha^{25} - 1)$ Substitute alpha^25 = frac8081$\alpha^{25} = \frac{80}{81}$: f(alpha) = left(left(frac8081right)^frac125right)^2000 cdot left(frac8081 - 1right)$$f(\alpha) = \left(\left(\frac{80}{81}\right)^{\frac{1}{25}}\right)^{2000} \cdot \left(\frac{80}{81} - 1\right)$$ f(alpha) = left(frac8081right)^80 cdot left(-frac181right)$$f(\alpha) = \left(\frac{80}{81}\right)^{80} \cdot \left(-\frac{1}{81}\right)$$ ### Step 2: Structuring into the Given Format f(alpha) = frac80^8081^80 cdot (-81)^-1 = 80^80 cdot (-81)^-81$$f(\alpha) = \frac{80^{80}}{81^{80}} \cdot (-81)^{-1} = 80^{80} \cdot (-81)^{-81}$$ Comparing this with (80)^80(n)^-81$(80)^{80}(n)^{-81}$, we immediately get n = -81$n = -81$. ### Pattern Recognition When evaluating exponents in the form x^A - x^B$x^A - x^B$, the stationary point occurs at x = (B/A)^1/(A-B)$x = (B/A)^{1/(A-B)}$. Plugging this back into the original function seamlessly factors out (B/A)^B cdot (B/A - 1)$(B/A)^B \cdot (B/A - 1)$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Applications of Derivatives

### Related Formula f'(x) < 0 text for strictly decreasing function$$f'(x) < 0 \text{ for strictly decreasing function}$$ g'(x) = 0 text to find critical points for extrema$$g'(x) = 0 \text{ to find critical points for extrema}$$ ### Core Logic For f(t) = frac|t + 1|t^2$f(t) = \frac{|t + 1|}{t^2}$, graph evaluation reveals decreasing behavior.

For t in (-2, -1)$t \in (-2, -1)$, the function is decreasing. Thus, the interval is (-2, -1)$(-2, -1)$. Given the interval is (2alpha, alpha)$(2\alpha, \alpha)$, this implies 2alpha = -2 Rightarrow alpha = -1$2\alpha = -2 \Rightarrow \alpha = -1$. ### Step 1: Finding function g(x) g(x) = 2 log_e(x - 2) - x^2 + 4x + 1 quad (x > 2)$$g(x) = 2 \log_e(x - 2) - x^2 + 4x + 1 \quad (x > 2)$$ g'(x) = frac2x - 2 - 2x + 4$$g'(x) = \frac{2}{x - 2} - 2x + 4$$ g'(x) = frac2 - 2x(x - 2) + 4(x - 2)x - 2$$g'(x) = \frac{2 - 2x(x - 2) + 4(x - 2)}{x - 2}$$ g'(x) = frac2 - 2x^2 + 4x + 4x - 8x - 2 = frac-2x^2 + 8x - 6x - 2$$g'(x) = \frac{2 - 2x^2 + 4x + 4x - 8}{x - 2} = \frac{-2x^2 + 8x - 6}{x - 2}$$ g'(x) = frac-2(x^2 - 4x + 3)x - 2 = frac-2(x - 3)(x - 1)x - 2$$g'(x) = \frac{-2(x^2 - 4x + 3)}{x - 2} = \frac{-2(x - 3)(x - 1)}{x - 2}$$ ### Step 2: Determining Maxima For x > 2$x > 2$, critical point is x = 3$x = 3$. Testing signs around x = 3$x = 3$: For 2 < x < 3$2 < x < 3$, g'(x) > 0$g'(x) > 0$ (Increasing) For x > 3$x > 3$, g'(x) < 0$g'(x) < 0$ (Decreasing) Maxima occurs at x = 3$x = 3$. ### Step 3: Calculating Local Maximum Value g(3) = 2 log_e(3 - 2) - (3)^2 + 4(3) + 1$$g(3) = 2 \log_e(3 - 2) - (3)^2 + 4(3) + 1$$ g(3) = 2 log_e(1) - 9 + 12 + 1$$g(3) = 2 \log_e(1) - 9 + 12 + 1$$ g(3) = 0 + 4 = 4$$g(3) = 0 + 4 = 4$$ ### Pattern Recognition Connecting nested parameter values (alpha$\alpha$) between independent functions ( f(t)$f(t)$ and g(x)$g(x)$) demands extreme care on interval boundaries. Derivative sign analysis effortlessly identifies maximums on restricted domains. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Application of Derivatives

### Related Formula textFor x geq 1, quad ln x geq 0 text and x-1 geq 0$$\text{For } x \geq 1, \quad \ln x \geq 0 \text{ and } x-1 \geq 0$$ textFor 0 < x < 1, quad ln x < 0 text and x-1 < 0$$\text{For } 0 < x < 1, \quad \ln x < 0 \text{ and } x-1 < 0$$ ### Core Logic Break the function into piecewise intervals based on the critical point x=1$x=1$. f(x) = begincases ln x - (x - 1) & x geq 1 \\ -ln x - (-(x - 1)) & 0 < x < 1 endcases$$f(x) = \begin{cases} \ln x - (x - 1) & x \geq 1 \\ -\ln x - (-(x - 1)) & 0 < x < 1 \end{cases}$$ f(x) = begincases ln x - x + 1 & x geq 1 \\ -ln x + x - 1 & 0 < x < 1 endcases$$f(x) = \begin{cases} \ln x - x + 1 & x \geq 1 \\ -\ln x + x - 1 & 0 < x < 1 \end{cases}$$ ### Step 1: Checking Differentiability (Statement I) Find the derivative f'(x)$f'(x)$: f'(x) = begincases frac1x - 1 & x > 1 \\ -frac1x + 1 & 0 < x < 1 endcases$$f'(x) = \begin{cases} \frac{1}{x} - 1 & x > 1 \\ -\frac{1}{x} + 1 & 0 < x < 1 \end{cases}$$ Check left-hand derivative (LHD) and right-hand derivative (RHD) at x = 1$x = 1$: textRHD = f'(1^+) = frac11 - 1 = 0$$\text{RHD} = f'(1^+) = \frac{1}{1} - 1 = 0$$ textLHD = f'(1^-) = -frac11 + 1 = 0$$\text{LHD} = f'(1^-) = -\frac{1}{1} + 1 = 0$$ Since textLHD = textRHD$\text{LHD} = \text{RHD}$, f(x)$f(x)$ is differentiable at x=1$x=1$ (and clearly everywhere else in its domain). Thus, Statement (I) is TRUE. ### Step 2: Checking Monotonicity (Statements II and III) Check for x > 1$x > 1$: f'(x) = frac1x - 1 < 0 text for all x > 1$f'(x) = \frac{1}{x} - 1 < 0 \text{ for all } x > 1$. Therefore, f$f$ is decreasing in (1, infty)$(1, \infty)$. Statement (III) is TRUE. Check for 0 < x < 1$0 < x < 1$: f'(x) = 1 - frac1x = fracx-1x < 0 text for all 0 < x < 1$f'(x) = 1 - \frac{1}{x} = \frac{x-1}{x} < 0 \text{ for all } 0 < x < 1$. Therefore, f$f$ is also decreasing in (0, 1)$(0, 1)$. Statement (II) is FALSE. Since f$f$ is decreasing in both intervals, it is monotonically decreasing across (0, infty)$(0, \infty)$. ### Step 3: Final Conclusion Only Statement (I) and Statement (III) are true. ### Pattern Recognition Adding or subtracting two modulus functions that change sign at the exact same critical point often smooths out the sharp corner, leading to a perfectly differentiable curve at that node. Always compute LHD and RHD to confirm. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Application of Derivatives Class 12 Maths: Limits, Continuity and Differentiability