Let x = -1$x = -1$ and x = 2$x = 2$ be the critical points of the function f(x) = x³ + ax² + b ₑ |x| + 1, x ≠ 0$f(x) = x^3 + ax^2 + b \log_e |x| + 1, x \neq 0$ . Let m$m$ and M$M$ respectively be the absolute minimum and the absolute maximum values of f$f$ in the interval [-2, -(1)/(2)]$\left[-2, -\frac{1}{2}\right]$ . Then |M + m|$|M + m|$ is equal to (Take ₑ 2 = 0.7$\log_e 2 = 0.7$ ):

Related Formula

For monotonicity, g'(x) > 0$g'(x) > 0 \implies$ increasing, and g'(x) < 0$g'(x) < 0 \implies$ decreasing.

Core Logic

Given g(x) = f(( x - 1)² + a - 1)$g(x) = f((\tan x - 1)^2 + a - 1)$. Differentiating w.r.t. x$x$:

Step 1: Analyze the Sign of the Derivative

We know f''(x) > 0 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 ( x - 1)² + a - 1$(\tan x - 1)^2 + a - 1$. Since ( x - 1)² ≥ 0$(\tan x - 1)^2 \geq 0$, it is strictly positive for x ≠ (π)/(4)$x \neq \frac{\pi}{4}$ in the given interval. Thus, ( x - 1)² + a - 1 ≥ a - 1 f'(( 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 ( x - 1)$(\tan x - 1)$ because 2 ² x > 0$2\sec^2 x > 0$. Case 1: x in (0, (π)/(4))$x \in \left(0, \frac{\pi}{4}\right)$ Here x < 1 x - 1 < 0$\tan x < 1 \implies \tan x - 1 < 0$. So, g'(x) < 0 g(x)$g'(x) < 0 \implies g(x)$ is decreasing.

Case 2: x in ((π)/(4), (π)/(2))$x \in \left(\frac{\pi}{4}, \frac{\pi}{2}\right)$ Here x > 1 x - 1 > 0$\tan x > 1 \implies \tan x - 1 > 0$. So, g'(x) > 0 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)$(u-k)$.

Chapter Mix

Class 12 Maths: Application of Derivatives

Related Formula

Core Logic

Given f(x) = x²⁰²⁵ - x²⁰⁰⁰$f(x) = x^{2025} - x^{2000}$ on x in [0,1]$x \in [0,1]$.

Differentiate to find critical points:

Step 1: Identifying the Minima

The critical points are x = 0$x = 0$ and x = ((2000)/(2025))(1)/(25) = ((80)/(81))(1)/(25)$x = \left(\frac{2000}{2025}\right)^{\frac{1}{25}} = \left(\frac{80}{81}\right)^{\frac{1}{25}}$. Let this root be α$\alpha$.

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(α) = α²⁰²⁵ - α²⁰⁰⁰ = α²⁰⁰⁰(α²⁵ - 1)$f(\alpha) = \alpha^{2025} - \alpha^{2000} = \alpha^{2000}(\alpha^{25} - 1)$

Substitute α²⁵ = (80)/(81)$\alpha^{25} = \frac{80}{81}$:

Step 2: Structuring into the Given Format

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 · (B/A - 1)$(B/A)^B \cdot (B/A - 1)$.

Chapter Mix

Class 12 Maths: Applications of Derivatives

Related Formula

Core Logic

For f(t) = (|t + 1|)/(t²)$f(t) = \frac{|t + 1|}{t^2}$, graph evaluation reveals decreasing behavior.

Step 1: Finding function g(x)

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

Pattern Recognition

Connecting nested parameter values (α$\alpha$) 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.

Chapter Mix

Class 12 Maths: Application of Derivatives

Related Formula

Core Logic

Break the function into piecewise intervals based on the critical point x=1$x=1$.

Step 1: Checking Differentiability (Statement I)

Find the derivative f'(x)$f'(x)$:

Check left-hand derivative (LHD) and right-hand derivative (RHD) at x = 1$x = 1$:

Since LHD = RHD$\text{LHD} = \text{RHD}$, 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) = (1)/(x) - 1 < 0 for all x > 1$f'(x) = \frac{1}{x} - 1 < 0 \text{ for all } x > 1$. Therefore, f$f$ is decreasing in (1, ∞)$(1, \infty)$. Statement (III) is TRUE.

Check for 0 < x < 1$0 < x < 1$: f'(x) = 1 - (1)/(x) = (x-1)/(x) < 0 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, ∞)$(0, \infty)$.

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.

Chapter Mix

Class 12 Maths: Application of Derivatives Class 12 Maths: Limits, Continuity and Differentiability

Related Formula

Core Logic

First, extract f(x)$f(x)$ from the integral equation: f(x) = 1 - 2x + e^x ∫₀x e-t f(t) dt$f(x) = 1 - 2x + e^x \int_{0}^{x} e^{-t} f(t) dt$ Divide by e^x$e^x$: e-x f(x) = e-x(1 - 2x) + ∫₀x e-t f(t) dt$e^{-x} f(x) = e^{-x}(1 - 2x) + \int_{0}^{x} e^{-t} f(t) dt$ Differentiate with respect to x$x$ using Leibniz rule: e-xf'(x) - e-xf(x) = e-x(-2) - e-x(1-2x) + e-xf(x)$e^{-x}f'(x) - e^{-x}f(x) = e^{-x}(-2) - e^{-x}(1-2x) + e^{-x}f(x)$ Cancel e-x$e^{-x}$: f'(x) - f(x) = -2 - 1 + 2x + f(x)$f'(x) - f(x) = -2 - 1 + 2x + f(x)$ f'(x) - 2f(x) = 2x - 3$f'(x) - 2f(x) = 2x - 3$

Execution

Solve the linear differential equation (dy)/(dx) - 2y = 2x - 3$\frac{dy}{dx} - 2y = 2x - 3$: Integrating factor IF = e∫ -2 dx = e-2x$IF = e^{\int -2 dx} = e^{-2x}$. Solution: y · e-2x = ∫ e-2x(2x - 3) dx$y \cdot e^{-2x} = \int e^{-2x}(2x - 3) dx$. Using integration by parts, it resolves to y = 1 - x$y = 1 - x$. So f(x) = 1 - x$f(x) = 1 - x$.

Now, analyze g'(x)$g'(x)$: g(x) = ∫₀x ((1 - t) + 2)¹⁵ (t - 4)⁶ (t + 12)¹⁷ dt$g(x) = \int_{0}^{x} ((1 - t) + 2)^{15} (t - 4)^{6} (t + 12)^{17} dt$ g'(x) = (3 - x)¹⁵ (x - 4)⁶ (x + 12)¹⁷ = -(x - 3)¹⁵ (x - 4)⁶ (x + 12)¹⁷$g'(x) = (3 - x)^{15} (x - 4)^{6} (x + 12)^{17} = -(x - 3)^{15} (x - 4)^{6} (x + 12)^{17}$.

Find critical points where g'(x) = 0$g'(x) = 0$: x = 3, 4, -12$x = 3, 4, -12$. Analyze sign changes of g'(x)$g'(x)$ around critical points using the wavy curve method. The leading coefficient is negative. For x = -12$x = -12$ (odd power 17), g'(x)$g'(x)$ changes from negative to positive ⇒$\Rightarrow$ Local Minima (p = -12$p = -12$). For x = 3$x = 3$ (odd power 15), g'(x)$g'(x)$ changes from positive to negative ⇒$\Rightarrow$ Local Maxima (q = 3$q = 3$). For x = 4$x = 4$ (even power 6), sign does not change.

Calculate |p + q|$|p + q|$: |p + q| = |-12 + 3| = |-9| = 9$|p + q| = |-12 + 3| = |-9| = 9$.

Pattern Recognition

Differentiating an integral equation involving ex-t$e^{x-t}$ always immediately yields a first-order linear differential equation. Once f(x)$f(x)$ is found, Newton-Leibniz on g(x)$g(x)$ converts it to a standard Wavy-Curve extrema problem.

Chapter Mix

Class 12 Maths: Application of Derivatives Class 12 Maths: Differential Equations