Let f(x) = (x + 3)² (x - 2)³, x in [-4, 4]$f(x) = (x + 3)^2 (x - 2)^3, x \in [-4, 4]$ . If M$M$ and m$m$ are the maximum and minimum values of f$f$ , respectively in [-4, 4]$[-4, 4]$ , then the value of M - m$M - m$ is:

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Core Logic

The vertices of the triangle are (0, 0)$(0, 0)$, (x, y)$(x, y)$, and (-x, y)$(-x, y)$. The base of the triangle lies along the horizontal line segment joining (-x, y)$(-x, y)$ and (x, y)$(x, y)$. Length of base = 2x$= 2x$. The height of the triangle from (0,0)$(0,0)$ to the line segment is y$y$. So, Area Δ = (1)/(2) (2x) (y) = xy$\Delta = \frac{1}{2} (2x) (y) = xy$ (Assuming x > 0$x > 0$).

Step 1: Setting up the area function

Substitute y = -2x² + 54$y = -2x^2 + 54$ into the area function:

Step 2: Differentiating to find maximum area

To find the maximum area, take the derivative with respect to x$x$ and set it to zero:

Setting A'(x) = 0$A'(x) = 0$:

Since x$x$ represents half the base length, x = 3$x = 3$.

Step 3: Calculating maximum area

Substitute x = 3$x = 3$ back into the area function:

Pattern Recognition

For symmetric figures inscribed under a parabolic arch, the area function A(x) = x · y(x)$A(x) = x \cdot y(x)$ is standard. Differentiate and find the critical point directly.

Chapter Mix

Class 12 Maths: Application of Derivatives

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Rolle's Theorem: If f(x)$f(x)$ is continuous on [a, b]$[a, b]$, differentiable on (a, b)$(a, b)$, and f(a) = f(b)$f(a) = f(b)$, then there exists at least one c in (a, b)$c \in (a, b)$ such that f'(c) = 0$f'(c) = 0$.

Core Logic

Given f(x) = (1)/(2)[g(x) + g(2 - x)]$f(x) = \frac{1}{2}\left[g(x) + g(2 - x)\right]$. Differentiating f(x)$f(x)$ with respect to x$x$:

Step 1: Evaluating the derivative at given points

Evaluate f'((3)/(2))$f'\left(\frac{3}{2}\right)$:

Since g'((1)/(2)) = g'((3)/(2))$g'\left(\frac{1}{2}\right) = g'\left(\frac{3}{2}\right)$, we get:

Also evaluate f'((1)/(2))$f'\left(\frac{1}{2}\right)$:

Additionally, observe that f'(1) = (g'(1) - g'(1))/(2) = 0$f'(1) = \frac{g'(1) - g'(1)}{2} = 0$.

Step 2: Applying Rolle's Theorem

We have f'((1)/(2)) = 0$f'\left(\frac{1}{2}\right) = 0$, f'(1) = 0$f'(1) = 0$, and f'((3)/(2)) = 0$f'\left(\frac{3}{2}\right) = 0$. By Rolle's Theorem on f'(x)$f'(x)$ (though we are asked about f'(x)=0$f'(x)=0$ directly), we already found three distinct roots for f'(x) = 0$f'(x) = 0$: x = (1)/(2), 1, (3)/(2)$x = \frac{1}{2}, 1, \frac{3}{2}$. Notice that x = 1/2$x = 1/2$ and x=1$x=1$ are in (0, 2)$(0, 2)$, and x=3/2$x=3/2$ is in (0, 2)$(0, 2)$. Thus f'(x) = 0$f'(x) = 0$ for at least three values in (0, 2)$(0, 2)$, which satisfies the condition "at least two x$x$ in (0, 2)$(0, 2)$".

Pattern Recognition

Symmetry in function definition g(x) + g(a-x)$g(x) + g(a-x)$ guarantees a critical point at x = a/2$x = a/2$. Additional critical points arise due to the given condition on g'(x)$g'(x)$, creating a scenario perfectly suited for Rolle's implications.

Chapter Mix

Class 12 Maths: Application of Derivatives

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Core Logic

Given f(x) = ex³ - 3x + 1$f(x) = e^{x^3 - 3x + 1}$. Check monotonicity on (-∞, -1]$(-\infty, -1]$:

For x ≤ -1$x \le -1$, (x+1) ≤ 0$(x+1) \le 0$ and (x-1) < 0$(x-1) < 0$, making f'(x) ≥ 0$f'(x) \ge 0$. Hence f(x)$f(x)$ is strictly increasing. Since f$f$ is onto (a,b]$(a,b]$, the range is dictated by the domain boundaries:

Point P(2b + 4, a + 2) ≡ P(2e³ + 4, 2)$P(2b + 4, a + 2) \equiv P(2e^3 + 4, 2)$. Find distance from line x + e⁻³y - 4 = 0$x + e^{-3}y - 4 = 0$:

Chapter Mix

Class 12 Maths: Applications of Derivatives Class 11 Maths: Straight Lines

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Core Logic

Using Newton-Leibniz formula:

The critical points are roots of f'(x) = 0$f'(x) = 0$: x = 0, (1)/(2), 2, 3, 5$x = 0, \frac{1}{2}, 2, 3, 5$.

Step 1: Sign Change Analysis

Perform the Wavy Curve Method for x in (-1, ∞)$x \in (-1, \infty)$: At x=5$x=5$ (odd power 61): sign changes - → +$- \to +$ (Local Minima) At x=3$x=3$ (even power 12): sign doesn't change (Neither) At x=2$x=2$ (odd power 7): sign changes + → -$+ \to -$ (Local Maxima) At x=1/2$x=1/2$ (odd power 5): sign changes - → +$- \to +$ (Local Minima) At x=0$x=0$ (from e^x - 1$e^x - 1$, odd power 11): sign changes + → -$+ \to -$ (Local Maxima)

Step 2: Compute p and q

Local minima at x = 5, (1)/(2)$x = 5, \frac{1}{2}$. Local maxima at x = 2, 0$x = 2, 0$.

Step 3: Final Computation

Chapter Mix

Class 12 Maths: Application of Derivatives Class 12 Maths: Integrals