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

Using the quotient rule:

If f'(x) < 0$f'(x) < 0$, the function decreases.

Core Logic

Let us compute f'(x)$f'(x)$:

Step 1: Sign Assessment

Observe that x² + 16 > 0$x^2 + 16 > 0$ for all real x$x$ and the denominator (x² - 6x - 16)² > 0$(x^2 - 6x - 16)^2 > 0$ everywhere except at the excluded boundary asymptotes x = -2$x = -2$ and x = 8$x = 8$. Therefore:

Thus, the function decreases throughout its domain intervals:

Pattern Recognition

When the numerator of the derivative simplifies to a strictly negative or positive constant/expression (like -x² - 16$-x^2 - 16$), the function exhibits monotonic behavior over all continuous subsets.

Chapter Mix

Class 12 Mathematics: Application of Derivatives

Related Formula

Core Logic

Expand the given logarithmic inequality:

Divide both sides by (x - y)$(x - y)$. The inequality sign will behave differently based on whether (x - y)$(x - y)$ is positive or negative. Case 1: Let x > y$x > y$ (so x - y > 0$x - y > 0$).

Taking the limit as y → x^-$y \to x^-$:

Step 1: Analyzing the second case

Case 2: Let x < y$x < y$ (so x - y < 0$x - y < 0$). Dividing flips the inequality sign:

Taking the limit as y → x^+$y \to x^+$:

Step 2: Squeeze Theorem deduction

Since f$f$ is given as a differentiable function, both the left-hand and right-hand limits must yield the exact same derivative. Thus, it is sandwiched between the bounds:

Step 3: Evaluating the Summation

We need to compute Σn=1²⁰ f'((1)/(n²))$\sum_{n=1}^{20} f'\left(\frac{1}{n^2}\right)$. Substitute x = (1)/(n²)$x = \frac{1}{n^2}$ into the derivative:

Apply the summation:

Pattern Recognition

Symmetric functional inequalities bounded by identical structural forms always compress down to an equality via the Sandwich/Squeeze theorem. Create the difference quotient limit to extract the derivative directly.

Chapter Mix

Class 12 Maths: Application of Derivatives Class 11 Maths: Sequences and Series Class 11 Maths: Limits and Derivatives

Related Formula

Core Logic

Since f'(1)$f'(1)$, f''(2)$f''(2)$, and f'''(3)$f'''(3)$ are evaluated at specific points, they act strictly as numerical constants. Let's differentiate the function sequentially:

f'''(x) = 6$f'''(x) = 6$

Step 1: Finding Unknown Constants

From the third derivative, evaluate it at x=3$x=3$: f'''(3) = 6$f'''(3) = 6$

Next, evaluate the second derivative at x=2$x=2$:

Finally, evaluate the first derivative at x=1$x=1$:

Rearranging gives:

Step 2: Solving the System

Substitute f''(2) = 12 + 2f'(1)$f''(2) = 12 + 2f'(1)$ into the rearranged equation:

Now find f''(2)$f''(2)$:

Step 3: Final Output Calculation

We now have the complete explicitly defined first derivative equation:

Evaluate this at x=10$x=10$:

Pattern Recognition

Functional equations that contain derivatives evaluated at specific points are just polynomials with unknown coefficients. Treat those evaluations as scalar constants (A, B, C$A, B, C$), differentiate systematically, and plug the anchors back in to build a basic system of linear equations.

Chapter Mix

Class 12 Maths: Application of Derivatives Class 11 Maths: Limits and Derivatives

Related Formula

Intermediate Value Theorem (IVT): If a function is continuous and monotonic on [a,b]$[a,b]$, and f(a)$f(a)$ and f(b)$f(b)$ have opposite signs, there is exactly one root in (a,b)$(a,b)$.

Core Logic

Analyze the derivative of f(x)$f(x)$ on the interval [(1)/(2), 1]$[\frac{1}{2}, 1]$:

For x in [(1)/(2), 1]$x \in [\frac{1}{2}, 1]$, the minimum value of x²$x^2$ is (1)/(4)$\frac{1}{4}$.

Thus, f'(x) ≥ 0$f'(x) \ge 0$ for all x in [(1)/(2), 1]$x \in [\frac{1}{2}, 1]$, meaning the function is strictly increasing on this interval.

Now, check the boundary values:

Since f((1)/(2)) lt 0$f(\frac{1}{2}) \lt 0$ and f(1) gt 0$f(1) \gt 0$ and the function is strictly increasing, it must cross the x-axis exactly once. Statement (I) is correct.

Step 1: Evaluate Statement II

To find where f(x) = 0$f(x) = 0$:

Recognizing the trigonometric identity 4 ³θ - 3 θ = 3θ$4\cos^3\theta - 3\cos\theta = \cos 3\theta$, substitute x = θ$x = \cos\theta$ (which is valid since x in [1/2, 1]$x \in [1/2, 1]$ implies θ in [0, π/3]$\theta \in [0, \pi/3]$):

Hence, the root is x = (π)/(12)$x = \cos\frac{\pi}{12}$. Statement (II) is correct.

Pattern Recognition

The structural polynomial form 4x³ - 3x$4x^3 - 3x$ is an immediate trigger for the 3θ$\cos 3\theta$ substitution. Proving monotonicity first guarantees that the single trig root found is the only root in the constrained domain.

Chapter Mix

Class 12 Mathematics: Application of Derivatives Class 11 Mathematics: Trigonometric Functions

Related Formula

To find points where a curve intersects the x-axis, set y = 0$y = 0$.

Core Logic

First, evaluate m$m$, the number of roots for f(x) = 0$f(x) = 0$:

Plotting the exponential curve y = 2^x$y = 2^x$ against the parabola y = x²$y = x^2$:

• At x = 2$x = 2$, 2² = 4$2^2 = 4$ and 2² = 4$2^2 = 4$ (Intersection 1).
• At x = 4$x = 4$, 2⁴ = 16$2^4 = 16$ and 4² = 16$4^2 = 16$ (Intersection 2).
• For x lt 0$x \lt 0$, y=2^x$y=2^x$ decays to 0 while y=x²$y=x^2$ grows to infinity. They strictly cross exactly once at x = α$x = \alpha$ (where α ≈ -0.76$\alpha \approx -0.76$).

Hence, the total number of intersections is 3. Therefore, m = 3$m = 3$.

Step 1: Evaluate n for derivative roots

Now, evaluate n$n$, the number of roots for f'(x) = 0$f'(x) = 0$:

Set f'(x) = 0 ⇒ 2^x ln 2 = 2x$f'(x) = 0 \Rightarrow 2^x \ln 2 = 2x$Plotting the scaled exponential curve$ Plotting the scaled exponential curve $y = 2^x \ln 2against the line$ against the line $y = 2x:

• The line$:
  • The line $y=2xpasses through origin with slope 2.
  • The curve$ passes through origin with slope 2.
  • The curve $y = 2^x \ln 2has a y-intercept of$ has a y-intercept of $\ln 2(approx 0.693).

  Because exponential growth eventually overtakes any linear function, it crosses the line twice for$ (approx 0.693).

  Because exponential growth eventually overtakes any linear function, it crosses the line twice for $x > 0. Since the line goes negative for$. Since the line goes negative for $x < 0and the exponential stays positive, they do not cross on the negative axis. Hence, the total number of intersections is 2. Therefore,$ and the exponential stays positive, they do not cross on the negative axis. Hence, the total number of intersections is 2. Therefore, $n = 2.

  

  Step 2: Final Computation

  Sum the variables:

  $.

  

Step 2: Final Computation

Sum the variables: