Mastering Applications of Derivative Ex 6.2 Class 12 NCERT

Understanding Monotonicity: The Core Concepts

To master the applications of derivative ex 6.2 class 12 ncert, we must first understand what makes a function increasing or decreasing on an interval $I$.

1. Strictly Increasing Function: A function $f(x)$ is said to be strictly increasing on an open interval $(a, b)$ if for any two points $x_1, x_2 \in (a, b)$, $x_1 < x_2 \implies f(x_1) < f(x_2)$. Visually, the graph rises continuously from left to right.

1. Strictly Decreasing Function: A function $f(x)$ is strictly decreasing on $(a, b)$ if $x_1 < x_2 \implies f(x_1) > f(x_2)$. The graph falls continuously.

The Derivative Connection:
Calculus simplifies this analysis. Instead of comparing infinite point pairs, we check the sign of the first derivative, $f'(x)$:

• If $f'(x) > 0$ for all $x \in (a, b)$, then $f(x)$ is strictly increasing in $(a, b)$.
• If $f'(x) < 0$ for all $x \in (a, b)$, then $f(x)$ is strictly decreasing in $(a, b)$.
• If $f'(x) = 0$ for all $x \in (a, b)$, then $f(x)$ is a constant function.

Step-by-Step Method to Find Intervals of Monotonicity

1. Step 1: Differentiate the Function — Find the first derivative of the given function, $f'(x)$, using standard rules of differentiation (Product, Quotient, or Chain rule).
2. Step 2: Find Critical Points — Set $f'(x) = 0$ and solve for $x$. These values are the critical points that divide the real number line (or the given domain) into sub-intervals.
3. Step 3: Test the Intervals — Pick an arbitrary test point from each sub-interval and substitute it into $f'(x)$. Determine if $f'(x)$ is positive ($>0$) or negative ($<0$) in that interval.
4. Step 4: State the Final Intervals — Conclude which intervals are strictly increasing (where $f'(x) > 0$) and which are strictly decreasing (where $f'(x) < 0$).

Common Exam Traps & NCERT Guidelines

• Open vs. Closed Intervals: For strictly increasing/decreasing, NCERT solutions generally use open intervals $(a, b)$ where the derivative is strictly positive or negative. Do not include critical points where $f'(x) = 0$ inside strictly increasing intervals.
• Trigonometric Domain Bounds: Pay close attention to restricted domains like $[0, 2\pi]$ or $[0, \pi/2]$ in trigonometric problems. Do not solve for critical points outside the given domain limit.
• Sign Mistakes with Negative Coefficients: When factoring, if your leading coefficient is negative, remember that it reverses the signs of the sub-intervals when using the wavy-curve method.

Worked Examples from NCERT Exercise 6.2

• Example 1 (Polynomial Function): Find the intervals in which the function $f(x) = 2x^3 - 3x^2 - 36x + 7$ is strictly increasing or strictly decreasing.

  Step 1: Differentiate.
  $f'(x) = \frac{d}{dx}(2x^3 - 3x^2 - 36x + 7) = 6x^2 - 6x - 36$

  Step 2: Find critical points by setting $f'(x) = 0$.
  $6(x^2 - x - 6) = 0 \implies 6(x - 3)(x + 2) = 0$
  Thus, $x = 3$ and $x = -2$. These points divide the real line into three intervals: $(-\infty, -2)$, $(-2, 3)$, and $(3, \infty)$.

  Step 3: Test intervals.

  • For $(-\infty, -2)$, choose $x = -3$: $f'(-3) = 6(-3 - 3)(-3 + 2) = 6(-6)(-1) = 36 > 0$ (Strictly Increasing).
  • For $(-2, 3)$, choose $x = 0$: $f'(0) = 6(0 - 3)(0 + 2) = -36 < 0$ (Strictly Decreasing).
  • For $(3, \infty)$, choose $x = 4$: $f'(4) = 6(4 - 3)(4 + 2) = 36 > 0$ (Strictly Increasing).

  Step 4: Conclusion.
  The function is strictly increasing in $(-\infty, -2) \cup (3, \infty)$ and strictly decreasing in $(-2, 3)$.

• Example 2 (Trigonometric Function): Find the intervals in which $f(x) = \sin x + \cos x$ is strictly increasing or strictly decreasing for $0 \le x \le 2\pi$.

  Step 1: Differentiate.
  $f'(x) = \cos x - \sin x$

  Step 2: Find critical points.
  Set $f'(x) = 0 \implies \cos x - \sin x = 0 \implies \tan x = 1$
  In the domain $[0, 2\pi]$, $\tan x = 1$ at $x = \frac{\pi}{4}$ and $x = \frac{5\pi}{4}$.
  This divides the domain into three intervals: $[0, \frac{\pi}{4})$, $(\frac{\pi}{4}, \frac{5\pi}{4})$, and $(\frac{5\pi}{4}, 2\pi]$.

  Step 3: Test intervals.

  • In $[0, \frac{\pi}{4})$, choose $x = 0$: $f'(0) = \cos 0 - \sin 0 = 1 > 0$ (Strictly Increasing).
  • In $(\frac{\pi}{4}, \frac{5\pi}{4})$, choose $x = \frac{\pi}{2}$: $f'(\frac{\pi}{2}) = \cos \frac{\pi}{2} - \sin \frac{\pi}{2} = -1 < 0$ (Strictly Decreasing).
  • In $(\frac{5\pi}{4}, 2\pi]$, choose $x = \frac{3\pi}{2}$: $f'(\frac{3\pi}{2}) = \cos \frac{3\pi}{2} - \sin \frac{3\pi}{2} = 0 - (-1) = 1 > 0$ (Strictly Increasing).

  Step 4: Conclusion.
  The function is strictly increasing in $[0, \frac{\pi}{4}) \cup (\frac{5\pi}{4}, 2\pi]$ and strictly decreasing in $(\frac{\pi}{4}, \frac{5\pi}{4})$.

Practice Questions with Solutions

• Q: Show that the function $f(x) = e^{2x}$ is strictly increasing on $\mathbb{R}$.
  A: Step 1: Differentiate the given function with respect to $x$.
  $f'(x) = \frac{d}{dx}(e^{2x}) = 2e^{2x}$

  Step 2: Analyze the sign of the derivative.
  We know that the exponential function $e^y > 0$ for all real values of $y$. Therefore, $e^{2x} > 0$ for all $x \in \mathbb{R}$.
  Multiplying by $2$, we get $2e^{2x} > 0$ for all $x \in \mathbb{R}$.

  Step 3: Formulate the conclusion.
  Since $f'(x) > 0$ for all real numbers $x$, the function $f(x) = e^{2x}$ is strictly increasing on $\mathbb{R}$.
  Final answer: $f(x) = e^{2x}$ is strictly increasing on $\mathbb{R}$.

• Q: Find the intervals in which the function $f(x) = x^2 - 4x + 6$ is (a) strictly increasing, (b) strictly decreasing.
  A: Step 1: Differentiate the function.
  $f'(x) = 2x - 4$

  Step 2: Find the critical points.
  Set $f'(x) = 0 \implies 2x - 4 = 0 \implies x = 2$.
  This critical point divides the real line into two intervals: $(-\infty, 2)$ and $(2, \infty)$.

  Step 3: Test each interval.

  • For $(-\infty, 2)$, choose a test point $x = 0$. $f'(0) = 2(0) - 4 = -4 < 0$. Thus, $f(x)$ is strictly decreasing.
  • For $(2, \infty)$, choose a test point $x = 3$. $f'(3) = 2(3) - 4 = 2 > 0$. Thus, $f(x)$ is strictly increasing.

  Final answer: (a) Strictly increasing in $(2, \infty)$, (b) Strictly decreasing in $(-\infty, 2)$.

• Q: Prove that the function $f(x) = 3x + 17$ is strictly increasing on $\mathbb{R}$.
  A: Step 1: Find the first derivative.
  $f'(x) = \frac{d}{dx}(3x + 17) = 3$

  Step 2: Test the sign of $f'(x)$.
  Since $3$ is a positive constant, $f'(x) = 3 > 0$ for all $x \in \mathbb{R}$.

  Step 3: Conclusion.
  Since the derivative is strictly greater than zero everywhere, the function is strictly increasing.
  Final answer: $f(x)$ is strictly increasing on $\mathbb{R}$.

• Q: Find the intervals in which the function $f(x) = -2x^3 - 9x^2 - 12x + 1$ is strictly increasing or strictly decreasing.
  A: Step 1: Find $f'(x)$.
  $f'(x) = -6x^2 - 18x - 12$

  Step 2: Set $f'(x) = 0$ to find critical points.
  $-6(x^2 + 3x + 2) = 0 \implies -6(x + 1)(x + 2) = 0$
  Critical points are $x = -1$ and $x = -2$. These partition the real line into $(-\infty, -2)$, $(-2, -1)$, and $(-1, \infty)$.

  Step 3: Test the intervals.

  • For $(-\infty, -2)$, try $x = -3$: $f'(-3) = -6(-3+1)(-3+2) = -6(-2)(-1) = -12 < 0$ (Strictly decreasing).
  • For $(-2, -1)$, try $x = -1.5$: $f'(-1.5) = -6(-0.5)(0.5) = 1.5 > 0$ (Strictly increasing).
  • For $(-1, \infty)$, try $x = 0$: $f'(0) = -12 < 0$ (Strictly decreasing).

  Final answer: Strictly increasing in $(-2, -1)$ and strictly decreasing in $(-\infty, -2) \cup (-1, \infty)$.

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