Applications of Derivative Ex 6.5 Class 12 NCERT

Understanding Local and Absolute Extrema

To solve problems in Exercise 6.5, we must first distinguish between local and absolute maximum/minimum values. Let $f$ be a continuous function defined on an interval $I$.

1. Local Extrema: A point $c$ is a local maximum if $f(c) \ge f(x)$ for all $x$ in a small neighborhood around $c$. Conversely, $c$ is a local minimum if $f(c) \le f(x)$ in its neighborhood. These occur at critical points where $f'(c) = 0$ or where $f(x)$ is non-differentiable.
2. Absolute Extrema: This refers to the highest or lowest value of the function over the entire interval $[a, b]$. Unlike local extrema, absolute extrema can also occur at the boundary endpoints $a$ and $b$.

To find local extrema, we rely on two key tools: the First Derivative Test (tracking sign changes of $f'(x)$) and the Second Derivative Test (using the concavity of the curve).

The Second Derivative Test Protocol

1. Find the First Derivative — Differentiate the given function $f(x)$ with respect to $x$ to obtain $f'(x)$.
2. Locate Critical Points — Set $f'(x) = 0$ and solve the equation to find critical points $x = c_1, c_2, \dots$
3. Compute the Second Derivative — Differentiate $f'(x)$ again to obtain $f''(x)$.
4. Evaluate at Critical Points — Substitute $x = c$ into $f''(x)$. If $f''(c) < 0$, then $x = c$ is a local maximum. If $f''(c) > 0$, then $x = c$ is a local minimum. If $f''(c) = 0$, the test fails; revert to the First Derivative Test.

Avoid These Common Board Exam Traps

1. Show the Second Derivative Test: In word problems, students often find the critical point and directly write the answer. CBSE marking schemes reserve 1 to 1.5 marks for explicitly showing that $f''(x) < 0$ (for maximum) or $f''(x) > 0$ (for minimum). Do not skip this step!
2. Boundary Points for Closed Intervals: If the question specifies a closed interval $[a, b]$, you must evaluate $f(x)$ at the boundary points $a$ and $b$ as well as the critical points. The absolute maximum/minimum is the absolute largest/smallest of these values.
3. Dimensional Constraints: Ensure variables like radius ($r$) and height ($h$) are constrained to positive real values ($r > 0$, $h > 0$).

Exercise 6.5 High-Yield Solved Problems

• Q: Find local maximum and minimum values of the function: $f(x) = 2x^3 - 6x^2 + 6x + 5$. A: Step 1: Differentiate the function with respect to $x$. $f'(x) = 6x^2 - 12x + 6 = 6(x^2 - 2x + 1) = 6(x-1)^2$ Step 2: Set $f'(x) = 0$ to find critical points. $6(x-1)^2 = 0 \implies x = 1$ Step 3: Apply the derivative test. Note that $f''(x) = 12(x-1)$. At $x = 1$, $f''(1) = 0$. Since the Second Derivative Test fails, we use the First Derivative Test. Step 4: Check the sign of $f'(x) = 6(x-1)^2$ near $x = 1$. For $x < 1$, $f'(x) > 0$ because of the squared term. For $x > 1$, $f'(x) > 0$ as well. Since $f'(x)$ does not change sign as $x$ passes through $1$, the point $x = 1$ is a point of inflection. There are no local maximum or minimum values. Final answer: No local maximum or minimum; $x=1$ is a point of inflection.
• Q: Find the absolute maximum and minimum values of $f(x) = 3x^4 - 8x^3 + 12x^2 - 48x + 25$ on the interval $[0, 3]$. A: Step 1: Differentiate $f(x)$: $f'(x) = 12x^3 - 24x^2 + 24x - 48 = 12(x^3 - 2x^2 + 2x - 4) = 12[x^2(x-2) + 2(x-2)] = 12(x^2 + 2)(x-2)$ Step 2: Set $f'(x) = 0$. Since $x^2 + 2 = 0$ has no real roots, the only real critical point is $x = 2$, which lies in $[0, 3]$. Step 3: Evaluate $f(x)$ at critical point $x=2$ and endpoints $x=0, 3$: $f(0) = 3(0) - 8(0) + 12(0) - 48(0) + 25 = 25$ $f(2) = 3(16) - 8(8) + 12(4) - 48(2) + 25 = 48 - 64 + 48 - 96 + 25 = -39$ $f(3) = 3(81) - 8(27) + 12(9) - 48(3) + 25 = 243 - 216 + 108 - 144 + 25 = 16$ Step 4: Compare values. Absolute Maximum value is $25$ at $x = 0$. Absolute Minimum value is $-39$ at $x = 2$. Final answer: Absolute maximum is 25, absolute minimum is -39.
• Q: Show that of all the rectangles of given area $A$, the square has the smallest perimeter. A: Step 1: Let the sides of the rectangle be $x$ and $y$. Given area $A = xy \implies y = \frac{A}{x}$. Step 2: Express Perimeter $P$ as a function of $x$: $P = 2(x + y) = 2\left(x + \frac{A}{x}\right)$ Step 3: Find $\frac{dP}{dx}$ and set it to $0$: $\frac{dP}{dx} = 2\left(1 - \frac{A}{x^2}\right) = 0 \implies x^2 = A \implies x = \sqrt{A}$ (since side length $x > 0$) Step 4: Use the second derivative test to verify minimum: $\frac{d^2P}{dx^2} = 2\left(0 + \frac{2A}{x^3}\right) = \frac{4A}{x^3}$ Since $A > 0$ and $x > 0$, $\frac{d^2P}{dx^2} > 0$ at $x = \sqrt{A}$, verifying a local minimum. Step 5: Find $y$ when $x = \sqrt{A}$: $y = \frac{A}{\sqrt{A}} = \sqrt{A}$ Since $x = y$, the rectangle is a square. Final answer: Proved mathematically that a square minimizes perimeter for a given area.
• Q: Prove that the volume of the largest cone that can be inscribed in a sphere of radius $R$ is $\frac{8}{27}$ of the volume of the sphere. A: Step 1: Let $r$ and $h$ be the radius and height of the inscribed cone. Let $x$ be the distance from the center of the sphere of radius $R$ to the base of the cone. Thus, height $h = R + x$, and by Pythagoras' theorem, radius $r^2 = R^2 - x^2$. Step 2: Formulate the Volume $V$ of the cone: $V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (R^2 - x^2)(R + x)$ Step 3: Find $\frac{dV}{dx}$ and equate to $0$: $\frac{dV}{dx} = \frac{\pi}{3} [ (R^2 - x^2)(1) + (R+x)(-2x) ] = \frac{\pi}{3}(R+x)[(R-x) - 2x] = \frac{\pi}{3}(R+x)(R-3x)$ Setting $\frac{dV}{dx} = 0 \implies x = -R$ (not possible) or $x = \frac{R}{3}$. Step 4: Find $\frac{d^2V}{dx^2}$ to confirm maximum: $\frac{d^2V}{dx^2} = \frac{\pi}{3}[ (R-3x) - 3(R+x) ] = \frac{\pi}{3}(-2R - 6x)$ At $x = \frac{R}{3}$, $\frac{d^2V}{dx^2} = \frac{\pi}{3}(-2R - 2R) < 0$, which confirms maximum volume. Step 5: Calculate maximum volume $V_{max}$: $V_{max} = \frac{1}{3}\pi \left(R^2 - \frac{R^2}{9}\right)\left(R + \frac{R}{3}\right) = \frac{1}{3}\pi \left(\frac{8R^2}{9}\right)\left(\frac{4R}{3}\right) = \frac{8}{27} \left( \frac{4}{3}\pi R^3 \right) = \frac{8}{27} V_{sphere}$. Final answer: Proved that $V_{cone} = \frac{8}{27} V_{sphere}$.

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