NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.5

(i) Solution for $f(x) = (2x - 1)^2 + 3$

Step 1: Analyze the squared term. We know that for any real number $x$, the square of a term is always non-negative. $$(2x - 1)^2 \ge 0$$

Step 2: Add 3 to both sides of the inequality. $$(2x - 1)^2 + 3 \ge 0 + 3$$ $$f(x) \ge 3$$ The minimum value of $f(x)$ is 3. This occurs when $(2x-1)^2 = 0$, which is at $x=1/2$.

Step 3: Check for maximum value. Since $(2x-1)^2$ can be made arbitrarily large by choosing a large $x$, the function $f(x)$ can also be made arbitrarily large. Thus, it has no maximum value.

Final Answer: Minimum value = 3. Maximum value does not exist.

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(ii) Solution for $f(x) = 9x^2 + 12x + 2$

Step 1: Convert the quadratic function into vertex form by completing the square. $$f(x) = 9x^2 + 12x + 2 = (3x)^2 + 2(3x)(2) + (2)^2 - (2)^2 + 2$$ $$f(x) = (3x + 2)^2 - 4 + 2$$ $$f(x) = (3x + 2)^2 - 2$$

Step 2: Analyze the squared term. As before, $(3x+2)^2 \ge 0$. $$(3x + 2)^2 - 2 \ge 0 - 2$$ $$f(x) \ge -2$$ The minimum value of $f(x)$ is -2. This occurs at $3x+2 = 0$, or $x = -2/3$.

Step 3: Check for maximum value. The function is a parabola opening upwards, so it increases indefinitely. It has no maximum value.

Final Answer: Minimum value = -2. Maximum value does not exist.

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(iii) Solution for $f(x) = -(x - 1)^2 + 10$

Step 1: Analyze the squared term. $$(x - 1)^2 \ge 0$$

Step 2: Multiply by -1 (this reverses the inequality sign) and add 10. $$-(x - 1)^2 \le 0$$ $$-(x - 1)^2 + 10 \le 0 + 10$$ $$f(x) \le 10$$ The maximum value of $f(x)$ is 10. This occurs when $(x-1)^2 = 0$, or $x=1$.

Step 3: Check for minimum value. The function is a parabola opening downwards, so it decreases indefinitely. It has no minimum value.

Final Answer: Maximum value = 10. Minimum value does not exist.

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(iv) Solution for $g(x) = x^3 + 1$

Step 1: Analyze the behavior of the function as $x$ approaches infinity and negative infinity. As $x \to \infty$, $g(x) \to \infty$. As $x \to -\infty$, $g(x) \to -\infty$.

Step 2: Use calculus. Find the derivative. $$g'(x) = 3x^2$$ Setting $g'(x)=0$ gives $x=0$. However, $g'(x)$ is always non-negative ($3x^2 \ge 0$), meaning the function is always increasing (or stationary at $x=0$). It never turns around to form a maximum or minimum.

Final Answer: The function has no maximum or minimum value.

(i) Solution for $f(x) = |x + 2| - 1$

Step 1: The absolute value function $|a|$ has a minimum value of 0. So, for all real $x$, $$|x + 2| \ge 0$$

Step 2: Subtract 1 from both sides. $$|x + 2| - 1 \ge 0 - 1$$ $$f(x) \ge -1$$ The minimum value is -1, which occurs when $|x+2|=0$, i.e., at $x=-2$. The function is not bounded above, so there is no maximum value.

Final Answer: Minimum value = -1. Maximum value does not exist.

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(ii) Solution for $g(x) = -|x + 1| + 3$

Step 1: We know $|x + 1| \ge 0$. Multiplying by -1 reverses the inequality. $$-|x + 1| \le 0$$

Step 2: Add 3 to both sides. $$-|x + 1| + 3 \le 3$$ $$g(x) \le 3$$ The maximum value is 3, which occurs when $|x+1|=0$, i.e., at $x=-1$. The function is not bounded below, so there is no minimum value.

Final Answer: Maximum value = 3. Minimum value does not exist.

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(iii) Solution for $h(x) = \sin(2x) + 5$

Step 1: The range of the sine function is $[-1, 1]$. $$-1 \le \sin(2x) \le 1$$

Step 2: Add 5 to all parts of the inequality. $$-1 + 5 \le \sin(2x) + 5 \le 1 + 5$$ $$4 \le h(x) \le 6$$

Final Answer: Maximum value = 6, Minimum value = 4.

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(iv) Solution for $f(x) = |\sin(4x + 3)|$

Step 1: The range of $\sin(4x+3)$ is $[-1, 1]$. $$-1 \le \sin(4x + 3) \le 1$$

Step 2: The absolute value of any number in this range will be between 0 and 1. For example, $|-1|=1$, $|0|=0$, $|0.5|=0.5$, $|1|=1$. $$0 \le |\sin(4x + 3)| \le 1$$

Final Answer: Maximum value = 1, Minimum value = 0.

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(v) Solution for $h(x) = x + 1, x \in (-1, 1)$

Step 1: This is a linear function defined on an open interval. Let's check the values at the boundaries of the interval. As $x \to -1$ from the right, $h(x) \to -1 + 1 = 0$. As $x \to 1$ from the left, $h(x) \to 1 + 1 = 2$.

Step 2: Since the interval is open, the function never actually reaches the values 0 and 2. It gets arbitrarily close to them, but never attains them. Therefore, there is no absolute minimum or maximum value on this open interval.

Final Answer: No maximum or minimum value.

(i) Solution for $f(x) = x^2$

Step 1: Find the first derivative and critical points. $$f'(x) = 2x$$ Set $f'(x) = 0 \implies 2x = 0 \implies x = 0$.

Step 2: Use the second derivative test. $$f''(x) = 2$$ Since $f''(0) = 2 > 0$, $x=0$ is a point of local minimum.

Step 3: Find the local minimum value. Local minimum value = $f(0) = (0)^2 = 0$.

Final Answer: Local minimum at $x=0$, value = 0.

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(ii) Solution for $g(x) = x^3 - 3x$

Step 1: Find the first derivative and critical points. $$g'(x) = 3x^2 - 3 = 3(x^2 - 1)$$ Set $g'(x) = 0 \implies 3(x^2 - 1) = 0 \implies x = \pm 1$.

Step 2: Use the second derivative test. $$g''(x) = 6x$$ At $x=1$, $g''(1) = 6(1) = 6 > 0 \implies$ Local Minimum. At $x=-1$, $g''(-1) = 6(-1) = -6 < 0 \implies$ Local Maximum.

Step 3: Find the values. Local minimum value = $g(1) = (1)^3 - 3(1) = -2$. Local maximum value = $g(-1) = (-1)^3 - 3(-1) = -1 + 3 = 2$.

Final Answer: Local maximum at $x=-1$, value = 2. Local minimum at $x=1$, value = -2.

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(iii) Solution for $h(x) = \sin x + \cos x, 0 < x < \pi/2$

Step 1: Find the derivative. $$h'(x) = \cos x - \sin x$$ Set $h'(x) = 0 \implies \cos x = \sin x \implies \tan x = 1$. In the interval $(0, \pi/2)$, this occurs at $x = \pi/4$.

Step 2: Second derivative test. $$h''(x) = -\sin x - \cos x$$ At $x = \pi/4$, $h''(\pi/4) = -\sin(\pi/4) - \cos(\pi/4) = -1/\sqrt{2} - 1/\sqrt{2} = -2/\sqrt{2} = -\sqrt{2} < 0$. This indicates a local maximum.

Step 3: Find the value. Local maximum value = $h(\pi/4) = \sin(\pi/4) + \cos(\pi/4) = 1/\sqrt{2} + 1/\sqrt{2} = \sqrt{2}$.

Final Answer: Local maximum at $x=\pi/4$, value = $\sqrt{2}$.

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(iv) Solution for $f(x) = \sin^4 x + \cos^4 x, 0 < x < \pi/2$

Step 1: Find the derivative. $$f'(x) = 4\sin^3 x \cos x + 4\cos^3 x (-\sin x)$$ $$f'(x) = 4\sin x \cos x (\sin^2 x - \cos^2 x)$$ $$f'(x) = 2\sin(2x) (-\cos(2x)) = -\sin(4x)$$

Step 2: Find critical points. Set $f'(x) = 0 \implies -\sin(4x) = 0$. This means $4x = n\pi$. Since $0 < x < \pi/2$, we have $0 < 4x < 2\pi$. So, $4x = \pi \implies x = \pi/4$.

Step 3: Second derivative test. $$f''(x) = -4\cos(4x)$$ At $x = \pi/4$, $f''(\pi/4) = -4\cos(\pi) = -4(-1) = 4 > 0$. This indicates a local minimum.

Step 4: Find the value. Local minimum value = $f(\pi/4) = (\sin(\pi/4))^4 + (\cos(\pi/4))^4 = (1/\sqrt{2})^4 + (1/\sqrt{2})^4 = 1/4 + 1/4 = 1/2$.

Final Answer: Local minimum at $x=\pi/4$, value = 1/2.

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(v) Solution for $f(x) = x^3 - 6x^2 + 9x + 15$

Step 1: Find the derivative and critical points. $$f'(x) = 3x^2 - 12x + 9 = 3(x^2 - 4x + 3) = 3(x-1)(x-3)$$ Set $f'(x)=0 \implies x=1$ or $x=3$.

Step 2: Second derivative test. $$f''(x) = 6x - 12$$ At $x=1$, $f''(1) = 6(1) - 12 = -6 < 0 \implies$ Local Maximum. At $x=3$, $f''(3) = 6(3) - 12 = 6 > 0 \implies$ Local Minimum.

Step 3: Find the values. Local maximum value = $f(1) = 1 - 6 + 9 + 15 = 19$. Local minimum value = $f(3) = (3)^3 - 6(3)^2 + 9(3) + 15 = 27 - 54 + 27 + 15 = 15$.

Final Answer: Local maximum at $x=1$, value = 19. Local minimum at $x=3$, value = 15.

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(vi) Solution for $g(x) = x/2 + 2/x, x > 0$

Step 1: Find the derivative. $$g'(x) = 1/2 - 2/x^2$$ Set $g'(x)=0 \implies 1/2 = 2/x^2 \implies x^2 = 4$. Since $x>0$, we take $x=2$.

Step 2: Second derivative test. $$g''(x) = -2(-2x^{-3}) = 4/x^3$$ At $x=2$, $g''(2) = 4/(2)^3 = 4/8 = 1/2 > 0 \implies$ Local Minimum.

Step 3: Find the value. Local minimum value = $g(2) = 2/2 + 2/2 = 1 + 1 = 2$.

Final Answer: Local minimum at $x=2$, value = 2.

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(vii) Solution for $g(x) = 1/(x^2 + 2)$

Step 1: Find the derivative. $$g'(x) = \frac{-(2x)}{(x^2+2)^2}$$ Set $g'(x)=0 \implies -2x = 0 \implies x=0$.

Step 2: Second derivative test. $$g''(x) = \frac{-2(x^2+2)^2 - (-2x)(2(x^2+2)(2x))}{(x^2+2)^4} = \frac{-2(x^2+2) + 8x^2}{(x^2+2)^3} = \frac{6x^2-4}{(x^2+2)^3}$$ At $x=0$, $g''(0) = -4/8 = -1/2 < 0 \implies$ Local Maximum.

Step 3: Find the value. Local maximum value = $g(0) = 1/(0^2+2) = 1/2$.

Final Answer: Local maximum at $x=0$, value = 1/2.

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(viii) Solution for $f(x) = x\sqrt{1-x}, 0 < x < 1$

Step 1: Find the derivative using the product rule. $$f'(x) = (1)\sqrt{1-x} + x \cdot \frac{1}{2\sqrt{1-x}}(-1)$$ $$f'(x) = \sqrt{1-x} - \frac{x}{2\sqrt{1-x}} = \frac{2(1-x) - x}{2\sqrt{1-x}} = \frac{2 - 3x}{2\sqrt{1-x}}$$

Step 2: Find critical points. Set $f'(x)=0 \implies 2-3x=0 \implies x=2/3$.

Step 3: First derivative test. For $x < 2/3$ (e.g., $x=0.5$), $f'(x) > 0$. For $x > 2/3$ (e.g., $x=0.8$), $f'(x) < 0$. The sign changes from positive to negative, indicating a local maximum.

Step 4: Find the value. Local maximum value = $f(2/3) = (2/3)\sqrt{1-2/3} = (2/3)\sqrt{1/3} = \frac{2}{3\sqrt{3}}$.

Final Answer: Local maximum at $x=2/3$, value = $2/(3\sqrt{3})$.

(i) Solution for $f(x) = e^x$

Step 1: Find the derivative. $f'(x) = e^x$.

Step 2: Check for critical points. The exponential function $e^x$ is always positive for all real $x$. Therefore, $f'(x) = e^x \neq 0$ for any value of $x$. Since there are no critical points, the function has no maxima or minima.

Final Answer: $f(x) = e^x$ does not have a maximum or minimum.

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(ii) Solution for $g(x) = \log x$

Step 1: Find the derivative. The domain of $\log x$ is $x > 0$. $g'(x) = 1/x$.

Step 2: Check for critical points. For $x>0$, $g'(x) = 1/x$ is always positive and can never be zero. Since there are no critical points, the function has no maxima or minima.

Final Answer: $g(x) = \log x$ does not have a maximum or minimum.

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(iii) Solution for $h(x) = x^3 + x^2 + x + 1$

Step 1: Find the derivative. $h'(x) = 3x^2 + 2x + 1$.

Step 2: Check if the derivative can be zero. We analyze the quadratic $3x^2 + 2x + 1$. Its discriminant is $D = b^2 - 4ac = (2)^2 - 4(3)(1) = 4 - 12 = -8$.

Step 3: Since the discriminant is negative ($D < 0$) and the leading coefficient is positive ($a=3>0$), the quadratic $3x^2 + 2x + 1$ is always positive for all real $x$. Thus, $h'(x) \neq 0$ for any value of $x$. No critical points exist.

Final Answer: $h(x)$ does not have a maximum or minimum.

(i) Solution for $f(x) = x^3, x \in [-2, 2]$

Step 1: Find critical points. $f'(x) = 3x^2$. Set $f'(x)=0 \implies x=0$. This point lies in the interval $[-2, 2]$.

Step 2: Evaluate the function at the critical point and the endpoints of the interval. $$f(0) = 0^3 = 0$$ $$f(-2) = (-2)^3 = -8$$ $$f(2) = 2^3 = 8$$

Step 3: Compare the values. The largest value is the absolute maximum and the smallest is the absolute minimum. Maximum value is 8. Minimum value is -8.

Final Answer: Absolute maximum = 8 (at $x=2$), Absolute minimum = -8 (at $x=-2$).

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(ii) Solution for $f(x) = \sin x + \cos x, x \in [0, \pi]$

Step 1: Find critical points. $f'(x) = \cos x - \sin x$. Set $f'(x)=0 \implies \tan x = 1$. In $[0, \pi]$, this occurs at $x = \pi/4$.

Step 2: Evaluate the function at the critical point and endpoints. $$f(0) = \sin 0 + \cos 0 = 0 + 1 = 1$$ $$f(\pi/4) = \sin(\pi/4) + \cos(\pi/4) = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \sqrt{2} \approx 1.414$$ $$f(\pi) = \sin \pi + \cos \pi = 0 + (-1) = -1$$

Step 3: Compare the values. Maximum value is $\sqrt{2}$. Minimum value is -1.

Final Answer: Absolute maximum = $\sqrt{2}$ (at $x=\pi/4$), Absolute minimum = -1 (at $x=\pi$).

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(iii) Solution for $f(x) = 4x - \frac{1}{2}x^2, x \in [-2, 9/2]$

Step 1: Find critical points. $f'(x) = 4 - x$. Set $f'(x)=0 \implies x=4$. This point lies in $[-2, 9/2]$ since $9/2=4.5$.

Step 2: Evaluate at critical point and endpoints. $$f(-2) = 4(-2) - \frac{1}{2}(-2)^2 = -8 - 2 = -10$$ $$f(4) = 4(4) - \frac{1}{2}(4)^2 = 16 - 8 = 8$$ $$f(9/2) = 4(9/2) - \frac{1}{2}(9/2)^2 = 18 - \frac{81}{8} = \frac{144-81}{8} = \frac{63}{8} = 7.875$$

Step 3: Compare the values. Maximum value is 8. Minimum value is -10.

Final Answer: Absolute maximum = 8 (at $x=4$), Absolute minimum = -10 (at $x=-2$).

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(iv) Solution for $f(x) = (x-1)^2 + 3, x \in [-3, 1]$

Step 1: Find critical points. $f'(x) = 2(x-1)$. Set $f'(x)=0 \implies x=1$. This is one of the endpoints.

Step 2: Evaluate at critical point (which is an endpoint) and the other endpoint. $$f(-3) = (-3-1)^2 + 3 = (-4)^2 + 3 = 16 + 3 = 19$$ $$f(1) = (1-1)^2 + 3 = 0 + 3 = 3$$

Step 3: Compare the values. Maximum value is 19. Minimum value is 3.

Final Answer: Absolute maximum = 19 (at $x=-3$), Absolute minimum = 3 (at $x=1$).

Given: Profit function $p(x) = 41 - 72x - 18x^2$.

Step 1: Find the derivative to locate critical points. $$p'(x) = -72 - 36x$$

Step 2: Set the derivative to zero to find the value of $x$ that maximizes profit. $$-72 - 36x = 0 \implies 36x = -72 \implies x = -2$$

Step 3: Use the second derivative test to confirm it's a maximum. $$p''(x) = -36$$ Since $p''(x) = -36 < 0$, the profit is maximum at $x=-2$. (Assuming $x$ can be negative, which might not be realistic for units of a product, but we proceed mathematically).

Step 4: Calculate the maximum profit by substituting $x=-2$ into the profit function. $$p(-2) = 41 - 72(-2) - 18(-2)^2$$ $$p(-2) = 41 + 144 - 18(4) = 41 + 144 - 72 = 113$$

Final Answer: The maximum profit is 113. Note: The OCR for Q6 appears incorrect. The question in NCERT is $p(x) = 41-24x-18x^2$. Let's solve that one.

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Corrected Q6 from NCERT Book: $p(x) = 41 - 24x - 18x^2$

Step 1: $p'(x) = -24 - 36x$.

Step 2: $-24 - 36x = 0 \implies 36x = -24 \implies x = -24/36 = -2/3$.

Step 3: $p''(x) = -36 < 0$, confirming a maximum.

Step 4: Max Profit = $p(-2/3) = 41 - 24(-2/3) - 18(-2/3)^2 = 41 + 16 - 18(4/9) = 41 + 16 - 8 = 49$.

Final Answer (for correct question): The maximum profit is 49.

Given: $f(x) = 3x^4 - 8x^3 + 12x^2 - 48x + 25$ on the interval $[0, 3]$.

Step 1: Find the derivative and critical points. $$f'(x) = 12x^3 - 24x^2 + 24x - 48$$ $$f'(x) = 12(x^3 - 2x^2 + 2x - 4)$$ $$f'(x) = 12(x^2 + 2)(x - 2)$$ Set $f'(x) = 0 \implies 12(x^2 + 2)(x - 2) = 0$. Since $x^2 + 2 > 0$ for all real numbers, we get $x = 2$.

Step 2: Evaluate the function at the critical point and the endpoints of the interval ($x=0, 2, \text{and } 3$). $$f(0) = 3(0)^4 - 8(0)^3 + 12(0)^2 - 48(0) + 25 = 25$$ $$f(2) = 3(16) - 8(8) + 12(4) - 48(2) + 25 = 48 - 64 + 48 - 96 + 25 = -39$$ $$f(3) = 3(81) - 8(27) + 12(9) - 48(3) + 25 = 243 - 216 + 108 - 144 + 25 = 16$$

Final Answer: Maximum value = 25 (at $x=0$). Minimum value = -39 (at $x=2$).

Given: $f(x) = \sin 2x$ on the interval $[0, 2\pi]$.

Step 1: Find the derivative to locate critical points. $$f'(x) = 2\cos 2x$$ Set $f'(x) = 0 \implies \cos 2x = 0$. In the given interval, this happens when $2x = \pi/2, 3\pi/2, 5\pi/2, \text{and } 7\pi/2$. $$\implies x = \pi/4, 3\pi/4, 5\pi/4, 7\pi/4$$

Step 2: Evaluate to find where it attains the maximum value. We know the maximum value of the sine function is $1$. $$\sin 2x = 1$$ This occurs when $2x = \pi/2$ or $2x = 5\pi/2$. $$\implies x = \pi/4 \text{ or } x = 5\pi/4$$

Final Answer: $\sin 2x$ attains its maximum value at $x = \pi/4$ and $x = 5\pi/4$.

Given: $f(x) = \sin x + \cos x$

Step 1: Consider the interval $[0, 2\pi]$. Find the derivative to locate critical points. [cite: 403] $$f'(x) = \cos x - \sin x$$ [cite: 404] Set $f'(x) = 0 \implies \cos x = \sin x \implies \tan x = 1$. This gives $x = \pi/4$ and $x = 5\pi/4$.

Step 2: Use the second derivative test. $$f''(x) = -\sin x - \cos x$$ [cite: 407] At $x = \pi/4$: $f''(\pi/4) = -1/\sqrt{2} - 1/\sqrt{2} = -\sqrt{2} < 0$, which indicates a Maximum. At $x = 5\pi/4$: $f''(5\pi/4) = -(-1/\sqrt{2}) - (-1/\sqrt{2}) = \sqrt{2} > 0$, which indicates a Minimum.

Step 3: Calculate the maximum value. Maximum value $= f(\pi/4) = \sin(\pi/4) + \cos(\pi/4) = 1/\sqrt{2} + 1/\sqrt{2} = 2/\sqrt{2} = \sqrt{2}$.

Final Answer: The maximum value is $\sqrt{2}$.

Given: $f(x) = 2x^3 - 24x + 107$

Step 1: Find the derivative and critical points. $$f'(x) = 6x^2 - 24 = 6(x^2 - 4) = 6(x - 2)(x + 2)$$ Set $f'(x) = 0 \implies x = 2$ or $x = -2$.

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For the interval $[1, 3]$:

Step 2: The critical point in $[1, 3]$ is $x = 2$. Evaluate the function at the critical point and the endpoints ($x = 1, 2, 3$). $$f(1) = 2(1)^3 - 24(1) + 107 = 2 - 24 + 107 = 85$$ $$f(2) = 2(2)^3 - 24(2) + 107 = 16 - 48 + 107 = 75$$ $$f(3) = 2(3)^3 - 24(3) + 107 = 54 - 72 + 107 = 89$$

Final Answer for $[1, 3]$: Maximum value $= 89$ (at $x=3$).

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For the interval $[-3, -1]$:

Step 3: The critical point in $[-3, -1]$ is $x = -2$. Evaluate the function at the critical point and endpoints ($x = -3, -2, -1$). $$f(-3) = 2(-3)^3 - 24(-3) + 107 = -54 + 72 + 107 = 125$$ $$f(-2) = 2(-2)^3 - 24(-2) + 107 = -16 + 48 + 107 = 139$$ $$f(-1) = 2(-1)^3 - 24(-1) + 107 = -2 + 24 + 107 = 129$$

Final Answer for $[-3, -1]$: Maximum value $= 139$ (at $x=-2$).

Given: $f(x) = x^4 - 62x^2 + ax + 9$ has a maximum at $x = 1$.

Step 1: Find the derivative of the function. $$f'(x) = 4x^3 - 124x + a$$

Step 2: Since the function attains its maximum at $x = 1$, the first derivative at $x = 1$ must be zero. $$f'(1) = 0$$ $$4(1)^3 - 124(1) + a = 0$$ $$4 - 124 + a = 0$$ $$-120 + a = 0 \implies a = 120$$

Step 3: You can verify this using the second derivative test. $$f''(x) = 12x^2 - 124$$ At $x=1$, $f''(1) = 12(1)^2 - 124 = -112 < 0$, which confirms it is a maximum.

Final Answer: $a = 120$.

Given: $f(x) = x + \sin 2x$ on the interval $[0, 2\pi]$

Step 1: Find the derivative to locate critical points. $$f'(x) = 1 + 2\cos 2x$$ Set $f'(x) = 0 \implies 2\cos 2x = -1 \implies \cos 2x = -\frac{1}{2}$

Step 2: Find the values of $x$ in the given interval. We know $\cos 2x = -\frac{1}{2}$ when $2x = \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{8\pi}{3}, \frac{10\pi}{3}$. $$\implies x = \frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3}$$

Step 3: Evaluate the function at critical points and endpoints ($0$ and $2\pi$). $$f(0) = 0 + \sin 0 = 0$$ $$f\left(\frac{\pi}{3}\right) = \frac{\pi}{3} + \sin\left(\frac{2\pi}{3}\right) = \frac{\pi}{3} + \frac{\sqrt{3}}{2} \approx 1.99$$ $$f\left(\frac{2\pi}{3}\right) = \frac{2\pi}{3} + \sin\left(\frac{4\pi}{3}\right) = \frac{2\pi}{3} - \frac{\sqrt{3}}{2} \approx 1.22$$ $$f\left(\frac{4\pi}{3}\right) = \frac{4\pi}{3} + \sin\left(\frac{8\pi}{3}\right) = \frac{4\pi}{3} + \frac{\sqrt{3}}{2} \approx 4.99$$ $$f\left(\frac{5\pi}{3}\right) = \frac{5\pi}{3} + \sin\left(\frac{10\pi}{3}\right) = \frac{5\pi}{3} - \frac{\sqrt{3}}{2} \approx 4.22$$ $$f(2\pi) = 2\pi + \sin 4\pi = 2\pi + 0 = 2\pi \approx 6.28$$

Final Answer: Maximum value = $2\pi$ (at $x = 2\pi$). Minimum value = $0$ (at $x = 0$).

Step 1: Let the two numbers be $x$ and $y$. Given: $x + y = 24 \implies y = 24 - x$.

Step 2: The product is $P = xy$. Substitute $y$ to get a function of one variable. $$P(x) = x(24 - x) = 24x - x^2$$

Step 3: Find the derivative and critical points. $$P'(x) = 24 - 2x$$ Set $P'(x) = 0 \implies 24 - 2x = 0 \implies x = 12$.

Step 4: Use the second derivative test to confirm maximum. $$P''(x) = -2 < 0$$. This confirms a maximum.

Step 5: Find the second number. If $x=12$, then $y = 24 - 12 = 12$.

Final Answer: The two numbers are 12 and 12.

Given: $x + y = 35$, maximize the product $x^2y^5$.

Step 1: Express one variable in terms of the other. $$y = 35 - x$$ Let the product be $p$. $$p = x^2y^5 = x^2(35 - x)^5$$

Step 2: Differentiate with respect to $x$ and find critical points. $$\frac{dp}{dx} = 2x(35 - x)^5 + x^2 \cdot 5(35 - x)^4(-1)$$ $$\frac{dp}{dx} = x(35 - x)^4 [2(35 - x) - 5x]$$ $$\frac{dp}{dx} = x(35 - x)^4 (70 - 7x)$$ Set $\frac{dp}{dx} = 0 \implies 70 - 7x = 0 \implies x = 10$.

Step 3: Find the corresponding value of $y$. $$y = 35 - 10 = 25$$

Final Answer: The two numbers are $10$ and $25$.

Given: Sum = $16$, minimize the sum of their cubes.

Step 1: Let the two numbers be $x$ and $(16 - x)$. Let the sum of their cubes be $S$. $$S = x^3 + (16 - x)^3$$

Step 2: Find the derivative and critical points. $$\frac{dS}{dx} = 3x^2 + 3(16 - x)^2(-1)$$ $$\frac{dS}{dx} = 3x^2 - 3(16 - x)^2 = 3[x^2 - (16 - x)^2]$$ $$\frac{dS}{dx} = 3[x^2 - (256 - 32x + x^2)] = 3(32x - 256)$$ Set $\frac{dS}{dx} = 0 \implies 32x - 256 = 0 \implies x = 8$.

Step 3: Use the second derivative test. $$\frac{d^2S}{dx^2} = 3(32) = 96$$ Since $\frac{d^2S}{dx^2} = 96 > 0$, it indicates a Minimum.

Step 4: Find the second number. Second number = $16 - 8 = 8$.

Final Answer: The two numbers are $8$ and $8$.

Step 1: Let the side of the square cut from each corner be $x$ cm. The original side is 18 cm. After cutting the corners, the base of the box will be a square with side $(18 - 2x)$ cm. The height of the box will be $x$ cm. The dimensions of the box are: Length = $18-2x$, Breadth = $18-2x$, Height = $x$. The domain for $x$ is $0 < x < 9$.

Step 2: Write the volume $V$ as a function of $x$. $$V(x) = (18 - 2x)(18 - 2x)(x) = x(18 - 2x)^2$$

Step 3: Find the derivative and critical points. $$V'(x) = 1 \cdot (18-2x)^2 + x \cdot 2(18-2x)(-2)$$ $$V'(x) = (18-2x) [ (18-2x) - 4x ] = (18-2x)(18-6x)$$ Set $V'(x) = 0 \implies 18-2x=0$ or $18-6x=0$. This gives $x=9$ or $x=3$. Since $x=9$ is not in the domain $(0,9)$, we only consider $x=3$.

Step 4: Use the second derivative test. $$V''(x) = (-2)(18-6x) + (18-2x)(-6) = -36 + 12x - 108 + 12x = 24x - 144$$ At $x=3$, $V''(3) = 24(3) - 144 = 72 - 144 = -72 < 0$. This confirms a maximum.

Final Answer: The side of the square to be cut off should be 3 cm.

Given: Rectangular sheet = $45 \text{ cm} \times 24 \text{ cm}$.

Step 1: Let the side of the square to be cut off be $x$ cm.

Step 2: Determine the dimensions of the box. Length $= 45 - 2x$ Breadth $= 24 - 2x$ Height $= x$ The volume $V$ of the box is: $$V = (45 - 2x)(24 - 2x)x$$

Step 3: Find the derivative and set it to zero. $$\frac{dV}{dx} = (45 - 2x)(24 - 2x) + x[-2(24 - 2x) - 2(45 - 2x)]$$ $$\frac{dV}{dx} = (45 - 2x)(24 - 2x) - 2x(69 - 4x)$$ Setting $\frac{dV}{dx} = 0$ gives a quadratic equation. Solving this gives $x = 5$ or $x = 18$. Since $x = 18$ makes the breadth negative ($24 - 36 = -12$), it is not a physically possible dimension. So, $x = 5$ cm.

Step 4: Verify using the second derivative test. Calculating $\frac{d^2V}{dx^2}$ at $x = 5$ will yield a negative value ($\frac{d^2V}{dx^2} < 0$), confirming it is a maximum.

Final Answer: The side of the square to be cut off is $5$ cm.

Given: A rectangle inscribed in a fixed circle of radius $a$.

Step 1: Let the length of the rectangle be $x$ and the breadth be $y$. The diagonal of the rectangle will be the diameter of the circle, which is $2a$. Using Pythagoras theorem: $$x^2 + y^2 = (2a)^2 = 4a^2 \quad \text{...(i)}$$

Step 2: Write the Area $A$ as a function of one variable. $$A = xy$$ From equation (i), $y = \sqrt{4a^2 - x^2}$. $$A = x\sqrt{4a^2 - x^2}$$

Step 3: Differentiate with respect to $x$ to find critical points. $$\frac{dA}{dx} = 1 \cdot \sqrt{4a^2 - x^2} + x \cdot \frac{1}{2\sqrt{4a^2 - x^2}}(-2x)$$ $$\frac{dA}{dx} = \frac{4a^2 - x^2 - x^2}{\sqrt{4a^2 - x^2}} = \frac{4a^2 - 2x^2}{\sqrt{4a^2 - x^2}}$$ Set $\frac{dA}{dx} = 0 \implies 4a^2 - 2x^2 = 0 \implies 2x^2 = 4a^2 \implies x^2 = 2a^2 \implies x = a\sqrt{2}$.

Step 4: Find the corresponding value of $y$. From (i): $y^2 = 4a^2 - x^2 = 4a^2 - 2a^2 = 2a^2 \implies y = a\sqrt{2}$. Since $x = y = a\sqrt{2}$, the length and breadth are equal, proving the rectangle is a square.

Step 5: Verify using the second derivative test. Calculating $\frac{d^2A}{dx^2}$ at $x = a\sqrt{2}$ will be negative ($\frac{d^2A}{dx^2} < 0$), confirming maximum area.

Final Answer: The square has the maximum area among all rectangles inscribed in a circle.

Given: A cylinder with radius $r$, height $h$, and a given constant surface area $S$.

Step 1: Write the formulas for Surface Area and Volume. Total Surface Area, $S = 2\pi r^2 + 2\pi rh \quad \text{(constant)}$ Volume, $V = \pi r^2 h$

Step 2: Express $h$ in terms of $r$ and $S$. $$2\pi rh = S - 2\pi r^2 \implies h = \frac{S - 2\pi r^2}{2\pi r}$$

Step 3: Substitute $h$ into the volume formula and differentiate. $$V = \pi r^2 \left( \frac{S - 2\pi r^2}{2\pi r} \right) = \frac{r(S - 2\pi r^2)}{2} = \frac{Sr - 2\pi r^3}{2}$$ $$\frac{dV}{dr} = \frac{S - 6\pi r^2}{2}$$ Set $\frac{dV}{dr} = 0 \implies S - 6\pi r^2 = 0 \implies S = 6\pi r^2$.

Step 4: Substitute the value of $S$ back into the original surface area equation to find the relationship between $h$ and $r$. $$6\pi r^2 = 2\pi r^2 + 2\pi rh$$ $$4\pi r^2 = 2\pi rh$$ Dividing both sides by $2\pi r$: $$2r = h$$ Since $2r$ is the diameter, this means height = diameter.

Final Answer: For a given surface area, the volume of the cylinder is maximum when its height is equal to the diameter of its base.

Given: Volume $V = 100 \text{ cm}^3$, minimize the surface area.

Step 1: Express height in terms of radius. $$V = \pi r^2 h = 100 \implies h = \frac{100}{\pi r^2}$$

Step 2: Write the surface area function. Total Surface Area, $S = 2\pi r^2 + 2\pi rh$ $$S(r) = 2\pi r^2 + 2\pi r\left(\frac{100}{\pi r^2}\right) = 2\pi r^2 + \frac{200}{r}$$

Step 3: Differentiate and find critical points. $$\frac{dS}{dr} = 4\pi r - \frac{200}{r^2}$$ Set $\frac{dS}{dr} = 0 \implies 4\pi r = \frac{200}{r^2} \implies 4\pi r^3 = 200 \implies r^3 = \frac{50}{\pi} \implies r = \sqrt[3]{\frac{50}{\pi}}$$

Step 4: Find the height. $$h = \frac{100}{\pi r^2} = \frac{2 \cdot 50}{\pi r^2} = \frac{2(\pi r^3)}{\pi r^2} = 2r$$ So, $h = 2\sqrt[3]{\frac{50}{\pi}}$.

Final Answer: Radius = $\sqrt[3]{\frac{50}{\pi}}$ cm, Height = $2\sqrt[3]{\frac{50}{\pi}}$ cm.

Given: Wire length = $28$ m.

Step 1: Let the length of the piece for the circle be $x$ and for the square be $(28 - x)$. For the circle: Circumference $2\pi r = x \implies r = \frac{x}{2\pi}$. Area of the circle, $A_1 = \pi r^2 = \pi \left(\frac{x}{2\pi}\right)^2 = \frac{x^2}{4\pi}$. For the square: Perimeter $4s = 28 - x \implies s = \frac{28 - x}{4}$. Area of the square, $A_2 = s^2 = \left(\frac{28 - x}{4}\right)^2 = \frac{(28 - x)^2}{16}$.

Step 2: Total area $A = A_1 + A_2$. $$A = \frac{x^2}{4\pi} + \frac{(28 - x)^2}{16}$$

Step 3: Differentiate and set to zero. $$\frac{dA}{dx} = \frac{2x}{4\pi} + \frac{2(28 - x)(-1)}{16} = \frac{x}{2\pi} - \frac{28 - x}{8}$$ Set $\frac{dA}{dx} = 0 \implies \frac{x}{2\pi} = \frac{28 - x}{8} \implies 4x = \pi(28 - x) \implies 4x = 28\pi - \pi x$ $$x(4 + \pi) = 28\pi \implies x = \frac{28\pi}{4 + \pi}$$

Step 4: Find the length of the other piece. $$28 - x = 28 - \frac{28\pi}{4 + \pi} = \frac{112 + 28\pi - 28\pi}{4 + \pi} = \frac{112}{4 + \pi}$$

Final Answer: The length of the piece for the circle is $\frac{28\pi}{4 + \pi}$ m, and for the square is $\frac{112}{4 + \pi}$ m.

Given: A sphere of fixed radius $R$. We need to maximize the volume of the inscribed cone.

Step 1: Let the radius of the cone's base be $r$ and its height be $h$. Using Pythagoras theorem in the cross-section of the sphere: $$r^2 + (h - R)^2 = R^2 \implies r^2 + h^2 - 2hR + R^2 = R^2 \implies r^2 = 2hR - h^2$$

Step 2: Write the volume of the cone. $$V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (2hR - h^2)h = \frac{1}{3}\pi (2Rh^2 - h^3)$$

Step 3: Differentiate with respect to $h$ to find the maximum. $$\frac{dV}{dh} = \frac{1}{3}\pi (4Rh - 3h^2)$$ Set $\frac{dV}{dh} = 0 \implies h(4R - 3h) = 0$. Since $h \neq 0$, we get $h = \frac{4R}{3}$.

Step 4: Find the maximum volume. Substitute $h = \frac{4R}{3}$ back into the volume formula: $$V_{max} = \frac{1}{3}\pi \left( 2R\left(\frac{4R}{3}\right)^2 - \left(\frac{4R}{3}\right)^3 \right)$$ $$V_{max} = \frac{1}{3}\pi \left( 2R\left(\frac{16R^2}{9}\right) - \frac{64R^3}{27} \right) = \frac{1}{3}\pi \left( \frac{32R^3}{9} - \frac{64R^3}{27} \right)$$ $$V_{max} = \frac{1}{3}\pi \left( \frac{96R^3 - 64R^3}{27} \right) = \frac{1}{3}\pi \left( \frac{32R^3}{27} \right) = \frac{8}{27} \left( \frac{4}{3}\pi R^3 \right)$$

Final Answer: The maximum volume of the cone is $\frac{8}{27}$ times the volume of the sphere. Hence Proved.

Given: A cone with a constant volume $V$. We need to minimize the curved surface area $S$.

Step 1: Express height $h$ in terms of $V$ and $r$. $$V = \frac{1}{3}\pi r^2 h \implies h = \frac{3V}{\pi r^2}$$

Step 2: Write the curved surface area formula. Let's use $Z = S^2$ to avoid square roots for easier differentiation. $$S = \pi rl = \pi r \sqrt{r^2 + h^2}$$ $$Z = S^2 = \pi^2 r^2 (r^2 + h^2) = \pi^2 r^4 + \pi^2 r^2 h^2$$ Substitute $h$: $$Z = \pi^2 r^4 + \pi^2 r^2 \left( \frac{9V^2}{\pi^2 r^4} \right) = \pi^2 r^4 + \frac{9V^2}{r^2}$$

Step 3: Differentiate $Z$ to find the minimum. $$\frac{dZ}{dr} = 4\pi^2 r^3 - \frac{18V^2}{r^3}$$ Set $\frac{dZ}{dr} = 0 \implies 4\pi^2 r^3 = \frac{18V^2}{r^3} \implies 2\pi^2 r^6 = 9V^2$$

Step 4: Substitute $V = \frac{1}{3}\pi r^2 h$ back into the equation. $$2\pi^2 r^6 = 9 \left( \frac{1}{9}\pi^2 r^4 h^2 \right)$$ $$2\pi^2 r^6 = \pi^2 r^4 h^2$$ Divide by $\pi^2 r^4$: $$2r^2 = h^2 \implies h = \sqrt{2}r$$

Final Answer: Altitude $h = \sqrt{2}r$. Hence Proved.

Given: A cone with a constant slant height $l$. Let the semi-vertical angle be $\theta$.

Step 1: Express radius $r$ and height $h$ using trigonometry. $$r = l\sin\theta \quad \text{and} \quad h = l\cos\theta$$

Step 2: Write the volume formula in terms of $\theta$. $$V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (l\sin\theta)^2 (l\cos\theta) = \frac{1}{3}\pi l^3 \sin^2\theta \cos\theta$$

Step 3: Differentiate with respect to $\theta$ and find the critical point. $$\frac{dV}{d\theta} = \frac{1}{3}\pi l^3 \left[ (\sin^2\theta)(-\sin\theta) + (\cos\theta)(2\sin\theta\cos\theta) \right]$$ $$\frac{dV}{d\theta} = \frac{1}{3}\pi l^3 [2\sin\theta\cos^2\theta - \sin^3\theta]$$ Set $\frac{dV}{d\theta} = 0$: $$2\sin\theta\cos^2\theta - \sin^3\theta = 0$$ Divide by $\sin\theta\cos^2\theta$: $$2 - \tan^2\theta = 0 \implies \tan^2\theta = 2 \implies \tan\theta = \sqrt{2}$$

Step 4: Find the angle. $$\theta = \tan^{-1}\sqrt{2}$$

Final Answer: The semi-vertical angle is $\tan^{-1}\sqrt{2}$. Hence Proved.

Given: A cone with constant total surface area $S$.

Step 1: Express slant height $l$ in terms of $S$ and $r$. Total Surface Area, $S = \pi r^2 + \pi rl \implies l = \frac{S - \pi r^2}{\pi r}$

Step 2: Express volume $V$ and square it to simplify differentiation (let $Z = V^2$). We know $h^2 = l^2 - r^2$. $$h^2 = \left(\frac{S - \pi r^2}{\pi r}\right)^2 - r^2 = \frac{S^2 - 2S\pi r^2 + \pi^2 r^4 - \pi^2 r^4}{\pi^2 r^2} = \frac{S^2 - 2S\pi r^2}{\pi^2 r^2}$$ Now, $V^2 = \left(\frac{1}{3}\pi r^2 h\right)^2 = \frac{1}{9}\pi^2 r^4 h^2$ $$Z = V^2 = \frac{1}{9}\pi^2 r^4 \left(\frac{S^2 - 2S\pi r^2}{\pi^2 r^2}\right) = \frac{r^2}{9}(S^2 - 2S\pi r^2) = \frac{S^2 r^2 - 2S\pi r^4}{9}$$

Step 3: Differentiate $Z$ with respect to $r$. $$\frac{dZ}{dr} = \frac{2S^2 r - 8S\pi r^3}{9}$$ Set $\frac{dZ}{dr} = 0 \implies 2S^2 r = 8S\pi r^3 \implies 2S = 8\pi r^2 \implies S = 4\pi r^2$$

Step 4: Find the semi-vertical angle $\theta$. We know $\sin\theta = \frac{r}{l}$. Substitute $S = 4\pi r^2$ into the equation for $l$: $$l = \frac{4\pi r^2 - \pi r^2}{\pi r} = \frac{3\pi r^2}{\pi r} = 3r$$ $$\sin\theta = \frac{r}{3r} = \frac{1}{3} \implies \theta = \sin^{-1}\left(\frac{1}{3}\right)$$

Final Answer: The semi-vertical angle is $\sin^{-1}\left(\frac{1}{3}\right)$. Hence Proved.

Step 1: Let the point on the curve be $P(x,y)$. We are given $y = x^2/2$. So, $P$ is $(x, x^2/2)$. The fixed point is $A(0,5)$.

Step 2: The distance formula is $D = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$. Minimizing the distance $D$ is the same as minimizing the square of the distance, $Z = D^2$. $$Z = (x-0)^2 + (x^2/2 - 5)^2 = x^2 + (x^2/2 - 5)^2$$

Step 3: Differentiate $Z$ with respect to $x$. $$\frac{dZ}{dx} = 2x + 2(x^2/2 - 5) \cdot \frac{d}{dx}(x^2/2) = 2x + 2(x^2/2 - 5) \cdot x$$ $$\frac{dZ}{dx} = 2x + x(x^2 - 10) = 2x + x^3 - 10x = x^3 - 8x$$

Step 4: Find critical points by setting $\frac{dZ}{dx} = 0$. $$x^3 - 8x = 0 \implies x(x^2 - 8) = 0$$ This gives $x=0$ or $x^2=8 \implies x = \pm 2\sqrt{2}$.

Step 5: Find the corresponding $y$ values. If $x=0$, $y = 0^2/2 = 0$. Point is $(0,0)$. If $x = \pm 2\sqrt{2}$, $y = (2\sqrt{2})^2/2 = 8/2 = 4$. Points are $(2\sqrt{2}, 4)$ and $(-2\sqrt{2}, 4)$.

Step 6: Check which point is nearest by calculating $Z$ for each critical point. At $(0,0)$, $Z = 0^2 + (0-5)^2 = 25$. At $(\pm 2\sqrt{2}, 4)$, $Z = (\pm 2\sqrt{2})^2 + (4-5)^2 = 8 + (-1)^2 = 9$. The minimum distance corresponds to $Z=9$.

Final Answer: The points are $(2\sqrt{2}, 4)$ and $(-2\sqrt{2}, 4)$. Option (A) is correct.

Given: $y = \frac{1 - x + x^2}{1 + x + x^2}$

Step 1: Find the derivative using the quotient rule. $$\frac{dy}{dx} = \frac{(1 + x + x^2)\frac{d}{dx}(1 - x + x^2) - (1 - x + x^2)\frac{d}{dx}(1 + x + x^2)}{(1 + x + x^2)^2}$$ $$\frac{dy}{dx} = \frac{(1 + x + x^2)(-1 + 2x) - (1 - x + x^2)(1 + 2x)}{(1 + x + x^2)^2}$$

Step 2: Simplify the numerator to find critical points. Numerator: $(2x + 2x^2 + 2x^3 - 1 - x - x^2) - (1 + 2x - x - 2x^2 + x^2 + 2x^3)$ $= (2x^3 + x^2 + x - 1) - (2x^3 - x^2 + x + 1)$ $= 2x^2 - 2$ So, $\frac{dy}{dx} = \frac{2(x^2 - 1)}{(1 + x + x^2)^2}$

Step 3: Set $\frac{dy}{dx} = 0$. $$2(x^2 - 1) = 0 \implies x^2 = 1 \implies x = \pm 1$$

Step 4: Find the minimum value by substituting $x = 1$ and $x = -1$ into the original function. At $x = 1$: $y = \frac{1 - 1 + (1)^2}{1 + 1 + (1)^2} = \frac{1}{3}$ At $x = -1$: $y = \frac{1 - (-1) + (-1)^2}{1 + (-1) + (-1)^2} = \frac{1 + 1 + 1}{1 - 1 + 1} = 3$ The minimum value between $\frac{1}{3}$ and $3$ is $\frac{1}{3}$.

Final Answer: The correct option is (D).

Given: $y = [x(x - 1) + 1]^{1/3}$ on the interval $[0, 1]$.

Step 1: Let the function inside the bracket be $f(x)$. $$f(x) = x(x - 1) + 1 = x^2 - x + 1$$ So, $y = [f(x)]^{1/3}$. To maximize $y$, we need to maximize $f(x)$ because the cube root function is strictly increasing.

Step 2: Find the critical points of $f(x)$. $$f'(x) = 2x - 1$$ Set $f'(x) = 0 \implies 2x - 1 = 0 \implies x = \frac{1}{2}$.

Step 3: Evaluate $f(x)$ at the critical point and the endpoints of the interval ($0$ and $1$). At $x = 0$: $f(0) = 0^2 - 0 + 1 = 1$ At $x = 1$: $f(1) = 1^2 - 1 + 1 = 1$ At $x = \frac{1}{2}$: $f\left(\frac{1}{2}\right) = \left(\frac{1}{2}\right)^2 - \frac{1}{2} + 1 = \frac{1}{4} - \frac{1}{2} + 1 = \frac{3}{4}$

Step 4: The maximum value of $f(x)$ in this interval is $1$. Therefore, the maximum value of $y = (1)^{1/3} = 1$.

Final Answer: The correct option is (C).