Class 12-NCERT Solutions-Chapter-06-Application of Derivatives-Ex 6.3

Solution:

(i) \( f(x) = (2x - 1)^2 + 3 \)
Method: We know that \( (2x-1)^2 \ge 0 \) for all \( x \in \mathbb{R} \).
Therefore, \( f(x) \ge 0 + 3 \).
Minimum value: 3 (at \( 2x-1=0 \Rightarrow x=1/2 \)).
Maximum value: None (value can go to infinity).

(ii) \( f(x) = 9x^2 + 12x + 2 \)
\( f(x) = (3x)^2 + 2(3x)(2) + 4 - 2 = (3x+2)^2 - 2 \).
Since \( (3x+2)^2 \ge 0 \), \( f(x) \ge -2 \).
Minimum value: -2 (at \( x=-2/3 \)).
Maximum value: None.

(iii) \( f(x) = -(x - 1)^2 + 10 \)
Since \( (x-1)^2 \ge 0 \), we have \( -(x-1)^2 \le 0 \).
Adding 10 to both sides: \( -(x-1)^2 + 10 \le 10 \).
Maximum value: 10 (at \( x=1 \)).
Minimum value: None.

(iv) \( g(x) = x^3 + 1 \)
\( g'(x) = 3x^2 \). Setting \( g'(x) = 0 \Rightarrow x = 0 \).
Test: For \( x < 0 \), \( g'(x) > 0 \). For \( x > 0 \), \( g'(x) > 0 \).
Since the sign of \( g'(x) \) does not change, \( x=0 \) is a point of inflection.
Ans: No maximum or minimum value.

Solution:

(i) \( |x+2| \ge 0 \Rightarrow |x+2| - 1 \ge -1 \).
Min value: -1. Max value: None.

(ii) \( |x+1| \ge 0 \Rightarrow -|x+1| \le 0 \Rightarrow -|x+1|+3 \le 3 \).
Max value: 3. Min value: None.

(iii) We know \( -1 \le \sin(2x) \le 1 \).
Add 5: \( 4 \le \sin(2x) + 5 \le 6 \).
Max value: 6. Min value: 4.

(iv) \( -1 \le \sin(4x) \le 1 \).
Add 3: \( 2 \le \sin(4x) + 3 \le 4 \).
Taking modulus: \( 2 \le |\sin 4x + 3| \le 4 \).
Max value: 4. Min value: 2.

(v) \( h(x) = x + 1 \) on \( (-1, 1) \).
The function is strictly increasing. Since the interval is open, the function approaches 0 and 2 but never reaches them.
Ans: No maximum or minimum value.

Solution:

(i) \( f(x) = x^2 \)
\( f'(x) = 2x \). Set \( 2x = 0 \Rightarrow x = 0 \).
\( f''(x) = 2 \).
At \( x=0 \): \( f''(0) = 2 > 0 \).
Thus, \( x=0 \) is a point of Local Minima.
Local Min Value = \( f(0) = 0 \).

(ii) \( g(x) = x^3 - 3x \)
\( g'(x) = 3x^2 - 3 = 3(x^2 - 1) \). Set \( g'(x) = 0 \Rightarrow x = \pm 1 \).
\( g''(x) = 6x \).
At \( x=1 \): \( g''(1) = 6 > 0 \) (Local Minima). Value = \( 1-3 = -2 \).
At \( x=-1 \): \( g''(-1) = -6 < 0 \) (Local Maxima). Value = \( -1+3 = 2 \).

(iii) \( h(x) = \sin x + \cos x, 0 < x < \frac{\pi}{2} \)
\( h'(x) = \cos x - \sin x \). Set \( h'(x) = 0 \Rightarrow \tan x = 1 \Rightarrow x = \pi/4 \).
\( h''(x) = -\sin x - \cos x \).
At \( x=\pi/4 \): \( h''(\pi/4) = -(\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}) = -\sqrt{2} < 0 \).
Thus, \( x=\pi/4 \) is Local Maxima. Value = \( \sqrt{2} \).

(iv) \( f(x) = \sin x - \cos x, 0 < x < 2\pi \)
\( f'(x) = \cos x + \sin x \). Set \( f'(x)=0 \Rightarrow \tan x = -1 \).
\( x = \frac{3\pi}{4}, \frac{7\pi}{4} \).
\( f''(x) = -\sin x + \cos x \).
At \( x=\frac{3\pi}{4} \): \( f'' = -\frac{1}{\sqrt{2}} + (-\frac{1}{\sqrt{2}}) = -\sqrt{2} < 0 \) (Maxima).
At \( x=\frac{7\pi}{4} \): \( f'' = -(-\frac{1}{\sqrt{2}}) + \frac{1}{\sqrt{2}} = \sqrt{2} > 0 \) (Minima).
Max Value = \( \sqrt{2} \), Min Value = \( -\sqrt{2} \).

(v) \( f(x) = x^3 - 6x^2 + 9x + 15 \)
\( f'(x) = 3x^2 - 12x + 9 = 3(x^2 - 4x + 3) = 3(x-1)(x-3) \).
CP: \( x=1, 3 \).
\( f''(x) = 6x - 12 \).
At \( x=1 \): \( f''(1) = -6 < 0 \) (Maxima). Value = 19.
At \( x=3 \): \( f''(3) = 6 > 0 \) (Minima). Value = 15.

(vi) \( g(x) = \frac{x}{2} + \frac{2}{x} \)
\( g'(x) = \frac{1}{2} - \frac{2}{x^2} \). Set \( g'(x)=0 \Rightarrow x^2=4 \Rightarrow x=2 \) (since \( x>0 \)).
\( g''(x) = 0 - 2(-2x^{-3}) = \frac{4}{x^3} \).
At \( x=2 \): \( g''(2) = \frac{4}{8} > 0 \) (Minima).
Local Min Value = \( 1+1=2 \).

(vii) \( g(x) = \frac{1}{x^2 + 2} \)
\( g'(x) = \frac{-2x}{(x^2+2)^2} \). CP: \( x=0 \).
Using First Derivative Test:
\( x<0 \Rightarrow g'(x)>0 \) (Increasing).
\( x>0 \Rightarrow g'(x)<0 \) (Decreasing).
Sign change from + to - implies Local Maxima at \( x=0 \). Value = \( 1/2 \).

(viii) \( f(x) = x\sqrt{1-x} \)
\( f'(x) = 1\cdot\sqrt{1-x} + x\frac{-1}{2\sqrt{1-x}} = \frac{2(1-x)-x}{2\sqrt{1-x}} = \frac{2-3x}{2\sqrt{1-x}} \).
CP: \( 2-3x=0 \Rightarrow x=2/3 \).
Sign of \( f'(x) \): For \( x < 2/3 \), \( f'>0 \). For \( x > 2/3 \), \( f'<0 \).
Change from + to - implies Local Maxima.
Value = \( \frac{2}{3}\sqrt{1/3} = \frac{2}{3\sqrt{3}} \).

Solution:

(i) \( f'(x) = e^x \).
Since \( e^x > 0 \) for all real \( x \), \( f'(x) \) is never 0. No critical points. Thus, no local maxima or minima.

(ii) \( g'(x) = \frac{1}{x} \).
For the domain \( x > 0 \), \( g'(x) \neq 0 \). No critical points. No maxima or minima.

(iii) \( h'(x) = 3x^2 + 2x + 1 \).
Discriminant \( D = b^2 - 4ac = 4 - 12 = -8 < 0 \).
Since \( a=3 > 0 \) and \( D < 0 \), the quadratic \( 3x^2+2x+1 \) is always positive.
So \( h'(x) > 0 \) for all \( x \). The function is strictly increasing. No maxima or minima.

Solution:

(i) \( f'(x) = 3x^2 = 0 \Rightarrow x=0 \).
Evaluate at CP and endpoints:
\( f(-2) = -8 \).
\( f(0) = 0 \).
\( f(2) = 8 \).
Abs Max: 8. Abs Min: -8.

(ii) \( f'(x) = \cos x - \sin x = 0 \Rightarrow \tan x = 1 \Rightarrow x = \pi/4 \).
\( f(0) = 0 + 1 = 1 \).
\( f(\pi/4) = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \sqrt{2} \approx 1.414 \).
\( f(\pi) = 0 - 1 = -1 \).
Abs Max: \( \sqrt{2} \). Abs Min: -1.

(iii) \( f'(x) = 4 - x = 0 \Rightarrow x=4 \).
\( f(-2) = 4(-2) - 0.5(4) = -8 - 2 = -10 \).
\( f(4) = 16 - 0.5(16) = 16 - 8 = 8 \).
\( f(4.5) = 4(4.5) - 0.5(20.25) = 18 - 10.125 = 7.875 \).
Abs Max: 8. Abs Min: -10.

(iv) \( f'(x) = 2(x-1) = 0 \Rightarrow x=1 \).
\( f(-3) = (-4)^2 + 3 = 19 \).
\( f(1) = 0 + 3 = 3 \).
Abs Max: 19. Abs Min: 3.

Solution:

Given \( p(x) = 41 - 72x - 18x^2 \).

\( p'(x) = -72 - 36x \).

For critical point: \( -72 - 36x = 0 \Rightarrow 36x = -72 \Rightarrow x = -2 \).

Second Derivative Check: \( p''(x) = -36 \).

Since \( p''(-2) = -36 < 0 \), \( x=-2 \) is a point of local maxima.

Max Profit = \( p(-2) = 41 - 72(-2) - 18(4) = 41 + 144 - 72 = 113 \).

Ans: 113.

Solution:

Let \( f(x) = 3x^4 - 8x^3 + 12x^2 - 48x + 25 \).

\( f'(x) = 12x^3 - 24x^2 + 24x - 48 \).

\( = 12(x^3 - 2x^2 + 2x - 4) = 12[x^2(x-2) + 2(x-2)] \).

\( = 12(x-2)(x^2+2) \).

Setting \( f'(x)=0 \), real root is \( x=2 \) (since \( x^2+2=0 \) has no real roots).

Evaluate at critical point \( x=2 \) and endpoints \( x=0, 3 \):

• \( f(0) = 25 \).
• \( f(2) = 3(16) - 8(8) + 12(4) - 48(2) + 25 = 48 - 64 + 48 - 96 + 25 = -39 \).
• \( f(3) = 3(81) - 8(27) + 12(9) - 48(3) + 25 = 243 - 216 + 108 - 144 + 25 = 16 \).

Maximum Value: 25. Minimum Value: -39.

Solution:

Let \( f(x) = \sin 2x \). We know the maximum value of sine function is 1.

\( \sin 2x = 1 \).

General solution: \( 2x = 2n\pi + \frac{\pi}{2} \).

For \( x \in [0, 2\pi] \), \( 2x \in [0, 4\pi] \).

Possible values for \( 2x \): \( \frac{\pi}{2}, \frac{5\pi}{2} \).

Values for \( x \): \( \frac{\pi}{4}, \frac{5\pi}{4} \).

Ans: \( x = \frac{\pi}{4} \) and \( x = \frac{5\pi}{4} \).

Solution:

Let \( f(x) = \sin x + \cos x \).

\( f'(x) = \cos x - \sin x \). Set \( f'(x) = 0 \Rightarrow \tan x = 1 \Rightarrow x = \pi/4 \) (principal).

\( f''(x) = -\sin x - \cos x \).

At \( x=\pi/4 \), \( f'' < 0 \). Maxima.

Max Value = \( \sin(\pi/4) + \cos(\pi/4) = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \sqrt{2} \).

Solution:

\( f(x) = 2x^3 - 24x + 107 \).

\( f'(x) = 6x^2 - 24 = 6(x^2 - 4) \). Critical points: \( x = \pm 2 \).

Interval [1, 3]: Only \( x=2 \) is in interval.
\( f(1) = 2 - 24 + 107 = 85 \).
\( f(2) = 16 - 48 + 107 = 75 \).
\( f(3) = 54 - 72 + 107 = 89 \).
Max Value: 89.

Interval [-3, -1]: Only \( x=-2 \) is in interval.
\( f(-3) = -54 + 72 + 107 = 125 \).
\( f(-2) = -16 + 48 + 107 = 139 \).
\( f(-1) = -2 + 24 + 107 = 129 \).
Max Value: 139.

Solution:

Let \( f(x) = x^4 - 62x^2 + ax + 9 \).

\( f'(x) = 4x^3 - 124x + a \).

Since the function attains maximum at \( x=1 \), \( f'(1) \) must be 0.

\( 4(1)^3 - 124(1) + a = 0 \).

\( 4 - 124 + a = 0 \Rightarrow a - 120 = 0 \Rightarrow a = 120 \).

Solution:

\( f(x) = x + \sin 2x \).

\( f'(x) = 1 + 2\cos 2x \). Set \( f'(x)=0 \Rightarrow \cos 2x = -1/2 \).

\( 2x = 2\pi/3, 4\pi/3, 8\pi/3, 10\pi/3 \) (within \( [0, 4\pi] \)).

\( x = \pi/3, 2\pi/3, 4\pi/3, 5\pi/3 \).

Evaluating:
\( f(0) = 0 \).
\( f(\pi/3) = \pi/3 + \sqrt{3}/2 \).
\( f(2\pi/3) = 2\pi/3 - \sqrt{3}/2 \).
\( f(4\pi/3) = 4\pi/3 + \sqrt{3}/2 \).
\( f(5\pi/3) = 5\pi/3 - \sqrt{3}/2 \).
\( f(2\pi) = 2\pi \).

Max is \( 2\pi \) (approx 6.28). Min is 0.

Solution:

Let numbers be \( x \) and \( 24-x \).

Product \( P(x) = x(24-x) = 24x - x^2 \).

\( P'(x) = 24 - 2x \). Set \( P'(x)=0 \Rightarrow x=12 \).

Second Derivative Test: \( P''(x) = -2 < 0 \).
Thus, \( x=12 \) gives a Maximum.

Numbers are 12 and 12.

Solution:

Given \( x + y = 60 \Rightarrow x = 60 - y \).

Let function \( f(y) = (60 - y)y^3 = 60y^3 - y^4 \).

\( f'(y) = 180y^2 - 4y^3 = 4y^2(45 - y) \).

Set \( f'(y)=0 \Rightarrow y=45 \) (since \( y>0 \)).

\( f''(y) = 360y - 12y^2 \).
At \( y=45 \): \( f''(45) = 360(45) - 12(45)^2 = 45(360 - 540) < 0 \).

Thus, Maxima at \( y=45 \). \( x = 60 - 45 = 15 \).

Numbers are 15 and 45.

Solution:

\( x + y = 35 \Rightarrow x = 35 - y \).

Let \( P(y) = (35 - y)^2 y^5 \).

\( P'(y) = 2(35-y)(-1)y^5 + (35-y)^2(5y^4) \)
\( = y^4(35-y) [-2y + 5(35-y)] \)
\( = y^4(35-y) [175 - 7y] \).

Critical points: \( y=35 \) (product 0, rejected), \( y=25 \).

Sign Test for \( P'(y) \):
For \( y < 25 \), \( P' > 0 \). For \( y > 25 \), \( P' < 0 \).
Change from + to - implies Maxima.

So \( y=25, x=10 \).

Solution:

Let numbers be \( x \) and \( 16-x \).

Function \( S(x) = x^3 + (16-x)^3 \).

\( S'(x) = 3x^2 + 3(16-x)^2(-1) = 3x^2 - 3(256 - 32x + x^2) \)
\( = 3x^2 - 768 + 96x - 3x^2 = 96x - 768 \).

Set \( S'(x) = 0 \Rightarrow 96x = 768 \Rightarrow x = 8 \).

2nd Derivative: \( S''(x) = 96 > 0 \).
Therefore, Minima at \( x=8 \).

Numbers are 8 and 8.

Solution:

Let the side of the square to be cut off be \( x \) cm.

After cutting squares of side \( x \) from each corner, the dimensions of the box will be:

• Length \( L = 18 - 2x \)
• Breadth \( B = 18 - 2x \)
• Height \( H = x \)

Volume \( V = L \times B \times H = x(18 - 2x)^2 \).

Differentiating w.r.t \( x \):
\( V'(x) = 1 \cdot (18 - 2x)^2 + x \cdot 2(18 - 2x)(-2) \)
\( = (18 - 2x) [ (18 - 2x) - 4x ] \)
\( = (18 - 2x) (18 - 6x) \).

For critical points, set \( V'(x) = 0 \):
\( 18 - 2x = 0 \Rightarrow x = 9 \) (Not possible as \( 18-2(9)=0 \))
\( 18 - 6x = 0 \Rightarrow x = 3 \).

Second Derivative Test:
\( V''(x) = \frac{d}{dx} [ (18 - 2x)(18 - 6x) ] \)
\( = -2(18 - 6x) + (18 - 2x)(-6) \)
\( = -36 + 12x - 108 + 12x = 24x - 144 \).
At \( x = 3 \): \( V''(3) = 24(3) - 144 = 72 - 144 = -72 < 0 \).
Since \( V'' < 0 \), volume is maximum at \( x = 3 \).

Ans: The side of the square to be cut off is 3 cm.

Solution:

Let side of cut square be \( x \) cm.

Dimensions of box: Length \( = 45 - 2x \), Breadth \( = 24 - 2x \), Height \( = x \).

Volume \( V = x(45 - 2x)(24 - 2x) \)
\( V = x(1080 - 90x - 48x + 4x^2) \)
\( V = 4x^3 - 138x^2 + 1080x \).

\( V'(x) = 12x^2 - 276x + 1080 \).
Set \( V'(x) = 0 \):
\( 12(x^2 - 23x + 90) = 0 \)
\( 12(x - 18)(x - 5) = 0 \).
Possible values: \( x = 18 \) or \( x = 5 \).
Since breadth is 24, \( x \) cannot be 18 (as \( 24 - 2(18) < 0 \)). So, \( x = 5 \).

Second Derivative Test:
\( V''(x) = 24x - 276 \).
At \( x = 5 \): \( V''(5) = 24(5) - 276 = 120 - 276 = -156 < 0 \).
So, Volume is maximum at \( x = 5 \).

Ans: Side of square is 5 cm.

Solution:

Let the circle have radius \( R \) (constant). Let rectangle dimensions be \( 2x \) and \( 2y \).

From geometry: \( x^2 + y^2 = R^2 \Rightarrow y = \sqrt{R^2 - x^2} \).

Area \( A = (2x)(2y) = 4xy \).
To maximize \( A \), we can maximize \( Z = A^2 \) (to avoid square roots).
\( Z = 16x^2 y^2 = 16x^2 (R^2 - x^2) = 16(R^2 x^2 - x^4) \).

\( \frac{dZ}{dx} = 16(2R^2 x - 4x^3) \).
Set \( \frac{dZ}{dx} = 0 \Rightarrow 2x(R^2 - 2x^2) = 0 \).
\( x = 0 \) (rejected) or \( x^2 = R^2/2 \Rightarrow x = R/\sqrt{2} \).

Second Derivative Test:
\( \frac{d^2Z}{dx^2} = 16(2R^2 - 12x^2) \).
At \( x^2 = R^2/2 \): \( 16(2R^2 - 6R^2) = -64R^2 < 0 \).
Thus, Area is maximum when \( x = R/\sqrt{2} \).

When \( x = R/\sqrt{2} \), \( y = \sqrt{R^2 - R^2/2} = R/\sqrt{2} \).
Since \( x = y \), the length \( 2x \) equals breadth \( 2y \).
Therefore, the rectangle is a square.

Solution:

Let surface area be \( S \) (constant), radius \( r \), and height \( h \).

\( S = 2\pi r^2 + 2\pi rh \implies h = \frac{S - 2\pi r^2}{2\pi r} \).

Volume \( V = \pi r^2 h = \pi r^2 \left( \frac{S - 2\pi r^2}{2\pi r} \right) = \frac{r}{2}(S - 2\pi r^2) = \frac{Sr}{2} - \pi r^3 \).

Differentiating w.r.t \( r \):
\( \frac{dV}{dr} = \frac{S}{2} - 3\pi r^2 \).
Set \( \frac{dV}{dr} = 0 \Rightarrow \frac{S}{2} = 3\pi r^2 \Rightarrow S = 6\pi r^2 \).

Substitute \( S \) back into height equation:
\( h = \frac{6\pi r^2 - 2\pi r^2}{2\pi r} = \frac{4\pi r^2}{2\pi r} = 2r \).

Since \( h = 2r \), height is equal to diameter.

Check Maxima: \( \frac{d^2V}{dr^2} = -6\pi r < 0 \). Maxima confirmed.

Solution:

Given Volume \( V = \pi r^2 h = 100 \Rightarrow h = \frac{100}{\pi r^2} \).

Surface Area \( S = 2\pi r^2 + 2\pi r h \).
Substitute \( h \): \( S = 2\pi r^2 + 2\pi r \left( \frac{100}{\pi r^2} \right) = 2\pi r^2 + \frac{200}{r} \).

\( \frac{dS}{dr} = 4\pi r - \frac{200}{r^2} \).
Set \( \frac{dS}{dr} = 0 \Rightarrow 4\pi r = \frac{200}{r^2} \Rightarrow r^3 = \frac{50}{\pi} \).
\( r = \sqrt[3]{\frac{50}{\pi}} \).

Second Derivative Test:
\( \frac{d^2S}{dr^2} = 4\pi + \frac{400}{r^3} \).
Since \( r > 0 \), \( S'' > 0 \). Thus, Surface Area is minimum.

Dimensions:
\( r = \left( \frac{50}{\pi} \right)^{1/3} \) cm.
\( h = \frac{100}{\pi r^2} = \frac{2(50)}{\pi r^2} = \frac{2(\pi r^3)}{\pi r^2} = 2r = 2 \left( \frac{50}{\pi} \right)^{1/3} \) cm.

Solution:

Total length = 28 m. Let length for square be \( x \). Then length for circle is \( 28 - x \).

Square: Perimeter \( 4s = x \Rightarrow \text{side } s = x/4 \). Area \( A_s = (x/4)^2 = x^2/16 \).

Circle: Circumference \( 2\pi r = 28 - x \Rightarrow r = \frac{28-x}{2\pi} \).
Area \( A_c = \pi r^2 = \pi \left( \frac{28-x}{2\pi} \right)^2 = \frac{(28-x)^2}{4\pi} \).

Total Area \( A = \frac{x^2}{16} + \frac{(28-x)^2}{4\pi} \).

\( \frac{dA}{dx} = \frac{2x}{16} + \frac{2(28-x)(-1)}{4\pi} = \frac{x}{8} - \frac{28-x}{2\pi} \).

Set \( \frac{dA}{dx} = 0 \Rightarrow \frac{x}{8} = \frac{28-x}{2\pi} \Rightarrow \pi x = 4(28-x) \).
\( \pi x = 112 - 4x \Rightarrow x(\pi + 4) = 112 \Rightarrow x = \frac{112}{\pi + 4} \).

Second Derivative Test:
\( \frac{d^2A}{dx^2} = \frac{1}{8} + \frac{1}{2\pi} > 0 \). Minima confirmed.

Dimensions:
Piece for Square = \( \frac{112}{\pi + 4} \) m.
Piece for Circle = \( 28 - \frac{112}{\pi + 4} = \frac{28\pi + 112 - 112}{\pi + 4} = \frac{28\pi}{\pi + 4} \) m.

Solution:

Let sphere radius be \( R \). Let cone height be \( h \) and radius \( r \).
Let the center of the sphere be O. Let the distance from O to the base of the cone be \( x \).
Then \( h = R + x \). Also \( r^2 + x^2 = R^2 \Rightarrow r^2 = R^2 - x^2 \).

Volume \( V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (R^2 - x^2)(R + x) \).

\( \frac{dV}{dx} = \frac{\pi}{3} [ (R^2 - x^2)(1) + (R + x)(-2x) ] \)
\( = \frac{\pi}{3} (R + x) [ R - x - 2x ] = \frac{\pi}{3} (R + x) (R - 3x) \).

Set \( \frac{dV}{dx} = 0 \Rightarrow x = R/3 \) (since \( x = -R \) implies height 0).

Second Derivative Test:
At \( x = R/3 \), derivative changes sign from + to -. Maxima.

Max Volume:
Height \( h = R + R/3 = 4R/3 \).
Radius \( r^2 = R^2 - R^2/9 = 8R^2/9 \).
\( V = \frac{1}{3}\pi \left( \frac{8R^2}{9} \right) \left( \frac{4R}{3} \right) = \frac{32\pi R^3}{81} \).

Compare to Sphere Volume \( V_s = \frac{4}{3}\pi R^3 \):
\( V = \frac{8}{27} \left( \frac{4}{3}\pi R^3 \right) = \frac{8}{27} V_s \). Proved.

Solution:

Let volume \( V \) be constant. \( V = \frac{1}{3}\pi r^2 h \Rightarrow h = \frac{3V}{\pi r^2} \).

Curved Surface Area \( S = \pi r l = \pi r \sqrt{r^2 + h^2} \).
Minimize \( Z = S^2 = \pi^2 r^2 (r^2 + h^2) = \pi^2 r^4 + \pi^2 r^2 h^2 \).

Substitute \( h \):
\( Z = \pi^2 r^4 + \pi^2 r^2 \left( \frac{9V^2}{\pi^2 r^4} \right) = \pi^2 r^4 + \frac{9V^2}{r^2} \).

\( \frac{dZ}{dr} = 4\pi^2 r^3 - \frac{18V^2}{r^3} \).
Set \( \frac{dZ}{dr} = 0 \Rightarrow 2\pi^2 r^6 = 9V^2 \).

Substitute \( V = \frac{1}{3}\pi r^2 h \) into equation:
\( 2\pi^2 r^6 = 9 \left( \frac{1}{9}\pi^2 r^4 h^2 \right) = \pi^2 r^4 h^2 \).
\( 2r^2 = h^2 \Rightarrow h = \sqrt{2}r \).

Second Derivative Test:
\( \frac{d^2Z}{dr^2} = 12\pi^2 r^2 + \frac{54V^2}{r^4} > 0 \). Minima confirmed.

Hence, altitude is \( \sqrt{2} \) times radius.

Solution:

Let slant height \( l \) be constant. Let semi-vertical angle be \( \alpha \).
Radius \( r = l \sin \alpha \), Height \( h = l \cos \alpha \).

Volume \( V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (l \sin \alpha)^2 (l \cos \alpha) \)
\( V = \frac{1}{3}\pi l^3 (\sin^2 \alpha \cos \alpha) \).

Differentiating w.r.t \( \alpha \):
\( \frac{dV}{d\alpha} = \frac{\pi l^3}{3} [ 2\sin\alpha\cos\alpha \cdot \cos\alpha + \sin^2\alpha(-\sin\alpha) ] \)
\( = \frac{\pi l^3}{3} \sin\alpha (2\cos^2\alpha - \sin^2\alpha) \).

Set \( \frac{dV}{d\alpha} = 0 \):
\( \sin\alpha \neq 0 \), so \( 2\cos^2\alpha = \sin^2\alpha \).
\( \tan^2\alpha = 2 \Rightarrow \tan\alpha = \sqrt{2} \Rightarrow \alpha = \tan^{-1}\sqrt{2} \).

Second Derivative Test:
Analysis shows \( V'' < 0 \) at this point. Maxima confirmed.

Solution:

Let Total Surface Area \( S \) be constant. \( S = \pi r l + \pi r^2 \).
\( l = \frac{S - \pi r^2}{\pi r} \).

Volume \( V = \frac{1}{3}\pi r^2 h \).
\( V^2 = \frac{1}{9}\pi^2 r^4 (l^2 - r^2) \).
Substitute \( l \):
\( V^2 = \frac{1}{9}\pi^2 r^4 \left[ \left( \frac{S}{\pi r} - r \right)^2 - r^2 \right] \)
\( V^2 = \frac{1}{9}\pi^2 r^4 \left[ \frac{S^2}{\pi^2 r^2} - \frac{2S}{\pi} \right] = \frac{1}{9} ( S^2 r^2 - 2S\pi r^4 ) \).

Differentiating w.r.t \( r \):
\( \frac{d(V^2)}{dr} = \frac{1}{9} ( 2S^2 r - 8S\pi r^3 ) \).
Set to 0: \( 2S^2 r = 8S\pi r^3 \Rightarrow S = 4\pi r^2 \).

Now, we need \( \sin \alpha = \frac{r}{l} \).
From \( S = \pi r l + \pi r^2 \), substitute \( S = 4\pi r^2 \):
\( 4\pi r^2 = \pi r l + \pi r^2 \Rightarrow 3\pi r^2 = \pi r l \Rightarrow l = 3r \).
\( \sin \alpha = \frac{r}{3r} = \frac{1}{3} \).
\( \alpha = \sin^{-1}(1/3) \).

Answer: (A)

Solution:

Let point be \( (x, y) \). Distance squared \( D = (x-0)^2 + (y-5)^2 = x^2 + (y-5)^2 \).
Substitute \( x^2 = 2y \):
\( D(y) = 2y + (y-5)^2 = 2y + y^2 - 10y + 25 = y^2 - 8y + 25 \).

\( D'(y) = 2y - 8 \).
Set \( D'(y) = 0 \Rightarrow y = 4 \).

If \( y = 4 \), \( x^2 = 2(4) = 8 \Rightarrow x = \pm 2\sqrt{2} \).
Point is \( (2\sqrt{2}, 4) \).

Answer: (D)

Solution:

Let \( y = \frac{1 - x + x^2}{1 + x + x^2} \).
\( \frac{dy}{dx} = \frac{(-1+2x)(1+x+x^2) - (1-x+x^2)(1+2x)}{(1+x+x^2)^2} \).
Numerator: \( (2x^3 + 2x^2 + 2x - x^2 - x - 1) - (2x^3 - 2x^2 + 2x + x^2 - x + 1) \)
\( = (2x^3 + x^2 + x - 1) - (2x^3 - x^2 + x + 1) \)
\( = 2x^2 - 2 = 2(x^2 - 1) \).

Critical points: \( x = \pm 1 \).
At \( x = 1 \): \( y = \frac{1-1+1}{1+1+1} = \frac{1}{3} \).
At \( x = -1 \): \( y = \frac{1+1+1}{1-1+1} = 3 \).
Minimum value is \( \frac{1}{3} \).

Answer: (C)

Solution:

Let \( g(x) = x^2 - x + 1 \).
\( g'(x) = 2x - 1 \). Critical point \( x = 1/2 \).
Values in \( [0, 1] \):
\( g(0) = 1 \).
\( g(1) = 1 \).
\( g(1/2) = 1/4 - 1/2 + 1 = 3/4 \).
Maximum value of \( g(x) \) is 1.
Therefore, Max value of \( [g(x)]^{1/3} \) is \( (1)^{1/3} = 1 \).