QBManager

Application of Derivatives

85618 The altitude for a right circular cone of minimum volume circumscribed about a sphere of radius $r$ is

1

2 r

2

3 r

3

5 r

4

4 r

Explanation:

(D) : Let the radius
Of cone be $R$ and
Height be $h$ .
In $△ ABC$ ,
$\sin θ = \frac{r}{h - r}$
In $△ ADE$ ,
$\sin θ = \frac{R}{\sqrt{h^{2} + R^{2}}}$
Equating (i) and (ii), we get
$\frac{r}{h - r} = \frac{R}{\sqrt{h^{2} + R^{2}}}$
Squaring both sides, we get
$r^{2} (h^{2} + R^{2}) = R^{2} (h - r)^{2}$
$\Rightarrow R^{2} = \frac{r^{2} h}{h - 2 r}$
Volume of cone is given by $V = \frac{1}{3} π R^{2} h$
$\Rightarrow V = \frac{1}{3} π (\frac{r^{2} h}{h - 2 r}) h$
[Using (iii)]
So, $\frac{d V}{d h} = \frac{π r^{2}}{3} [\frac{(h - 2 r) 2 h - h^{2} (1)}{(h - 2 r)^{2}}]$
$\Rightarrow h (h - 4 r) = 0$
Now, for minimum or maximum volume
$\frac{dV}{dh} = 0 \Rightarrow h (h - 4 r) = 0$
$\Rightarrow h = 4 r [∵ h \neq 0]$
Also, ${(\frac{d^{2} V}{{dh}^{2}})}_{h = 4 r} > 0$
$∴$ Altitude $h = 4 r$ .

Shift-I

Application of Derivatives

85621 The largest value of $y = 2 x^{3} - 3 x^{2} - 12 x + 5$ for $- 2 \leq x \leq 2$ occurs at $x$ is equal to

1 -2

2 -1

3 2

4 4

Explanation:

(B) : Given equation : $y = 2 x^{3} - 3 x^{2} - 12 x + 5$ ....(i)
Differentiate equation (i),
$\frac{d y}{d x} = 6 x^{2} - 6 x - 12$
For max value, put $(\frac{d y}{d x}) = 0$
$0 = 6 x^{2} - 6 x - 12 \Rightarrow x^{2} - x - 2 = 0$
$(x - 2) (x + 1) = 0 \Rightarrow x = - 1, 2$
Again differentiating equation (ii), we get
$\frac{d^{2} y}{{dx}^{2}} = 12 x - 6$
at $x = - 1 \Rightarrow \frac{d^{2} y}{{dx}^{2}} = 12 (- 1) - 6$
$\Rightarrow \frac{d^{2} y}{{dx}^{2}} = - 12 - 6 \Rightarrow \frac{d^{2} y}{{dx}^{2}} = - 18 (- ve)$
Therefore, $y$ is $max$ at $x = - 1$ .

BITSAT-2005

Application of Derivatives

85622 The function $f (x) = 2 x^{3} - 3 x^{2} - 12 x + 4$ , has

1 two points of local maximum

2 two points of local minimum

3 one maxima and one minima

4 no maxima or minima

Explanation:

(C) : $f (x) = 2 x^{3} - 3 x^{2} - 12 x + 4$
$\Rightarrow f^{'} (x) = 6 x^{2} - 6 x - 12 = 6 (x^{2} - x - 2)$
$= 6 (x - 2) (x + 1)$
For maxima and minima $f^{'} (x) = 0$
$∴ 6 (x - 2) (x + 1) = 0 \Rightarrow x = 2, - 1$
Now, $f^{''} (x) = 12 x - 6$
At $x = 2; f^{''} (x) = 24 - 6 = 18 > 0$
$∴ x = 2$ , local minima point
At $x = - 1; f^{''} (x) = 12 (- 1) - 6 = - 18 < 0$
$∴ x = - 1$ local max. point

BITSAT-2011

Application of Derivatives

85623 If $a^{2} x^{4} + b^{2} y^{4} = c^{4}$ , then the maximum value of xy i

1

\frac{c}{\sqrt{ab}}

2

\frac{c^{2}}{2 \sqrt{a b}}

3

\frac{c}{2 \sqrt{ab}}

4

\frac{c^{2}}{\sqrt{2 ab}}

Explanation:

(D) : If the sum of two positive quantities is a constant, then their product is maximum, when they are equal.
$∴ a^{2} x^{4} \cdot b^{2} y^{4}$ is maximum when
$a^{2} x^{4} = b^{2} y^{4} = \frac{1}{2} (a^{2} x^{4} + b^{2} y^{4}) = \frac{c^{4}}{2}$
$∴$ Maximum value of
$a^{2} x^{4} \cdot b^{2} y^{4} = \frac{c^{4}}{2} \cdot \frac{c^{4}}{2} = \frac{c^{8}}{4}$
Maximum value of $x y = {(\frac{c^{8}}{4 a^{2} b^{2}})}^{1 / 4} = \frac{c^{2}}{\sqrt{2 a b}}$

BITSAT-2013

Application of Derivatives

85618 The altitude for a right circular cone of minimum volume circumscribed about a sphere of radius $r$ is

1

2 r

2

3 r

3

5 r

4

4 r

Explanation:

(D) : Let the radius
Of cone be $R$ and
Height be $h$ .
In $△ ABC$ ,
$\sin θ = \frac{r}{h - r}$
In $△ ADE$ ,
$\sin θ = \frac{R}{\sqrt{h^{2} + R^{2}}}$
Equating (i) and (ii), we get
$\frac{r}{h - r} = \frac{R}{\sqrt{h^{2} + R^{2}}}$
Squaring both sides, we get
$r^{2} (h^{2} + R^{2}) = R^{2} (h - r)^{2}$
$\Rightarrow R^{2} = \frac{r^{2} h}{h - 2 r}$
Volume of cone is given by $V = \frac{1}{3} π R^{2} h$
$\Rightarrow V = \frac{1}{3} π (\frac{r^{2} h}{h - 2 r}) h$
[Using (iii)]
So, $\frac{d V}{d h} = \frac{π r^{2}}{3} [\frac{(h - 2 r) 2 h - h^{2} (1)}{(h - 2 r)^{2}}]$
$\Rightarrow h (h - 4 r) = 0$
Now, for minimum or maximum volume
$\frac{dV}{dh} = 0 \Rightarrow h (h - 4 r) = 0$
$\Rightarrow h = 4 r [∵ h \neq 0]$
Also, ${(\frac{d^{2} V}{{dh}^{2}})}_{h = 4 r} > 0$
$∴$ Altitude $h = 4 r$ .

Shift-I

Application of Derivatives

85621 The largest value of $y = 2 x^{3} - 3 x^{2} - 12 x + 5$ for $- 2 \leq x \leq 2$ occurs at $x$ is equal to

1 -2

2 -1

3 2

4 4

Explanation:

(B) : Given equation : $y = 2 x^{3} - 3 x^{2} - 12 x + 5$ ....(i)
Differentiate equation (i),
$\frac{d y}{d x} = 6 x^{2} - 6 x - 12$
For max value, put $(\frac{d y}{d x}) = 0$
$0 = 6 x^{2} - 6 x - 12 \Rightarrow x^{2} - x - 2 = 0$
$(x - 2) (x + 1) = 0 \Rightarrow x = - 1, 2$
Again differentiating equation (ii), we get
$\frac{d^{2} y}{{dx}^{2}} = 12 x - 6$
at $x = - 1 \Rightarrow \frac{d^{2} y}{{dx}^{2}} = 12 (- 1) - 6$
$\Rightarrow \frac{d^{2} y}{{dx}^{2}} = - 12 - 6 \Rightarrow \frac{d^{2} y}{{dx}^{2}} = - 18 (- ve)$
Therefore, $y$ is $max$ at $x = - 1$ .

BITSAT-2005

Application of Derivatives

85622 The function $f (x) = 2 x^{3} - 3 x^{2} - 12 x + 4$ , has

1 two points of local maximum

2 two points of local minimum

3 one maxima and one minima

4 no maxima or minima

Explanation:

(C) : $f (x) = 2 x^{3} - 3 x^{2} - 12 x + 4$
$\Rightarrow f^{'} (x) = 6 x^{2} - 6 x - 12 = 6 (x^{2} - x - 2)$
$= 6 (x - 2) (x + 1)$
For maxima and minima $f^{'} (x) = 0$
$∴ 6 (x - 2) (x + 1) = 0 \Rightarrow x = 2, - 1$
Now, $f^{''} (x) = 12 x - 6$
At $x = 2; f^{''} (x) = 24 - 6 = 18 > 0$
$∴ x = 2$ , local minima point
At $x = - 1; f^{''} (x) = 12 (- 1) - 6 = - 18 < 0$
$∴ x = - 1$ local max. point

BITSAT-2011

Application of Derivatives

85623 If $a^{2} x^{4} + b^{2} y^{4} = c^{4}$ , then the maximum value of xy i

1

\frac{c}{\sqrt{ab}}

2

\frac{c^{2}}{2 \sqrt{a b}}

3

\frac{c}{2 \sqrt{ab}}

4

\frac{c^{2}}{\sqrt{2 ab}}

Explanation:

(D) : If the sum of two positive quantities is a constant, then their product is maximum, when they are equal.
$∴ a^{2} x^{4} \cdot b^{2} y^{4}$ is maximum when
$a^{2} x^{4} = b^{2} y^{4} = \frac{1}{2} (a^{2} x^{4} + b^{2} y^{4}) = \frac{c^{4}}{2}$
$∴$ Maximum value of
$a^{2} x^{4} \cdot b^{2} y^{4} = \frac{c^{4}}{2} \cdot \frac{c^{4}}{2} = \frac{c^{8}}{4}$
Maximum value of $x y = {(\frac{c^{8}}{4 a^{2} b^{2}})}^{1 / 4} = \frac{c^{2}}{\sqrt{2 a b}}$

BITSAT-2013

Application of Derivatives

85618 The altitude for a right circular cone of minimum volume circumscribed about a sphere of radius $r$ is

1

2 r

2

3 r

3

5 r

4

4 r

Explanation:

(D) : Let the radius
Of cone be $R$ and
Height be $h$ .
In $△ ABC$ ,
$\sin θ = \frac{r}{h - r}$
In $△ ADE$ ,
$\sin θ = \frac{R}{\sqrt{h^{2} + R^{2}}}$
Equating (i) and (ii), we get
$\frac{r}{h - r} = \frac{R}{\sqrt{h^{2} + R^{2}}}$
Squaring both sides, we get
$r^{2} (h^{2} + R^{2}) = R^{2} (h - r)^{2}$
$\Rightarrow R^{2} = \frac{r^{2} h}{h - 2 r}$
Volume of cone is given by $V = \frac{1}{3} π R^{2} h$
$\Rightarrow V = \frac{1}{3} π (\frac{r^{2} h}{h - 2 r}) h$
[Using (iii)]
So, $\frac{d V}{d h} = \frac{π r^{2}}{3} [\frac{(h - 2 r) 2 h - h^{2} (1)}{(h - 2 r)^{2}}]$
$\Rightarrow h (h - 4 r) = 0$
Now, for minimum or maximum volume
$\frac{dV}{dh} = 0 \Rightarrow h (h - 4 r) = 0$
$\Rightarrow h = 4 r [∵ h \neq 0]$
Also, ${(\frac{d^{2} V}{{dh}^{2}})}_{h = 4 r} > 0$
$∴$ Altitude $h = 4 r$ .

Shift-I

Application of Derivatives

85621 The largest value of $y = 2 x^{3} - 3 x^{2} - 12 x + 5$ for $- 2 \leq x \leq 2$ occurs at $x$ is equal to

1 -2

2 -1

3 2

4 4

Explanation:

(B) : Given equation : $y = 2 x^{3} - 3 x^{2} - 12 x + 5$ ....(i)
Differentiate equation (i),
$\frac{d y}{d x} = 6 x^{2} - 6 x - 12$
For max value, put $(\frac{d y}{d x}) = 0$
$0 = 6 x^{2} - 6 x - 12 \Rightarrow x^{2} - x - 2 = 0$
$(x - 2) (x + 1) = 0 \Rightarrow x = - 1, 2$
Again differentiating equation (ii), we get
$\frac{d^{2} y}{{dx}^{2}} = 12 x - 6$
at $x = - 1 \Rightarrow \frac{d^{2} y}{{dx}^{2}} = 12 (- 1) - 6$
$\Rightarrow \frac{d^{2} y}{{dx}^{2}} = - 12 - 6 \Rightarrow \frac{d^{2} y}{{dx}^{2}} = - 18 (- ve)$
Therefore, $y$ is $max$ at $x = - 1$ .

BITSAT-2005

Application of Derivatives

85622 The function $f (x) = 2 x^{3} - 3 x^{2} - 12 x + 4$ , has

1 two points of local maximum

2 two points of local minimum

3 one maxima and one minima

4 no maxima or minima

Explanation:

(C) : $f (x) = 2 x^{3} - 3 x^{2} - 12 x + 4$
$\Rightarrow f^{'} (x) = 6 x^{2} - 6 x - 12 = 6 (x^{2} - x - 2)$
$= 6 (x - 2) (x + 1)$
For maxima and minima $f^{'} (x) = 0$
$∴ 6 (x - 2) (x + 1) = 0 \Rightarrow x = 2, - 1$
Now, $f^{''} (x) = 12 x - 6$
At $x = 2; f^{''} (x) = 24 - 6 = 18 > 0$
$∴ x = 2$ , local minima point
At $x = - 1; f^{''} (x) = 12 (- 1) - 6 = - 18 < 0$
$∴ x = - 1$ local max. point

BITSAT-2011

Application of Derivatives

85623 If $a^{2} x^{4} + b^{2} y^{4} = c^{4}$ , then the maximum value of xy i

1

\frac{c}{\sqrt{ab}}

2

\frac{c^{2}}{2 \sqrt{a b}}

3

\frac{c}{2 \sqrt{ab}}

4

\frac{c^{2}}{\sqrt{2 ab}}

Explanation:

(D) : If the sum of two positive quantities is a constant, then their product is maximum, when they are equal.
$∴ a^{2} x^{4} \cdot b^{2} y^{4}$ is maximum when
$a^{2} x^{4} = b^{2} y^{4} = \frac{1}{2} (a^{2} x^{4} + b^{2} y^{4}) = \frac{c^{4}}{2}$
$∴$ Maximum value of
$a^{2} x^{4} \cdot b^{2} y^{4} = \frac{c^{4}}{2} \cdot \frac{c^{4}}{2} = \frac{c^{8}}{4}$
Maximum value of $x y = {(\frac{c^{8}}{4 a^{2} b^{2}})}^{1 / 4} = \frac{c^{2}}{\sqrt{2 a b}}$

BITSAT-2013

Application of Derivatives

85618 The altitude for a right circular cone of minimum volume circumscribed about a sphere of radius $r$ is

1

2 r

2

3 r

3

5 r

4

4 r

Explanation:

(D) : Let the radius
Of cone be $R$ and
Height be $h$ .
In $△ ABC$ ,
$\sin θ = \frac{r}{h - r}$
In $△ ADE$ ,
$\sin θ = \frac{R}{\sqrt{h^{2} + R^{2}}}$
Equating (i) and (ii), we get
$\frac{r}{h - r} = \frac{R}{\sqrt{h^{2} + R^{2}}}$
Squaring both sides, we get
$r^{2} (h^{2} + R^{2}) = R^{2} (h - r)^{2}$
$\Rightarrow R^{2} = \frac{r^{2} h}{h - 2 r}$
Volume of cone is given by $V = \frac{1}{3} π R^{2} h$
$\Rightarrow V = \frac{1}{3} π (\frac{r^{2} h}{h - 2 r}) h$
[Using (iii)]
So, $\frac{d V}{d h} = \frac{π r^{2}}{3} [\frac{(h - 2 r) 2 h - h^{2} (1)}{(h - 2 r)^{2}}]$
$\Rightarrow h (h - 4 r) = 0$
Now, for minimum or maximum volume
$\frac{dV}{dh} = 0 \Rightarrow h (h - 4 r) = 0$
$\Rightarrow h = 4 r [∵ h \neq 0]$
Also, ${(\frac{d^{2} V}{{dh}^{2}})}_{h = 4 r} > 0$
$∴$ Altitude $h = 4 r$ .

Shift-I

Application of Derivatives

85621 The largest value of $y = 2 x^{3} - 3 x^{2} - 12 x + 5$ for $- 2 \leq x \leq 2$ occurs at $x$ is equal to

1 -2

2 -1

3 2

4 4

Explanation:

(B) : Given equation : $y = 2 x^{3} - 3 x^{2} - 12 x + 5$ ....(i)
Differentiate equation (i),
$\frac{d y}{d x} = 6 x^{2} - 6 x - 12$
For max value, put $(\frac{d y}{d x}) = 0$
$0 = 6 x^{2} - 6 x - 12 \Rightarrow x^{2} - x - 2 = 0$
$(x - 2) (x + 1) = 0 \Rightarrow x = - 1, 2$
Again differentiating equation (ii), we get
$\frac{d^{2} y}{{dx}^{2}} = 12 x - 6$
at $x = - 1 \Rightarrow \frac{d^{2} y}{{dx}^{2}} = 12 (- 1) - 6$
$\Rightarrow \frac{d^{2} y}{{dx}^{2}} = - 12 - 6 \Rightarrow \frac{d^{2} y}{{dx}^{2}} = - 18 (- ve)$
Therefore, $y$ is $max$ at $x = - 1$ .

BITSAT-2005

Application of Derivatives

85622 The function $f (x) = 2 x^{3} - 3 x^{2} - 12 x + 4$ , has

1 two points of local maximum

2 two points of local minimum

3 one maxima and one minima

4 no maxima or minima

Explanation:

(C) : $f (x) = 2 x^{3} - 3 x^{2} - 12 x + 4$
$\Rightarrow f^{'} (x) = 6 x^{2} - 6 x - 12 = 6 (x^{2} - x - 2)$
$= 6 (x - 2) (x + 1)$
For maxima and minima $f^{'} (x) = 0$
$∴ 6 (x - 2) (x + 1) = 0 \Rightarrow x = 2, - 1$
Now, $f^{''} (x) = 12 x - 6$
At $x = 2; f^{''} (x) = 24 - 6 = 18 > 0$
$∴ x = 2$ , local minima point
At $x = - 1; f^{''} (x) = 12 (- 1) - 6 = - 18 < 0$
$∴ x = - 1$ local max. point

BITSAT-2011

Application of Derivatives

85623 If $a^{2} x^{4} + b^{2} y^{4} = c^{4}$ , then the maximum value of xy i

1

\frac{c}{\sqrt{ab}}

2

\frac{c^{2}}{2 \sqrt{a b}}

3

\frac{c}{2 \sqrt{ab}}

4

\frac{c^{2}}{\sqrt{2 ab}}

Explanation:

(D) : If the sum of two positive quantities is a constant, then their product is maximum, when they are equal.
$∴ a^{2} x^{4} \cdot b^{2} y^{4}$ is maximum when
$a^{2} x^{4} = b^{2} y^{4} = \frac{1}{2} (a^{2} x^{4} + b^{2} y^{4}) = \frac{c^{4}}{2}$
$∴$ Maximum value of
$a^{2} x^{4} \cdot b^{2} y^{4} = \frac{c^{4}}{2} \cdot \frac{c^{4}}{2} = \frac{c^{8}}{4}$
Maximum value of $x y = {(\frac{c^{8}}{4 a^{2} b^{2}})}^{1 / 4} = \frac{c^{2}}{\sqrt{2 a b}}$

BITSAT-2013