QBManager

Application of Derivatives

85802 If the curves $2 x = y^{2}$ and $2 x y = K$ intersect perpendicularly, then the value of $K^{2} i$

1

2 \sqrt{2}

2 2

3 8

4 4

Explanation:

(C) : Given,
Differentiating equation (i) and (ii) with respect to $x$ , we get,
$2 = 2 y \frac{d y}{d x} \Rightarrow \frac{d y}{d x} = \frac{1}{y}$
$m_{1} = \frac{1}{y} ($ Where, $m =$ slope $)$
$2 xy = k$
$2 [x \cdot \frac{dy}{dx} + y \cdot 1] = 0$
$∴ \frac{dy}{dx} = - \frac{y}{x} \Rightarrow m_{2} = - \frac{y}{x}$
We know that,
$m_{1} \times m_{2} = 1$
$\frac{1}{y} (- \frac{y}{x}) = - 1$
$x = 1$
From equation (ii),
$y = \frac{k}{2 x}$
Putting the value of $y$ in equation (i),
$2 x = {(\frac{k}{2 x})}^{2}$
$2 x = \frac{k^{2}}{4 x^{2}}$
$8 x^{3} = k^{2}$
$at x = 1$
$k^{2} = 8$

Karnataka CET-2020

Application of Derivatives

85803 The value of $\sqrt{24.99} i$

1 4.999

2 5.001

3 4.899

4 4.897

Explanation:

(A) : Given,
The value of $\sqrt{24.99}$
Let, $f^{'} (x) : \sqrt{x} \Rightarrow f^{'} (x) = \frac{1}{2 \sqrt{x}}$
Here, $x = 25$ and $Δ x = 0.01$
Now, $f (x + Δ x) - f (x) = f^{'} (x) Δ x$
$f (24.99) = \frac{1}{2 \sqrt{25}} (- 0.01) + \sqrt{25}$
$f (24.99) = 5 - 0.001$
$f (24.99) = 4.999$

Karnataka CET-2019

Application of Derivatives

85804 If a circular plate is heated uniformly, its area expands $3 c$ times as fast as its radius, then the value of $c$ when the radius is 6 units, $i$

1

4 π

2

2 π

3

6 π

4

3 π

Explanation:

(A) : Let A square units is the area measured when the radius is $r$ units.
$Area A = π r^{2}$
Differentiate both side with respect to ' $x$ '
$\begin{aligned} (i) & \frac{dA}{dx} = 2 π \frac{dr}{dx} \\ \frac{dA}{dx} = 3 c \frac{dr}{dx} \end{aligned}$
From equation (i) we get,
$3 \frac{c \cdot dr}{dx} = 2 π r \frac{dr}{dx}$
$3 c = 2 π r$
$c = \frac{2}{3} π r$
$c = \frac{2}{3} π (6)$
$c = 4 π$

BITSAT-2010

Application of Derivatives

85805 The fuel charges for running a train are proportional to the square of the speed generated in miles per hour and costs $Math input error$ per hour at 16 miles per hour. The most economical speed if the fixed charges i.e. salaries etc. amount to $Math input error$ per hour $i$

1 10

2 20

3 30

4 40

Explanation:

(D) : Let the speed of the train be $v$ and distance to be covered be $s$ so that total time taken is $s / v$ hours. Cost of fuel per hour $= {kv}^{2}$ ( $k$ is constant)
Also $48 = k {.16}^{2}$ by given condition
$∴ k = \frac{3}{16}$
$∴$ Cost of fuel per hour $\frac{3}{16} v^{2}$
Other charger per hour are 300.
Total running cost,
$C = (\frac{3}{16} v^{2} + 300) \frac{s}{v} = \frac{3 s}{16} v + \frac{300 s}{v}$
$\frac{dC}{dv} = \frac{3 s}{16} - \frac{300 s}{v^{2}} = 0 \Rightarrow v = 40$
$\frac{d^{2} C}{{dv}^{2}} = \frac{600 s}{v^{3}} > 0$
$∴ v = 40$ results in minimum running cost

BITSAT-2014

Application of Derivatives

85806 If $0 < x < \frac{π}{2}$ , then

1

\tan x < x < \sin x

2

x < \sin x < \tan x

3

\sin x < \tan x < x

4 None of these

Explanation:

(D) : Let us assume the functions $f (x)$ and $g (x)$ given by,
$f (x) = \tan x - x$
And, $g (x) = x - \sin x$ for $0 < x < \frac{π}{2}$
Now, $f^{'} (x) = \sec^{2} x - 1$ and $g^{'} (x) = 1 - \cos x$
$f^{'} (x) > 0 and g^{'} (x) > 0, \forall x \in (0, \frac{π}{2})$
$f^{'} (x) > f (0)$ and $g^{'} (x) > 0, \forall x \in (0, \frac{π}{2})$
$\tan x - x > 0$ and $x - \sin x > 0$
$\tan x > x$ and $x > \sin x$ $\sin x < x < \tan x$

BITSAT-2015

Application of Derivatives

85802 If the curves $2 x = y^{2}$ and $2 x y = K$ intersect perpendicularly, then the value of $K^{2} i$

1

2 \sqrt{2}

2 2

3 8

4 4

Explanation:

(C) : Given,
Differentiating equation (i) and (ii) with respect to $x$ , we get,
$2 = 2 y \frac{d y}{d x} \Rightarrow \frac{d y}{d x} = \frac{1}{y}$
$m_{1} = \frac{1}{y} ($ Where, $m =$ slope $)$
$2 xy = k$
$2 [x \cdot \frac{dy}{dx} + y \cdot 1] = 0$
$∴ \frac{dy}{dx} = - \frac{y}{x} \Rightarrow m_{2} = - \frac{y}{x}$
We know that,
$m_{1} \times m_{2} = 1$
$\frac{1}{y} (- \frac{y}{x}) = - 1$
$x = 1$
From equation (ii),
$y = \frac{k}{2 x}$
Putting the value of $y$ in equation (i),
$2 x = {(\frac{k}{2 x})}^{2}$
$2 x = \frac{k^{2}}{4 x^{2}}$
$8 x^{3} = k^{2}$
$at x = 1$
$k^{2} = 8$

Karnataka CET-2020

Application of Derivatives

85803 The value of $\sqrt{24.99} i$

1 4.999

2 5.001

3 4.899

4 4.897

Explanation:

(A) : Given,
The value of $\sqrt{24.99}$
Let, $f^{'} (x) : \sqrt{x} \Rightarrow f^{'} (x) = \frac{1}{2 \sqrt{x}}$
Here, $x = 25$ and $Δ x = 0.01$
Now, $f (x + Δ x) - f (x) = f^{'} (x) Δ x$
$f (24.99) = \frac{1}{2 \sqrt{25}} (- 0.01) + \sqrt{25}$
$f (24.99) = 5 - 0.001$
$f (24.99) = 4.999$

Karnataka CET-2019

Application of Derivatives

85804 If a circular plate is heated uniformly, its area expands $3 c$ times as fast as its radius, then the value of $c$ when the radius is 6 units, $i$

1

4 π

2

2 π

3

6 π

4

3 π

Explanation:

(A) : Let A square units is the area measured when the radius is $r$ units.
$Area A = π r^{2}$
Differentiate both side with respect to ' $x$ '
$\begin{aligned} (i) & \frac{dA}{dx} = 2 π \frac{dr}{dx} \\ \frac{dA}{dx} = 3 c \frac{dr}{dx} \end{aligned}$
From equation (i) we get,
$3 \frac{c \cdot dr}{dx} = 2 π r \frac{dr}{dx}$
$3 c = 2 π r$
$c = \frac{2}{3} π r$
$c = \frac{2}{3} π (6)$
$c = 4 π$

BITSAT-2010

Application of Derivatives

85805 The fuel charges for running a train are proportional to the square of the speed generated in miles per hour and costs $Math input error$ per hour at 16 miles per hour. The most economical speed if the fixed charges i.e. salaries etc. amount to $Math input error$ per hour $i$

1 10

2 20

3 30

4 40

Explanation:

(D) : Let the speed of the train be $v$ and distance to be covered be $s$ so that total time taken is $s / v$ hours. Cost of fuel per hour $= {kv}^{2}$ ( $k$ is constant)
Also $48 = k {.16}^{2}$ by given condition
$∴ k = \frac{3}{16}$
$∴$ Cost of fuel per hour $\frac{3}{16} v^{2}$
Other charger per hour are 300.
Total running cost,
$C = (\frac{3}{16} v^{2} + 300) \frac{s}{v} = \frac{3 s}{16} v + \frac{300 s}{v}$
$\frac{dC}{dv} = \frac{3 s}{16} - \frac{300 s}{v^{2}} = 0 \Rightarrow v = 40$
$\frac{d^{2} C}{{dv}^{2}} = \frac{600 s}{v^{3}} > 0$
$∴ v = 40$ results in minimum running cost

BITSAT-2014

Application of Derivatives

85806 If $0 < x < \frac{π}{2}$ , then

1

\tan x < x < \sin x

2

x < \sin x < \tan x

3

\sin x < \tan x < x

4 None of these

Explanation:

(D) : Let us assume the functions $f (x)$ and $g (x)$ given by,
$f (x) = \tan x - x$
And, $g (x) = x - \sin x$ for $0 < x < \frac{π}{2}$
Now, $f^{'} (x) = \sec^{2} x - 1$ and $g^{'} (x) = 1 - \cos x$
$f^{'} (x) > 0 and g^{'} (x) > 0, \forall x \in (0, \frac{π}{2})$
$f^{'} (x) > f (0)$ and $g^{'} (x) > 0, \forall x \in (0, \frac{π}{2})$
$\tan x - x > 0$ and $x - \sin x > 0$
$\tan x > x$ and $x > \sin x$ $\sin x < x < \tan x$

BITSAT-2015

Application of Derivatives

85802 If the curves $2 x = y^{2}$ and $2 x y = K$ intersect perpendicularly, then the value of $K^{2} i$

1

2 \sqrt{2}

2 2

3 8

4 4

Explanation:

(C) : Given,
Differentiating equation (i) and (ii) with respect to $x$ , we get,
$2 = 2 y \frac{d y}{d x} \Rightarrow \frac{d y}{d x} = \frac{1}{y}$
$m_{1} = \frac{1}{y} ($ Where, $m =$ slope $)$
$2 xy = k$
$2 [x \cdot \frac{dy}{dx} + y \cdot 1] = 0$
$∴ \frac{dy}{dx} = - \frac{y}{x} \Rightarrow m_{2} = - \frac{y}{x}$
We know that,
$m_{1} \times m_{2} = 1$
$\frac{1}{y} (- \frac{y}{x}) = - 1$
$x = 1$
From equation (ii),
$y = \frac{k}{2 x}$
Putting the value of $y$ in equation (i),
$2 x = {(\frac{k}{2 x})}^{2}$
$2 x = \frac{k^{2}}{4 x^{2}}$
$8 x^{3} = k^{2}$
$at x = 1$
$k^{2} = 8$

Karnataka CET-2020

Application of Derivatives

85803 The value of $\sqrt{24.99} i$

1 4.999

2 5.001

3 4.899

4 4.897

Explanation:

(A) : Given,
The value of $\sqrt{24.99}$
Let, $f^{'} (x) : \sqrt{x} \Rightarrow f^{'} (x) = \frac{1}{2 \sqrt{x}}$
Here, $x = 25$ and $Δ x = 0.01$
Now, $f (x + Δ x) - f (x) = f^{'} (x) Δ x$
$f (24.99) = \frac{1}{2 \sqrt{25}} (- 0.01) + \sqrt{25}$
$f (24.99) = 5 - 0.001$
$f (24.99) = 4.999$

Karnataka CET-2019

Application of Derivatives

85804 If a circular plate is heated uniformly, its area expands $3 c$ times as fast as its radius, then the value of $c$ when the radius is 6 units, $i$

1

4 π

2

2 π

3

6 π

4

3 π

Explanation:

(A) : Let A square units is the area measured when the radius is $r$ units.
$Area A = π r^{2}$
Differentiate both side with respect to ' $x$ '
$\begin{aligned} (i) & \frac{dA}{dx} = 2 π \frac{dr}{dx} \\ \frac{dA}{dx} = 3 c \frac{dr}{dx} \end{aligned}$
From equation (i) we get,
$3 \frac{c \cdot dr}{dx} = 2 π r \frac{dr}{dx}$
$3 c = 2 π r$
$c = \frac{2}{3} π r$
$c = \frac{2}{3} π (6)$
$c = 4 π$

BITSAT-2010

Application of Derivatives

85805 The fuel charges for running a train are proportional to the square of the speed generated in miles per hour and costs $Math input error$ per hour at 16 miles per hour. The most economical speed if the fixed charges i.e. salaries etc. amount to $Math input error$ per hour $i$

1 10

2 20

3 30

4 40

Explanation:

(D) : Let the speed of the train be $v$ and distance to be covered be $s$ so that total time taken is $s / v$ hours. Cost of fuel per hour $= {kv}^{2}$ ( $k$ is constant)
Also $48 = k {.16}^{2}$ by given condition
$∴ k = \frac{3}{16}$
$∴$ Cost of fuel per hour $\frac{3}{16} v^{2}$
Other charger per hour are 300.
Total running cost,
$C = (\frac{3}{16} v^{2} + 300) \frac{s}{v} = \frac{3 s}{16} v + \frac{300 s}{v}$
$\frac{dC}{dv} = \frac{3 s}{16} - \frac{300 s}{v^{2}} = 0 \Rightarrow v = 40$
$\frac{d^{2} C}{{dv}^{2}} = \frac{600 s}{v^{3}} > 0$
$∴ v = 40$ results in minimum running cost

BITSAT-2014

Application of Derivatives

85806 If $0 < x < \frac{π}{2}$ , then

1

\tan x < x < \sin x

2

x < \sin x < \tan x

3

\sin x < \tan x < x

4 None of these

Explanation:

(D) : Let us assume the functions $f (x)$ and $g (x)$ given by,
$f (x) = \tan x - x$
And, $g (x) = x - \sin x$ for $0 < x < \frac{π}{2}$
Now, $f^{'} (x) = \sec^{2} x - 1$ and $g^{'} (x) = 1 - \cos x$
$f^{'} (x) > 0 and g^{'} (x) > 0, \forall x \in (0, \frac{π}{2})$
$f^{'} (x) > f (0)$ and $g^{'} (x) > 0, \forall x \in (0, \frac{π}{2})$
$\tan x - x > 0$ and $x - \sin x > 0$
$\tan x > x$ and $x > \sin x$ $\sin x < x < \tan x$

BITSAT-2015

Application of Derivatives

85802 If the curves $2 x = y^{2}$ and $2 x y = K$ intersect perpendicularly, then the value of $K^{2} i$

1

2 \sqrt{2}

2 2

3 8

4 4

Explanation:

(C) : Given,
Differentiating equation (i) and (ii) with respect to $x$ , we get,
$2 = 2 y \frac{d y}{d x} \Rightarrow \frac{d y}{d x} = \frac{1}{y}$
$m_{1} = \frac{1}{y} ($ Where, $m =$ slope $)$
$2 xy = k$
$2 [x \cdot \frac{dy}{dx} + y \cdot 1] = 0$
$∴ \frac{dy}{dx} = - \frac{y}{x} \Rightarrow m_{2} = - \frac{y}{x}$
We know that,
$m_{1} \times m_{2} = 1$
$\frac{1}{y} (- \frac{y}{x}) = - 1$
$x = 1$
From equation (ii),
$y = \frac{k}{2 x}$
Putting the value of $y$ in equation (i),
$2 x = {(\frac{k}{2 x})}^{2}$
$2 x = \frac{k^{2}}{4 x^{2}}$
$8 x^{3} = k^{2}$
$at x = 1$
$k^{2} = 8$

Karnataka CET-2020

Application of Derivatives

85803 The value of $\sqrt{24.99} i$

1 4.999

2 5.001

3 4.899

4 4.897

Explanation:

(A) : Given,
The value of $\sqrt{24.99}$
Let, $f^{'} (x) : \sqrt{x} \Rightarrow f^{'} (x) = \frac{1}{2 \sqrt{x}}$
Here, $x = 25$ and $Δ x = 0.01$
Now, $f (x + Δ x) - f (x) = f^{'} (x) Δ x$
$f (24.99) = \frac{1}{2 \sqrt{25}} (- 0.01) + \sqrt{25}$
$f (24.99) = 5 - 0.001$
$f (24.99) = 4.999$

Karnataka CET-2019

Application of Derivatives

85804 If a circular plate is heated uniformly, its area expands $3 c$ times as fast as its radius, then the value of $c$ when the radius is 6 units, $i$

1

4 π

2

2 π

3

6 π

4

3 π

Explanation:

(A) : Let A square units is the area measured when the radius is $r$ units.
$Area A = π r^{2}$
Differentiate both side with respect to ' $x$ '
$\begin{aligned} (i) & \frac{dA}{dx} = 2 π \frac{dr}{dx} \\ \frac{dA}{dx} = 3 c \frac{dr}{dx} \end{aligned}$
From equation (i) we get,
$3 \frac{c \cdot dr}{dx} = 2 π r \frac{dr}{dx}$
$3 c = 2 π r$
$c = \frac{2}{3} π r$
$c = \frac{2}{3} π (6)$
$c = 4 π$

BITSAT-2010

Application of Derivatives

85805 The fuel charges for running a train are proportional to the square of the speed generated in miles per hour and costs $Math input error$ per hour at 16 miles per hour. The most economical speed if the fixed charges i.e. salaries etc. amount to $Math input error$ per hour $i$

1 10

2 20

3 30

4 40

Explanation:

(D) : Let the speed of the train be $v$ and distance to be covered be $s$ so that total time taken is $s / v$ hours. Cost of fuel per hour $= {kv}^{2}$ ( $k$ is constant)
Also $48 = k {.16}^{2}$ by given condition
$∴ k = \frac{3}{16}$
$∴$ Cost of fuel per hour $\frac{3}{16} v^{2}$
Other charger per hour are 300.
Total running cost,
$C = (\frac{3}{16} v^{2} + 300) \frac{s}{v} = \frac{3 s}{16} v + \frac{300 s}{v}$
$\frac{dC}{dv} = \frac{3 s}{16} - \frac{300 s}{v^{2}} = 0 \Rightarrow v = 40$
$\frac{d^{2} C}{{dv}^{2}} = \frac{600 s}{v^{3}} > 0$
$∴ v = 40$ results in minimum running cost

BITSAT-2014

Application of Derivatives

85806 If $0 < x < \frac{π}{2}$ , then

1

\tan x < x < \sin x

2

x < \sin x < \tan x

3

\sin x < \tan x < x

4 None of these

Explanation:

(D) : Let us assume the functions $f (x)$ and $g (x)$ given by,
$f (x) = \tan x - x$
And, $g (x) = x - \sin x$ for $0 < x < \frac{π}{2}$
Now, $f^{'} (x) = \sec^{2} x - 1$ and $g^{'} (x) = 1 - \cos x$
$f^{'} (x) > 0 and g^{'} (x) > 0, \forall x \in (0, \frac{π}{2})$
$f^{'} (x) > f (0)$ and $g^{'} (x) > 0, \forall x \in (0, \frac{π}{2})$
$\tan x - x > 0$ and $x - \sin x > 0$
$\tan x > x$ and $x > \sin x$ $\sin x < x < \tan x$

BITSAT-2015

Application of Derivatives

85802 If the curves $2 x = y^{2}$ and $2 x y = K$ intersect perpendicularly, then the value of $K^{2} i$

1

2 \sqrt{2}

2 2

3 8

4 4

Explanation:

(C) : Given,
Differentiating equation (i) and (ii) with respect to $x$ , we get,
$2 = 2 y \frac{d y}{d x} \Rightarrow \frac{d y}{d x} = \frac{1}{y}$
$m_{1} = \frac{1}{y} ($ Where, $m =$ slope $)$
$2 xy = k$
$2 [x \cdot \frac{dy}{dx} + y \cdot 1] = 0$
$∴ \frac{dy}{dx} = - \frac{y}{x} \Rightarrow m_{2} = - \frac{y}{x}$
We know that,
$m_{1} \times m_{2} = 1$
$\frac{1}{y} (- \frac{y}{x}) = - 1$
$x = 1$
From equation (ii),
$y = \frac{k}{2 x}$
Putting the value of $y$ in equation (i),
$2 x = {(\frac{k}{2 x})}^{2}$
$2 x = \frac{k^{2}}{4 x^{2}}$
$8 x^{3} = k^{2}$
$at x = 1$
$k^{2} = 8$

Karnataka CET-2020

Application of Derivatives

85803 The value of $\sqrt{24.99} i$

1 4.999

2 5.001

3 4.899

4 4.897

Explanation:

(A) : Given,
The value of $\sqrt{24.99}$
Let, $f^{'} (x) : \sqrt{x} \Rightarrow f^{'} (x) = \frac{1}{2 \sqrt{x}}$
Here, $x = 25$ and $Δ x = 0.01$
Now, $f (x + Δ x) - f (x) = f^{'} (x) Δ x$
$f (24.99) = \frac{1}{2 \sqrt{25}} (- 0.01) + \sqrt{25}$
$f (24.99) = 5 - 0.001$
$f (24.99) = 4.999$

Karnataka CET-2019

Application of Derivatives

85804 If a circular plate is heated uniformly, its area expands $3 c$ times as fast as its radius, then the value of $c$ when the radius is 6 units, $i$

1

4 π

2

2 π

3

6 π

4

3 π

Explanation:

(A) : Let A square units is the area measured when the radius is $r$ units.
$Area A = π r^{2}$
Differentiate both side with respect to ' $x$ '
$\begin{aligned} (i) & \frac{dA}{dx} = 2 π \frac{dr}{dx} \\ \frac{dA}{dx} = 3 c \frac{dr}{dx} \end{aligned}$
From equation (i) we get,
$3 \frac{c \cdot dr}{dx} = 2 π r \frac{dr}{dx}$
$3 c = 2 π r$
$c = \frac{2}{3} π r$
$c = \frac{2}{3} π (6)$
$c = 4 π$

BITSAT-2010

Application of Derivatives

85805 The fuel charges for running a train are proportional to the square of the speed generated in miles per hour and costs $Math input error$ per hour at 16 miles per hour. The most economical speed if the fixed charges i.e. salaries etc. amount to $Math input error$ per hour $i$

1 10

2 20

3 30

4 40

Explanation:

(D) : Let the speed of the train be $v$ and distance to be covered be $s$ so that total time taken is $s / v$ hours. Cost of fuel per hour $= {kv}^{2}$ ( $k$ is constant)
Also $48 = k {.16}^{2}$ by given condition
$∴ k = \frac{3}{16}$
$∴$ Cost of fuel per hour $\frac{3}{16} v^{2}$
Other charger per hour are 300.
Total running cost,
$C = (\frac{3}{16} v^{2} + 300) \frac{s}{v} = \frac{3 s}{16} v + \frac{300 s}{v}$
$\frac{dC}{dv} = \frac{3 s}{16} - \frac{300 s}{v^{2}} = 0 \Rightarrow v = 40$
$\frac{d^{2} C}{{dv}^{2}} = \frac{600 s}{v^{3}} > 0$
$∴ v = 40$ results in minimum running cost

BITSAT-2014

Application of Derivatives

85806 If $0 < x < \frac{π}{2}$ , then

1

\tan x < x < \sin x

2

x < \sin x < \tan x

3

\sin x < \tan x < x

4 None of these

Explanation:

(D) : Let us assume the functions $f (x)$ and $g (x)$ given by,
$f (x) = \tan x - x$
And, $g (x) = x - \sin x$ for $0 < x < \frac{π}{2}$
Now, $f^{'} (x) = \sec^{2} x - 1$ and $g^{'} (x) = 1 - \cos x$
$f^{'} (x) > 0 and g^{'} (x) > 0, \forall x \in (0, \frac{π}{2})$
$f^{'} (x) > f (0)$ and $g^{'} (x) > 0, \forall x \in (0, \frac{π}{2})$
$\tan x - x > 0$ and $x - \sin x > 0$
$\tan x > x$ and $x > \sin x$ $\sin x < x < \tan x$

BITSAT-2015

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