QBManager

Application of Derivatives

85297 The function $f (x) = \frac{x}{\log x}$ increases on the interval

1

(0, \infty)

2

(0, e)

3

(e, \infty)

4 None of these

Explanation:

(C) : Given
$f (x) = \frac{x}{\log x}$
$f (x)$ is defined for $x > 0$
$f^{'} (x) = \frac{\log x \times 1 - x \frac{1}{x}}{(\log x)^{2}} \Rightarrow f^{'} (x) = \frac{\log x - 1}{(\log x)^{2}}$
$∴ f^{'} (x) > 0$
$\log x > 1$
$\log_{e} x > \log_{e} e$
$x > e$
$∴ x \in (e, \infty)$

[JCECE-2011]

Application of Derivatives

85298 The function $f (x) = \tan x - x$

1 always increases

2 always decreases

3 never increases

4 sometimes increases and sometimes decreases

Explanation:

(A) : Given,
$f (x) = \tan x - x$
On differentiating w.r.t. to $x$ we get -
$f^{'} (x) = \sec^{2} x - 1$
$f^{'} (x) > 0 \forall x \in R$
So, $f^{'} (x)$ is always increases.

[JCECE-2018]

Application of Derivatives

85299 The function $f (x) = 2 - 3 x$ is :

1 increasing

2 decreasing

3 neither decreasing nor increasing

4 none of the above

Explanation:

(B): If $f (x)$ is a function, it will be increasing or decreasing if $f^{'} (x) > 0 f^{'} (x) < 0$ .
We have,
$f (x) = 2 - 3 x$
On differentiating w.r.t. to $x$ we get -
$f^{'} (x) = - 3 < 0$
$∴$ Function is decreasing for every value of $x$ .
Alternate solution -
Let $y = f (x) = 2 - 3 x$
$y + 3 x = 2$
$\frac{x}{2 / 3} + \frac{y}{2} = 1$

It is clear from the figure that for increasing the value of $x$ from $- \infty$ to $\infty$ , we will get the decreasing value of $y$ from $\infty$ to $- \infty$ .
$∴$ It is decreasing function.

Application of Derivatives

85300 A point is in motion along the curve $12 y = x^{3}$ . The $x$ -coordinate, changes faster than $y$ coordinate, if

1

x \in [- 2, 2]

2

x \in [- \infty, 2] \cup [2, \infty]

3

x \in [- 2, 2]

4

x \in [- \infty, - 2] \cup [2, \infty]

Explanation:

(A) : Given,
$x$ -coordinate, changes faster than $y$ coordinate.
So, $\frac{dy}{dt} > \frac{dx}{dt}$
$∵ 12 y = x^{3}$
$\frac{12 dy}{dt} = 3 x^{2} \frac{dx}{dt}$
$\frac{d y}{d t} = \frac{x^{2}}{4} \frac{d x}{d t}$
From equation (i),
$\frac{x^{2}}{4} \frac{d x}{d t} > \frac{d x}{d t}$
$x^{2} > 4$
$x^{2} - 4 > 0$
$(x - 2) (x + 2) > 0$
$∴ x \in [- 2, 2]$

CG PET-2010

Application of Derivatives

85297 The function $f (x) = \frac{x}{\log x}$ increases on the interval

1

(0, \infty)

2

(0, e)

3

(e, \infty)

4 None of these

Explanation:

(C) : Given
$f (x) = \frac{x}{\log x}$
$f (x)$ is defined for $x > 0$
$f^{'} (x) = \frac{\log x \times 1 - x \frac{1}{x}}{(\log x)^{2}} \Rightarrow f^{'} (x) = \frac{\log x - 1}{(\log x)^{2}}$
$∴ f^{'} (x) > 0$
$\log x > 1$
$\log_{e} x > \log_{e} e$
$x > e$
$∴ x \in (e, \infty)$

[JCECE-2011]

Application of Derivatives

85298 The function $f (x) = \tan x - x$

1 always increases

2 always decreases

3 never increases

4 sometimes increases and sometimes decreases

Explanation:

(A) : Given,
$f (x) = \tan x - x$
On differentiating w.r.t. to $x$ we get -
$f^{'} (x) = \sec^{2} x - 1$
$f^{'} (x) > 0 \forall x \in R$
So, $f^{'} (x)$ is always increases.

[JCECE-2018]

Application of Derivatives

85299 The function $f (x) = 2 - 3 x$ is :

1 increasing

2 decreasing

3 neither decreasing nor increasing

4 none of the above

Explanation:

(B): If $f (x)$ is a function, it will be increasing or decreasing if $f^{'} (x) > 0 f^{'} (x) < 0$ .
We have,
$f (x) = 2 - 3 x$
On differentiating w.r.t. to $x$ we get -
$f^{'} (x) = - 3 < 0$
$∴$ Function is decreasing for every value of $x$ .
Alternate solution -
Let $y = f (x) = 2 - 3 x$
$y + 3 x = 2$
$\frac{x}{2 / 3} + \frac{y}{2} = 1$

It is clear from the figure that for increasing the value of $x$ from $- \infty$ to $\infty$ , we will get the decreasing value of $y$ from $\infty$ to $- \infty$ .
$∴$ It is decreasing function.

Application of Derivatives

85300 A point is in motion along the curve $12 y = x^{3}$ . The $x$ -coordinate, changes faster than $y$ coordinate, if

1

x \in [- 2, 2]

2

x \in [- \infty, 2] \cup [2, \infty]

3

x \in [- 2, 2]

4

x \in [- \infty, - 2] \cup [2, \infty]

Explanation:

(A) : Given,
$x$ -coordinate, changes faster than $y$ coordinate.
So, $\frac{dy}{dt} > \frac{dx}{dt}$
$∵ 12 y = x^{3}$
$\frac{12 dy}{dt} = 3 x^{2} \frac{dx}{dt}$
$\frac{d y}{d t} = \frac{x^{2}}{4} \frac{d x}{d t}$
From equation (i),
$\frac{x^{2}}{4} \frac{d x}{d t} > \frac{d x}{d t}$
$x^{2} > 4$
$x^{2} - 4 > 0$
$(x - 2) (x + 2) > 0$
$∴ x \in [- 2, 2]$

CG PET-2010

Application of Derivatives

85297 The function $f (x) = \frac{x}{\log x}$ increases on the interval

1

(0, \infty)

2

(0, e)

3

(e, \infty)

4 None of these

Explanation:

(C) : Given
$f (x) = \frac{x}{\log x}$
$f (x)$ is defined for $x > 0$
$f^{'} (x) = \frac{\log x \times 1 - x \frac{1}{x}}{(\log x)^{2}} \Rightarrow f^{'} (x) = \frac{\log x - 1}{(\log x)^{2}}$
$∴ f^{'} (x) > 0$
$\log x > 1$
$\log_{e} x > \log_{e} e$
$x > e$
$∴ x \in (e, \infty)$

[JCECE-2011]

Application of Derivatives

85298 The function $f (x) = \tan x - x$

1 always increases

2 always decreases

3 never increases

4 sometimes increases and sometimes decreases

Explanation:

(A) : Given,
$f (x) = \tan x - x$
On differentiating w.r.t. to $x$ we get -
$f^{'} (x) = \sec^{2} x - 1$
$f^{'} (x) > 0 \forall x \in R$
So, $f^{'} (x)$ is always increases.

[JCECE-2018]

Application of Derivatives

85299 The function $f (x) = 2 - 3 x$ is :

1 increasing

2 decreasing

3 neither decreasing nor increasing

4 none of the above

Explanation:

(B): If $f (x)$ is a function, it will be increasing or decreasing if $f^{'} (x) > 0 f^{'} (x) < 0$ .
We have,
$f (x) = 2 - 3 x$
On differentiating w.r.t. to $x$ we get -
$f^{'} (x) = - 3 < 0$
$∴$ Function is decreasing for every value of $x$ .
Alternate solution -
Let $y = f (x) = 2 - 3 x$
$y + 3 x = 2$
$\frac{x}{2 / 3} + \frac{y}{2} = 1$

It is clear from the figure that for increasing the value of $x$ from $- \infty$ to $\infty$ , we will get the decreasing value of $y$ from $\infty$ to $- \infty$ .
$∴$ It is decreasing function.

Application of Derivatives

85300 A point is in motion along the curve $12 y = x^{3}$ . The $x$ -coordinate, changes faster than $y$ coordinate, if

1

x \in [- 2, 2]

2

x \in [- \infty, 2] \cup [2, \infty]

3

x \in [- 2, 2]

4

x \in [- \infty, - 2] \cup [2, \infty]

Explanation:

(A) : Given,
$x$ -coordinate, changes faster than $y$ coordinate.
So, $\frac{dy}{dt} > \frac{dx}{dt}$
$∵ 12 y = x^{3}$
$\frac{12 dy}{dt} = 3 x^{2} \frac{dx}{dt}$
$\frac{d y}{d t} = \frac{x^{2}}{4} \frac{d x}{d t}$
From equation (i),
$\frac{x^{2}}{4} \frac{d x}{d t} > \frac{d x}{d t}$
$x^{2} > 4$
$x^{2} - 4 > 0$
$(x - 2) (x + 2) > 0$
$∴ x \in [- 2, 2]$

CG PET-2010

Application of Derivatives

85297 The function $f (x) = \frac{x}{\log x}$ increases on the interval

1

(0, \infty)

2

(0, e)

3

(e, \infty)

4 None of these

Explanation:

(C) : Given
$f (x) = \frac{x}{\log x}$
$f (x)$ is defined for $x > 0$
$f^{'} (x) = \frac{\log x \times 1 - x \frac{1}{x}}{(\log x)^{2}} \Rightarrow f^{'} (x) = \frac{\log x - 1}{(\log x)^{2}}$
$∴ f^{'} (x) > 0$
$\log x > 1$
$\log_{e} x > \log_{e} e$
$x > e$
$∴ x \in (e, \infty)$

[JCECE-2011]

Application of Derivatives

85298 The function $f (x) = \tan x - x$

1 always increases

2 always decreases

3 never increases

4 sometimes increases and sometimes decreases

Explanation:

(A) : Given,
$f (x) = \tan x - x$
On differentiating w.r.t. to $x$ we get -
$f^{'} (x) = \sec^{2} x - 1$
$f^{'} (x) > 0 \forall x \in R$
So, $f^{'} (x)$ is always increases.

[JCECE-2018]

Application of Derivatives

85299 The function $f (x) = 2 - 3 x$ is :

1 increasing

2 decreasing

3 neither decreasing nor increasing

4 none of the above

Explanation:

(B): If $f (x)$ is a function, it will be increasing or decreasing if $f^{'} (x) > 0 f^{'} (x) < 0$ .
We have,
$f (x) = 2 - 3 x$
On differentiating w.r.t. to $x$ we get -
$f^{'} (x) = - 3 < 0$
$∴$ Function is decreasing for every value of $x$ .
Alternate solution -
Let $y = f (x) = 2 - 3 x$
$y + 3 x = 2$
$\frac{x}{2 / 3} + \frac{y}{2} = 1$

It is clear from the figure that for increasing the value of $x$ from $- \infty$ to $\infty$ , we will get the decreasing value of $y$ from $\infty$ to $- \infty$ .
$∴$ It is decreasing function.

Application of Derivatives

85300 A point is in motion along the curve $12 y = x^{3}$ . The $x$ -coordinate, changes faster than $y$ coordinate, if

1

x \in [- 2, 2]

2

x \in [- \infty, 2] \cup [2, \infty]

3

x \in [- 2, 2]

4

x \in [- \infty, - 2] \cup [2, \infty]

Explanation:

(A) : Given,
$x$ -coordinate, changes faster than $y$ coordinate.
So, $\frac{dy}{dt} > \frac{dx}{dt}$
$∵ 12 y = x^{3}$
$\frac{12 dy}{dt} = 3 x^{2} \frac{dx}{dt}$
$\frac{d y}{d t} = \frac{x^{2}}{4} \frac{d x}{d t}$
From equation (i),
$\frac{x^{2}}{4} \frac{d x}{d t} > \frac{d x}{d t}$
$x^{2} > 4$
$x^{2} - 4 > 0$
$(x - 2) (x + 2) > 0$
$∴ x \in [- 2, 2]$

CG PET-2010

Explanation:

Explanation:

Explanation:

Explanation:

Question Bank Series

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation: