QBManager

Application of Derivatives

85274 The function $f (x) = \log x - \frac{2 x}{x + 2}$ is increasing for all

1

x \in (0, \infty)

2

x \in (- \infty, 1)

3

x \in (- 1, \infty)

4

x \in (- \infty, 0)

Explanation:

(A) : Given,
$f (x) = \log x - \frac{2 x}{x + 2}$
On differentiating both sides w. r. t. to $x$
$f^{'} (x) = \frac{1}{x} - [\frac{2 x - (x + 2) \cdot 2}{(x + 2)^{2}}]$
$f^{'} (x) = \frac{1}{x} - [\frac{2 x - 2 x - 4}{(x + 2)^{2}}]$
$f^{'} (x) = \frac{1}{x} + \frac{4}{(x + 2)^{2}}$
$∵ f^{'} (x) \geq 0$
$\frac{1}{x} + \frac{4}{(x + 2)^{2}} \geq 0 = \frac{(x + 2)^{2} + 4 x}{x (x + 2)^{2}} \geq 0$
$\frac{x^{2} + 8 x + 4}{x (x + 2)^{2}} \geq 0 \Rightarrow ∴ x \in (0, \infty)$

MHT CET-2020

Application of Derivatives

85275 For every value of $x$ , the function $f (x) = \frac{1}{a^{x}}, a > 0$ is

1 constant

2 increasing

3 neither increasing nor decreasing

4 decreasing

Explanation:

(D) : Given function,
$f (x) = \frac{1}{a^{x}}$
$f (x) = a^{- x}, a > 0$
$f^{'} (x) = - a^{- x} \log a$
$f^{'} (x) = \frac{- \log a}{a^{x}}$
$f^{'} (x) < 0 \forall x \in R, a > 0$
$∴ f (x)$ is decreasing for every value of $x$ .

MHT CET-2020

Application of Derivatives

85276 The function $f (x) = (x + 2) e^{- x}$ is

1 decreasing in

(- \infty, - 1)

and increasing in

(- 1, \infty)

2 decreasing for all

x

3 increasing in

(- \infty, - 1)

and decreasing in

(- 1, \infty)

4 increasing for all

x

Explanation:

(C) : Given,
$f (x) = (x + 2) e^{- x} \Rightarrow f^{'} (x) = e^{- x} - e^{- x} (x + 2)$
$f^{'} (x) = - e^{- x} [x + 1]$
For increasing function-
$- e^{- x} (x + 1) > 0$ or $e^{- x} (x + 1) < 0$
$e^{- x} > 0$ or $(x + 1) < 0$
$x \in (- \infty, \infty)$ and $x \in (- \infty, - 1)$
Therefore, $x \in (- \infty, - 1)$
The function is increasing in $(- \infty, - 1)$
For the decreasing function-
$- e^{- x} (x + 1) < 0$ or $e^{- x} (x + 1) > 0, x \in (- 1, \infty)$
Hence,
The function is decreasing in $(- 1, \infty)$

MHT CET-2020

Application of Derivatives

85277 The function $f (x) = x^{3} - 3 x$ is

1 decreasing in

(- \infty, - 1) \cup (1, \infty)

and increasing in

(- 1, 1)

2 decreasing in

(0, \infty)

and increasing in

(- \infty, 0)

3 increasing in

(- \infty, - 1) \cup (1, \infty)

and decreasing in

(- 1, 1)

4 increasing in

(0, \infty)

and decreasing in

(- \infty, 0)

Explanation:

(C): Given function,
$f (x) = x^{3} - 3 x$
$f^{'} (x) = 3 x^{2} - 3$
$f^{'} (x) = 3 (x + 1) (x - 1)$
Here,
$f^{'} (x) > 0$ in $(- \infty, - 1) \cup (1, \infty)$ and $f^{'} (x) < 0$ in $(- 1, 1)$ $∴ f (x)$ is increasing in $(- \infty, - 1) \cup (1, \infty)$ and decreasing in $(- 1, 1)$ .

MHT CET-2019

Application of Derivatives

85274 The function $f (x) = \log x - \frac{2 x}{x + 2}$ is increasing for all

1

x \in (0, \infty)

2

x \in (- \infty, 1)

3

x \in (- 1, \infty)

4

x \in (- \infty, 0)

Explanation:

(A) : Given,
$f (x) = \log x - \frac{2 x}{x + 2}$
On differentiating both sides w. r. t. to $x$
$f^{'} (x) = \frac{1}{x} - [\frac{2 x - (x + 2) \cdot 2}{(x + 2)^{2}}]$
$f^{'} (x) = \frac{1}{x} - [\frac{2 x - 2 x - 4}{(x + 2)^{2}}]$
$f^{'} (x) = \frac{1}{x} + \frac{4}{(x + 2)^{2}}$
$∵ f^{'} (x) \geq 0$
$\frac{1}{x} + \frac{4}{(x + 2)^{2}} \geq 0 = \frac{(x + 2)^{2} + 4 x}{x (x + 2)^{2}} \geq 0$
$\frac{x^{2} + 8 x + 4}{x (x + 2)^{2}} \geq 0 \Rightarrow ∴ x \in (0, \infty)$

MHT CET-2020

Application of Derivatives

85275 For every value of $x$ , the function $f (x) = \frac{1}{a^{x}}, a > 0$ is

1 constant

2 increasing

3 neither increasing nor decreasing

4 decreasing

Explanation:

(D) : Given function,
$f (x) = \frac{1}{a^{x}}$
$f (x) = a^{- x}, a > 0$
$f^{'} (x) = - a^{- x} \log a$
$f^{'} (x) = \frac{- \log a}{a^{x}}$
$f^{'} (x) < 0 \forall x \in R, a > 0$
$∴ f (x)$ is decreasing for every value of $x$ .

MHT CET-2020

Application of Derivatives

85276 The function $f (x) = (x + 2) e^{- x}$ is

1 decreasing in

(- \infty, - 1)

and increasing in

(- 1, \infty)

2 decreasing for all

x

3 increasing in

(- \infty, - 1)

and decreasing in

(- 1, \infty)

4 increasing for all

x

Explanation:

(C) : Given,
$f (x) = (x + 2) e^{- x} \Rightarrow f^{'} (x) = e^{- x} - e^{- x} (x + 2)$
$f^{'} (x) = - e^{- x} [x + 1]$
For increasing function-
$- e^{- x} (x + 1) > 0$ or $e^{- x} (x + 1) < 0$
$e^{- x} > 0$ or $(x + 1) < 0$
$x \in (- \infty, \infty)$ and $x \in (- \infty, - 1)$
Therefore, $x \in (- \infty, - 1)$
The function is increasing in $(- \infty, - 1)$
For the decreasing function-
$- e^{- x} (x + 1) < 0$ or $e^{- x} (x + 1) > 0, x \in (- 1, \infty)$
Hence,
The function is decreasing in $(- 1, \infty)$

MHT CET-2020

Application of Derivatives

85277 The function $f (x) = x^{3} - 3 x$ is

1 decreasing in

(- \infty, - 1) \cup (1, \infty)

and increasing in

(- 1, 1)

2 decreasing in

(0, \infty)

and increasing in

(- \infty, 0)

3 increasing in

(- \infty, - 1) \cup (1, \infty)

and decreasing in

(- 1, 1)

4 increasing in

(0, \infty)

and decreasing in

(- \infty, 0)

Explanation:

(C): Given function,
$f (x) = x^{3} - 3 x$
$f^{'} (x) = 3 x^{2} - 3$
$f^{'} (x) = 3 (x + 1) (x - 1)$
Here,
$f^{'} (x) > 0$ in $(- \infty, - 1) \cup (1, \infty)$ and $f^{'} (x) < 0$ in $(- 1, 1)$ $∴ f (x)$ is increasing in $(- \infty, - 1) \cup (1, \infty)$ and decreasing in $(- 1, 1)$ .

MHT CET-2019

Application of Derivatives

85274 The function $f (x) = \log x - \frac{2 x}{x + 2}$ is increasing for all

1

x \in (0, \infty)

2

x \in (- \infty, 1)

3

x \in (- 1, \infty)

4

x \in (- \infty, 0)

Explanation:

(A) : Given,
$f (x) = \log x - \frac{2 x}{x + 2}$
On differentiating both sides w. r. t. to $x$
$f^{'} (x) = \frac{1}{x} - [\frac{2 x - (x + 2) \cdot 2}{(x + 2)^{2}}]$
$f^{'} (x) = \frac{1}{x} - [\frac{2 x - 2 x - 4}{(x + 2)^{2}}]$
$f^{'} (x) = \frac{1}{x} + \frac{4}{(x + 2)^{2}}$
$∵ f^{'} (x) \geq 0$
$\frac{1}{x} + \frac{4}{(x + 2)^{2}} \geq 0 = \frac{(x + 2)^{2} + 4 x}{x (x + 2)^{2}} \geq 0$
$\frac{x^{2} + 8 x + 4}{x (x + 2)^{2}} \geq 0 \Rightarrow ∴ x \in (0, \infty)$

MHT CET-2020

Application of Derivatives

85275 For every value of $x$ , the function $f (x) = \frac{1}{a^{x}}, a > 0$ is

1 constant

2 increasing

3 neither increasing nor decreasing

4 decreasing

Explanation:

(D) : Given function,
$f (x) = \frac{1}{a^{x}}$
$f (x) = a^{- x}, a > 0$
$f^{'} (x) = - a^{- x} \log a$
$f^{'} (x) = \frac{- \log a}{a^{x}}$
$f^{'} (x) < 0 \forall x \in R, a > 0$
$∴ f (x)$ is decreasing for every value of $x$ .

MHT CET-2020

Application of Derivatives

85276 The function $f (x) = (x + 2) e^{- x}$ is

1 decreasing in

(- \infty, - 1)

and increasing in

(- 1, \infty)

2 decreasing for all

x

3 increasing in

(- \infty, - 1)

and decreasing in

(- 1, \infty)

4 increasing for all

x

Explanation:

(C) : Given,
$f (x) = (x + 2) e^{- x} \Rightarrow f^{'} (x) = e^{- x} - e^{- x} (x + 2)$
$f^{'} (x) = - e^{- x} [x + 1]$
For increasing function-
$- e^{- x} (x + 1) > 0$ or $e^{- x} (x + 1) < 0$
$e^{- x} > 0$ or $(x + 1) < 0$
$x \in (- \infty, \infty)$ and $x \in (- \infty, - 1)$
Therefore, $x \in (- \infty, - 1)$
The function is increasing in $(- \infty, - 1)$
For the decreasing function-
$- e^{- x} (x + 1) < 0$ or $e^{- x} (x + 1) > 0, x \in (- 1, \infty)$
Hence,
The function is decreasing in $(- 1, \infty)$

MHT CET-2020

Application of Derivatives

85277 The function $f (x) = x^{3} - 3 x$ is

1 decreasing in

(- \infty, - 1) \cup (1, \infty)

and increasing in

(- 1, 1)

2 decreasing in

(0, \infty)

and increasing in

(- \infty, 0)

3 increasing in

(- \infty, - 1) \cup (1, \infty)

and decreasing in

(- 1, 1)

4 increasing in

(0, \infty)

and decreasing in

(- \infty, 0)

Explanation:

(C): Given function,
$f (x) = x^{3} - 3 x$
$f^{'} (x) = 3 x^{2} - 3$
$f^{'} (x) = 3 (x + 1) (x - 1)$
Here,
$f^{'} (x) > 0$ in $(- \infty, - 1) \cup (1, \infty)$ and $f^{'} (x) < 0$ in $(- 1, 1)$ $∴ f (x)$ is increasing in $(- \infty, - 1) \cup (1, \infty)$ and decreasing in $(- 1, 1)$ .

MHT CET-2019

Application of Derivatives

85274 The function $f (x) = \log x - \frac{2 x}{x + 2}$ is increasing for all

1

x \in (0, \infty)

2

x \in (- \infty, 1)

3

x \in (- 1, \infty)

4

x \in (- \infty, 0)

Explanation:

(A) : Given,
$f (x) = \log x - \frac{2 x}{x + 2}$
On differentiating both sides w. r. t. to $x$
$f^{'} (x) = \frac{1}{x} - [\frac{2 x - (x + 2) \cdot 2}{(x + 2)^{2}}]$
$f^{'} (x) = \frac{1}{x} - [\frac{2 x - 2 x - 4}{(x + 2)^{2}}]$
$f^{'} (x) = \frac{1}{x} + \frac{4}{(x + 2)^{2}}$
$∵ f^{'} (x) \geq 0$
$\frac{1}{x} + \frac{4}{(x + 2)^{2}} \geq 0 = \frac{(x + 2)^{2} + 4 x}{x (x + 2)^{2}} \geq 0$
$\frac{x^{2} + 8 x + 4}{x (x + 2)^{2}} \geq 0 \Rightarrow ∴ x \in (0, \infty)$

MHT CET-2020

Application of Derivatives

85275 For every value of $x$ , the function $f (x) = \frac{1}{a^{x}}, a > 0$ is

1 constant

2 increasing

3 neither increasing nor decreasing

4 decreasing

Explanation:

(D) : Given function,
$f (x) = \frac{1}{a^{x}}$
$f (x) = a^{- x}, a > 0$
$f^{'} (x) = - a^{- x} \log a$
$f^{'} (x) = \frac{- \log a}{a^{x}}$
$f^{'} (x) < 0 \forall x \in R, a > 0$
$∴ f (x)$ is decreasing for every value of $x$ .

MHT CET-2020

Application of Derivatives

85276 The function $f (x) = (x + 2) e^{- x}$ is

1 decreasing in

(- \infty, - 1)

and increasing in

(- 1, \infty)

2 decreasing for all

x

3 increasing in

(- \infty, - 1)

and decreasing in

(- 1, \infty)

4 increasing for all

x

Explanation:

(C) : Given,
$f (x) = (x + 2) e^{- x} \Rightarrow f^{'} (x) = e^{- x} - e^{- x} (x + 2)$
$f^{'} (x) = - e^{- x} [x + 1]$
For increasing function-
$- e^{- x} (x + 1) > 0$ or $e^{- x} (x + 1) < 0$
$e^{- x} > 0$ or $(x + 1) < 0$
$x \in (- \infty, \infty)$ and $x \in (- \infty, - 1)$
Therefore, $x \in (- \infty, - 1)$
The function is increasing in $(- \infty, - 1)$
For the decreasing function-
$- e^{- x} (x + 1) < 0$ or $e^{- x} (x + 1) > 0, x \in (- 1, \infty)$
Hence,
The function is decreasing in $(- 1, \infty)$

MHT CET-2020

Application of Derivatives

85277 The function $f (x) = x^{3} - 3 x$ is

1 decreasing in

(- \infty, - 1) \cup (1, \infty)

and increasing in

(- 1, 1)

2 decreasing in

(0, \infty)

and increasing in

(- \infty, 0)

3 increasing in

(- \infty, - 1) \cup (1, \infty)

and decreasing in

(- 1, 1)

4 increasing in

(0, \infty)

and decreasing in

(- \infty, 0)

Explanation:

(C): Given function,
$f (x) = x^{3} - 3 x$
$f^{'} (x) = 3 x^{2} - 3$
$f^{'} (x) = 3 (x + 1) (x - 1)$
Here,
$f^{'} (x) > 0$ in $(- \infty, - 1) \cup (1, \infty)$ and $f^{'} (x) < 0$ in $(- 1, 1)$ $∴ f (x)$ is increasing in $(- \infty, - 1) \cup (1, \infty)$ and decreasing in $(- 1, 1)$ .

MHT CET-2019

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